Teaching is my passion and I have been teaching since I was in engineering college. I believe in "Learning" and not in "Mugging" After taking a sabbatical from my professional career I have been associated with an institute for past 3 years where I teach maths & sciences.
My approach is to first identify the need of the student because each student has different learning ability and learning style. • Develop a plan along with the students so that he/she enjoys the learning. • Break the topics into chunks which will follow a logical order with a clear learning goal. • Each learning sessions will be interactive where I will use audio-visuals, charts, videos, and illustrations. • Key concepts will be repeated several times so that it builds the foundation of the student • Subject-wise, chapter/topic wise discussion, and solution in the session • Regular test analysis of student performance and regular feedback to parents on the progress of the students My teaching style will Stop Mugging and help the student to start Learning
Class 9 - 10 | Mathematics Physics Chemistry All Boards | INR 500 / Hour |
Test Preparation | CMAT | INR 1000 / Hour |
MBA & BBA | Management Subjects | INR 1000 / Hour |
Nominal Value (par value) of the share is Rs. 150. Nominal Value of 1000 share owned by the person= 150 X 1000 = Rs. 150,000.00 Comapny declared 15% dividend. Dividend amount = (15/100)X150X1000=Rs 22,500.00 (Dividend is calculated on nominal value and that is the sum received by the investor) Let the cost of each share purchased be "s".( the purchase price of a share is normally higher/different than nominal value) The person earned 12% profit on his investment amount= 1000s ( s X 1000 share) Then (12/100)X 1000s= 22,500.00 (since profit, = sum received from an investment) Therefore s= (22,500x100)/(12X1000)= Rs 187.50. Ans: Cost of each share bought was Rs 187.50
Raju invested Rs, 24,000.00 for buying share quoted at Rs 48.00 Therefore Raju purchased (24,000/48) =500 shares. The face value of each share is Rs. 60 10% dividend on each share =(10/100)X60=Rs 6 Therefore total dividend earned on 500 shares =Rs 500X6=Rs. 3,000.00 Ans: Raju's income is Rs 3,000.00
Amount qualified for paying interest @4% in June is Rs 4,500.00 Simple Interest = (P x R x T)/100 where Principal = P, Rate = R% per annum (p.a.) and Time = T years. (i) I = 4500X4X1/(100X12) since T =1 month=1/12 year Interest for June = Rs 15.00 (ii) Interest for July = 8500X4X1/(100X12)= Rs 28.33
Maturity value of a recurring deposit is calculated by the following formula M = ( R x [(1+r)n - 1 ] ) / (1-(1+r)-1/3) Where, M = Maturity value R = Monthly Installment r = Rate of Interest (i) / 400 (Annual interest rate converted into quarterly rate) n = Number of Quarters (Assuming the compounding is done quarterly) 7725= (R* ((1+7/400)^8 - 1 ) ) / (1-(1+7/400)-1/3) R=299.18 (nearest approximation) Ans: Monthly instalment will be Rs 299.18
Titration of Na2CO3 using phenolphthalein as indicator involves the reaction: Na2CO3 + HCl = NaHCO3 + NaCl so 1 Mole Na2CO3 consumes 1 Mole HCl or "x" ml 0.05 M Na2CO3 soln consumes "x" ml 0.05 M HCl soln. so A = x. Titration of Na2CO3 using methyl orange as indicator involves the reactions: Na2CO3 + 2HCl = H2CO3 + 2NaCl and also NaHCO3 + HCl = H2CO3 + NaCl so 1 Mole Na2CO3 consumes 2 Moles of HCl and 1 Mole NaHCO3 consumes 1 Mole of HCl so "x" ml 0.05 M Na2CO3 soln consumes "2x" ml 0.05 M HCl soln. and "x" ml 0.05M NaHCO3 soln consumes "x" ml 0.05 M HCl soln. (as the compound contains both Na2CO3 and NaHCO3 in equimolar proportion) hence B= 2x+x=3x. TherforeB/A=3x/x=3:1
employee-1: Let the monthly income be 3x and monthly expenditure be 5y employee-2 Let the monthly income be 5x and monthly expenditure be 9y employee-1: Savings= 3x-5y (income minus expenditure) employee-2: Savings= 5x-9y (income minus expenditure) Therefore 3x-5y=5x-9y=300-(I) (both saves Rs 300 per month) By solving we get 2x=4y or x=2y. Substituting value of x in terms of y in eq-1 we get 3(2y)-5y=300 or y=300. Therefore x=600 Ans: Monthly income of first employee=3x=3(600) =Rs 1800.
Given P(x)=x3 - 6x2 + ax - 6 and Q(x)=x3 + (5-a)x2+12x + 8 leave the same remainder when divided by x - 2. By remainder theorem, if f(x) is divided by (x − a) then the remainder is f(a) Here when P(x) and Q(x) are divided by (x − 2) the remainders are p(2) and Q(2) respectively. Also given that P(2) = Q(2) → (1) Put x = 2 in both P(x) and Q(x) [Zero of the linear polynomial (x-2)=2] Hence equation (1) becomes, 2^3-6(2)^2+a(2)-6= 2^3+(5-a)(2)^2+12(2)-8 or 2a-22=44-4a or 6a=66 or a=11 Ans: Value of a is 11
Electrostatic (repulsion) force between these two charges "Q" and "Q-q" has magnitude F=k(q)(Q−q)/r2=−k(q2−Qq)/r2 here Q and r are constant so for maximum force we have to differentiate the force "F" w.r.t "q" dF/dq = k/r2 [Q -2q] For maximum force dF/dq=0 Therefore Q-2q=0 or 2q=Q Hence the repulsion force will be maximum when q/Q =1/2
Molecular mass of water =2×1+1×16=18g=2×1+1×16=18g For 178.2g of water nA=9.9 Molecular mass of glucose =6×12+12×1+6×16=180g=6×12+12×1+6×16=180g For 180g of glucose nB=0.1 XB=0.1/(0.1+9.9)=0.01 XA=0.99 For lowering of vapour pressure, By using the formula for colligative property P=P0AXA=P0A(1−XB) P=760(1−0.01) =760−7.6 =752.4 torr
The distance bird has to travel is equal to the length of the train.i.e 175 m. Since the bird is flying parallel and opposite to the direction of the train the relative velocity will be the velocity of train plus velocity of bird = 72+18=90 km/hr=25m/s. Time taken by the bird to cross the trian= 175m/25m/s= 7 secs
The force used by John on the door was 50N, at a distance 1.0m from the pivot point;i.e. hinges of the door. (since the door is 2m wide and he opens it by applying the force at the middle of the door) Since John hit the door perpendicular to its plane, the angle between the door and the direction of force was 90 degrees. Since ţ(torque) = r x F = r F sin(90) where r is the distance of the point of application of force from the pivot point, then the torque on the door was: ţ = (1.0m) (50N) sin(90) = 50 N m Farther from the hinge, less force is required to open the door. So if John opens the door at the edge then he needs to apply minimum force. F1Xr1=F2xr2 or (1.0m) (50N)=F2X (2.0m) or F2=25N The minimum force need to be applied for opening the door is 25N
Pipes A and B can fill the tank in 12 minutes and 16 minutes respectively. Therefore, in 1 minute, pipe A can fill 1/12 parts of the tank and pipe B can fill 1/16 parts of the tank Pipe C can empty the tank in 8 minutes. Therefore, in 1 minute, pipe C can empty 1/8 parts of the tank Net part filled by Pipes A,B,C together in 1 minute= 1/12+1/16-1/8=1/48 When all the taps are opened at the same time, 1/48 parts of the tank can be filled in 1 minute. i.e., the whole tank can be filled in 48 minutes
The centripetal force acting on the stone. F = ma = mv2/r = [0.2 x 10^2] / 1 = 20 N
The colourless salt (H) is either ammonium nitrate(NH4NO3) or ammonium nitrite (NH4NO2) When boiled with NaOH it will produce ammonia gas -NH3 Case-1 NH4NO3+NaOH→NaNO3+NH3↑+H2O When zinc is added to the same solution further ammonia will be produced 7NaOH+NaNO3+4Zn→4Na2ZnO2+NH3↑+2H2O Case-2 NH4NO2+NaOH→NaNO2+NH3↑+H2O 3Zn+5NaOH+NaNO2→3Na2ZnO2+NH3↑+H2O
1 g/L = 1000 ppm; 1 ppm = 0.001 g/L 100 gm water=100ml of water .025 gm chlorine in 100 ml water = .0025gm chlorine per litre= 2.5 ppm concentration of chlorine will be 2.5 ppm
Number of matches won by the team in June: 30% of 60 = 18 After a phenomenal winning streak the average is 50% Assuming the team played x matches continuously and won all of them to achieve an average winning percentage of 50%. So total matches played by the team = (60+x) Total matches won by the team = (18+x) Total matches won = 50% of Total matches played Therfore (18+x)/(60+x)=0.5 (50% in fraction)=1/2 or 36+2x = 60+x or x = 24 The team has won 24 games
The mass of the bullet,m1 =20g = 0.020 kg Mass of the block. m2 =4.0 kg Initial velocity of the bullet before collision, u1 =600 m/s Initial velocity of the block, u2 = 0 Velocity of the bullet after collision,v1 Velocity of the block after collision,v2 According to the conservation of linear momentum, m1u1+m2u2=m1v1+m2v2 0.02×600+4x (0)=0.02v1+4v2 --------- (eq1) Since the block rises by 0.2 m after collision and by applying law of conservation of energy, the kinetic energy of the bullet before hitting is equal to change in the potential energy of the block plus the kinetic energy of the bullet after the collision Here, (v2)^2-(u2)^2=2gh (change in the potential energy of the block) or v2= √2x10x0.2, v2=2m/s By substituting value of v2 in eq-1 we get ∴ 0.02×600=0.02v1+4×2 ⇒ 0.02v1=12−8 ⇒ v1=4/.02 ⇒v1=200m/s The velocity of the bullet when it comes out is 200m/s
Let the volume of the round bottomed flask be V1 Then, the volume of air inside the flask at 27° C is also V1(since no liquid added in the flask) Now, T1 = 25°C = 298° K T2 = 327° C = 600° K According to Charles’s law, V1/T1 = V2/T2 => V2 =V1T2/T1 = V1(600/298) = 2.013 V1 Therefore, volume of air expelled out = 2.013 V1 – V1 = 1.013 V Hence, percentage of air expelled out = (1.013)/(2.013)x100 = 50.32%
0.73 g of HCl=0.73/36.5=0.02 mole of HCl=20 millimoles of HCl 200 mL of the solution needs 20 millimoles of HCl for coagulation Thus, 1000 mL of the solution will require 5x 20 millimoles of HCl for complete coagulation = 100 millimoles so, by definition, flocculation value of HCl for the given colloid= 100
Present population of the town =240,000 ∴ The population after 6 years will be 240,000 x 90x90/(100x100)= 194,400 Let the population before 6 years be 'p" ∴ px90x90/(100x100)=240,000, p=296,297 population before 6 years was 296,297
Refraction at first surface u=∞;μ1=4/3; μ2=1; Radius of curvature=+2 Using equation 1/v1−(4/3)/(−∞)=(1−4/3)/(+2) V1=−3(2) V=−6 Negative sign indicates that the image is virtual and is on the same side as the object at a distance 6 mm from the first surface For refraction at second surface, the first image serves as the object which is at a distance of 6 mm +4 mm = 10 mm from the second surface μ2=4/3; μ1=1 u=−10; Radius of curvature=−2 ∴(4/3)/v2−1/(−10)=(4/3−1)/(−2) ∴4/3v2−1/(−10)=(4/3−1)/(−2) V2=-5 The final image is at a distance 5 mm from the second surface towards the left of the second surface. i.e. the final image is formed at a distance of 1 mm from the first surface
According to Einstein's explanation of photoelectric emission, a photon of energy 'E' performs two operations: 1. Removes the electron from the surface of metal 2. Supplies some part of energy to move photo electron towards anode Since minimum amount of energy to remove electron from a surface is equal to work function, we can write Einstein equation as: Energy Supplied = Energy Consumed in ejecting an electron + maximum Kinetic energy of electron E= work function (W) + KEmax, So we can rewrite the formula as hν−W=1/2mv2(max) =eVo Here h is Plank's constant, ν is the frequency of the incident radiation, W is the work function, m is the mass of the emitted electron, v(max)is the speed of the emitted electron, e is the charge of the electron and Vo is the stopping potential and λ is the wavelength of the incident radiation. Using the wavelength of incident radiation the equation can be written as hc/λ−W=eVo or, hc/λ=W + eVo In the first case with wavelength 1λ and stopping potential 3V, we get, hc/1λ =W+ e(3V)-----(1) In the second case wavelength is 2λ and stopping potential 1v, we get, hc/2λ =W+ e(V) -----(2) Multiplying second equation by 2 we get, hc/λ=2W+2eV --------(3) Now, Substituting (2) in(1) , we get, 2W+ 2eV = W + 3 eV; or W=eV Substituting it in eq-3 we get, hc/λ=2W+2W; hc/λ = 4W But we can write Work function W = hc/λ0 where λ0= threshold wavelength hc/λ= 4 hc/λ0 or λ0= 4λ Threshold wavelength is 4λ
Length of the rod, L = 0.45 m Mass suspended by the wires, m = 50 g = 50 × 10 – 3 kg Acceleration due to gravity, g = 10.0 m/s2 Current in the rod flowing through the wire, I = 5.0 A B= magnitude of magnetic field (Tesla, T) Tension in the wire will be zero only when magnetic field (B) is equal and opposite to the weight of the rod, (ignoring weight of the wire), ∴ BIL=mg or B= mg/IL= (50 × 10 – 3 × 10)/ (5 × 0.45) = 0.22T The magnetic field should be 0.22T
Time taken by Harpreet to cover 128 km distance = t motion + t rest= 128/16 + 5(8)= 8 hrs + 40 minutes= 8 2/3 hour Harpreet will take 5 breaks of 8 minutes each to cover 128 km
COPcooling=Q/W Q is the heat removed from the cold reservoir. W is the work done, COP is the co-efficient of performance W= Q/COP= 2000/4= 500 calories Work done by electric motor is 500 calories
Pascal's principle, for the two pistons, implies F1A2=F2A1 where A1 and A2 are cylinder areas. A1= πr2 and A2=π(4r)2 F1= F2A1/A2 Now F2=mg= 100 x 9.8 (assuming g=9.8 m/s2) or F1= (100 x 9.8 x πr2)/16πr2 = 61.25 N
The block slides off when the normal force is zero; By applying the formula mv2/R= mg cosθ or v2/R=gcosθ or cosθ= 5x5/5x10 or cosθ=1/2 θ=60 The angle made by the radius vector with the vertical will be 60 degree
If the ellipse intersects the hyperbola orthogonally, their foci will coincide equations of the ellipse & the hyperbola be x2/a2+y2/b2=1and x2/A2−y2/B2=1 respectively. Rewriting the hyperbola equation 2x^2-2y^2=1 in the x^2/A^2 form we get A^2=1/2 or A=1/√2 The eccentricity of the hyperbola E=√(A2+B2)/A=√2 and the foci will be (ae,0) for a parabola and (AE,0) for a hyperbola. As the point coincdes, ae=AE=1 e= 1/√2 (e is reciprocal of E) ∴ a= √2 or a^2=2 For ellipse b2=a2(1−e2) or b=1 substituting the values we get ellipse equation as x^2/2 + y^2/1=1
Let A(at1^2, 2at1) and B(at2^2, 2at2) be two distinct points on a parabola y^2=4x Since the value of a=1 for the given parabola coordinates of A and B will be (t1^2, 2t1) and (t2^2, 2t2) Since the axis of the parabola touches a circle of radius r having AB as the diameter, A and B are also lying on the circle and must satisfy circle equation. Coordinates for the centre of the circle will be ((t1^2+t2^2)/2, t1+t2)) Since the axis of the parabola touches the circle, radius of the circle will be y coordinate of the centre of the circle ∴ t1+t2=±r Slope of the line joining AB will be (2t2-2t1)/(t2^2-t1^2)=2(t2-t1)/(t2-t1)(t2+t1) =2/(t2+t1)= ±2/r
%error=(exact value-approximate value)/exact value X 100
Percentage error in voltage (100 ±5)= |(100±5)|/100X100 = 5% Percentage error in current (10 ±0.2)= |(10 ± 10.2)|/10X 100 =2% (discard any negative sign and take the absolute value) R= V/I Since R is a function two variables, the percentage error will be the sum of the percentage error of each variable
Percentage error in resistance is 5%+2%=7%
Let the number of students in the class be "s"
Number of students passed =0.75s (75% of s), Nuber of students failed =0.25s
Number of boys passed = 0.45 x 0.75s ( since 55% of passed students are girls)
Number of girls failed = 0.65 x 0.25s (since 35% of failed students are boys)
% of passed boys over failed girls = ( 0.45 x 0.75s)/(0.65 x 0.25s)x100=27/13X100= 207.7%
Targeted Delivery of Government subsidies and benefits (like adhaar linked transfer) will eliminate the ghost (claiming benefits under a false identity) and duplicates (same person claiming multiple benefits). Since the entire process of transfer, in an Aadhaar-enabled system, each and every transaction can be authenticated. Suppose a beneficiary goes to a fair price shop to buy his monthly quota of ration. If the shop has a biometric authentication device connected to the entitlement database, every transaction at the shop will end up benefiting the intended beneficiaries. All that the beneficiary has to do is to authenticate whether the person is the intended beneficiary with the help of unique biometrics like finger prints or iris. Since the benefit is linked to Aadhaar number, no person with a false identity or same person cannot get benefits for the same schemes multiple times.
Since it is a better welfare delivery mechanism it will reduce the cost of the subsidy and Government will have more money to spend on other projects like improving railways infrastructure or developing ports
Magnetic flux Ф= B.A ( magnetic field X Area)= Bπa^2, (a is the radius of the coil)= btπa^2 (B=bt given)
Induced emf ϵ=- NdФ/dt=-NAdB/dt=-bπa^2 (since number of turns N=1)
But ϵ= IR hence IR- InR=bπa^2 Therefore IR(1-n)=bπa^2 or IR= bπa^2/(1-n)
dФ= Ф1-Ф2= -bπa^2/(1-n)- (bπa^2)/2=- bπa^2 [1/(1-n)-1/2]=-bπa^2(1-n)/2(1+n)
Electric field , ϵ2πa= -bπa^2(1-n)/2(1+n) or ϵ=-ba(1-n)/(1+n) Since cross sectional radius a= R and Resistance is one-half is 10 times another half, n=10
ϵ=9/11Rb
The velocity of the centre of mass will be zero. The initial velocity of the particles were zero as they were at rest so momentum will be zero.By the law of conservation of momentum, the final momentum will also be zero. as mass cannot be zero, final velocity will be zero
Magnetic flux Ф= B.A ( magnetic field X Area)= Bπa^2, (a is the radius of the coil)= btπa^2 (B=bt given)
Induced emf ϵ=- NdФ/dt=-NAdB/dt=-bπa^2 (since number of turns N=1)
But ϵ= IR hence IR- InR=bπa^2 Therefore IR(1-n)=bπa^2 or IR= bπa^2/(1-n)
dФ= Ф1-Ф2= -bπa^2/(1-n)- (bπa^2)/2=- bπa^2 [1/(1-n)-1/2]=-bπa^2(1-n)/2(1+n)
Electric field (E)= ϵ/L(Length), Eπa= -bπa^2(1-n)/2(1+n) or E=-ba(1-n)/2(1+n) Since cross sectional radius a= R and Resistance is one-half is 10 times another half, n=10
E=9/22Rb
The family will be as follows K (Grandmother/Female) L(male) is K's son M(female) is L's wife K& L has son N(male) and daughter P (female). O (female) is N's wife and Q (female) is N's sister. So M becomes the mother-in law of K's only daughter-in-law "O". K and N are the only males So O is the daughter-in-law of M
EMF induced in the secondary is e2= -MdI/dt= -(2)x10/(0.1)=200 V Induced emf in the secondary will cause a change in the flux through N loops of wire changes by dФ in a time dt,
e2=-NdФ/dt or dФ (change in flux) = e2Xdt/N= 200X(0.1)/500=0.04 wb
Change in flux in secondary will be 0.04wb
The cell cycle is divided into two major phases- Interphase and mitosis. The chromosomal DNA is duplicated during a sub portion of interphase known as the "S", or synthesis, phase.
EMF induced in the secondary is e2= -MdI/dt= -(2)x10/(0.1)=200 V Induced emf in the secondary will cause a change in the flux through N loops of wire changes by dФ in a time dt,
e2=-NdФ/dt or dФ (change in flux) = e2Xdt/N= 200X(0.1)/500=0.04 wb
Change in flux in secondary will be 0.04wb
From Archimedes principle, we know m-m apparent = dliq X v (v=volume of liquid). In the given problem Mass of balloon filled with water is M+m ∴ M+m -mapparent= 1XM (since density of water is 1 and volume of liquid in the balloon will be M)
m apparent = m+m-M=m Apparent mass of balloon with water in it when completely immersed in water will be m(loss of weight of balloon = weight of water displaced
Nt/N0=(1/2)^t/T1/2=(1/2)^1/2=1/√2
where Nt is the number of undisintegrated atoms after time t and N0 is the initial number of atoms
Work done in an electric field W=q ΔV where q is the charge and ΔV is the electric potential
W=-q(V1-V2)
The electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation: V = kQ/r, where k is a constant with a value 1/4πϵ
Potential at the centre of ring 1 is V1= Q1/kR + Q2/k √2R ( The distance from charge Q2 is the hypotenuse of right angle triangle with base=height=R) ∵potential of two charges must be added
Potential at the centre of ring 2 is V2= Q2/kR + Q1/k √2R
∴ W= q(V2-V1)= q/k[(Q2/R + Q1/√2R ) – (Q1/R- Q2/√2R)]= q/kR[(Q2-Q1)-1/√2(Q2-Q1)]=q/K√2R(Q2-Q1)( √2-1)
W = q/4 π ϵ√2R(Q2-Q1)( √2-1)
d(distance) = SXt (speed x time)
The distance traveled by the ball is 120 X 3 =360 m If it moves at 100m/s it will take 360/100=3.6 sec to hit the same object
When the bomb is dropped, the horizontal component of the velocity Vh= 500 m/s
The vertical component of initial velocity Uv=0 and the vertical component of final velocity
Vv=Uv+gt= 0+ 10(10)=100 m/s
Let the angle of the bomb with horizon be Ф ∴ tan Ф=vertical velocity/horizontal velocity= Vv/Vh=100/500=1/5
Ф= tan^-1(1/5)
Length of steel wire=length of copper wire=L Dimater of steel wire=2X Diameter of copper wire
∴ Area of cross-section of steel wire =4 X Area of cross section of copper wire As=4Ac
Change in length of steel wire ΔLs Change in length of copper wire = ΔLc
Force applied in both cases = F
Young modulus of steel wire Ys= (F/As)(L/ ΔLs)
Young modulus of copper wire Yc= (F/Ac)(L/ ΔLc)
Ys/Yc=2/1=( FXLXAcX ΔLc)/(FXLXAsX ΔLs)= ΔLc/4 ΔLs
∴ ΔLs/ ΔLc=1/8 Elongation ratio steel:copper=1:8
The small circle with centre Q will touch the big circle at B and M (M will be the centre of the big circle) Radius = AB/2
Radius of small circle with centre Q=AB/4
Efficiency= Power output/power input X 100= (11x90x100)/(220X5)= 90%
H = I2Rt If voltage and t is constant then H α R where R will vary with length and cross section area.
So the heat will be doubled if length and the radius of the wire is doubled
From Bernoulli's theorem, we know
1/2V^2=(P1-P2)/ρ where V is the velocity of fluid( water in this case) P1 & P2 are the pressures and ρ is the density of the fluid
V^2= 2(4.5 X 10^5-4.0 X 10^5)/10^3 m/s = 2 x 0.5 x 10^2=10^2 m/s
∴ V=10 m/s
From Ideal Gas Law we know PV = nRT or n = PV/RT
p1= 3 atm V1=1.65 L T1= (273+220) =493K P2=0.7 atm, T2= (273+110)=383K, V2= 1.0L
percent nitorgen to be released= (n1-n2)/n1X100=(1-n2/n1)X100
= [1- (0.7X1X493)/(3X1.65X383)]X100= 81.7%
Equation of the line QR= 2x+y=3 or y= -2x+3 so the slope of the line will be m1 = (-2)
In the Δ PQR, PQ=PR ∴ <PQR=<PRQ=45 Since the angle between QR and PQ=45
Tan 45= (m1-m2)/(1+m1m2) where m2= slope of PQ
1= (-2)-m2/1+(2)m2 or m2=3 and, Point P(2,1) is common to both PQ and PR
Eqn of line PQ= y-1/(x-2)=3 or 3x-y-5=0
Since PQ and PR are perpendicular to each other m2xm3=-1 (m3 is the slope of PR)
m3= -1/3 Eqn of line PR= y-1/x-2=-1/3 or 2x+3y-7=0
∴ Joint eqn of PQ and PR will be (3x-y-5)( 2x+3y-7 )=0 Multiplying it we get
6x^2-3y^2+7xy-31x-8y+35=0 which will represent the pair of lines PQ and PR
q1--------d--------------q2 .
A B
The electric potential energy of the system V1=q1q2/kd When the distance between the charges are doubled the new potential will be V2= q1q2/k2d Change in the potential ΔV= V1-V2= q1q2/k(1/d-1/2d)= q1q2/2kd
At 20°C, Length of copper rod= Measure in the scale=L
At 21°C Lenght of copper rod will be L+ ΔL= L+ LαΔT = L+Lα (ΔT=21°-20° =1) α is the coefficient of linear expansion of copper
At 21°C Length of the steel scale will be L+ ΔL= = L +Lβ where β is coefficient of linear expansion of steel
Considering the value of α at 20°C, as 17X10^-6 and β as 12 X 10^-6
At 21°C, the ratio of the length of the copper rod to the measure of the scale will be 1.000017/1.000012=1.000005
So the scale will measure the steel rod as 1.000005L
Floor to ceiling distance of the elevator= 2.7m
Acceleration of the elevator = 1.2 m/s^2 upwards
Relative acceleration of the bolt = (g+a) = 9.8+1.2 = 11m/s^2. ( the acceleration has to be added when the elevator is going up because it is accelerating against downward gravitational pull)
Let time t to reach the distance of 2.7 m
U = 0 s = 2.7m , a = 11 m/s^2
From the second equation
S = ut+1/2 at^2
S = ½ *11* t^2
5.4 = 11 t^2
t = 0.7 sec time taken to hit the floor will be 0.7s
The heat of neutralization between a strong acid and a strong base is -57.62kJ/mol
1 M of NaOH will be exactly neutralised with 1 M of HCl.
∴ 0.1M of NaOH will neutralize 0.1 M of HCl to give 57.62kJ x 0.1 = 5.76 KJ of heat
When a charged particle is moving through a magnetic field, a magnetic force acts on it. –This force has a maximum value when the charge moves perpendicularly to the magnetic field lines. –This force is zero when the charge moves along the field lines. So when current is passed through the coil a magnetic field will be induced in the coil. The direction of the force can be found by applying the "Right hand rule" and the intensity of the field can be calculated from Bio-Savart's Law
For walking 1 km the man needs 150KJ of energy. To walk 5 kms he needs 150X5=750 KJ energy.
If only 30% of energy is available for muscular work, the total energy needed will be 750/30X100= 2500KJ
1 mole of glucose produces 3000 KJ. Then for producing 2500 KJ, 5/6 mole of glucose =5/6X180 =150 gms of glucose will be needed (assuming molecular weight of glucose=180 gms)
The length of the pendulum is "l" and mass "g". suspended vertically in an electric field E and carrying a charge q
Gravitational force downward on the pendulum =mg Electric force upward on the pendulum = Eq
Since the electrostatic force is less than gravitational force the net force downward on the pendulum = (mg-Eq)
Net acceleration of the pendulum (mg-Eq)/m= (g-Eq/m)
Time period T =2π√(l/(g-Eq/m)
V1/T1=V2/T2 V1/(27+273)=1.18V1/(27n+273) or V1/300= 1.18V1/(27n+273)
or 27n + 273 = 354 or 27n =81 or n= 3
Height of the bird in air=X
Depth of fish in water =Y
The fish is viewing the bird in air from a denser medium water of refractive index μ (assume as the data is missing).
If the bird is viewed normally by the fish then
μ= (Apparent depth in air)/(Real depth in air) or Apparent depth in air= μX
Since the fish is Y m below, it will see the bird at a distance(μX +Y) from it
Percentage of Iron = 0.335% molecular weight of haemoglobin
Weight of Fe present = (0.335/100) ×16716= 55.9 gm
Molecular weight of approx 55.9 Therefore one molecule of iron weighs 55.9gm Therefore, 55.9/55.9 = 1 Molecular weight of iron= atomic weight of iron
Hence, 1 atom of iron is present in one molecule of haemoglobin.
From Law of mass action, we know Rate of reaction
R1=k[NO]^2[O2] where k is the rate constant.
When the volume of the vessel is doubled, the concentration of the reactants will be halved.
New rate R2= k(1/2[NO]^2)(1/2[O2}= 1/8(R1)
By doubling the volume, the rate of reaction will become 1/8th of the previous rate
Mass of trolley m=2kg., angular frequency ω= 10 rads/sec
When the turntable is rotated, the length of the spring becomes the radius of the circle along which the trolley moves. So r= 40 cm=0.4m
When the turntable is rotated, the tension in the spring is equal but opposite in direction to the centripetal force
∴ T=mv^2/r= mrω^2= 2X 0.4 X (10)^2 =80N
Extension produced in this spring =40-35=5 cm =,05m
Force constant of spring = Force/extension = 80/.05 N/m=1.6 X 10^3 N/m
The time period of the pendulum in the stationary lift is T1=2 π √l/g where l is the length of the pendulum and g is the acceleration due to gravity
When the lift start accelerating upwards
From Newton's 2nd Law we know (ΣF=ma) acting on the pendulum. The overall acceleration of the pendulum is upward (with the lift). So ma is positive (upward). The only external forces acting on the pendulum are the force of gravity acting down (-W=-mg) and the supporting Normal Force FN upward on the pendulum
. So ΣFN=ma(net)=-ma –mk (k=acceleration of the lift)_
∴a=−(g+k) In the problem it is given k= g/3
a=g+g/3=4g/3 ignoring the direction of the force
Let the new time period be T2 when the pendulum is accelerating with lift
T1/T2= √ [g/(4g/3)] or T2=T1√3g/4g or T2= √3/2T1
The potential at any point inside the sphere is the same as the potential at the surface ie 10V
When the ball is at h the PE1 of the ball is mgh
When the ball is pushed with a velocity v it reaches the bottom with KE1=1/2mv^2
Then by inertia the ball travels upto R when its PE2= mgR and the velocity =0 KE2=0
From law of conservation of energy
PE1+KE1=PE2+KE2 or mgh+1/2mv^2=mgR+0 solving it we get V= √2g(R-h)
Solution enclosed
Solution enclosed
As the car is moving at a constant velocity, its acceleration is zero. According to Newton’s second of motion the net force F=ma (mass of the car X acceleration of the car) Since a=0
Net force acting on the arc is zero.
We can find the electric field created by a point charge by using the equation E = kQ / r2 . (where K= 9 X 10^9 N m2/C2)
and Q=ne ( where n is the number of electrons and e is the charge of one electron = 1.6 X10^-19 C)
Therefore n= Er2./Ke=( 0.036 X 0.1 X 0.1)/(9 X 10^9 X 1.6 X10^-19 )= 2.5 X 10^5
For the above circuit, we get a balanced Wheatstone bridge, because 3Ώ/4Ώ=6Ώ/8Ώ. Hence the point R and S will be at the same potential. Hence no current will flow through the resistance 7Ώ. The network is therefore equivalent to two resistances (3+4)Ώ and (6+8)Ώ joined in parallel. Equivalent resistance R will be 1/R=1/7+1/14
∴ R= 14/3= 4.67Ώ
solution attached
Since the number that will divide 90 leaving a reminder 2 and 107 leaving a render 3, the number should be the H.C.F of the numbers (90-2)=88 and (107-3) = 104
88= 8 X 11
104 = 8 X 13
HCF= 8
∴The greatest number which divides 90 and 107 leaving remainder 2 and 3 respectively is 8.
solution
From Newton's law, we can derive Force, F=mg=GmM/r2 (where g is the acceleration due to gravity, G is universal gravitation constant, m is mass of the body and M is mass of earth)
∴ g= GM/r^2 Hence g/gnew= r^2/rnew^2 eqn -1
Since the radius of earth is reduced by 1%, rnew =0.99r
and rnew^2=(0.99r)^2=0.98r^2
Putting it in eqn 1 we get g/gnew=r^2/0.98r^2, gnew=gX(1.02)
∴ Percentage change in g = (1.02g-g)/gX100= 2%
Acceleration due to gravity will change by 2% for 1% reduction in earth's radius
Centripetal Force F=mω2r m=5kg, ω=2 rad/s r=1m Substituting the values we get
F=5X4X1= 20N
solution
The separation between two consecutive bright and dark fringes is same and is called the fringe width β
β = λD/d where λ is the wavelength of the light, D is the slit screen distance and d is the distance between the slits.
So fringe width is directly proportional to wavelength λ and the slit screen distance D and inversely proportional to distance d between the slits
Keeping all other factors constant
if the yellow light is replaced with red light since λr > λy (wavelength of red light greater than the wavelength of yellow light) the fringe width will increase
t1/2 of A= 20 mins and B= 40 mins
t=80 =4tA=2tB
After 80 mins, number of nuclei of A remaining is N0/2^4=N0/16 (where No is the initial number of nuclei of A, B)
After 80 mins, number of nuclei of B remaining is N0/2^2=N0/4
∴ Ratio of nuclei remaining of A and B = 1:4
The question needs to be corrected as "Find locus of midpoint of chord of the circle x2 + y2 = 4 subtending a right angle at the origin" Solution attached
Wrong quadratic equation roots are (-2) and (-15) and b= 17
Sum of the roots (-2) + (-15)= 17/a or a=1 Product of the roots are (-2) X (-15) =30 =c/a ∴ c=30
For correct equation b=13 Sum of the roots =-13/1=-13 Product of the roots = 30/1=30
So the roots are (-13+/-√(13^2-4(1)(30))/2(1) = (-13 + 7)/2 and (-13-7)/2 = (-3) and (-10)
Roots of correct equations are -3 and -10
When the food packet is dropped the vertical component of its velocity is zero and horizontal component is 720 km/hr.
Let the time taken by the food to reach ground be "t"
dvert = ut +1/2gt^2 or 396.6= (o)t + 1/2(9.8)t^2 or t^2= (396.9 X 2)/9.8= 81 ot t=9
dhor = vhor X t= (720 X 9 X 10^3)/3600= 1800
The food packet will take 9 secs to reach the ground and horizontal range is 1800 m
D is the correct option
The options need to be checked in the Question. Option A should be 0.0218 mol/dm3
Option C is the correct option
The race course will be an ellipse with flag posts as its foci. Major Axis a= 1/2 X 10 =5, Minor axis b= 3 (By applying Pythagoras theorem)
Area of the path enclosed =π X5 X3=15π=47.12 m2
Option C is the correct option
Option A is the correct option
Cannot be answered as the circuit diagram is missing
Option B is the correct option
By applying the theorem of parallel axes moment of Inertia "I" of a body is I=Ig + MR^2= where Ig is the moment of inertia about an axis passing through the centre of mass but perpendicular to the plane of the body plus MR^2 where M is the mass of the body and R is the distance between the axis and the centre of mass
I = 2 + 1X2^2=6 kgm2
Hence option B is the correct option
Out of pack of 52 cards one king and one queen can be drawn by(4C1)∗(4C1)=16 ways (assuming the order doesn't matter)
Out of Out of pack of 52 cards two cards can be drawn by 52C2= 1326 ways
Probability of getting a king and a queen will be 16/1326=8/663 ways
Hence option D is the correct option
solution
ignore earlier solution as the wrong file was attached Correct solution
solution
The direction of the magnetic field is out of the plane of the figure.
solution
Option B. It is an instrumental error due to wrong design or wrong calibration of the thermometer
Average weight of the group D cannot be determined as we do not know the average weight of each student. Data insufficient
Hence option D is the correct option
solution
Flux Ф through the sphere is given by the relation Ф = q/ε0 where q is the charge and
ε0 is the permittivity of the free space
The total flux through the closed sphere is independent of the radius of the sphere .
So the ratio of flux of sphere A and B will be 1 as the flux will be equal
The question has an error. The Charge q= 20 micro coloumb and not 20 C Then option B will be the correct option
Please find the solution
The instantaneous power for a pure capacitative circuit will be (VmXIm/2)Xcos(90)-(VmXIm/2)Xcos(2wt+90)
Average power will be zero
Please see the working for the details
Amount = 0.03 X 60 =1.8 g
Option A
Temperature coefficient μ = KT+10/KT
μ=2, K27= 10^-3
2=10^-3/K17 Hence K17= 10^-3/2= 5X10^-4
Option B
Buoyant Force (Force with which the water is pushing up the body) = weight in air - Weight in water = 5N-2N=3N
Option C
As the satellite is orbiting around earth (considering earth's orbit as circular rather than elliptical)
its centripetal force must be equal to Gravitational Force
Centripetal Force = mRoω^2= GMm/Ro^2 (ω= angular velocity of the satellite and G is the universal gravitational constant) so ω= √(GM/Ro^3)
Angular momentum of the satellite, J= mωRo^2= m√(GMRo)
A man with heterozygous blood group A will have the genotype AO married to a woman with heterozygous blood group B with genotype BO
A O
B AB BO
O AO OO
Chances of the first child having AB =1/4= 25%
Sara can finish the work in 12 Days. In I day she does 1/12 parts of the work
Bula can finish the work in 18 Days. In I day she does 1/18 parts of the work
Kamal can finish the work in d Days. In I day she does 1/d parts of the work
They together can complete in 4 days. In I day they can do 1/4 parts of the work
Hence 1/12+1/18+1/d=1/4 so 1/d= 1/4-1/12-1/18= 1/9
Hence Kamal can complete the work in 9 days Option A
For any element: Number of Protons = Atomic Number
Number of Electrons = Number of Protons = Atomic Number
Number of Neutrons = Mass Number - Atomic Number
Atomic mass =number of protons +number of neutrons
In 6C12,number of Protons,Electrons: 6
Number of Neutrons: 12-6=6
Mass of neutron is halved and the electron is doubled. Mass of electron is negligible and hence it will not contribute to any change
New atomic mass = 6/3(N) +6(P)=9 Hence atomic mass will reduce by 12-9=3
Reduction is 3/12X100= 25%
Option D
4f subshell has seven orbitals for a total of 14 electrons
ms, the spin quantum number defines the spin state of each electron. Since there are only two allowed values of spin, thus there can only be two electrons per orbital. The values of m=± 1/2 ie. 1 electron will have +1/2 value per orbitalHence 7 orbitals will have 7 electrons with ms +1/2
Option D
Option D 6 electrons refer to the attached explanation
Electrons in a particular subshell (such as s, p, d, or f) are defined by values of ℓ (0, 1, 2, or 3).
For 4s orbital mℓ=0
Option B
Wavelength of ray's in increasing order of λ
ɣ ray < X-ray <UV <IR <Microwave<Radio wave
Hence Option A i.e Radiowave will have maximum wavelength
Angular velocity ω=θ/t Where θ=2π ؞ ω1/ω2=θ/t 1/θ/t 2=t2/t1 Since t1=t2 time taken are same
ω1:ω2= 1:1 ie same angular velocity
Electrons can act like wave and particle known as wave-particle duality. Heisenberg uncertainty principle states that the exact position and momentum of an electron cannot be simultaneously determined because of wave-particle duality.
Hence option C
Aufbau principle states that the electrons will first occupy the orbitals that have the lowest energy
The order in which the electrons should be filled is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p…..
The order of the increasing energy of the orbitals can also be calculated using (n+l) rule.
If the value of n+l for two orbitals is equal then the orbital with lower value of n will have lower energy level, hence electron will first occupy it
Option A
The principal quantum number, n, refers to the energy level an electron placed in and to the size of the orbital
Option A
Option D 7:32
Ratio of oxygen: nitrogen = 1: 4= 1gm:4gm
Molecular weight of oxygen=32 and Nitrogen =28. 32 gm of Oxygen OR 28 gm of Nitrogen will contain N molecules
1gm Oxygen contains N/32 molecules of oxygen whereas 4 gms of nitrogen will contain 4N/28 molecules of nitrogen
Ratio of molecules of oxygen: molecules of nitrogen= N/32/4N/28= 7:32
Option C
Reason: The nucleus of an atom has a relatively small diameter compared with that of the atom. So most of the space in an atom is empty. Hence the most of alpha particles can pass straight through the metal foil
Option B the same in both the balloons
Reasons: Avogadro's principle states that at the same temperature and pressure equal volumes of all gases contain the same number of molecules. Both balloons are at same temperature and pressure and unit volume(=1)
(a) Force of friction = ½ ma = ½ (10 X 3.27) = 16.35N
(b).Hence, the work done is 0.
(c) The maximum allowed angle for rolling without slipping for cylinder rolling down the plane,θmax = tan-1(3µ)
= tan-1(3X0.25) = 36.86degree
Please refer to the solution below
λ = 10√(h/m)
Yes, the man will be able to land on the next building
Initial velocity of the car U= 126 km/hr = 126 X1000/(60X60)= 35 m/s
Since the car is stopping, the final velocity= 0
(i) V^2-U^2=2aS (S=200 m given) a= (35)^2/2(200)= -3.06 m/s2
Retardation of the car = 3.06 ms-2
(ii) V=U+at 0= 35 -3.06t or t= 35/3.06 = 11.42 s
The car will take 11.44 s to stop
Solution
Solution
solution
solution
initial velocity of the packet , u = 4.9 m/s final velocity = v m/s height, h = 245 m g = 9.8 m/s² time taken = t seconds h = ut + 0.5gt² [ since u is upward it is taken as negative) 245=-4.9t + 0.5(9.8)t² ⇒245 = -4.9t + 0.5×(9.8)×t² ⇒ 245 =- 4.9t + 4.9t² ⇒ 4.9t² -4.9t -245 =0 ⇒ t² -t -50 =0 Solving it, we get t = 7.59s 0R t = -6.59s (discarded since time can't be negative)
v = u + gt = -4.9 +9.8(7.59) = -4.9 +74.38 = 69.48 m/s So final velocity is 69.48 m/s downward
solution
Option C is the correct answer
Let nA and nB be the frequency of the organ pipes A and B respectively.
We have, nA= v/4lA and nB= v/4lB where v=vlocity of sound in air, lA is the length of the air column resonating with pipe A and lB is the length resonating with pipe B.
and nA/nB= lB/lA = 51/50 (given) eqn .....(1)
Since both pipes together produce 5 beats per second, nA-nB = 5 eqn......(2)
By solving eqn (1) and (2) we get nA=255 Hz and nB=250 Hz
Hence option D is the correct answer
Let the volume of alcohol in the mixture be 4x liters. Since Alcohol: water= 4:3, there will be 3x liters of water in the mixture.
Now 5 liters of water is added and the new ratio becomes 4:5
4x/(3x+5)= 4/5 by solving we get x= 5/2.
Volume of alcohol in the mixture = 4(5/2)= 10 liters. Hence option A is the correct option
Let Tushar's speed be x km/hr and Ashish be y km/hr.
To cover 40 kms. Tushar took 40/x hrs and Asish took 40/y hrs.Since Tushar took 2 hours more than Ashish,
40/x - 40/y = 2 eqn ....(1)
If Tushar doubles his speed, then he took 40/2xhours to cover the same distance. Since he took 1 hour less than Ashish,
40/y-40/2x=1 eqn.....(2) Solving the eqns we get
x=20/3 = 6.67 and y=10. Therefore Tushar's speed is 6.67 kms/hr
Let A = amount, P = principal, r= rate of interest compounded per annum n=time in years
By using compound interest formula we get 2A= A(1+r)^4 or (1+r)^4=2 eqn...(1)
Let the amount become 8 times in 't' years Therefore 8A=A(1+r)^t or (1+r)^t= 8 =2^3....eqn(2)
(1+r) = 2^1/4= 2^3/t ∴ 1/4= 3/t or t =12
Option D is the correct answer
Cannizzaro reaction is a chemical reaction of benzaldehyde with potassium hydroxide to produce benzyl alcohol
2 C6H5CHO + KOH → C6H5CH2OH (benzyl alcohol) + C6H5COOK
Option B
According to the Einstein’s Photoelectric equation
Ek=hν−Ф Ek is the kinetic energy of the electron, h is Planck's constant, ν is the frequency and Ф is the stopping potential
Compare it with equation of a straight line y = mx + c
Ek and v (frequency) are variables. The slope of the line m is plank’s constant (h).
Thus, the slope is same for all metals and independent of the intensity of radiation.
The correct option is D.
Atomizers use Bernoulli's principle.The pressure inside an atomizer is relatively high. At one end of the horizontal tube is a squeeze-pump which causes air to flow through it. Squeezing of the bulb over the fluid creates a low-pressure area due to the higher speed of the air, As the pressure toward the top of the bottle is reduced, the perfume flows along the tube, drawn from the higher pressure area at the bottom which subsequently draws the fluid up.
The voltage of the capacitorC1 is V0= q1/C1 and the voltage of the uncharged capacitor C2 is 0 as its charge is 0.
When the two capacitors are electrically connected then the charge will flow from higher potential to lower potential till the difference of potential ceases to exist. Let the common potential of C1 and C2 be V. Assuming no loss charge during the charge sharing process, Total charge before connection = Total charge after connection
i.e C1V0 + 0 = C1V + C2V = (C1+C2)V or V= Vo C1/(C1+C2)
Final potential V= Vo C1/(C1+C2)
Since the Earth is flattened at the poles and bulges at the equator, an oblate spheroid the distance from Earth's center to sea level is roughly 21 kilometers (13 miles) greater at the equator than at the poles. Since the value of g varies inversely with the square of the earth's radius, the shorter radius at the pole will result in a greater value of g.
The second reason is that at latitudes nearer the Equator, the outward centrifugal force produced by Earth's rotation is larger than at polar latitudes. This counteracts the Earth's gravity to a small degree – up to a maximum of 0.3% at the Equator – and reduces the value of g at the equator.
So as we move from equator to pole value of g will increase. Hence option C is the correct option
It is caused by thin film interference which depends on the thickness of the film. Option C is the correct option
The pulmonary artery carries blood from the right ventricle of the heart to the lungs for oxygenation. Its name is derived from the Spanish word 'pulmon' meaning lungs
option A
Option D Molasses (a by-product of the refining of sugar beets and sugar cane) with very high sugar concentration is used in industrial production of baker's yeast because yeast needs sugar for growth. Molasses contains a mixture of sucrose, fructose, and glucose, (different forms of sugar) which are readily fermentable and the yeast hydrolyzes molasses to fermentable sugar which is required for its growth.
solution
Option C is the correct answer
Total age of all students =22×21 Let the age of the teacher be t years Total age of all students + Age of the teacher =21×22 + t ؞ The new average with teacher is (21×22 + t)/23
and the problem states that the new average with the teacher is increased by 1 i.e new average = 21+1=22
(21×22 + t)/23 = 22 or t= 44
Age of the teacher is 44 years
Let the intensities of the light sources be I1 and I2 and amplitude be a1 and a2
The intensity of a wave is proportional to the square of its amplitude i.e. I =a^2
؞ I1/I2 = (a1/a2)^2 = 100/1 (given) a12=100 or a1=10 (a2)^2= 1 or a^2=1
Imax/Imin= (a+1)2/(a-1)2= (10+1)2/(10-1)2= 11^2/9^2=121/81
Ratio of Imax:Imin= 121:81
The question seems to be wrong as the question states "number grow by x during the year" where it should be number grow by x% during the year
Let us assume the value of x to be 10%. Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million. i.e., 0.1 million sheep have died in 2001. In terms of the percentage of the number of sheep alive at the beginning of 2001, it will be (0.1/1.1)*100 % = 9.09%. Hence x > y. option A will be the correct answer
However, if we take increase was by x number (as mentioned in the problem) and decrease by y number then
The number of sheep in the herd at the beginning of the year 2001 (end of 2000) will be 1 million + x = (1,000,000+x)
In 2001, the numbers decrease by y and at the end of the year the number sheep in the herd = 1 million. then y number of sheep died
In that case 1,000,000+x-y =1,000,000 or x=y
Then option B will be the correct answer
Correct answer is B Luminal. Chemical name for Luminal is Phenobarbitol; Phenylethylbarbiturate; Phenobarbituric acid . Phenobarbital is a barbituric acid derivative that acts as a nonselective central nervous system depressant and falls under the group of hypnotic.
Phenacetin is a pain relieving and fever relieving drug and it falls under the group called NSAID (Non-steroidal anti-inflammatory drug)
Quinine belongs to a class of drugs known as antimalarials. It is used for the treatment of malaria.
Penicillin is a group of antibiotics used for treating infections caused by susceptible gram-positive bacteria like Streptococci, Staphylococci, Clostridium, and Listeria genera
Volume of HCl V1=35.4 ml Normality of HCl =N1
Molar mass of NaOH = 40 gm Normality of NaOH is N2= 0.275/40 N. Let the volume of NaOH be V2= 1000 ml
N1V1=N2V2 or N1X 35.4 = 0.275 X 1000/40 or N1= (0.275 X100)/(35.4X40) =0.194 N
Option C- 0.194 N is the correct answer
If each wager is for half the money remaining, each win lead to multiplying the amount by 1.5 times and each loss multiplying the amount by 0.5 times.
the person won three times and lost three times. So we need to multiply the initial amount by 1.5 thrice and the loss by 0.5 times thrice in any order.
So the final amount will be Rs. 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5) = Rs 27 (ina ll caes)
The loss will be 64-27 =37
option B will be the correct option
Mass of moon = M, Mass of earth =81M Distance between earth and moon be d =60R, R=radius of earth
Let y be the distance from the moon where the gravitational force will be zero
At y, the gravitational pull by earth will be equal to that of moon
GM/y^2= 81GM/(d-y)^2 or 81y^2= (d-y)^2 or d-y=9y or d=10y.
So 60R=10y or y=6R
Option C is the correct option
(4,p)X______________________Y (0,q) Slope,3/4 = (p-q)/(4-0) or p-q=3
Option C is the correct answer
OF2 dissolves in water but does not give any oxyacid solution. OF2 + H2O → 2HF + O2
Option B is the correct answer
P1V1/T1=P2V2/T2 P1=P, p2= 2 atms V2=2V1 since the concentration was halved.
T1=327+273=600 and T2= 427+273=700 P1V1/600=2(2V1)/700 or p1= 6X4/7=24/7
Option B is the correct answer
Ferric hydroxide Fe(oH)3 is a positively charged solution. To coagulate we need a negatively charged solution.
By Herdy-Schulz rule -Coagulating power of an electrolyte is directly proportional to the valency of the active ions (ions causing coagulation).
Similarly, to coagulate a positive solution, such as Fe(oH)3 the coagulating power of different anions has been found to decrease in the order : [Fe(CN)6]4- > [CrO4] 2- > Cl-
Option D is the correct answer
800 gm of solution X contains 40% solute. ie it contains 40/100*800=320 gm of solute X
After colling 100 gm of solute is precipitated. Therefore 220 gms of solute X in the solution
Percentage of the solute X in remaining solution is 220/700*100 = 31.4%
Option A is the correct answer
Option B is the correct answer
H2I+2BOH⇒2BI+2H2O=ΔH=−285.77kJ/mol
H+OH–⇒H2O;ΔH==−56.07kJ/mol
The heat of ionization or the energy of dissociation in this reaction is equal to (–285.77 + 56.07) kJ/mol =-229.7 KJ/mol
Option B
option B is the correct answer
Option B is the correct answer
solution
Case 1 Probability of picking up a white marble = 5/15=1/3
Case-2 x white marbles were added.
Probability of selecting white marble = (5+x)/(15+x)=3/5 or 25+5x= 45+3x or x=10
Option D
The number set of two dice adding to 10 or less than 10 are as follows
(1,1]) (1.2).......(1,6)
(2,1), (2.2)....(2,6)
(3,1), (3,2)....(3,6)
(4,1), (4,2)....(4,6)
(5,1) (5,2)....(5,5)
(6,1), (6,2)...(6,4)
P= 33/36=11/12
Option B
The problem is misprinted. The problem should be " if α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) =2ac/(a² - c²) " and not 2ac/(2a² - c²)
Solution
a tan θ + b sec θ = c (given)
Rearranging the given equation we get, atan(θ) - c = -bsec(θ) Squaring both sides, a²tan²θ - 2ac*tan(θ) + c² = b²sec²θ Then Using the identity, sec²θ = 1 + tan²θ and simplifying the above, it reduces to: (a² - b²)tan²θ - 2actan(θ) + (c² - b²) = 0 The above is a quadratic in tan(θ), which has two roots - tan(α) & tan(β) So applying properties of quadratic equation, Sum of roots: tan(α) + tan(β) = 2ac/(a² - b²) -------- (1) and Product of roots: tan(α)*tan(β) = (c² - b²)/(a² - b²) -------- (2) Therefore, tan(α + β) = {tan(α) + tan(β)}/{1 - tan(α)*tan(β)} Substituting the values from (1) & (2) above, tan(α + β) = [2ac/(a² - b²)]/[1 - (c² - b²)/(a² - b²)] = 2ac/(a² - c²) Thus, tan(α + β) = 2ac/(a² - c²)
Let nC(r-1) = 45, nCr = 120 and nC(r+1) = 210
=> nCr / nC(r-1) = 120/45=8/3 and nC(r+1)/nCr = 210/120=7/4
nCr / nC(r-1)=[ ni /(r)i (n-r)i ]/[ni /(r-1)i (n-r+1)i] = (n-r+1)/r = 8/3 and (n-r)/(r+1) = 7/4
=> 4n -11r =7… eq-1 and 3n - 11r =- 3… eq-2
Subtracting 2nd eqn. from the 1st,
n = 10 and r = 3
To reach the path represented by the equation 6x - 7y+ 8=0 in the shortest possible time, the family has to travel through the path which is perpendicular to the path 6x - 7y + 8=0 and passes through the intersection of the paths whose equations are 2x - 3y - 4 =0 and 3 x + 4y -5 =0
To get the point of intersection of the two lines: 2x - 3y - 4 =0 and 3 x + 4y -5 =0, we need to solve the equation which will give the the point of intersection,O (x1,y1)
2x1 - 3y1 - 4 =0 ....line1
3x1 + 4y1 -5 =0......line 2
x1=31/17 and y1=(-2/17)
The slope of the given line(line-3) is 6x - 7 y + 8=0 , m1=6/7 The slope of the required line(shortest path) will be m2= -7/6 ,ince the shortest path will be perpendicular to line 3 and product of slopes of perpendicular lines are -1
The equation of the required line is : (y - y1)/(x - x1)=- 7/6
⇒ [y-(-2/17)]/(x -31/7)= (-7/6)
⇒ (17y+2)/(17x-31)=(-7/6)
⇒ 119x +102y= 205
Hence the equation of the path they should follow is 119x +102y- 205=0
Option D is the correct answer
Let the one end of the diameter be (x,y). The center (2,-5) is the midpoint of the diameter, whose other end is (-2,-3)
By using the midpoint formula
x+(-2)/2=2 and y+(-3)/2=-5 => x=6 and y = -7
Coordinates of the other end of the diameter is (6,-7)
Option A is the correct answer
solution
Solution
Option A is the correct answer
Option B is the correct answer
Option B is the correct answer
Option D is the correct answer
Solution
Solution
Option C is the correct answer
Solution
Let AB= x and AC=y
Tan30 = x/2a => 1/√3= x/2a => x=2a/√3
Tan60=y/a => √3 =y/a => y= a√3
AB/BC = x/(y-x) => (2a/√3)/(√3a-2a/√3) = 2:1
Option D is the correct answer
Let h be the height of the tower and x be the length of the shadow when the sun is at 60°
tan60 = h/x => x= h/√3 and tan 30= h/(x+25) => 1/√3 = h/(x+25) => 3h = h + 25√3 => 2h= 25√3
=> h = 25√3/2
Sorry the wrong solution got attached please see the correct solution
Option C is the correct option
x2 + y2 + z2 = 8R2 + 2r2
The wheel moves at 140 rev/min. In 1 hour it is 140 X 60 rev.
In I rev the wheel will travel 2πr cm = 2X 22/7X 35 cm = 2X 22/7X 35 X10-5 kms
In 1 hour it will travel 140 X 60 X2X 22/7X 35 X10-5 = 18,480 X 10-3 Kms = 18.48 Kms
Option B is the correct answer
P will divide the line segment CD in 2:3 ratio and y=6
If all the three dices are rolled simultaneously, they can fall in 6^3 =216 ways. If the same number has to appear in all the three dices, then the possible combinations are as follows.
(1,1,1), (2,2,2), (3.3,3), (4,4,4), (5,5,5) and (6,6,6)= 6 combinations
Probability of getting the same number on three dices are 6/216=1/36
Hence option B is the correct answer
Tangents drawn from an external point to a circle are always equal.
The quadrilateral, ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. Therefore A,B,C and D are external points from which tangents AS=AP=5, BP=BQ, CQ=CR=3 and DR=DS are drawn
BC=7. => BQ + CQ =BC=7 => BQ= 7-3=4 ∴ BP=4
AB=x, => AP +BP = x => 5+4 =x => x=9
Option B is the correct answer
When a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.
The diameter of the circle is 28cm. Area of the circle is 22/7X14X14 sqcm= 22X28 sq. cm.
The area of the square is 28 X28 sq. cm
∴ The are inscribed between square and circle is 28X28-22X28= 28(28-22) =28X6= 168 sq. cm
Option D is the correct answer
Bhairav (B) has given 256 meters(m) lead to Akshyay(A)
B will cover 1600 m, A will cover 1600-256=1344m. Let the time taken be t to complete the race and speed of B and A be Vb and Va
∴ 1344/Va=1600/Vb => Va/Vb= 1600/1344= 21/25......(i)
Similarly, B Covers 1600m when C covers 1600-64= 1536 m
∴ 1536/Vc= 1600/Vb => Vb/Vc= 1600/1536= 25/24
∴ Va:Vb:Vc= 21:25:24 => Va:Vc=21:24
∴ Speed of A is Va=21x and speed of C Vc=24x
Time taken by C to complete one and a half mile race= (1600X1.5)=2400m be T
T= 2400/24x=100/x. In the same time A will run 100/x*21x=2100 m
∴ The winner will be Chinmay C by a lead of 2400-2100=300m= 300/1600 mile =3/16mile
Option B is the correct answer
The number of visitors in January was 960.
The average number of visitors January-July is 870
∴ Total number of visitors January-July is 870 X 7= 6,090.
∴ Total number of visitors February-July =6,090-960= 5,130
Actual number of visitors in August was 827
∴ Total number of visitors February -August =5,130+827= 5,957
∴ Average number of visitors February-August= 5,957/7=851
Hence Option B is the correct answer
The question has missing data. It needs to specify "how long the pipe was opened every day?"
Assuming it is opened for 1 hour every day, then the solution will be as follows
In 1 hour P will fill up 1/12th of the tank and in 1 hour Q will fill up 1/18th of the tank.
In 2 days (1/12 + 1/18)= 5/36 parts of the tank will be filled.
The tank will be filled in 36/5*2= 72/5=14.4 days
Option D is the correct answer.
Option B is the correct answer
Option B
We know that acceleration due to gravity, g on the surface of the Earth: g = GM/r2
Where M is the mass of the Earth, r the radius of the Earth (or distance between the center of the Earth and point on the surface of the earth), and G is the gravitational constant. i.e. g ∝ 1/r2
The acceleration due to gravity at a height h above the surface of the earth is g'
g/g' = (r+h)2/r2 => g' = gr2/(r+h)2 => g' = g(6400)2/6432)2 = 0.99g
Option A is the correct answer
The problem needs to be corrected as "If the number of water bottles sold is 1/3rd of the number of cold drinks sold, Help me to find the total number of cold drinks sold".
Solution
Let the number of cold drinks sold be '3x', then the number of water bottles sold will be 1/3*3x=x
Let the price of each bottle of cold drink be 'c' and each water bottle be 'w'
Total collection from the sales of cold drinks and water botlle= (3cx+wx)
60% of the collection is from cold drinks(given)
∴ 60/100*(3cx+wx) = 3cx => 10cx= 6cx+2wx => 4cx = 2wx => w = 2c
Had 30 more cold drinks being sold, its contribution would have been 80%
∴ (3x+30)c = 80/100[(3x+30)c + wx] => 15cx + 150c = 12cx + 120c + 4wx
=> 15cx + 150c = 12cx + 120c + 4(2c)x => 5cx = 30c => x = 6
∴ Number of cold drinks sold, 3x = 3(6) =18
Option C is the correct answer
Let the body be projected at an angle θ to the horizontal with a velocity 'u'. The vertical height attained by the body is given by H= u2(Sinθ)2/2g Now it is given that H1=H2
∴ (u1)2(Sinθ1)2/2g = (u2)2(Sinθ2)2/2g => u1/u2 = Sinθ2/Sinθ1 => u1/u2 = sin60/sin45 = (√3 /2)/(1/√2)
=> u1/u2 = (√3/√2) =√(3/2)
Ratio of their velocities will be √3:√2
The given options are misprinted and hence none of the optionis correct
Let the radius of each semi-circle be 'r'. Then the side of the square will be 2r
From the given figure we can calculate, 4+r+2r+r+2=18
=> 4r = 12 => r =3
∴ Area of shaded semi-circles will be 4*πr2/2 = 2πr2= 2*(3.14)*9= 56.57 sq. cm
and area of the square is 18*18 =324 sq.cm
∴ Area of the unshaded region is (324-56.57) = 267.43 sq.cm
Hence Option D is the correct answer
Solution
Option B is the correct answer
Cylinder
Diameter= 6m Radius r = 3m Height h1= 5m. Volume of cylinder V1= πr2h1 = (22/7)*3*3*5 m3 = 141.43m3
Conical lid
Diameter= 6m Radius r = 3m Height h2= 2m. Volume of cone V2= 1/3πr2h2= (1/3)*(22/7)*3*3*2= 18.86 m3
Volume of the solid = Volume of Cylinder(V1)+ Volume of conical lid (V2)= 141.43 + 18.86 = 160.29 m3
hence Option D is the correct answer
The height of the cylinder h=8 cm. Total surface area=528 cm2 Let r be the radius of the cyclinder
Total surface area of the cylinder is (2 π * r2 + 2 π * r * h) = 2 π r(r+h)
2 π r(r+h) =528 (given) => 2*(22/7)*r(r+8) = 528 => r2 +8r = 84 => r2 +8r - 84 = 0 => (r+ 14)(r-6) = 0
=> r = 6 OR r = -14 (discarded as radius can't be negative)
؞ Volume of the cylinder V= π * r2*h
The cylinder is melted and recasted into 'n' number of spheres of radius r2=2 cm
Volume of n spheres = 4/3*π * r3*n . Euating the volume of cylinder and spheres we get
π *6*6*8 = 4/3 *π*2*2*2*n => n = 27
27 spheres can be obtained
Hence option D is the correct option
Volume of hollow cylinder V = π(R2-r2)h where R= radius of the outer surface and r= radius of the inner surface
Volume of each cone v= 1/3π*(r1)2 *h2 where r2 = radius of the cone and h2=height of the cone
Let the cylinder be melted into 'n' number of cones. Then by equating the volume of hollow cylinder and cone we get,
π(R2-r2)h = n* 4/3π*(r1)3 => n = (100-64)X5X3/(3*3*5) => n = 12
Option B is the correct answer
The ratio of the radius and height of the cone is 3:4. Let the radius be 3x and height be 4x
Volume of the cone V = 1/3πr2h => 1/3(3.14)(3x)2(4x)= 301.44 => 1/3*3.14*9*4x3 = 301.44
=> x3 = 8 => x=2
Curved suraface area of the cone =πrl (l= slatn height) , l = √(r2+h2) = √(9x2+16x2)= 5x
πrl= 3.14*3x*5x= 3.14*15*4 = 188.4 cm2
Option D is the correct answer
First Train: Speed u1= 15 m/s. Retardation a1= (-1) m/s Final velocity =0; Distance travelled before stopping be s1
SecondTrain: Speed u2= 20 m/s. Retardation a1= (-1) m/s Final velocity =0; Distance travelled before stopping be s2
0 -(u1)^2 = 2 (-1) s1 => -(15)^2 = -2s1 => s1 = 225/2 = 112.5 m
0 -(u2)^2 = 2 (-1) s2 => -(20)^2 = -2s2 => s2 = 400/2 = 200 m
The distance between them after they stop = [500-(112.5+200)] = 187.5 m
Option C is the correct answer
=>
Velocity is the change in the position of a body with time
The height y (vertical distance) along the y-axis is given by y = (8t – 5t2) So we can get the vertical velocity by differentiating it over time. i.e. dy/dt =d/dt(8t-5t2) = 8-10t. At t=0, Vy=dy/dt = 8-10(0) = 8
The horizontal distance 'x' is given by x=6t ؞ Horizontal velocity Vx = dx/dt = d/dt(6t) = 6
The velocity (vector) with which the projectile is projected is 8i + 6j = √ 8^2+6^2 = √ 100 = +/- 10.
Since velocity cannot be negative Option C -10m/s is the correct answer
Option C is the correct answer
For the X- direction, Final velocity Vx, Initial velocity Ux,= 8 m/s, Acceleration a= 2 m/s2
Vx =Ux + at => Vx = 8 + 2t . therefore, velocity vector in x-direction Vx = (8 + 2t)i........(i)
For the Y- direction, Final velocity Vy, Initial velocity Uy,= -15 m/s, Acceleration a= 0
Vy =-15 + (o)t => Vy = -15 therefore, velocity vector in x-direction Vy = -15j ....(ii)
The total velocity vector at time t will be V = Vx +Vy = (8 + 2t)i. + (-15)j.) = √ ((8 + 2t)2 + (-15)2)
and at t=0 it will be √ (64 + 225) = √289 = 17 m/s
Option A is the correct answer
Resistance of a wire is given by R=ρ.L/A, where ρ is the specific resistance of the wire,
A is area of cross-section of the wire, and L is length of the wire and If V is volume of the wire
Volume V = A*L => A = V / L, Then ρ = A R / L = V R / L²
Hence R=ρ. L² /V …….(i)
If d is density and m is the mass of the wire, then R=ρ. L²d /m
Let mass of aluminum be mA and that of copper be mc
∴ Ratio of mass of aluminum: mass of copper = mA /mc = (ρA.LA²dA)/ (ρc.Lc²dc)
But LA= Lc , ρA=2 ρC , and dc = 3dA
∴ mA /mc = (2/1)* (1/3) = 2/3
Hence option B is the correct answer
Let the curve be represented by the function, y = f(x)=3x4 −16x3 +24x2 +37
Differentiating the function we get
dy/dx = 12x3 −48x2+48x For maxima and minima dy/dx = 0 ⟹12x3 −48x2+48x =0 ⟹12x(x2−4x+4)=0 ⟹ x(x−2)2=0
⟹x=0, 2 The second derivative d2y/dx2 will give the critical points d2y/dx2 =36x2−96x+48 at x=0, d2y/dx2 =48 since d2y/dx2 =48>0 it means that x=0 is local minima. but at x=2, d2y/dx2 =0 so we can't apply second derivative test.
, we can only apply first derivative test.
⟹ dy/dx is negative for all x < 0 and positive for all x > 0
Therefore y=f(x) has only one minimum (extremum) at x=0
Hence option B is the correct answer
OptionB
solution
Option C is the correct answer
Option A is the correct answer
According to Faraday's law of electrolysis, as the current passed will be same,
Weight of silver deposited/equivalent weight of silver = Weight of Hydrogen liberated/equivalent weight of hydrogen
=> 1.08/108 = Weight of Hydrogen liberated / 1 => Weight of Hydrogen liberated = 0.01 gm
Now, 2 gms of hydrogen occupies 22,400 mL at STP
Therefore, 0.01 gm of hydrogen will occupy (22,400 X 0.01) /2 = 112 mL at STP
Option B is the correct answer
Initial concentration of A was [A]0 = 4[b]0, 4 times the initial concentration of B; Half-life of A is 5 minutes.
A's concentration will decrease as follows 4[b]0 => 2[b]0=> [b]0 => 0.5[b]0 after each half-life
B's concentration will decrease as follows [b]0 => 0.5[b]0 Half-life of B is 15 minutes.
Hence after one-half life of B = 3 half-lives of A = 15 munites concentration of A and B will be equal.
Option A is the correct answer
The velocity of an incompressible liquid is inversely proportional to the area of the circular pipe.
A1V1 = A2V2 where A is the area of the cross-section of the pipe and V is the velocity of the liquid.
V2 = (A1V1)/A2 = [π(3)^2/π(1.5)^2] X 2 = 8 m/s
Hence option C is the correct answer
(i) Actual length of the bacteria is 5/50,000= 0.0001 cms
(ii) Length after 20,000 enlargement will be 2 cms
Option D is the correct answer
There are total 3+4+5 =12 points. To form a Δ we need to choose 3 points, except the ones which are collinear.
Side AB has 3, side BC has 4 and side AC has 5 collinear points.
hence the number of Δ s formed are 12C3- 3C3-4C3-5C3 = 220 -1 -4 -10 = 205
Option D is the correct answer
Insufficient data. Cannot be solved
Body-1 => mass= m1=40 kg, velocity u1= 4 m/s
Body-2 => mass= m2=60 kg, velocity u2= 2 m/s
For perfectly inelastic collision, Loss in kinetic energy = (1/2)m1m2/(m1+m2)(u1-u2)^2 = 0.5 40*60(40+60)*(4-2)^2
= 48J
hence, option B is the correct answer
Use approximation to convert the decimals to nearest whole numbers
(11.92)^2 + (16.01)^2 = ?^2 × (3.85)^2 ͌ (12)^2 + (16)^2 = ?^2 × (4)^2 ͌ 144 + 256/16= 25 = 5^2
? = 5
Option D is the correct answer
option A
The equation for photon energy is E = hc/λ i.e E α 1/λ => Eaλa = Ebλb => Eb = 2.1 X 10-18 *λ/3λ = 0.7 X 10-18
= 7.0 X 10-19 J
option B is the correct answer
C.P of VCR = C.P of TV = 8,000. 4% loss on VCR. S.P. of VCR =96% of Rs. 8,000 = Rs. 7,680.
8% gain on TV. S.P. of TV =108% of Rs. 8,000 = Rs. 8640.
Total S.P. = 16,320 Total C.P. 8,000 + 8,000 = 16,000
Profit/Loss % = (S.P. - C.P.)/C.P/ X 100 = 2%
The shopkeeper made 2% profit on entire transaction
C.P = 15,500. Repairs =Rs 450. Total cost = Rs. (15,500 + 450) = Rs. 15,950. Sold for 15% profit
S.P = 1.15 X 15,950 = 18,342.5
Selling price of the article = Rs.18,342.50
mass of bullet= m1 = 10gms = 0.01 kg. initial velocity of bullet = u1= 1000m/s. final velocity of bullet v1=400m/s
mass of block= m2 = 5 kg. initial velocity of block = 0. final velocity of block v2
By applying conservation of momentum, we get m1u1= m1v1 + m2v2 => v2 = m1/m2*(u1-v1)= (0.01/5)X 600.
The height traveled by the block h = V2^2/2g ( g=9.8 m/s2) = (0.01X 600/5)^2/(2X9.8) = 0.073 m
Option B is the correct answer
Option C is the correct answer
Let A's share at present be a and B's share at present be b
The problem states that a(1+20/100)^3 = b(1+20/100)^5 => a/b = (120/100)^5-3 => a/b = (6/5)^2 => a/b = 36/25
؞ a= 36/(36+25)*5500 = 36/61*5500= 3246. b= 25/(36+25)*5500= 2254.
hence A's present share should be Rs 3246 and B's share Rs 2254.
None of the options is correct.
Also, note that if A's share in 3 years has to be equal to B's share in 5 years with the same rate of interest, then A's present share has to be greater than B's present share. In all the options except option number A, B's present share is given as more than that of A's present share and hence cannot be correct
Option C i.e. velocity of river's water= 3km/hr is the correct answer
Let |z1| = |1 + √3i|^5 and |z2| = |1 - √3i|^5. (1 + √3i)^5 + 1 - √3i)^5 = (√(1 + 3)^5 + √(1 + 3)^5 = 2^5 + 2^5 = 64
Hence option D is the correct answer
solution
When we multiply decimals, the decimal point is placed in the product so that the number of decimal places in the product is the sum of the decimal places in the factors.
Let’s compare two multiplication problems that look similar: 2.04 X 1.94 and 2.04 and 1.95
2.04 X 1.94 = 3.9576 Add 2 decimals of 2.04 and 2 decimals of 1.94 = 4 decimal places.
however, if the last digit of the product of the last two digit is 0, then we need to reduce one decimal place as in case of 2.04 X 1.95 = 3.9780 but the answer can omit the final 0 after decimal places.
hence in the above problem without actually multiplying we can say that number of decimal place will be 5+3 =8 but the product of last digits 8 and 5 being 40, ending with 0, we can omit the 0 and reduce the number of decimal places by 1
Hence it will be 8-1=7 option C
Radius of disc 'r' =20 cm and angular velocity ω = 20 rad/s
Then linear velocity of point A which is on the rim, v = rω = 20*20 = 400 cms/s = 4 m/s
option D
The velocity of the ultrasound wave, v, is the product of its wavelength, λ , and its frequency, f.
Wavelength of a wave = velocity of the wave / frequency of the wave
λ = v/f, v = 1.6 km/s = 1.6 X 10^3 m/s, f = 3.2MHz = 3.2 X 10^6 Hz
λ = (1.6* 10^3)/f(3.2*10^6) = 0.5*10^-3 m = 0.5 mm
Option B is the correct answer
Slope of the graph is m= -5/6 m/s2 per sec. Intercept c = 5 m/s2.By using the eqn. y =mx + c , we can see that the acceleration varies with time as a= -5/6t + 5 where 't' is the time.
a= dv/dt = -5/6t + 5 Integrating ⌠dv (from 0 to v) = ⌠(-5/6t +5) dt (from 0 to t), we get
=> v = -5/12t^2 +5t From the graph, we can see that acceleration is decreasing with time and becomes 0 at t=6s. when v will be vmax. Using t=6 we get vmax = -5/12(36) + 5(6) = -15 + 30 =15 m/s
Hence option B is the correct answer
The question cannot be answered because the location of point D is not specified. Only if AOD is a straight line with O as the center of the circle, then BD = DC = a/√3 Option E
Option E is the correct answer (Proof follows)
OE is perpendicular to AB ؞ <BOE = 90o. But EOC = 60o ؞<BOC =30o
In Δ AOE, AO=EO (both are radius) <AOE =90o ؞ <OAE = <AEO = 45o
In Δ BOC, BO=CO (both are radius) <AOE =90o <BOC = 30o ؞ <OCB = <OBC = 75o
Now in Δ ABD, <BAD=45o and <ABD=75o ؞ <ADB = 180-(45+75) = 60o
؞ <EDB + <BOC = 60o + 30o = 90o
Option A is the correct answer
option C is the correct answer
Option D
Option C
Q_____a________q______a________Q
Let the charge q be placed at the midpoint of the line joining Q and Q
For a system of charges to be in equilibrium, the net force on all the three charges must be zero. F=0 Hence the charge q must be negative so that forces due to the other two charges will act in opposite direction.
The force on one of the charge Q by the other charge Q is KQ^2/(2a)^2 and the force on the same charge due to q is KQq/a^2
F=KQ^2/4a^2+ KQq/a^2 = 0 ⇒ q = -Q/4 The magnitude of the force will be Q/4. Hence option B is correct
Option C is the correct answer
Option D
Time taken by the bullet to cover to cover 100 meter horizontally = d/v = 100/500 = 0.2 s.
In 0.2 secs, the bullet will fall vertically by the distance h =gt^2/2 = 10*0.04/2 =0.2 m = 20 cms
Option D is the correct answer
Option C
The magnitude of the magnetic field (B)at distance R from the center of the wire produced by current (I) in a wire of infinite length is B = µoI/2πR .where µo is the magnetic permeability constant, If all other conditions remain unchanged, the strength of the magnetic field B varies inversely as distance from the wire
؞B1R1=B2R2 ⇒ B2= B*5/20 =B/4
Hence, option B is the correct answer
Option D is the correct answer
By solving the equations (i) and (ii) we get x=9 and y=-2 since 9 > 0 >-2; hence option A is the correct answer
We know F= mg = GmM/R^2 ⇒ g = GM/R^2 (where M is the mass of the body and R = radius of the body
But M= ρV ( where ρ= density of the body and V=volume of the body= 4/3πR^3) ؞ g = 4/3πGρR^3/R^2
؞g α R*ρ ⇒ ge/gp= Re*ρe/Rp*ρp ( ge and gp are respective gravitation on earth and planet, Re and Rp are their respective radii and ρe and ρp are respective densities) But ρe = ρp as densities are equal
ge/gp= Re/Rp ⇒ gp= ge*Rp/Re = g* 0.2R/R =0.2g
Hence option A is the correct answer
Option B is the correct answer
x^3-25 = 50X3-50=> x^3= 25+ 100 =125 => x= (125)^1/3 = 5
y^2 - 3√729= 5+(121/11) => y^2 = 9 + 16 =25 => y=5
؞ x=y Option E
Total number of cards =52
Total number of jacks of black color= 2 (jack of clubs and jack of spades)
P(getting a black of Jack) = 2/52=1/26
Hence option D is the correct answer
sin(90-θ) = cosθ and cos(90-θ)=sinθ
LHS of the expression sin(90-θ).cos(90-θ) = cosθ.sinθ
RHS of the expression tanθ/sec^2θ= (cos^2θ*sinθ)/cosθ = cosθ.sinθ
؞ LHS = RHS option D is the correct answer
Let 80-x = θ Therefore, sin(80-x) = sinθ = cos(90-θ) = cos(90-(80-x))= cos(10+x)
option C is the correct answer
Let the radius of the bigger circle be R and smaller circle radius be r.
؞π(R^2+r^2)= 500.5= 1001/2 ..(i) and R-r=3.5=7/2..(ii)
=> 22/7(R^2+r^2)=1001/2=> (R^2+r^2)= 637/4 from (ii) R = r+7/2 => R^2 = r^2 + 7r + 49/4 putting it in (i) we get
r^2 + 7r + 49/4 +r^2 = 637/4 => 2r^2 + 7r+ 49/4 = 637/4 => 8r^2 +28r-588 =0 => 2r^2 -14r +21r-147=0 => r=7, -21/2
Since radius can't be negative, discard-21/2 ؞ r=7 R= 7 +7/2=21/2. R/r = (21/2)/7= 21/14=3/2
hence option B is the correct answer
Area of the quadrant = π r2/4 (r=10cm) = 25π cm2; Area of the semi-circle drawn on the hypotenuse of the
isosceles Δ = π R2/2(R= 5√2 cm) =25π cm2 (since hypotenuse of the issocles Δ of side 10 cm = 10√2)
Area of the issoceles Δ =1/2*10*10=50 cm2
؞ Area of unshaded region = area of semi-circle- (area of quadrant- area of isosceles Δ) = 25-25+50 = 50cm2
hence option D is the correct answer
Area of the sheet = 28X28 = 784 cm2. Radius of each circle is 28/4= 7cm. Area of each square = π 7^2 = 49π= 154 cm2
Area of remaining sheet = 784 - 4(154)= 784-616= 168 cm2
Hence option B is the correct answer
Diameter AB of the circle =13 cm. Radius = 13/2 cms. Area of the circle = (13/2)^2π= 42.25π
Side AB = 13 and BC=5 Now ABC is a right angle Δ because the angle subtended by diameter on the circumference of the circle is a right angle. Therefore AC= √(169-25) = 12, Then area of the Δ ABC = 1/2*12*5 = 30 cm2.
؞ Area of the shaded region = (42.25π- 30)cm2. Hence option B is the correct answer
Let the equation of the line be y = mx+c where m is the slope of the line and c is the y-intercept. Since the line passes through the point (13,-11) and y-intercept= 3(given), -11=m*13+3 => m=-14/13 Substituting the value of m in the line equation we get, y=-14x/13+3 => 14x+13y-39=0 will be the equation of the line.
None of the options is correct. The equation of the line should be 14x+13y-39=0
ABCO is a square. Therefore, diagonals AB=OB=√2 cm. OB is the radius of the circle. Therefore the circumference of the circle= 2π*√2= 2√2π cms. Hence option C is the correct answer
The perimeter of the square = 44 cms. ؞ Each side of the square = 44/4 = 11 cms. ؞ Area of the square =11*11= 121 cm2
The circumference of the circle with radius r= 44. 2πr = 44 => πr =22 =>22r/7= 22 => r =7 Area of the circle= 49π= 49*22/7= 154 cm2
Difference between the area of circle and that of square = 154- 121=33 cm2. Hence Option A is the correct answer
Area of the square ABCD= 7*7= 49 cm2. Area of the circle with radius 7/2 (radius = half of the side of the square)
= 22/7*(7/2)(7/2) = 77/2 cm2. Area of the shaded region = 0.5*(area of square-area of circle) = 0.5 *(49-77/2)
= 0.5*(98-77)/2= 0.5*21/2 =5.25 cm2 Option A is the correct answer
Option A
Ve = √(2GM/R) where Ve = escape velocity at earth's surface, G universal gravitational constant, m= mass of earth and
R=radius of earth=distance between the center of masse of earth and the body. Now 3V> ve
As R→ R∞, 1/R∞ → 0 so 3V> ve → 0 Hence the speed at infinity will be 0
Option A is the correct answer
Radius of bigger circles = R = 20 cms and smaller circles = r =10 cms.(Diameter of smaller circle= radius of bigger circle)
Shaded region = 3 smaller semi-circles + 1 bigger semi-circle - 1 smaller semicircle
Perimeter of shaded region = 3π(10) + π(20) - π(10) = 40π = 125.71 cms
Area of the shaded region = 3π(100)/2 + π/2(400-100) = 150π + 150π = 300π = 942.85cm2
Hence option B is the correct amnswer
Option C is the correct answer
Lenght of train-A = 100 m and length of train-B= 120 m. Speed of train-A = 50 kmph and speed of train- B = x kmph
If train-A has to cross the train-B the distance traveled by A will be the sum of the length of the trains A and B= 100+120=220m = 0.22km Sice the trains are traveling in the opposite directions, the relative speed of train A will be the sum of the speed of A and B = (50 + x) kmph. Since time = distance/speed , the time to cross the train- B is 6 secs= 1/3600 hr.
6/3600= 0.22/(50+x) => 6(50 + x) = 22*3600/100= 50+x = 22*6 => x = 132-50 => x=82 kmph
Option B is the correct answer
Option D is the correct answer
option E- Cyprus Mistry
1_2_3_4_5_6_7_8_9_10_11_12_13_14_15_16_17_18
Before, 5th fm right → B P → 6th from left
After P B
Bimal is fifth from left and Prakash 6th from right. When they interchange Bimal is thirteenth from left. i.e. 13th from right = 6th from left => there are 13+6-1 =18 positions. Hence after interchange, Prakash is 14th from the right.
Hence option B is the correct answer
option B
A ______30___________C____10_____B
Distance AC=30 Kms, AB=40 kms Let speed of boat P be 'x' speed of boat Q be 'y' and speed of stream be 'z'
Relative speed of P upstream = x-y and speed of Q upstream z-y. Relative speed of P downstream = x+y and speed of Q upstream z+y. Now P travels from A to B = 40 kms upstream in 10 hours and B to A downstream in 5 hours.
time =distance/speed Therfore, x-y= 40/10 =4..(i) and x+y = 40/5=8 ...(ii) solving (i) and (ii) we get x= 6 and y=2
Speed of boat P = 6 kms/hr and speed of stream = 2 kms/hr and Speed of boat Q = z-2= 30/5=6 => z= 8 kms/hr
Now, P starts from A and Q starts from C and Pand Q will meet at D after t hours.
t = 30/(Speed of P upstream+speed of Q downstream) = 30/(4+10) = 15/7
In 15/7 hours P will travel 15/7*4= 60/7 kms upstream from A., = 40-60/7 =220/7 kms from B.
Therefore P and Q will meet at 220/7= 31.4 kms form B
Hence option A is the correct answer
K ____M
D____ N
Either statement specifies the direction of K w.r.t N
Hence option C is the correct answer
Speed of A = 36 km/hr. Therefore, Distance traveled by A in 2 mins = 36/60*2= 12/10kms = 1200 meters
Speed of B = 15 m/s. Therefore, Distance traveled by B in 2 mins = 15*120= 1800 meters
Distanc ebetween them after 2 mins = 1800-1200 = 600 meters
Option C
Let the <ACB= θ. tanθ= 450/778.5 =0.5773 Therefore θ= tan-1(0.5773) =30o
Option B is the correct answer
Price of W =0.21/6= $0.035/oz;
Price of X =0.48/16= $0.03/oz
Price of Y=0.56/20= $0.028/oz
Price of Z0.56/20= $0.03/oz
Y is least expensive per ounce Option C
Let Reena's age be x years and her daughter's be y years. x=8y
Eight years from now, (x+8)/(y+8)= 10/3 => 3x+24 =10y+80 => 24y+24= 10y+80 => y=4. Therefore, x = 8*4=32
Reena's present age is 32 years
Optiopn A is the correct answer
15p^2 - 26p +8 = 0=> 15p^2 - 20p -6p +8=0 => 5p(3p-4)-2(3p-4) => p=2/5 OR p=4/3
. 15p^2-2q-1=0 , When p=2/5, q= 1/2[15(4/25)-1] = 7/10 Therefore p<q
When p=4/3, q= 1/2[15(16/9)-1] = 77/6 Therefore p<q
Option D is the correct answer
If average age is increased by 2 years, total age is increased by 16 years
Total age of the two women = (35+45)+16 = 96 years Average of the two women is 96/2 = 48 years
Option A
Solution
solution
solution
N is the LCM of first 100 natural numbers.
Now 101 is a prime number, so 101 and N do not have any common factors. Hence their LCM = 101XN
102= 2X51, since 2<100 and 51<100, 102 divides N
103 is a prime number, no common factors between 103 and N
104= 52X2; since 2<100 and 52<100, 105 divides N
105 = 21X5; since 5<100 and 21<100, 105 divides N
Hence, LCM of first 105 natural number is NX101X103
Option B is the correct answer
solution
Let the sum borrowed by A be x and by B be y
Interest paid by A = IA= x*4*5/100 and Interest paid by B = IB= y*5*4/100
Since interest paid by A is double that of B's; IA=2(IB) => x/5 = 2y/5 => x = 2y
x + y =30,000 => 2y + y = 30,000 => y = 10,000. Therefore, x = 20,000
Sum borrowed by A is Rs. 20,000.00
Option B
Due to stoppage the bus travels (45-36) = 9 kms less.
Time taken for traveling 9 kms = 9/45*60= 12 mins
hence the stoppage is 12 mins Option B
In 100 gms of alloy-1(A1) - Gold = 20 gms and Silver = 80 gms
In x gms of alloy-2 (A2) - Gold= 40/100*x = 0.4x and Silver = 60/100*x = 0.6x
After mixing, ratio of Gold: silver = (20+0.4x)/(80+0.6x)= 32/68 => x= 150
150 grams of second alloy was mixed with 100 grams of first alloy
Option E is the correct answer
approximate value of cube root to3√926.158 is 3√1000. 3√1000 = 10
approximate value of cube of (12.99)3 is 13^3 = 2197
approximate value of cube of 74.98% of 679.8 = 75% of 680 = 3/4*680 = 510
3√926.158 + (12.99)3 + 74.98% of 679.8 = 10+2197+510= 2717= (52.12)^2
?= 52.12 Option B
Option B Missing number is 4576
numbers 4569 4571 4576 4586 4603 4629 4666
Difference 2 5 10 17 26 37
Difference 3 5 7 9 11
solution
The period of the sine and cosine function is 2π because the sine wave repeats every 2π units.
The period of k*cos(ax+b) or k*sin(ax+b) is 2π/a
Therefore, the period of cos(10t+1) = 2π/10 = π/5 and period of sin(4t-1) = 2π/4= π/2
The fundamental period of g(t) = 2cos(10t + 1) + sin(4t − 1) will be the LCM of the fundamental periods for each of the componenet term in x(t) i.e. LCM of π/5 and π/2 is π
Option B is the correct answer
1249.69 ÷ (49.94) + 7(17/41) of 942.99 + 69.87% of 800 = 1250/50 + 304*943/41 + 70/100*800=25 + 6992 + 560 = 7577
Option C is the correct answer
Mona traveled 3 kms south and then turned right and traveled 5 kms in the west direction and again turned right and traveled 7 kms in North direction
Now she is traveling in North direction. Hence option C is the correct answer
In the initial solution let the volume of A be 3x and that of B be 2x. When 10 litres of solution is taken out, 6 liters of A and 4 liters of B is taken out. Then 10 liters of B is added.
Therefore, (3x-6)/(2x-4+10)=1/2 => 3x-6 = x + 3 => 2x=9 => x= 9/2 => 5x = 45/2 = 22.5
Volume of initial solution = 22.5 liters Hence Option E is the correct answer
William share: John's share = 7:9 Total amount deposited $ 72,000. William's share = $7/16*72,000= $31,500
John's share = $40,500. William withdrew $ 7,500 and John withdrew $ x.
(31,500-7,500)/(40,500- x) = 4/5 => 24,000/(40,500-x) = 4/5 => x = 40,500-30,000 = $10,500
Option C is the correct answer
Option D is the correct answer
Option D
The magnetic field intensity at any given point is specified by both a direction and a magnitude (or strength); is represented by a vector field.
Option C:
According to the most accurate caesium atomic clock in the world, 1 second is the time that elapses during 9,192,631,770 cycles of the radiation produced by the transition between two levels of the caesium 133 atom
Let the time taken by the nth tap to fill up the tank be tn. The efficiency of the nth tap is equal to the sum of the efficiencies of all previous taps. i.e. the time taken by the 3rd tap= sum of the time taken by 1st and 2nd tap.
t3= t2+t1.
t4= t3+t2+t1= (t2+t1) + (t2+t1) = 2(t2+t1)
t5= t4+t3+t2+t1= 2(t2+t1)+(t2+t1) + (t2+t1) = 4(t2+t1)= 2^2(t2+t1).
؞tn= 2^(n-3)(t2+t1). the 8th tap takes 48 hours to fill the tank. ؞ t8= 2^(8-3)(t2+t1) ⇒ 2^5(t2+t1)= 48
⇒ 2^5(t2+t1) = 48
When tap number 1to9 are working together to fill the tank it will be equal to the time taken by 10th tap
2^(10-3)(t2+t1) = 2^7(t2+t1) = 2^2*t^5(t2+t1) = 4*48 = 192
Option C is the correct answer
Option C is the correct answer.
Earth (mass m) rotating around the Sun (mass M; M >> m) in a circular orbit of radius r with velocity v. Then, by applying Newton's law of gravitation and the second law of motion, we can write
Gravitational force = mass × centripetal acceleration
i.e. GMm/r2 = m(v2/r) ... (i)
or, v2 = GM/r ... (ii)
The time period of rotation, T, of earth around the sun is given by,
T = 2πr/v => Squaring both sides we get, T^2= 4(πr)^2/v^2 = 4(πr)^2/GM/r = 4π^2r^3/GM .... (iii)
=> M = 4π^2r^3/GT^2 => M = K* r^3/t^2 where K is a constant equals to 4π^2/G
Hence mass of sun M is proportional to r^3/t^2
i) 12p2+11p+2=0 => 12p2+3p+8p +2 = 0 => 3p(4p+1) + 2(4p+1) =0 => (3p+2)(4p+1)=0 => p = -2/3 OR p=-1/4
ii) 27q2+3q−10=0=> 27q2+18q -15q −10=0 => 9q(3q+2) -5q(3q+2) = 0 =>(9q-5)(3q+2) = 0 => q = 5/9 OR q = -2/3
p= -2/3, -1/4
q= -2/3, 5/9
Option E
Since O is the incenter, OP is the bisector of the < QPR. Therefore <QPS= 1/2*60o =30o
Then < QRS= 30o ( Chord QS will subtend equal angle at the circumference)
and <QOR= 2*60o = 120o (angle subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.)
Option B is the correct answer
Let N1, N2, N3 be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3 and remainders R1, R2, R3… respectively.
Now, Remainder of (N1X N2XN3)/D => Remainder of (R1XR2XR3)/D.
Therefore, Remainder of (14X 21X1423X1425)/12 => (2X9X7X5)/12 => (18 X35)/12 => (6X9)/12 => 6
Option D
Since we do not know the average weight of the class or average weight of each group, we cannot determine average weight of the students in group-D
Hence option D is the correct answer
W→→→→→→→E
--------x--------> +ve
Train A moves from west to east positive direction. So the relative velocity of A w.r.t to ground is Vae = +54 km/hr
Train B moves from east to west negative direction. So the relative velocity of B w.r.t to ground is Vbe = -90 km/hr
Therefore, relative velocity of B w.r.t A is Vba = Vbe + Vea = Vbe - Vae = (-90) - (+54) = -144 km/hr
= - 144*1000/(60*60) m/s = -40 m/s
Option D
option B
solution
solution
solution
A = P(1+r/n)^nt where A=amount, P= principal, r= rate of interest in decimal, n= number of times interest compounded per year, t= time in years
A = 36000(1+0.16/2)^2*3/2 = 36000 (1.08)^3 = 36000*1.25972 = 45,349.63
Therefore, interest = Rs. A - P = 45,349.63 - 36000= Rs. 9,349.63
Option A
solution
Option C aircraft: jet
Stratosphere is a layer of the atmosphere that means it is a type of atmosphere. It is analogous to aircraft: jet, where aircraft is the category and jet, is a type of aircraft.
Nimbus: cloud is not applicable because the order is category: type whereas Nimbus is not the category but it is a type of cloud. Other options do not satisfy category:type relation
Option C
Each shares (of Rs 20/-) gives 8% dividend = 8/100*20 = Rs 8/5
The man can earn Rs 8/5 from 1 shares
؞ He can earn Rs 1,000 from 1/8/5*1000 = 5000/8 = 625 shares.
Now each Rs. 20 shares is selling at Rs 40. ؞ To buy 635 shares he need to invest Rs 40 * 625 = Rs 25,000
Option D
Let the height of the building be 'h' and and OQ be 'b' QR = a (given)
In Δ POQ, tanx = h/b and in Δ POR tany = h/(b+a) Therefore h= btanx...(i) and h= (a+b)tany ...(ii)
Equating (i) and (ii) we get btanx = (a+b)tany ⇒ btanx = atany + btany ⇒ btanx - btany = atany ⇒ atany = b(tanx - tany)
⇒ b= atany/(tanx-tany) putting the value of b in (i) we get
height of the building, h= tanxtany(tanx-tany)
Option D speed. It has only magnitude and no direction, hence it is a scalar quantity.
option D
Scientific notation is the method of writing, or of displaying real numbers as a decimal number between 1 and 10 followed by an integer power of 10. In scientific notation, all numbers are written in the form of a⋅10ba⋅10b (a times ten raised to the power of b). eg 5X6X100,000 = 3,000,000 = 3.0 million.
In scientific notation, we can write it as = 3 X10^6
Total number of consumers n(AUB) =1000 Consumers who like product A, n(A) =720 Consumers who like product B , n(B) =450 Number of consumers who like both A and B (A∩B).
n(AUB) = n(A) + n(B) -(A∩B). ⇒ (A∩B) = n(A) + n(B) - n(AUB)
⇒ 720 +450 -1000 =1170-1000=170
Least number of people who liked both products is 170
option A is the correct answer
Maturity value of a recurring deposit is given by M =
R [ (1+i)n – 1] ——————– 1- (1+i) -1/3 Where, M = Maturity value R = Monthly Installment r = Rate of Interest (i) / 400 (Annual interest rate converted into quarterly rate) n = Number of Quarters (Assuming the compounding is done quarterly) By inputting the values given we get M= Rs21,244.00 Hence Option C is the closest to the correct answer
solution
option B
Significant figures are the digits which give us useful information about the accuracy of a measurement
Rule 1: Non-zero digits are always significant.e.g. The number 22.22 has FOUR significant figures because all of the digits present are non-zero.
Rule 2: Any zeros between two significant digits are significant.The number 11.2 has THREE significant figures because all of the digits present are non-zero. e.g.3021 has FOUR significant figures. The zero is between a 3 and a 2.
Rule 3: A final zero or trailing zeros in the decimal portion ONLY are significant.e.g There are FOUR significant figures in 32.00. It is different from 32 It's important to understand that "zero" does not mean "nothing." Zero denotes actual information, just like any other number. we cannot tag on zeros that aren't certain to belong there.
Rule 4: Leading zeros are NOT significant. e.g The number 0.0032 also has TWO significant figures. All of the zeros are leading.
solution
solution
option A
When the stone is dropped, the height fallen by the stone can be calculated from the formula S= ut + 1/2at^2 where s=h=height(distance), a=g =9.8 m/s2 (acceleration due to gravity), t= time and initial velocity u =0 Therefore,
h = 1/2gt^2....(i),Now error in measurement = |+/-(observed value-actual value)|.= dx. By differentiating eqn.(i) we get,
dh = 1/2*g*2tdt => dh = gtdt = 9.8*2*0.1 (since t=2, dt=0.1)
=> dh = 1.96 m Hence Option D
Since the cylinder is completely hollow and open on both sides, the curved(lateral) surface area CSA = 2πrh( r=radius of the cylinder and h=15 cm,height of the cylinder) 2πrh = 2500(given) => r = 2500/2πh
Volume of the cylinder = πr^2h = π*15 *(2500/2πh)^2 = 25*25*10^4/(4*15π) = 3.315727 X 10^4 = 33157.2 cm3
Hence option A
Solution
solution
Let the velocity of sound be v and the velocity of the train be u. u =v/20 (given)
Since the blast is blown at interval of 1 second, the actual frequency of blast f=1/t = 1/1 =1 Since the train is approaching the stationary observer the apparent frequency of the blast f"= v*f/(v-u) = v*1(v-v/20) = 20/19
Apparent time interval between the blast t' = 1/f" = 19/20 s Option C is the correct answer
solution
(i) The value of dimensionless constants cannot be determined by this method. e.g. π(pi)
(ii) This method cannot be applied to equations involving exponential and trigonometric functions.
(iii) It cannot be applied to an equation involving more than three physical quantities.
(iv) It can check only whether a physical relation is dimensionally correct or not. It cannot tell whether the relation is absolutely correct or not. For example, applying this technique s =ut + (1/4) at2 is dimensionally correct whereas the correct relation is s = ut + (1/2) at2
Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight and is measured by the angle or semi-angle of inclination between those two lines
To measure large distances, such as the distance of a planet or a star from Earth, astronomers use the principle of parallax. As the Earth orbits the sun, a nearby star will appear to move against the more distant background stars. Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position. The star's apparent motion is called stellar parallax.
There is a simple relationship between a star's distance and its parallax angle:
d = 1/p
The distance d is measured in parsecs and the parallax angle p is measured in arcseconds.
Distance of the star is 1 lightyear (ly). 1 ly = distance traveled by light in 1 year = 3X10^8x365X24X60X60= 9.46X10^15m
I parsec(pc) = 3.08 X10^16 m Therefore, 4.29 ly = (4.29*9.46 X10^15)/(3.08 X10^16) = 1.317 ≈ 1.32 parsec
Let θ be the parallax which the star (named Alpha Centauri) at a distance r of 1.32 parsec when viewed from two locations on earth six months apart. Then θ = d/r where d is the diameter of the earth's orbit = 2 A.U=2X1.5X10^11 (since radius of the earth's orbit = 1 AU = 1.5 X 10^11 m)
θ = d/r = arc/radius = 2 AU/1.32 parsec = 1.52
The question cannot be answered because the value of angular diameter is not given.
Distance of Jupiter from the earth D, 824.7 million km = 824.7 X 10^6 km = 8.247 X 10^9 m
Angular diameter = θ Using the relation θ = D/d where d = diameter of jupiter
d = θD = 824.7 X 10^9 θ
Time taken by quasar light to reach earth = 3 billion years = 3.0 X 10^9 years = 3.0 X 10^9 X 365 X 24 X 60 X60
= 9.46 X 10^16 sec Speed of light = 3 X 10^8 m/s
Therefore, distance between earth and sun, d =s*t = 9.46 X 10^16 X 3 X 10^8 = 28.38 X 10^24 m = 2.8 X 10^22 km
Least count = pitch/number of divisions in circular scale. Therefore, when the number of divisions in the circular scale is increased, least count will decrease and accuracy will increase (theoretically) but it is practically impossible to increase the number of divisions beyond a certain limit since the human eye cannot take accurate measurement beyond a certain limit. hence the accuracy cannot be increased beyond a limit
Magnification m = observed length (x)/actual length(y) m =100, x =3.5 mm y =?
y = 3.5/100 = 0.035 mm. Therefore, estmated thickness of human hair = 0.035 mm = 35 micrometre
The problem states that speed of light in vacuum = 1 unit/sec (unity)
Distance =speed x time = 1 unit/sec x (8*60+20) sec = 500 units
Distance between the earth and sun is 500 units
n= real depth/apparent depth where n= refractive index of the medium
assuming refractive index of glass = 1.5 real depth = 3cm (thickness of the slab = 3cm)
therefore, apparent depth = 3/1.5 = 2 cm
Initial height of the microscope was x + 3 where x is the height of the microscope above the glass slab when no glass slab was there. Now after the glass slab is placed the new height will be x+2. So the microscope to moved by (x+3)-(x+2) = 1 cm down Option D
Focal length of the lens f = +15 cm, image distance = +20 cm (Real image on a screen) Object distance,u= ?
from lens formula we know, 1/v-1/u =1/f => 1/u = 1/v-1/f = 1/20 - 1/15 => 1/u = (3-4)/60 =-1/60
Therefore, u = -60cm The object should be placed at 60 cms from the lens
Option D
The question has a mistake. If we assume the answer choices are correct, the correct equation should be
x= 40+12t-t^3 and not x= 40t+12t-t^3 When t=0, x = 40+12(0)-(0)^3 = 40m i.e. the particle has travelled 40 m at time zero
Velocity =dx/dt. When the particle comes to rest velocity will be zero. i.e. dx/dt=0 But, dx/dt = 12 -3t^2
Therefore, 12-3t^2=0 => 3t^2 = 12 => t^2 =4 => t=+/-2 (ignore -2 as time cannot be negative) Therefore, x = 40+12(2)-2^3=56m i.e. after 2 secs the body has travelled 56 m therefore, the body has travelled 56-40 =16 m before coming to rest Option D is the correct answer
The climate change in momentum can be calculated from the area under the force-time graph
For 0 → 2 secs, ΔP1= 1/2(2)*6 = +6 Ns For 2 → 4sec, ΔP2 = 2(-3) = -6 Ns For 4 →8 sec, ΔP3 = 4(3) = =12 Ns.
So total change in the momentum from 0→8 sec = ΔP1+ΔP2+ΔP3 = 6-6+12 = 12 Ns
Option C
Let the distance be 2d. Time taken to cover the first half of the journey, t1= distance/speed=d/40 hr and time taken to cover the second half of the journey, t2= d/60. Therefore, average speed of the car = total distance/total time= 2d/(d/40+d/60) = 2d*120/5 = 48 kmph
OptionA
Displacement x=(t-2) where t=time. When t=0, x= 0-2 = -2 So the distance travelled = 2m in the backward direction
When t=4, x = (4-2) = 2m, distance travelled = 2m in forward direction. therefore in first 4 secs the distance travelled = 2+2=4m Option A
Option A Speed
An object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed. While the speed of the object is constant, its velocity is changing because at every instant direction is changing
Option C acceleration
The equation v = u + at is applicable only when acceleration 'a' is constant. If acceleration is changing the equation cannot be used for calculating velocity(v)
Option A
The precise answer should be "At the highest point, vertical velocity will be zero and the direction of acceleration (due to gravity) is always vertically downward. However, at the highest point, the ball will have a horizontal velocity which will be constant throughout the path but no horizontal acceleration"
Option A
When a body freely falls under gravity (with no other forces acting on it, no air resistance) it is a uniformly accelerated motion where the velocity increases over time at a constant rate
The maximum speed will be the area under the a-t graph.
Maximum speed = area of the graph= (1/2*base of traiangle* height of triangle=1/2*11*10 = 55 m/s
Hence, option B is the correct answer
The height attained by each stone can be calculated from the formula h = ut -1/2gt^2 where 'u' is the initial velocity, 't' is the time of flight and 'g' is the acceleration due to gravity (- sign because acceleration is vertically down whereas the motion is vertically up). So the height attained by each stone will depend only on initial velocity and time of flight and it is independent of mass. When both stones are thrown simultaneously with same initial velocity, they will attain equal height in equal time.
Hence option C
Let the height of the tower be H. Time 't' taken by the body to reach the ground can be calculated from the relation
t = √(2H/g) where g is acceleration due to gravity Since the time is independent of mass, both bodies will take the same time to reach the ground
Hence option B
The speed of car-1 is 18 km/h and the speed of car-2 is 27 km/h. Therefore, the relative speed of the cars = 18+27=45 km/h (since they are moving in opposite direction) The cars are separated by 36 km. So the time taken by the cars to meet is (distance /relative speed) = 36/45 = 4/5 hour. The bird is flying at a speed of 36 km/h from the first car.
In 4/5 hours the bird will travel 4/5*36= 144/5 = 28.8 km. The bird will cover 28.8 km when the two cars will meet.
Hence option A
( a ) Possessing a positive value of acceleration has to be speeding up. The statement is false eg. -ve----|---- → +ve
Suppose when the car moves left to right it is +ve direction. Now if the car moves right to left and speeding up it will have a negative speed and negative acceleration, yet the car is speeding up
(b)) moving with a constant speed must have zero acceleration. This statement is also false.
eg. An object moving in uniform circular motion is moving in a circle with constant speed but the body is accelerating because the direction of the velocity vector is changing.
(c) with zero speed at an instant may have non-zero acceleration at that instant. This statement is true.
eg When a ball is dropped from a building it starts falling with zero speed but non-zero acceleration due to gravity
(d) possessing zero speed could have a non-zero velocity. This statement is false. When the speed is zero, the magnitude of velocity is also zero
So the correct option should be false-false-true-false option D
When the cars apply brakes, the distance traveled by each car before stopping be s1 and s2 respectively.
Since the stopping velocity is zero, the stopping distance can be calculated from the equation v^2- u^2 = 2(-a)s => s=u^2/2a s1 = (70*70*10^6)/(2*5*60*60*60*60) = 37.80 m and s2 = (60*60*10^6)/(2*5*60*60*60*60) = 27.77 m
So s1+s2 = 37.80 + 27.77 = 65.57 < 80. Since the minimum distance between the car should be greater than
65.57 m (here it is 80 m) the collision can be averted.
Hence Option A
Option C. The umbrella should be held at an angle θ=tan-1(1/2) opposite to the velocity of rain relative to the child
option A Speed remains constant during uniform circular motion
option D
Option D
Except for muscle, rest all are connective tissue. Muscle tissue is a class by itself which is distinctly different from connective tissue
The position of the particle is given by x=at^3 The velocity (say v) of the particle at any instant is given by differentiation of its position x with respect to time as: dx/dt = v Therefore, dx/dt = v = 3at^2
there is a typo error in the options as both options B and C are showing v=a.3/t^2
One of the options should be v=a.3t^2
The height (s=stories) fallen by the orange in time (t)is given by s = ut +1/2gt^2 where u=0, initial velocity of the orange and g= acceleration due to gravity. In 1 sec after dropping the orange s = (0*1) + 1/2 g(1)^2 => s = g/2 => g=2s
When t=2, the number of stories fallen,s2 = 1/2g(2)^2 => s2 = 1/2(2s)(4) =4s
Therefore, 2 secs after the orange is dropped it has fallen 4 stories. Hence Option D
Adipose tissue, or fat, is an anatomical term for loose connective tissue composed of adipocytes.
Hence adipose tissue is a type of connective tissue. Option B
Endothelium refers to cells that line the interior surface of blood vessels. It is a thin layer of simple squamous cells called endothelial cells. Hence Option C
Option B
Option C
Microsporum belongs to the kingdom of fungi which is the same kingdom of Rhizopus
Ascaris, a roundworm belongs to the kingdom Animalia, Wuchereria also belong to the kingdom Animalia
Stem has four main functions.
1. to support the leaves, flowers, and fruits. 2. to transport minerals and water to the leaves where they can be converted into usable products by photosynthesis 3. transport the synthesized food from leaves to other organs of the plant 4. storing food and water. In some plants, stems are adapted for specialized functions.e.g. Thorns are woody, sharply pointed branch stems
Option D kinetic friction
Option A
Option D
Option C Force = m*a = m*v/t (mass x velocity = momentum)
Option C 1 dyne = 10^(-5) N
Option C momentum
Option B Even if we stop paddling due to inertia of motion we will continue to move forward upto a certain distance as friction will finally make us stop
Option C decrease in friction.
When it rains, the water forms a film between the road surface and the footwear and reduce the friction between the two surfaces and if the friction is not enough to hold the grip, we tend to slip
Option A
When we want to remove dust from the blanket, we strike it using a rod because of which the blanket moves but the dust due to inertia of rest try to be in the same position and due to gravity now falls down
Option B Calcium and Phosphorus
Option D cellulose It is the most abundant organic compound on earth. It is a type of carbohydrate, containing chains of glucose rings. It provides strength and rigidity and forms the cell walls of plants. Cellulose is the primary constituent of wood, making this organic compound the most abundant one on the surface of the Earth
Option D The porphyrin ring of chlorophyll, has a magnesium atom at its center, is the part of a chlorophyll molecule that absorbs light energy.
Option D For storage, glucose needs to be converted into insoluble compound called glycogen in animals and as starch in plants.
Option C: Approximately 24 hours
Option A
Normal body cells of diploid organisms (other than sexual reproductive cells) always contain pairs of homolog chromosomes. All these cells are genetically identical to the zygote (the product resulting from fertilization) from which they descend. One set comes from mother side and another set from father side. However, because of meiotic "reshuffling" (= crossing-overs), a different pool of genes is present in the inherited chromosomes than in the cells of the father and the mother.
Sexual reproductive cells have only one sample of each chromosome.
Option D Karyokinesis is the division of the nucleus, and chromosome is contained in the nucleus
Option D In animals mitosis is only seen in diploid (having two sets of chromosomes) somatic ( any cell of living organism other than reproductive cells) cells
Option B
Interphase is the interval in the cell cycle between two cell divisions when the individual chromosomes cannot be distinguished. It was formerly called the resting phase. However, interphase does not describe a cell that is merely resting; rather, the cell is actively living and preparing for later cell division, so the name was changed.
Option A
Option B
In the S phase (synthesis phase) during mitosis DNA is replicated. It is also a critical point during interphase when the cell duplicates its chromosome and ensures its systems are ready for cell division.
Option A M stands for mitosis
Option C G0 phase is an inactive phase, also called as quiescent phase of the cell cycle. Cells in G0 stage remain metabolically active but do not proliferate unless called to do so depending on the requirements of the organism
Option D Onion root tip cell has 16 chromosomes in each cell.
Phospholipids are most abundant in all biological membranes. A phospholipid molecule is constructed from four components: fatty acids, a platform to which the fatty acids are attached, a phosphate, and an alcohol attached to the phosphate. Phospholipids may contain glycerol, a 3- carbon alcohol, or sphingosine, a more complex alcohol. Phospholipids derived from glycerol are called phosphoglycerides. whereas those derived from sphingosine, are called Sphingomyelin. The third group is called Glycolipids-sugar containing lipids.
Archaeal Membranes Are Built from Ether Lipids with Branched Chains
Cholesterol is a lipid with a structure quite different from that of phospholipids. It is a steroid, built from four linked hydrocarbon rings.It is especially abundant in the membranes of those cells, where it helps maintain the integrity of the membranes, and plays a role in facilitating cell signaling-- meaning the ability of your cells to communicate with each other.
Option A Phenylalanine and tryptophan are aromatic amino acids. containing an aromatic ring. Rest B,C & Dare aliphatic
amino acids
Option A The molecular formula of arachidonic acid is C20H32O2
The question has an error. Except for lysine, all others are neutral amino acids. Lysine is a basic amino acid with isoelectric point 9.59
Valine has an aliphatic hydrophobic side chain whereas tyrosine and tryptophan have aromatic hydrophobic side chain
Isoelectric points are 5.96, 5.66 and 5.89 respectively
Option A Prophase -I i is typically the longest phase of meiosis. During this phase homologous chromosomes pair and exchange DNA (homologous recombination).
Option C Oils have low melting point. Oils are unsaturated. The double bonds make it harder to form a nice, compact crystal so the oils melt at a lower temperature.
option C
When a vehicle can go round the curved road at a reasonable speed without skidding, the sufficient centripetal force is generated by raising the outer edge of the road a little above the inner edge. It is called Banking of Roads.
Option C
Option D Newton's first Law- The law of inertia (off course we need to assume there are no interplanetary forces acting on the aircraft)
Net force on the body = mass of the body x acceleration of the body. If net force =0, acceleration=0 because mass cannot be 0 Option B or C
Option A Newton's first law- Law of inertia.
When the tablecloth is pulled the chinaware because of inertia of rest tends to stay in its place
option B solution
Answer
Option C Cell Plate
Plant cells divide by a cell plate that eventually becomes the cell wall whereas animal cells divide by a cleavage furrow.
Option B
In the absence of mitogens, most cells exist in G0. The G0 phase is a period in the cell cycle in which cells exist in a quiescent state where the cell is neither dividing nor preparing to divide
Option A
During Anaphase, except in meiosis-I, the number of chromosomes changes (by doubling) when sister chromatids are separated into chromosome
During the S-phase (synthesis phase) of the interphase during which DNA synthesis occurs During this phase, amount of DNA per cell doubles but the chromosome number remains the same but they exist in a structure that looks like an X shape, each chromosome contains two sister chromatid. These chromatids are genetically identical
Option A It will have 4 chromosomes in at the end of the anaphase-II.
Option C Formation of contractile ring and phragmoplast
Option D
Plant shows mitotic divisions in both haploid and diploid cells whereas in animals mitotic division is seen only in diploid somatic cells
Option C No cell can live without protein
Option A
Meiosis-II is a equational division similar to mitosis
Option A
A grown-up cell divides when a maximum size is attained and which disturb the nucleus-cytoplasm ratio (N:C ratio), also known as karyoplasmic index or Kern-Plasma ratio
Option A Ampere
Option B Current will not flow
The flow of current between two charged bodies depend on the electric potential of the bodies. Current will flow only if there is a difference in the potential of the two bodies.It is analogous to heat flowing between two bodies. Heat will flow only if there is a difference in the temperature of the two bodies. Heat will not flow if both bodies are at same temperature.
Let √2 be 'p' and √3 be 'q' p = √2 = √(200/100) and 'q' = √(300/100) We need to find a rational number 'r' between p and q. Therefore, √(200/100) < r < √(300/100) So if we square it, then 200/100 <r2 <300/100
Let us take perfect squares between 200 and 330 like 225 and 256. By definition, rational numbers are
p/q, where q= not equal to zero. Now the rational numbers r1= √225/100= 15/10 = 1.5 and r5 = √256/100 = 16/10 =1.6 so we have found two rational numbers . Let the other rational numbers be r2, r3.......r8 . So r2= (1.5 +1.6)/2 = 1.55 = 155/100 r3 = (1.5 + 1.55)/2 = 1.525 =1525/1000 Therefore, r1= 15/10, r2= 155/100. r3= 1525/1000, r5= 16/10
If we continue like this, we can easily find 10 ratioanl numbers between √2 and √3
The zeros of a graph ( a graph is the pictorial representation of the function) are the points where the graph of the function crosses the x-axis.
In a) the graph crosses x-axis at 3 points. Hence number of zeros =3 In b) number of zeros=1
In c) number of zeros = 2 and in d) number of zeros = nil
If a number has to end with digit 0, the number must be a multiple of 10. i,e the number should be (10)^n where n can be 1, 2.... any positive integer. So the number must have prime factors 2 and 5.
Now, 6^n has prime factors as 2 and 3 but not 5. Therefore, 6^n cannot be divided by 10 hence, 6^n cannot end with digit 0
It has to be proved by contradiction. Proof
To find the maximum numbers of columns we need to take HCF of 616 and 32
616 = 2x2x2x7x11 and 32 = 2x2x2x2x2 Therefore HCF = 8
Hence they can march in maximum 8 columns
The plane will fly in a straight line in a tangential direction to the line which was holding the plane
Initial velocity u 2.0 m/s and final velocity v = 3.5 m/s time t =10s mass = 3kg.
Using the relation, v = u +at we get, a = (v-u)/t = (3.5-2.0)/10 = 1.5/10 =0.15 Therefore, Force F = ma = 3*0.15 = 0.45N acting on the direction of the motion. Hence Option C
Aristotle asserted that a body was in its natural state when it was at rest
Option A Their inertia will also be 1:5 as inertia is measured in terms of mass
When the bull pulls a cart by a force say F, the cart also pulls the bull with an equal force F in the opposite direction and the net force is zero. But the bull is also pushing the ground at an angle by a force say F1. Now the ground also pushes the bull with a force of same magnitude F1 but in opposite direction. The vertical component of the force balances with the weight of the bull and the cart but the unbalanced horizontal component of the force (if greater than frictional force) makes the bull move forward.
Hence option A
The nucleus maintains the integrity of genes and controls the activities of the cell by regulating gene expression.So it is called the control center of the cell.
When we remove the skin of the potato tuber, we remove the periderm, a cortical protective layer of many roots and stems
Option B nutrition -autotrophic or heterotrophic. The other criteria include cell structure (complex eukaryote or simple prokaryote) thallus organization, mode of reproduction and phylogenetic relationships
Option B Posterior side
Option A Vitamins do not provide any calories
Option B Myology is the study of the muscular system, including the study of the structure, function and diseases of muscle
Option A A tendon is a tough, flexible band of fibrous connective tissue that connects muscles to bones.
Option B
In older children and adults, an elastic-like muscle at the entry to the stomach closes like a valve to prevent liquids from being pushed back up. This valve or sphincter is called lower oesophageal sphincter or cardiac sphincter. In babies, however, this valve or sphincter isn't fully effective until between 6 and 12 months of age. Since it isn't fully developed yet, the valve is easily pushed back by the contents of the stomach - resulting in regurgitation or spitting up.
Boluses of food from pharynx is conveyed into esophagus by Upper Esophageal Sphincter (UES)
Option B
Of the various conventional treatments that are available, surgical removal of the gallbladder is the most widely used. The most commonly used surgical technique is known as laparoscopic cholecystectomy. The doctor makes several small incisions in the abdomen, then uses special pencil-thin instruments to remove the gallbladder.
Option B
DNA (Deoxyribose Nucleic Acids) are digested by enzymes called nucleases which are secreted by pancreas and succus entericus (intestinal juice)
Intervertebral disc is found in the vertebral column of___mammals.
Option B
Cardiac muscles are striated, single nucleated and involuntary
Option B Cardiac sphincter is much less effective in infants
OPTION C
Option D
People who are lactose intolerant have trouble digesting lactose, a sugar naturally found in milk and dairy products. This is because they do not have enough of an enzyme called lactase to break down lactose. Since milk is the principal source of dietary calcium, calcium supplements can help people with lactose intolerance meet their daily requirements of calcium and other important nutrients.
Option D
The liver and muscle can only store a small amount of glycogen, however, so if we continue to take in more carbohydrates than needed, then our body will then convert excess carbohydrates to fat.
Option C
The muscles located in the inner wall of alimentary canals are smooth muscles which are non-striated, involuntary, shaped like almonds (tapered ends), one nucleus per cell.
Option B
The primary cause of gum disease is plaque, which is a sticky film made up of mostly bacteria.The bacteria associated with periodontal diseases are predominantly gram-negative anaerobic bacteria and may include P. gingivalis (Porphyromonas gingivalis)
Option D
Actin was discovered by Straub (1942).
Option A Carbon dioxide
Normal pH of urine is about 6.0
Option C
Dialysis fluid does not contain urea so that urea moves across the partially permeable membrane, from the blood to the dialysis fluid, by diffusion
Option C Acetyl CoA + Co2
Pyruvic acid undergoes oxidative decarboxylation to form Acetyl Co, NADH and CO2
CH3C(=O)C(=O)O(pyruvate) + HSCoA + NAD+ → CH3C(=O)SCo(Aacetyl-CoA) + NADH + CO2
Option B
Succinyl CoA + Pi + NDP ↔ Succinate + CoA + NTP
where Pi denotes inorganic phosphate, NDP denotes nucleoside diphosphate (either GDP or ADP), and NTP denotes nucleoside triphosphate (either GTP or ATP)
Option D
32-34 ATP, H2O, NAD+, FAD are the end products of oxidative phosphorylation
Option A Glycogen
Total ATP from one molecule of glucose = 30
Total ATP from one molecule of glycogen= 31
The above graph is an acceleration graph which gives the change in acceleration over time.
For an acceleration graph, the slope = change in acceleration/change in time = Δa/Δt (where a = acceleration, t=time) The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.
Therefore area of the graph = Δv But we also know that a= Δv/Δt Hence, Δv= a*Δt = 1/2basexheight = 0.5*11*10=55 m/s. Hence the change in the velocity = 55 m/s during the time interval t=0 → t=11. Therefore the maximum velocity(V) = final velocity = initial velocity (V0)+ change in velocity (Δv) But V0= 0 ( as the body started from rest although it was having a positive acceleration)
So maximum velocity V = V0 + 55 = 55 m/s Option B is the corrrrect answer
Option D
Both birds and mammals are warm-blooded, which means they can maintain a constant body temperature and do not need to rely on an external heat source to stay warm.
Option A
Platypus is oviparous as it is an egg-laying mammal
Rest three are viviparous mammals
Average speed = 2/(1/40+1/60) = 2*120/5= 48 kmph
Option A
option D Charas, Ganja and marijuana are all sourced from Cannabis plant
Option B is correct
Morphine still is the drug of choice for the treatment of severe acute or chronic pain.Morphine is also frequently injected to prepare for surgery and it can also be given during the operation for the suppression of nociceptive stimuli.
Although barbiturates like sodium thiopental and amobarbital are used in an effort to obtain information from subjects who are unable or unwilling to provide it otherwise, these drugs have not been proven to cause consistent or predictable enhancement of truth-telling
Chewing tobacco does the reverse- it increases heart rate
Cocaine is never prescribed to a patient.
Option D
Valium (benzodiazepine) is a CNS depressant which is used to treat anxiety
Amphetamine is a neuro stimulant. Hashsis is an illegal drug and connot be prescribed whereas morphine is a narcotic drug which is used under controlled conditions to releive moderate to severe pain
Option D Multinucleate condition
A cell normally contains only one nucleus; under some conditions, however, the nucleus divides but the cytoplasm does not. This produces a multinucleate cell (syncytium) such as occurs in skeletal muscle fibres.
Option A
Colchicine, an alkaloid drug extracted from the autumn crocus is used in doubling chromosome
Option B
Phytochromes (is a pigment found in plants) are a class of photoreceptor that plants use to detect light. They are sensitive to light in the red and far-red region of the visible spectrum and can be classed as either Type I, which are activated by far-red light, or Type II that are activated by red light.
Option B
Cryopreservation means the storage of germplasm at very low temperatures by bringing the plant cell and tissue cultures to a zero metabolism or non-dividing state
Option A
Oxaloacetic acid -C4H4O5
Option D
Carotenoid pigments exhibit strong light absorption in the blue portion of the visible spectrum (400 to 500 nm).
Option A By the Nitrogen atom of the pyrole ring
Option C
Organic matters are decomposed by microorganisms, mainly bacteria, fungi and protozoa
Option D
Cocaine is a powerfully addictive stimulant drug made from the leaves of the coca plant native to South America.
Option A
Prophase ,metaphase,anaphase, telophase
Option B ATP
ATP is formed in the first part of photosynthesis, (light reaction). In the second part of the cycle, known as dark reactions (or Calvin Cycle), which happens in the stroma of the plant, ATP (and NADPH) combine with CO2 and H2O to make glucose. The energy required in the dark reaction comes from ATP (and NADPH)
Option D
All of the above
Option A
Triticale is an intergeneric hybrid of wheat (Triticum) and rye (Secale)
Option D
Mutations can be induced in a variety of ways, such as by exposure to ultraviolet or ionizing radiation or chemical mutagens or gamma-rays
Option A
Since the nucleus of dinoflagellates (a type of algae) resemble both the ‘eukaryotes and ‘prokaryotes’, it is called ‘mesokaryotic’ nucleus. Unlike a eukaryotic cell, during cell division, in all dinoflagellates, the nuclear envelope remains intact and nucleolus does not disappear, chromosomes are permanently condensed even during interphase.
Option B
The first stable compound in the Calvin cycle is a 3-carbon compound called 3-phosphoglyceric acid (PGA). Hence Calvin cycle is also called C-3 cycle
Etiolation in plants occurs when they are grown in either partial or a complete absence of light. The condition is characterized by long, weak stems with smaller, sparser leaves due to longer internodes, with a pale yellow or bleached white, chlorotic colour due to lack of chlorophyll.
option D
Conversion of phosphoglycolate to glycolate takes place in chloroplast. 2-phosphoglycolate is converted into glycolate by the enzyme phosphoglycolate phosphatase
option D All of the above is the application of germplasm
Option A
Explant culture is an in-vitro technique to organotypically culture cells from a piece or pieces of tissue or organ removed from a plant or animal.
Option D Vitamin B12
Option D
Penicillium is a genus of ascomycetous fungi of major importance in drug production. Members of the genus produce penicillin, a molecule that is used as an antibiotic, which kills or stops the growth of certain kinds of bacteria inside the body.
Option D
Biotechnology deals with techniques of using live organisms OR enzymes from organisms to produce products and processes that are useful to humans. e.g. A small piece of circular DNA called a plasmid is extracted from the bacteria or yeast cell. A small section is then cut out of the circular plasmid by restriction enzymes, 'molecular scissors'. The gene for human insulin is inserted into the gap in the plasmid. Researchers return the plasmid to the bacteria and put the “recombinant” bacteria in large fermentation tanks. There, the recombinant bacteria begin producing human insulin.
Option A
Restriction enzyme, also called restriction endonuclease, is a group of enzyme belonging to the class nuclease, that cleaves DNA at specific sites along the molecule.
Option D DNA fingerprinting can be used for all of the above
Option C
Herbert Boyer (1936-) and Stanley N. Cohen (1935-) developed recombinant DNA technology although many other scientists made important contributions to the new technology as well.
Option B recombinant DNA technique
Option A
In green plants, manganese plays a very important role in the photosynthesis process by splitting the water molecule to liberate oxygen. It is absorbed in the form of manganous ions (Mn²⁺). It also plays important role in respiration and nitrogen fixation (but a majority of plant species able to form nitrogen-fixing root nodules are in the legume family) whereas photosynthesis happens in all green plants. hence option A is a better choice
Turgor pressure is the force within the cell that pushes the plasma membrane against the cell wall. It is the turgor pressure in the plant cells which helps the plants to be erect.
Wall pressure is the pressure exerted on the contents of a plant cell by the cell wall that is equal in force and opposite in direction to the turgor pressure.
Option C
Detailed steps in recombinant DNA technology
(a) Selection and isolation of DNA segment of an insert which is to be cloned
(b) Selection of suitable cloning vector, a self-replicating DNA molecule, into which the DNA insert is to be integrated. most commonly used are plasmids and bacteriophages
(c) Fragmentation of the DNA by Restriction EndonucleaseThese are the enzymes that produce internal cuts (cleavage) in the strands of DNA, only within or near some specific sites called recognition sites/recognition sequences/ restriction sites or target sites.
(d) Then Exonuclease enzymes are used to removes nucleotides from the ends of a nucleic acid molecule.
(e) Then DNA ligase enzymes are used to join two fragments of DNA by synthesizing the phosphodiester bond
(f) Then DNA polymerases are used to synthesize a new complementary DNA strand of an existing DNA or RNA template
(g) The recombined DNA molecule is introduced into a suitable host. This process of entry of rec DNA into the host cell is called transformation
(h) Selection of transformed host cells (or recombinant cells) are those host cells which have taken up the recDNA molecule. In this step, the transformed cells are separated from the non-transformed cells by using various methods making use of marker genes
(i) Expression and Multiplication of DNA insert in the host is to be ensured that the foreign DNA inserted into the vector DNA is expressing the desired character in the host cells.
Option B
Triose Phosphate Isomerase (TPI) and Aldolase
option A nitrogen
Option A
Phycobilins are light-capturing bilins found in cyanobacteria and in the chloroplasts of red algae. Phycobilins consist of an open chain of four pyrrole rings (tetrapyrrole) and are structurally similar to the bile pigment bilirubin,
Option D
All of the above happens in cryopreservation
Option D
Saccharomyces cerevisiae (baker's yeast) is used in commercial production of ethanol
Option C Snail
By applying the rule, the sum of two sides of a triangle must be greater than the third side, we get 10+12 > x =>
22 > x => x <22 and 10+x >12, => x > 12-10 =>x > 2 So x can take integer values between 3 and 21
But the problem states that x is the side of an acute-angled triangle. Therefore each angle < 90.
If we take one leg of the triangle as 10 and the other as 12, and the hypotenuse as x, then 12^2 + 10^2 < x^2 (since the angle has to be less than 90) i.e x^2 <244. So possible values of x^2 can be 9,16,25,36,49, 64,1,100,121,144,169, 196, 225 (13 values) The other possibility is x & 10 are the legs and 12 is the hypotenuse.
Then, x^2 +10^2 > 12^2 => x^2 > 144-100 => x^2 >44 ( Then x cannot take the values 9, 16, 25 & 36). Therefore, possible integer values of x satisfying the above conditions are 13-4 = 9 values. Hence Option C is the correct option
Data (or graph) on sales of the branches missing. The problem cannot be answered
Let the breadth of the room be 'b' Therefore, area of the room = 18b m2.
Let 'l' be the length of the carpet. Therfore l = total cost of carpet/ rate of carpet per metre= 810/4.5
Since the carpet covers the room completeley, area of room = area of carpet.
Therefore, 18b = (75/100)*(810/4.5) => b = (75*810)/(18*4.5*100) = 7.5 m Option B
We know that the circumcentre (R) of a triangle is related to the sides of a triangle ABC, with sides a,b & c by the formula
2R = a/sinA=b/sinB= c/sinC. Now R = a=c (given)
2a= a/sinA => sinA = 1/2 therfore <A=30o since <A= <C=30o (AB=AC, given), then <B = 180-2*30 = 120= 2π/3
Option D
Let the distance be 'd' Time taken for going school t1 = d/3 and time taken for coming back t2= d/2
Total time t1 + t2 = 5 Therfore d/3 + d/2 = 5 => 5d/6 = 5 => d=6 Option C
Let C.P = Rs. 100. profit = 22.5% Therefore S.P = Rs. 100 + 22.5/100*100 = Rs. 122.5
When SP is 122.5, profit is 22.5 When SP is Rs. 392, profit = 392*22.5/122.5 = Rs 72.0 Option C
Let the present age of son be x years. Man's age is x + 24. After 2 years man's age will be x+26 and son's age will be x+2.
x+26 = 2(x+2) => x= 22 Therfore, present age of son =22 years. Option B
Option A Sulpher deficiency results in younger leaves turning yellow first, sometimes followed by older leaves. Small, stunted plants are produced. This should not be confused with nitrogen deficiency, which appears first on older leaves and then spreads to the whole plant.
Proabability of geting prize = 10C1/35C1= 10/35 = 2/7 Option C
Investment of Murugan(M):Prasanna(P):Arun(A) = Rs. 8,000:4,000:8,000= 8:4:8
Profit of Prasanna = (Profit X Prasanna's investment X No. of manths he invested)/(M*8 + P*8 + A*6)
= 4005*4*8/(8*8+ 4*8 + 8*6) = 4005*32/144= 890. Option A
option D
Factors affecting transpiration are light, temperature, humidity, wind and soil water
5 men can be selected from 8 men in 8C5 ways and 6 women from 10 women can be selected in 10C6 ways.
therfore, the committee of 5 men and 6 women can be selected in 8C5 X 10C6 ways = 56 X 210 = 11,760 ways.
hence option E
Option D Normal diastolic pressure in healthy individual is 80 mm of Hg whereas normal systolic pressure is 120 mm of Hg
Option B
On June 13, 2013, in the case of the Association for Molecular Pathology v. Myriad Genetics, Inc., the Supreme Court of the United States ruled that human genes cannot be patented in the U.S. because DNA is a "product of nature." The Court decided that because nothing new is created when discovering a gene, there is no intellectual property to protect, so patents cannot be granted. Prior to this ruling, more than 4,300 human genes were patented. The Supreme Court's decision invalidated those gene patents, making the genes accessible for research and for commercial genetic testing.
option D
Option A
Relative velocity of bodies is given by,vAB = vA−vB, where vAB is the velocity of A respect to B,vAand vB is the Velocity of A and B respectively with reference to an observer stationary on the ground
If we assign positive sign to one direction, the body mocing in the opposite direction will be negative.
In the problem, velocity of body-1 is V and body-2 is U. Therefore relative velocity of body-1w.r.t body-2 is
V12 = V-U (since both bodies are moving in same direction) Option D
Radius of the cyliners arein the ratio of 3;2. Let the radius of cylinder-1 be 3r and the cylinder-2 be 2r. and the height be
h1 and h2. Ratio of volume V1:V2= 9:7 Therfore π(3r)2*h1/π(2r)2h2= 9/7 => 9h1/4h2=9/7 => h1/h2= 4/7
Option B
solution
Solution
solution
Length of the rectangle, l = 5m44cm = 544 cm and breadth , b= 3m74cm =374 cm. If we cut the rectangle into 'x' pieces of square without any piece of rectangle left over, then x must be the HCF(highest common factor) of 544 and 374. By dividing x by the length of the rectangle we get 544/34 =16 and by divising along the breadth of the rectangle we get, 374/34=11. So the least number of sqaures that can be cut from the rectangle = 16X11 =176. Option A
solution
Let the average score of the batsamn in 16 innings be 'x'. Therefore, total score in 16 innings is 16x. In the 17th innings he made 87 runs.Hence, the total score in 17 innings = 16+87 and the average score in 17 innings is (16x+87)/17.Since, the average has increased by 3 runs, the new average is x+3. Therfore (16x+87)/17 = x+3 =>16x+87= 17x+51 => x= 36. Hence the average after 17 innings will be 36+3=39 runs
Let the numbers of students in the class be 'n'. The average marks got increased by 1/2 due to the wrong-entry. Total increase = 1/2*n =n/2. Since 63 was wrongly entered as 83, difference = 20 marks. Then,n/2=20 => n=20*2=40
There are 40 students in the class
Let the arera of the larger part be 'a' and smaller part be 'b'. Then a+b=700 and a-b=1/5*(a+b)/2 => a-b = 1/5*350 =>
a-b=70. Solving, the equations we get 2a=770 => a=385. Then b= 700-385=315 Smaller part is 315 hectares
If we draw 2 balls at random then at least 1 red ball can be selected by 2 ways, E1 & E2..
E1= 1 ball is red and other ball is green. E2 = both balls are red. Number of ways 1 red ball and 1 green ball can be selected is 6C1X 9C1 ways Total number of ways by which any 2 balls can be selected is 9+6C2=15C2. Probability of E1 = 6C1X 9C1/15C2= 54/105. Number of ways 2 red balls can be selcted is 6C2 ways.Probability of E2 = 6C2/15C2= 15/105. Therefore, P(E1) + P(E2) = 54/105+15/105= 69/105= 23/35.
Option C
6 machines can 270 botttles in 1 minute. 1 machine can produce 270/6 bottles in 1 min. 10 machines can produce 270*10/6 bottles in 1 minute. Therefore, 10 machines can produce in 4 minutes, 270*10*4/6 = 1,800 bottles Option A
Height of flagstaff:Length of shadow of flagstaff = Height of building:Length of shadow of building.
Therefore, Height of the building= Height of flagstaff X Length of shadow of building./Length of shadow of flagstaff = 17.5*28.75/40.25 = 12.5 m
The possible combinations in the set {0,1,2....10}, whose sum and diffrence will be multiples of 4 are {0,4}, (0,8}, {6,2}, {8,4}, {10,2} and {10,6}is 6. Number of ways any 2 numbers can be picked up from the set of 11 numbers in 11C2 ways = 55 ways. Therefore, the probaliites of picking up the numbers whose sum aswell as differencein multiple of 4 is, 6/55.
Option A
An ellipse with its centre at origin will have it foci as (c,0) and (-c,0). and having a,b as the semi-major and semi-minor axis. eccentricity of the ellipse, e = c/a=1/2(given) a=2c directrix of the ellipse,x = a2/c=-4(given)
4c2/c=-4 => c=-1; a=-2, b2=a2-c2= 4-1=3; b=√3. Then the equation of the ellipse will be x2/(-2)2+y2/3=1=> x2/4 +y2/3 =1. The equation of the normal to the ellipse at the point (x1,y1) is given by a2x/x1-b2y/y1=a2-b2 (x1=1, y1=3/2) given. Then the equation of the normal will be 4x/1-3y/3/2=4-3 => 4x-2y=1 Option B
Person Hours/day No of days
39 5 12
1 5 12X39
1 1 12 X 39 X 5
30 6 (12 X 39 X 5)/(30X6) = 13 days Option D
Let it make n number of rounds. Since the lenght of the rope will be same in both cases, we can write the equation,
70*2π*14= n* 2π*20 ⟹ n = 7*14/2 = 49 rounds oprion A
The arithmetic mean of two numbers a and b, a > b > 0, is (a+b)/2 and gemoteric mean is √ab. As per the problem,
(a+b)/2=5√ab. => a+b = 2*5√ab => Squaring both sides we get, (a+b)^2 = 100ab and (a-b)^2 = (a+b)^2-4ab= 100ab-4ab => (a-b)^2 = 96ab => a-b = √96 √ab Therefore, a+b/a-b = 10√ab/ √96 √ab = 10/96 = 5/2√6
Actual distance = 80.5 X 6.6/0.6 = 80.5 X 11 = 885.5 kms Option A
Sundays in the given month will be on 1st, 8th, 15th, 22nd and 29th i.e. 5 sunday. Total of sunday = 510*5=2550 Total of other days = 240X25 = 6,000. Total 6000+2550= 8550. Average number of visitors per day = 8550/30=285
solution
The problem needs to be corrected as "distance from the origin is √5" and not the distance from the straight is
solution
solution
Equation of the circle passing through the points of intersection of the circles S1= x2+y2+3x+7y+2p -5=0 and
S2=x2+y2+2x+2y-p2 =0 is S1+ λS2=0 where λ ≠ (-1) is an arbitrary real number. The circle passes through the point (1,1) Therefore, 1+1+3+7+2p-5+ λ(1+1+2+2-p2)=0 => 7+2p = λ(p2-6) =>
λ = 7+2p/(p2-6) Now p2 can take any real value except 6, ie. p cannot take value +√6 or -√6 Option C
Horizontal distance travelled by the particle is x= 4sin6t. Therefore, horizontal velocity = vx= dx/dt= 24cos6t.
Vertical distance travelled is y= 4(1-cos6t). Vertical velocity vy= dy/dt= - (-24sin6t)= 24sin6t. Then the resultant velocity
v= √(vx2+vy2) = √(24^2*(sin^2(6t)+cos^2(6t)) = 24*1 = 24 m/s. Therefore distasnce travelled by the particle in 1/4 secs = 24*1/4= 6m Option B
Volume of the iron ball (spherical) coated with ice, V= 4/3πr^3.....(i) where r = (radius of the ball + thickness of ice). The rate of melting of ice at any instant, (dr/dt) can be found out by differentiating the eqn.(i) dV/dt = 4/3*3πr^2dr/dt
=> dV/dt = 4πr^2dr/dt . r=(10+5)=15cm, dV/dt= 50 cm3/min. Therefore, dr/dt = 50/(4π*15*15)= 1/18π cm/min
Thickness of ice decrease @1/18π cm/min
There are four variables- Men(m), hours/day(h), No of days(d) and amount of job(w).
The relationship among the vraibles are as follows. mἀ1/d(inverse proportion) m ἀ 1/h (inverse proportion)
m ἀ w(direct proportion) Therefore, m = Kw/d*h, where K is a constant. hence, m1*d1*h1/w1 =m2*d2*h2/w2 =>
104X30X8/(2/5)= m2X26X9/(3/5) (since, in 30 days 2/5th work was completed and 1-2/5=3/5th work has to be completed in 56-30=26 days ) Therefore, m2= 104X30X8X3/26X9X2 = 160. Hence, 160-104= 54 additional men needed to complete the work within 56 days working 9 hours/day
By definition stress is force per unit area. S= F/A (whereas F=mg, m, being the mass and g is grav. constant) Then S= mg/A= V*ῥ*g/A (V= volume,ῥ=density)= L^3*ῥ*g/L^2 = L*ῥ*g (where L = linear dimensions, height, breadth and depth)
Therefore, S1/S2= L1*ῥ*g /L2*ῥ*g = L1/L2 => S2= S1*L2/L1= S1*9L1/L1 => S2=9S1. Option A
solution
solution
Dispalcement of a particle under constant (non-zero) acceleration is gven by the relation s= s0+udt+1/2dt2 (s0=initial displacement, u=initial velocity, t=time and a=constant non-zero acceleration). Comparing it with the equation y=a+bx+cx2 we get the relation between displacement and time is a quadratic function and therfore the curve is a parabola. Option B
solution
Average weight of all boys in the class = (16*50.25 + 8*45.15)/(16+8)= 11165.2/24 =48.55 kg.
Sale in the 6th month should be Rs. 6500*6-(6435+6927+6855+7230+6562) = Rs. 4,991
Morphine is an an opioid (also called a narcotic) analgesic, prescribed to post-sugical patients not responding to NSAIDS(Non-steroidal anti-inflammatories). Morphine is highly addictive(habbit forming). So the patient (cousin) is suffering from morphine withdrawl symptom. Had he continued with the same medication he would have become addicted to morphine which could have finally led to morphine dependence, i.e he cannot live without daily dose of morphine.
The dose and the frequency of the morphine should be reduced gradually over a period of 1-2 weeks, However dose should not be reduced more than fifty-percent at a time and slowly tapering-off of the dose will help him to overcome morphine addiction
Lack of the enzyme ADA(Adenosine deaminsae) can cause severe immunodeficiency. It is caused by mutation in the genes of x-chromosome.
The preferred treatment is bone marrow/stem cell transplantation from an HLA-identical healthy sibling or close relative.
Second option is Enzyme replacement therapy (ERT) with polyethylene glycol-modified bovine adenosine deaminase (PEG-ADA)
The permanent cure is introducing ADAcDNA into cells at an early embryonic stage
We know distance travelled s= ut + 0.5at^2, where u=initial velcity=0; as body started from rest, a=constant acceleration, t=time . S1= (0)t+0.5a(10)^2= 50a. Volocity after initial 10 seconds,V= u +at = 0 + 10a =10a
Distance traveled in the last 10 secs, S2= v(10) + 0.5a(10)^2= 10a(10) + 50a = 150a = 3S1 Option B
solution
Some data is missing in the problem. The equation of the straight line should be xcosα+ysinα=2 (alpha is missing)
solution
solution
solution
solution
Let the unit digit be y and ten's digit be x. Then the number is 10x+y. and x+y = 12 (given)....(i)
When the digits are reversed, the new number is 10y+x Now, 10y+x= 10x+y+36 (given) = > 9y-9x= 36
= > y-x =4 ....(ii) Solving (i) and (ii) we get,y=8, x=4 so the original number is 48. Option C
Let A be the set of familieshaving at least one car and B be the set of families having at least two cars.
Then the number of families having exactly one car is A'= A-B= 200-60=140
13/n+18/n+29/n= 1/n(13+18+29)= 60/n. For 60/n to be an integer, n must be a factor of 60 i.e 1,2,3,4,5,6,10,12,15,20, 30,60 There are 12 positive integers
3n+9n+11n>200 = > 23n>200 n>200/23 = > n>8 and n < or = 20 Between 9 and 20 (both inclusive) there are 12 integers
Vsound = √ɣRT/M where ɣ= adiabatic constant , R gas constant, M molecular mass of gas, T absolute tjemperature of the gas Therefore,. Vsound is directly proportional to sqare root of absolute temeperature of air Option D
Sound wave frequency 20 - 20,000 Hertz Radio wave frequency 20 kHz to 300 GHz Therfore, radio wave has higher frequency than sound wave. Option B
T = 1 / f = 1/20 = 0.05 secs Option D
Option D
Option C
4% blanking means 96% the visible trace time = 0.96*2.0 = 1.92 ms Option A
calling rate = number of calls per subscriber = 10,000/9,000= 10/9 Option B
Option D
Relation between total power and carrier power is given by Pt = Pc (1 + m 2 /2) where Pt =Total power Pc =Carrier power m =Modulation index Pt= 400(1+(0.75)^2/2) = 512.5 W Option B
The elementary charge, usually denoted as e (or q) is 1.6×10−19 C Since the body has positive charge, there is a deficiency of 8 * 10-19/1.6 * 10-19 = 5.0 electrons Option B
Circumference of circle A (whose radius is r)=1(4/7)= 11/7*4a (a being the side of a square). Area of the square = 784.
Therefore, a2= 784, = > a = 28. 2*22/7r= 11/7 *4*28 (given) r = 28 Diameter of circle B is 0.5*28=14. rb= 7.
Area of circle B = 22/7*7*7= 22*7= 154 sq cm
Let the number of blue marbles be 'b' and green marbles be 'g'. P(g) = 2/3. Therefore, (gC1)/(24C1)=2/3 = > gC1= 16.
Then g=16 Therefore, b= 24-16=8 There are 8 blue marbles
xdy-ydx+y2dx=0 = > xdy = (y-y2)dx = > dy/(y-y2) = dx/x = > Integrating we get ln|y| - ln |y-1| = ln x + lnC
= > ln|y/(y-1)| = lnxC = > y/(y-1) = Cx
From 960 litres of pure milk, 48 litres is removed and replaced by water = > fraction of milk will become (1-48/960) = 19/20 of orginal. If the process is repeated 'n' times, fraction of milk will become (19/20)^n. Since the process is repeated thrice, amount of pure milk will become (19/20)^3* 960 = 823.08 ltr. Option D
Let the principal amount be Rs. 100. C.I. Amount = 100(1+0.1)^2 = 121. If C.I. = S.I., then, 121 = 100+100*(10/100)*t
= > 10t= 121-100 = > t = 21/10 = 2.1 Option C
Solution
Let E1 and E2 be the event of drawing a bicycle from plant 1 & 2 respectively Let S be the event of drawing a standard bicycle. P(E1)= 60/100 =0.6 P(E2)= 40/100 = 0.4 P(S/E1)= 80/100 =0.8 P(S/E2)= 90/100 =0.9 Since events E1 & E2 are mutually exclusice, P(standard bicycle produced by plant-2) = P(E2)* P(S/E2)/[P(E1)*P(S/E1) + P(E2)*P(S/E2)]= 0.36/0.84 = 3/7
solution
Face value of the bill = Rs. 7,65.00 and discounted value of the bill = Rs 7,497.00 Period = 73 days (difference 18/05/2003 and 8/03/2003, both days inclusive) Dsicount amount = 7650-7497= 153. Rate of interest = (Interest X100)/[PrincipalXtime(yrs)](153*365*100)/(7650*73) = 10%. The banker charged 10% interest
total no of eggs=16. No of rotten eggs=5, no of good eggs =16-5 =11. Probability of one good egg and one rotten egg drawn is (11C1 X 5C1)/16C2 = 11 X 5 X2/16 X 15 = 11/24. Option A
solution
Let the cost price of each bed be x. Total cost price of two beds = 2x. S.P of bed-1 = 1.25x and S.P of bed-2
= (1.25x- 6596) Overall profit is 8% Then 1.25x + 1.25x- 6596 = 1.08*(2x) = > 2.5x -2.16x = 6596 = > 0.34x= 6596,
= > x = 6596*100/34 = 19,400. Cost of each bed is Ts 19,400/- Option D
solution
The gradiant of the curve x2=12y is dy/dx. 2xdx =12dy = > dy/dx = 2x/12 = x/6. Now the grandiant of the curve is equal to the gradiant of the tangent to the curve x=y+k. Rewriting the equation we get y = x-k. So the gradiant of the line is 1 (since coefficient of x=1) Therefore, x/6=1 = > x=6 and 12y =36, y=3. Then k = 6-3=3. Therfore, K=3 and point of contact is (6,3)
The given euation dy/dx= xy + x + y +1 can be rewritten as dy/dx = xy+y +y+1 = > dy/dx = y(x+1) +1(y+1) = >
dy/dx = (x+1)(y+1) Therefore, ʃ dy/(y+1)= ʃ (x+1)dx By integrating we get ln |y+1| = x2/2 + x +c = >
e^(x2/2+x+c) = y+1 = > y = e^(x2/2+x+c) - 1
Let the number of units produced is 'x'. Total Cost for producing 'x' units is Y = 900 + 3x + x2/100. Then average cost for producing x units will be Y/x = 900/x+3 + x/100 By differentiating we get dy/dx= -900/x2 + 1/100 If average cost is minimum, dy/dx =0 ...(i) and d2y/dx2 = +1800/(x^3)+1/100 >0 From (i) we get -900/x2 + 1/100 = 0 = > 900/x2 = 1/100
= > x2 = 90000 = > x =300 Hence the average cost will be minimum when for x=300 units.
Probability of picking a coin from either of the two bags is 1/2. So the probability of getting a silver coin in purse1= 1/2*2/6=1/6. Probability of getting a silver coin from 2nd purse is 1/2*4/7 = 2/7. So probability of geeting a silver coin from one of the two purses is 1/6+2/7= 19/42. Similarly probability of getting a golod coin is 1-19/42= 23/42.
The relation between t and distance x is t=ax2+bx....(i) Instantenous velocity = dx/dt and nstantenous acceleration= d2x/dt2. By differentiating eqn...(i) we get, dt = a(2x)dx + bdx = > dt = (2ax +b)dx = > Therefore, Instantenous velocity = dx/dt = 1/(2ax+b) ... (ii) By futher differentiating we get d2x/dt2 = d/dx((2ax+b)-1) = > (-1/(2ax+b)2)2a = >
Instantenuous accln. d2x/dt2 = -2a/(2ax+b)2*2a(dx/dt) = -2a/(2x+b)2*1/(2ax+b) = -2a/(2ax+b)3
Option B Triangular law of vector addition ( same as parallelogram law of vector addition) helps to resolve vectors into a single resulatant vector
General equation of a pair of straight lines is ax²+2hxy+by²+2gx+2fy+c=0. Now the given eqn may be rearranged as
kx² + 2(-½)xy+(-1)y² +2(-3/2)x+2(3/2)y+0=0 ... eqn (i) = > a=k, h= -½ b= -1 g= -3/2 f= 3/2 c=0 Eqn..(i) will represent a pair of straight lines only if abc + 2fgh – af² – bg² – ch² = 0 then k(-1)(0) +2(3/2)(-3/2)(-½) -K(3/2)²-(-1)(-3/2)²-(0)(-½)²=0
= > 0+9/4 -k(9/4) +9/4 -0 =0 = > k(9/4) = 2(9/4) = > k = 2
solution
Solution
Solution
Option B Vector quantity has both magnitude and direction while scalar quantity has only magnitude
Option A A unit vector is a vector of magnitude (length)= 1, also called a direction vector
Option A Vector addition is commutative (does not depend on order of displacement) and also associative
Option C
|A+B|=|A-B| (given) Square the magnitude of the vectors. (A+B)² = (A-B)² = > A²+B² +2AB = A²+B²-2AB = > 4A.B=0
= > A.B =0 = |A| X |B| X cos(90) [cos(90)=0] Hence |A| and |B| are perpendicular Option B
The rank co-efficient,r = 1 - 6∑d²/n(n²-1) r=0.2, n=10 (given) 0.5 = 1 -6∑d²/10*(100-1) = > ∑d² = 0.5*990/6= 82.5
Corrected ∑d² = 82.5 - 3² +7² = 82.5-9+49 = 122.5. Then corrected, r = 1-6*122.5/10*99 = 1-.7424 = 0.2576.
The correct coefficient of rank of correlation is 0.2576.
Solution
C |D E| F |G H| I |J K| L 4 9 (5) 16(7) 25(9) O |P Q| R |S T| U |V W| X
The next term should be L25O
Option B It came from the Italian word "banco" means bench because italian merchants made deals to lend and borrow money sitting on a bench
Option B (RBI)
4/2=2 12/4=3 48/12 =4 A |BC|D |EF|G |HI|J 11 13 (2) 17(4) 23(6)
Next term is 48J23
A B C D E FG H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9101112 13141516 17181920 21222324 2526
Difference between the letters of the next term in the series is 5 I-D=5, J-E=5, N-I=5
Next term in series is STQ
Z X V T R P ....N A D G J M P... S 1 2 6 21 88 445
(1+1)*1 (2+1)*2 (6+1)*3 (21+1)*4 (88+1)*5 .....(445+1)*6
Next term in the series N2676S
t1 t2 t3 t4 t5 t6 t7
2 8 20 48 92 188 380 t2-t1 = 6 t3-t2 =12 t4-t3 =24 t5-t4= 48 t6-t5 =96 t7-t6 = 192
t2= t1 + (t2-t1)= 2 +6 t3= t2 + (t3-t2) =8+12 ..... tn = tn-1 + (tn-tn-1) t7 = 188+ 192=380
Hence t4 = t3+ t4-t3 = 20 +24 =44 and not 48 Hence 48 is the wrong term
120 99 80 63 48 Difference between each consecutive terms are 21 19 17 15.... 13 48-13 =35.
Hence the next term in the series is 35 and the series will be 120 99 80 63 48 35
No. K2, also known as Mount Godwin-Austen is the second highest mountain in the world, after Mount Everest. It is located on the China–Pakistan border
Kanchenjunga is the is the third highest mountain in the world, and lies partly in Nepal and partly India (Sikkim state)
100:50 : : 30:a => a= 15 Fourth proportional is 15
Let hte time taken by car A be 12t and car B be 15t Since they cover same distance VA*12t =VB*15t =>
VB = 60*12t/15t= 48 km/hr
4^3.5/2^5 = 2^2*3*5/2^5 = 2^30/2^5 = 2^25 Ratio= 2^25
LCM of 12,25,45 and 60 =900. The smallest 5 digit number is 1,0000. Divide 10000/900 = 11.11. Take the next higher interger =12. Multiply 900 by 12= 10,800 the smallest 5-digit number perfectly divisible by 12,25,45 and 60
48-38=10, 60-50=10, 72-62=10, 108-98=10,110-100=10 The difference is 10 in all the cases.
So the required number is (LCM of 48,60,72,108,110)-10 = 23,760-10 =23,750
Dimension of the room is 15.17 m=1517 cm and 9.02m =902 cm. The size of the square tile will be
1517,X902/(HCF of 1517,902)² = 1,368,334/41² = 814cm =8.14m Largest possible tile should be 8.14m
Two numbers are in the ratio of 5:6. LCM = HCF X (Product Ratio) =5*5*6 = 150
LCM= HCFX (Product ratio) = HCF*3*4. LCMXHCF= 10,800 = > HCF² = 10,800/3*4 = 900. = > HCF = 30
Then the numbers are 3*30 = 90 and 4*30=120. Sum of the numbers are 120+90=210
HCF of two numners is 32. So the numbers must be multiple of 32. Let the numbers be x= 32a and y=32b
x+y=192 => 32a+32b=192 a+b =192/32 =6. The possible combinations which can give 6 are as follows. (1,5) (2,4) (3,3) (4,2) (5,1) So 5 such pairs can be formed
A number is divisible by ( only when sim of the digits of the number is divisible by 9.
187 → 1+8+7 =25 not divisible by 9. 648 → 6+4+8 = 18 divisible by 9 543 → 5+4+3 =10 Not divisible by 9
837 → 8+3+7 =18, divisible by 9 Therfore, only two numbers among above numbers are divisible by 9
If two ingredients are mixed in a ratio, quantity of cheaper/quantity of dearer= [(C.P of dearer)- Mean price]/[Mean price - (C.P of cheaper)] Using the same a nalogy for the above problem, we get 42 27 33 y x
Let the numbers be 3x and 2x. Their LCM= 6x. 6x=240 (given) x=40. The smaller number is 2(40)=80
dividend = divisor × quotient + remainder 1265 = d* 84 +5 => 84d =1260 => d=1260/84=15. The required number is 15.
The product of two prime number is always equal to the LCM of these two number. Hence LCM=493
The smalles number divisible by all odd numbers upto 15 = LCM of (1,3,5,7,9,11,13,15) = 45,045.
Statement -1 gives the C.P of each calculator and S.P of each calculator known. Hence we can profit from Statement -1
Statement -II gives the profit in % and S.P is known. So we can find the Profit.
Hence either statement -1 or statement -II alone is sufficient to answer the question. Option D
From 1 we get the number of days men will take to complete the work. Either of II or III gives the number of days women will tke to complete hte work.
Combining I and either II or III (II or III gives the same info) we get work of 1 man = work of 2 woman. Hence the question can be answered
Statement I is not sufficient to answer the question whereas statement-II is sufficient to answer the question. Option B
First generation antihistamines includes diphenhydramine (Benadryl), carbinoxamine (Clistin), clemastine (Taregyl), chlorpheniramine (Cadistin-Zydus)), and brompheniramine (Brovex, Bromaphen). These are older antihistamines and these are non- selective H1 blocker which can cross blood-brain barrier and blocks cholinergic receptors also. These agents also commonly have action at α-adrenergic receptors and/or 5-HT receptors and can produce sedation.
Rate of interest in scheme A =8% and B=9%
Every Re. 1, invested in A after 2 years will become 1*1.08*1.08= 1.1664
And if invested in B it will become 1*1.09*1.09= 1.1881.
If we convert it to simple interest, S.I for A =16.64/2= 8.32% and for B=18.81/2= 9.405%
Total interest on Rs. 27,000 for 2 years is Rs. 4818.3
Effective S.I= 4818.3/27,000*100/2=8.923%, which is the weighted average of the schemes A&B.
By applying allegation rule we get,
(B)9.405 8.32(A)
8.923
0.603 : 0.487
Therefore, amount invested = 0.482/(0.482+0.603)*27,000= 11,944 (round off to) Rs.12,000
Since the 4 strips has to be arranged in a square, each side of the square will be √(4X45X5) = 30 cm
So the area of the 4 strips is 30X30= 900 cm2
M=3 litres N= 5 litres. Divided into 4 equal parts. so in each part M=(1/4)*(3/8) and N= (1/4)*(5/8) Now 'x' amount of M is added to one part. Therefore, (1/4)*(3/8)+x/(1/4)*(5/8)= 6/5 => 3+4x/5= 6/5 => 4x=3 => x=3/4. (3/4 litres M is added)
New ratio of M/N= (3+3/4)/5= 15/20 = 3/4 Hence M:N = 3:4
A/B= 7/17 and B/C= 7/17 Therefore A/B/C = 7*7/17*7/17*17 = 49/119/289 (By taking LCM of the values of B)
But A=4900(given) therfore C= 28,900
Let the price of the bike be Rs 100. If the price is increased by 25% the new price will be 100+25/100*100=Rs 125.
To restore the original price, the price must be decreased by Rs 25 which is 25/125*100= 20%
Average speed = 1/[1/4/50+1/3/60+{1-(1/4+1/3)/100}]=1/[1/200+1/180+5/1200]= 1/[53/3600]= 3600/53=67.92 km/hr
Biggest number = 7(1+2+5+7)*135= 7/15*135= 63
By aligation method
280 ↘ ↙ 240
x
7 ↙ ↘ 9 => 9/7 = (280-x)/(x-240) => 16x= 4120 => x= 4120/16= 27.50
Average price of per kg mixed rice is Rs. 27.50
In mixture -1, water part = 1/2 and in mixture-II, water part is 2/5. After mixing, water part in the new mixture will be
1/2 2/5
x = water part => 2= (0.5-x)/(x-0.4) => 2x-0.8 = 0.5-x => 3x= 1.3 => x= 13/30 Therfore milk part will be (30-13)/30
20 : 40 =17/30 Therfore in the new mixture ratio of water:milk= 13;17
Let S.P. of both types of rice (R1 & R2) be Rs.100 per kg. On R1, 10% profit and on R2 10% loss incurred.
C.P of R1= (100/1.1)=Rs.90.9/kg and C.P of R2 = (100/0.9)=Rs.111.11./kg.
S.P of 1 Kg of R1 + 1 kg of R2 = 90.9+111.11= Rs. 202.00 Profit = 200-202 = -2 (loss of Rs 2)
% loss 2/202*100 = 0.99 approx 1%
Let the C.P of the watch be Rs 100. Sold for 15% profit. Therfore S.P =115. If she had purchased for 10% less the then
C.P(new) = Rs. 90 and sold for 20% profit. S.P(new) = 1.2*0.9 =108 Therfore he she should have sold for Rs (115-108)=
Rs 7 less.But actually she sold for Rs 3.5 less(1/2 of the assumed value). Therefore, actaul price of the watch was Rs 50(1/2 of the assumed value)
S.P= C.P ( goods sold at cost price.) Let C.P and S.P of 1 kg(=1000 gms) of goods be Re 1.
When Harrison picks the wrong weight, he sells 1100 gms but takes price for only 1000 gms.
When he picks up the wrong weight his C.P(wrong),will be Rs 1.10 but S.P still be Rs 1.0
Probability of picking wrong weight is 3/5 and right weight is 2/5.
So his average cost price (x)can be found out by applying allegation technique.
1.1 1.0
x
3/5 2/5 2/5:3/5= (1.1-x):(x-1.0) => 2x-2 = 3.3- 3x => 5x = 5.3 => x= 5.3/5
Average C.P = 5.3/5 and S.P=1, Loss% = (1-5.3/5)/(5.3/5)*100=> 0.3*100/5.3 =5.66%
Let S.P= C.P = Rs .100. Since the seller is making profit, he is selling less quantity for every 1 kg.
Since he is making 28% profit, his actaul cost price is Rs 100/1.28 = Rs .78.125.
For Rs. 100, he can buy 1000 gms. For Rs 78.125, he buys 78.125/100*1000= 781.25 gms.
Therfore he uses 781.25 gms weight for every Kg
Let the speed of the man in still water be x km/hr and time be t hours
Relative Speed downstream = x+3 and upstream = x-3.
Since the downstream distance is double the upstream distance
(x+3)*t = 2(x-3)*t => x+3 = 2x-6 => x=9.
Speed of the man in still water is 9 km/hr
36 30
x
15 25 C.P of the mixture (x) =>25/15= (36-x)/(x-30) => 5(x-30) = 3(36-x) => 5x-150 = 108-3x =>
x=258/8 = 32.25 Therfore, S.P. =1.1X32.25(10% profit over C.P)= Rs.35.47
He should sell I kg of mixture at Rs 35.47
For Alloy-1, C:Z=4:3 C= 4/7 In 14 kg of alloy-1, 8 kg copper For alloy-2, C:T =9:5 In 14 kg of alloy-2, 5 kg tin. When equal weights of two alloys are mixed together, in 28 kg of new alloy, 5 kg of tin. Therfore, in 1 kg of new alloy, 5/28 kg tin.
Initial velocity 'u'=0, final velocity 'v'= 8 m/s. acceleration ='a' Acceleration time 't1'= 2 s Uniform speed travel time 't2'=3
Diispalcement during acceleration ='s' and during uniform speed = s1.
v=u + at => 8 = 0 + a(2) => a= 8/2 = 4m/s² s= ut +0.5 at² => s= 0 + 0.5*4*2²= 8 m S1= 8*3=24m
Total dispalcement = s+s1= 8+24=32m
Ice at zero degree needs latent heat of 334 joules per gram of ice to convert into water at same temperature ( zero degree)
Since ice can absorb this heat from the enivironment, it has more cooling effectiveness compared to water at same temperature
T(°C) = T(K) - 273.15 => T(K)= 60+273.15= 333.15°K
Post your requirement in LearnPick