Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles. All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O.

D. All of above

A. Cry

Acrylic paint is a fast-drying paint made of pigment suspended in acrylic polymer emulsion. Acrylic paints are water-soluble, but become water-resistant when dry. Depending on how much the paint is diluted with water, or modified with acrylic gels, media, or pastes, the finished acrylic painting can resemble a watercolor or an oil painting, or have its own unique characteristics not attainable with other media.Acrylic paint is typically used for crafting, or in art classes in schools because it does not require any chemicals, and rinses away with just water. It also is less likely to leave a stain on clothes than oil paint.

John Osborne

Joseph Conrad

C. Significant decrease in platelets count. In classical dengue fever, it usually starts to fall by 3 days, most significant by 5 days and comes back to normal (if no complications arise)by 11 days.

D. Chloroplast. Conversion of phosphoglycolate to glycote takes place in chloroplast. Photorespiration is the uptake of O2and release of CO2in light that results from the biosynthesis of glycolate in chloroplast and subsequent metabolism of glycolic acid in the same leaf cell through other two cell organelles (peroxisome and mitochondria).

D. Violet and blue The visible spectrum of solar radiations are primarily absorbed by carotenoids of the higher plants are violet and blue.

C. toxin is inactive: In bacteria, the toxin is present in an inactive form, called prototoxin, which gets converted into active form when it enters the body of an insect.

A.1D

(A+B)-(B+C)=12, A-C=12 Therefor, C is younger than A by 12 years....

ratio sooraj vimal 1- 6 7 diff 1 4+ 7 8 diff 1 multiply them with others diff 1- 6:7 4+ 7:8 take diff of anyofthem 7-6=1 8-7=1 actully -1+4=5 years is diff multiply all answers with 5 1- 30 35 5+ 35 40 so vimal now is35+1=36 or40-4=36

Let Syam=x Sunil = y , According to question, x/y= 240.......(1) 2y-x= 44........(2). x=240/y.. Solving eq. (1)&(2)... 2y-240/y=4. y2-120=2y. y2-2y-120=0. y2-12y+10y-120=0 Y(y-12)+10(y-12). (y-12) (y+10). Y= 12, -10.. Therefor... Y= 12 years. So, Sunil's age would be 12years

D. 12 years Solution...Let the present age of the man and his wife be 4 x 4x and 3 x 3x respectively. After 4 4 years this ratio will become 9 : 7 97 ⇒ ( 4 x + 4 ) : ( 3 x + 4 ) = 9 : 7 ⇒ 7 ( 4 x + 4 ) = 9 ( 3 x + 4 ) ⇒ 28 x + 28 = 27 x + 36 ⇒ x = 8 4x43x497 74x493x4 28x2827x36 x8 Present age of the man = 4 x = 4 × 8 = 32 4x4832 Present age of his wife = 3 x = 3 × 8 = 24 3x3824 Assume that they got married before t t years. Then, ( 32 − t ) : ( 24 − t ) = 5 : 3 ⇒ 3 ( 32 − t ) = 5 ( 24 − t ) ⇒ 96 − 3 t = 120 − 5 t ⇒ 2 t = 24 ⇒ t = 24 2 = 12

Present age of Denis = 55 years Present age of Rahul = 55 − 22 = 33 Let present age of Ajay = x x Then, present age of Rahul = x − 6 18 x618 x − 66 18 = 33 ⇒ x − 6 = 3 × 18 = 54 ⇒ x = 54 + 6 = 60

Explanation: Let age of the son before 8 8 years = x x Then, age of Kamal before 8 8 years ago = 4 x 4x After 8 8 years, Kamal will be twice as old as his son ⇒ 4 x + 16 = 2 ( x + 16 ) ⇒ x = 8 4x162x16 x8 Present age of Kamal = 4 x + 8 = 4 × 8 + 8 = 40 4x848840

Let the present age the son = x Then, present age of the father = 3 x + 3 Given that, three years hence, father's age will be 10 10 years more than twice the age of the son ⇒ ( 3 x + 3 + 3 ) = 2 ( x + 3 ) + 10 ⇒ x = 10 3x332x310 x10 Father's present age = 3 x + 3 = 3 × 10 + 3 = 33

C. 30. Explanation: Let present age of the elder person = x x and present age of the younger person = x − 16 x16 ( x − 6 ) = 3 ( x − 16 − 6 ) ⇒ x − 6 = 3 x − 66 ⇒ 2 x = 60 ⇒ x = 60 2 = 30

Density of lake water = 1.25 g mL-1 Amount of Na+ ion in one kg of water = 92 g Atomic mass of Na+ = 23 g mo1-1 Let V ml of lake water be taken. Let take mass of solvent(water) = 1 Kg Then mass of Na+ ions = 92 g Number of moles of = 92/23 = 4 (molar mass of Na is 23g/mol) Plug the values we get Molality = 4 /1 = 4m Solution molality of Na + ions = 4m

Let the ages of father and son 10 years ago be 3x and x years respectively. Then, (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40 x = 20. Required ratio = (3x + 10) : (x + 10) = 70 : 30

Mother's age when Sobha's brother was born = 36 years. Father's age when Sobha's brother was born = (38 + 4) years = 42 years. Required difference = (42 - 36) years = 6 years.

A. Data inadequate... Explanation: B- A = A- C (B+ C) = 2A Now given that, (B + C) = 50 So, 50 = 2Aand therefore A= 25. Here we know the value(age) of A (25), but we don't know the age of B Therefore, (B-A) cannot be determined.

C. 40. Let the mother's present age be x years. Then, the person's present age = 2 x years. 5 2 x + 8 = 1 (x + 8) 5 2 2(2x + 40) = 5(x + 8) x = 40.

The basicity of amines depends on the +I effect of the alkyl groups. With increase in alkyl group the +I effect will increase which will increase the ease of donation of lone pair electron. Amine will accept a proton and from cation will be stabilized in water by salvation (by hydrogen bonding).better the salvation by hydrogen bonding higher will be the basic strength. Hence, alkyl amine is more basic than ammonia CH3NH2 > NH3

Let their present ages be 4x, 7x and 9x years respectively. Then, (4x - 8) + (7x - 8) + (9x - 8) = 56 20x = 80 x = 4. Their present ages are 4x (a)= 16 years, 7x(b) = 28 years and 9x(c) = 36 years respectively.

Answer : 7.30gcm−37.30gcm3 In a body-centred cubic lattice ,atoms touch each other along the body-diagonal of the cube.Hence 4r=3–√a4r3a or r=3–√a4=(1.732)(287pm)4=r3a41.732287pm4124.27124.27pm. The expression of density is ρ=Na3(MNA)ρNa3MNA Substituting the values,we get ρ=2(287×10−10cm)3(51.99gmol−16.023×1023mol−1)ρ22871010cm351.99gmol16.0231023mol1=7.30

D. 32 years FFather's age was 28 28 years when my sister was born. My sister's age was 4 4 years when my brother was born. Therefore, father's age was 28 + 4 = 32 32 years when my brother was born.

For total miscible liquids, Ptotal = Mol. fraction of A × pA0 + Mol. fraction of B × pB0 No. of moles of A = 28/140 Liquid B is water. its mass is (100−28), i.e., 72. No. of moles of B = 72/18 Total number of moles = 0.2 + 4.0 = 4.2 Given, Ptotal = 160 mm pB0 = 150 mm So 160 = 0.2/4.2 × pB0 + 4.0/4.2 × 150 pA0 = 17.15×4.2/0.2 = 360.15 mm

Answer: Option A Explanation: After 6 6 years, Shekhar's age will be 26 26 years Therefore, Present age of Shekhar = 26 − 6 = 20 26620 Let present age of Shobha = x x Then, 20 x = 4 3 x = 20 × 3 4 = 15

Answer: Option C Explanation: Given that, six years ago, the ratio of the ages of Vimal and Saroj = 6 : 5 65 Hence we can assume that age of Vimal six years ago = 6 x 6x age of Saroj six years ago = 5 x 5x After 4 4 years, the ratio of their ages = 11 : 10 1110 ⇒ ( 6 x + 10 ) : ( 5 x + 10 ) = 11 : 10 ⇒ 10 ( 6 x + 10 ) = 11 ( 5 x + 10 ) ⇒ 5 x = 10 ⇒ x = 10 5 = 2 Saroj's present age = ( 5 x + 6 ) = 5 × 2 + 6 = 16

Option D Explanation: Let the present ages of son and father be x and (60 -x) years respectively. Then, (60 - x) - 6 = 5(x - 6) 54 - x = 5x - 30 6x = 84 x = 14. Son's age after 6 years = (x+ 6) = 20 years..

Let age of Kiran and Bineesh be 7 x 7x and 9 x 9x respectively 7 x = 9 x − 7 x = 7 2 = 3.5 7x9x7 x723.5 Kiran's age = 7 x = 7 × 3.5 = 24.5

Answer: Option A Explanation: We are given with length and area, so we can find the breadth. as Length * Breadth = Area => 20 * Breadth = 680 => Breadth = 34 feet Area to be fenced = 2B + L = 2*34 + 20 = 88 feet

Answer: Option B Explanation: Consider a square plot as shown above. Let length of each side = 1 1 Then, length of the diagonal = √ 1 2 + 1 2 = √ 2 12122 Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2 112 Distance travelled if walked diagonally = BD = √ 2 = 1.41 21.41 Distance saved = 2 − 1.41 = 0.59 21.410.59 Percent distance saved = 0.59 2 × 100 = 0.59 × 50 ≈ 30 %

Answer: D) 50% increase Explanation: Let original length = x and original breadth = y. Original area = xy. New length = x/2 and New breadth=3y New area = Increase in area = New area-original area = Increase % =50 %

Let original length = x and original width = y Decrease in area will be =xy−(80x100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%

Answer: Option C Explanation: Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 x2 - 100x + 291 = 0 (x - 97)(x - 3) = 0 x = 3.

100 cm is read as 102 cm. A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 - A1) = [(102)2 - (100)2] = (102 + 100) x (102 - 100) = 404 cm2. Percentage error = 404 x 100 % = 4.04%

A circular loop behaves like a dipole with one surface North pole and One surface south pole. The lines force emerges from N and will meet at S So net flux through X−YXY plane is zero.

Perimeter = Distance covered in 8 min. = 12000 x 8 m = 1600 m. 60 Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2.

Ans : 885.5 km Solution: .6 cm in map = actual distance of 6.6 km 1 cm in map =6.6/.6 km 80.5 cm in map =80.5x6.6/.6 km = 885.5 km

price of 357 apples = Rs.1517.25price of 1 apple = Rs. 1517.25357price of 49 dozens apples = Rs.(49×12×1517.25357)≈Rs. 2500

Answer : Option B Explanation : Given that fort had provision of food for 150 men for 45 days Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days Remaining men after 10 days = 150 - 25 = 125 Assume that after 10 days,the remaining food is sufficient for 125 men for x days More men, Less days (Indirect Proportion) ⇒Men150:125}::x:35 ⇒150×35=125x⇒6×35=5x⇒x=6×7=42⇒The remaining food is sufficient for 125 men for 42 days

Answer: Work done = 5 8 Balance work = 1 - 5 = 3 8 8 Let the required number of days be x. Then, 5 : 3 = :: 10 : x 5 x x = 3 x 10 8 8 8 8 x = 3 x 10 x 8 8 5 x = 6.

Let the required number of bottles be x. More machines, More bottles (Direct Proportion) More minutes, More bottles (Direct Proportion) Machines 6 : 10 :: 270 : x Time (in minutes) 1 : 4 6 x 1 x x = 10 x 4 x 270 x = (10 x 4 x 270) (6) x = 1800.

The hostel had provisions for 250 men for 40 days. This means the total number of meals available were 250 * 40 = 10000. If 50 men left the hostel, the number of men left = 250 - 50 = 200. Now, the number of days food will last = 10000 / 200 = 50. Hence, now the food will last for 50 days.

Total meals available for 500 men= 500*27= 13500. Meals spent= 500*3= 1500. Remaining meals= 13500-1500= 12000. No of days food will last= 12000/800= 15 days.

Ans:- B 2.5 kW Power = word done in one second Now here motor in one second does work in 2 parts i) Lifting of water 10L water = 10 kg as density of water is 1Kg/L so Work = Change in PE = mgh = 10 * 10 * 20 =2kW ii) Now this water is thrown hence given kinetic energy which is equal to work done by the motor = 0.5 * 10 * 100 =0.5 kW adding the two we get the answer

Assume that initially garrison had provisions for x men for y daysSo, after 10 days, garrison had provisions for x men for (y-10) daysAlso, garrison had provisions for 4x5 men for y days (?x−x5=4x5)More men, Less days (Indirect Proportion)⇒Menx:4x5}::y:(y−10) ⇒x(y−10)=4xy5⇒(y−10)=4y5⇒5(y−10)=4y⇒5y−50=4y⇒y=50

tan (θ + β) = 1/2 and tan θ = 1/4 1 / 4 + tan β 1 − 1 / 4 tan β = 1 2 14tanβ114tanβ12 9 tanβ = 2 tan β = 2/9 The correct option is B.

p = 15/25 q = 10/25 n = 10 σ2 = npq = 12/5

Let M be the angle, the arc of the sector AOB makes with the center O. Let OA = OB = radius r. Given OA + Arc AB + OB = 20 --> 2r + rM = 20 --> M = (20-2r)/r. ..... (1) Area A of the sector = (M/2)rr = 10r-(r*r) ---------- From (1) dA/dr = 10 - 2r = 0 ----> r=5 Second derivative is negative for r =5 We get maximum area at r=5 B = 25 ----------- From (1)

Ans- D 485

D. 26 metric tonnes Let required amount of coal be x metric tonnes More engines, more amount of coal (direct proportion) If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption. If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption More rate of consumption, more amount of coal (direct proportion) More hours, more amount of coal(direct proportion) Hence we can write as enginesrate of consumptionhours9:813:148:13⎫⎭⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪::24:x ⇒9×13×8×x=8×14×13×24⇒3×8×x=8×6×13⇒3×x=6×13⇒x=2×13=26

Ans. C-7.8 mm

Decomposition reaction of H2O2 is: H 2 O 2 ⟶ H 2 O + 1 2 O 2 H2O2H2O12O2 2t½ = 50 mins t½ = 25 mins For first order reaction, t½ = 0.693/k k = 0.693/25 d [ H 2 O 2 ] d t = d [ O 2 ] d t × 2 dH2O2dtdO2dt2 d [ O 2 ] d t = 1 2 × 0.693 25 × 0.05 dO2dt120.693250.05 = 6.93 × 10–4

A- 50 days. Assume that initially garrison had provisions for x men for y daysSo, after 10 days, garrison had provisions for x men for (y-10) daysAlso, garrison had provisions for 4x5 men for y days (?x−x5=4x5)More men, Less days (Indirect Proportion)⇒Menx:4x5}::y:(y−10) ⇒x(y−10)=4xy5⇒(y−10)=4y5⇒5(y−10)=4y⇒5y−50=4y⇒y=50

D - 26 metric tonnes. Let required amount of coal be x metric tonnes More engines, more amount of coal (direct proportion) If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption. If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption More rate of consumption, more amount of coal (direct proportion) More hours, more amount of coal(direct proportion) Hence we can write as enginesrate of consumptionhours9:813:148:13⎫⎭⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪::24:x ⇒9×13×8×x=8×14×13×24⇒3×8×x=8×6×13⇒3×x=6×13⇒x=2×13=26

Let required number of hours be x More examiners, less hours(indirect proportion) More days, less hours(indirect proportion) More answer books, more hours(direct proportion) Hence we can write as examinersdaysanswer books9:412:302:1⎫⎭⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪::x:5 ⇒9×12×2×5=4×30×1×x⇒x=9×12×2×54×30×1=9×3×2×530=9×2×510=9

C- 1.818. Solution : A+B⇌C+DABCD 11111111 1−x1−x1+x1+x1x1x1x1x Kc=[C][D][A][B]KcCDAB =(1+x)2(1−x)21x21x2 =100100 => 1+x1−x1x1x=1010 => 1+x=10−10x1x1010x => 11x=911x9 x=911x911 [D]=1+x=1+911D1x1911 =1.818M

C-752.4. Molecular mass of water =2×1+1×16=18g2111618g For 178.2g178.2g water nA=9.9nA9.9 Molecular mass of glucose =6×12+12×1+6×16=180g612121616180g For 18g18g glucose nB=0.1nB0.1 XB=0.10.1+9.9)XB0.10.19.9=0.010.01 XA=0.99XA0.99 For lowering of vapour pressure , p=p0AXA=P0A(1−XB)ppA0XAPA01XB P=760(1−0.01)P76010.01 =760−7.67607.6 =752.4torr

D- -110.5.

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O22 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, what is the formula of the hydrocarbon? please get me out of this problem.

A-. 1.5 δ=i+e−AδieA 40=35+79−A403579A 40=114−A40114A A=114−40A11440 =74=r1+r274r1r2 From this we get, μ=1.5μ1.5 δmin

A. 0.01 Ω. Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt. If resistance of galvanometer is Rg and it gives full-scale deflection when current Ig is passed through it. Then, V = IgRg V = 1 mA × 100 Ω Let a shunt of resistance (Rs) is connected in parallel to galvanometer. If total current through the circuit is I. I = 10 A = Is + Ig V = IsRs = (I - Ig)Rs (I - Ig)Rs = IgRg (10 - 10-3)Rs = 100 × 10-3 Rs ≈ 0.01 Ω

For open end pipe in air, fundamental frequency: f = V/2L For the pipe closed at one end (dipped in water), fundamental frequency: f' = V/4L' L' = L/2 f' = V/2L = f

C 110. Originally let there be x men. Less men, More days (Indirect Proportion) (x-10) : x :: 100 :110 10x= 1100 x= 110

A- 885.5 km. Let the required actual distance be x km More scale distance, More actual distance(direct proportion) Hence we can write as scale distance.6:80.5}::6.6:x ⇒.6x=80.5×6.6⇒.1x=80.5×1.1⇒x=80.5×11=885.5

PRICE OF ONE APPLE = 1517.25/357 = 4.25 PRICE OF ONE DOZEN = 4.25 * 12 = 51 Price of 49 Dozens = 51*49 = Rs 2499/-

D-5:4.

D- 20 times nearer. Telescope creates a focused image of a distant object, so that the object looks nearer. Magnifying power of telescope, M = 20, implies that the tree appears 20 times nearer.

C- C, A, B, D.

D- 0.065 H. For the lamp with direct current, V = IR R = 8 Ω XL = Lω = L × (2πf) = L × 100π Z = √ 8 2 + ( L × ( 100 π ) ) 2 Z82L100π2 Z = V r m s I r m s = 220 10 ZVrmsIrms22010 484 = 64 + 104(π2L2) L = 0.065 H

B-42. Explanation : Given that fort had provision of food for 150 men for 45 days Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days Remaining men after 10 days = 150 - 25 = 125 Assume that after 10 days,the remaining food is sufficient for 125 men for x days More men, Less days (Indirect Proportion) ⇒Men150:125}::x:35 ⇒150×35=125x⇒6×35=5x⇒x=6×7=42⇒The remaining food is sufficient for 125 men for 42 days

Let BD be the lighthouse and A and C be the two points on ground. Then, BD, the height of the lighthouse = 60 m BAD = 45° , BCD = 60° tan 45 ° = BD BA ⇒ 1 = 60 BA ⇒ BA = 60 m ⋯ ( 1 ) tan45BDBA 160BA BA60 m1 tan 60 ° = BD BC ⇒ √ 3 = 60 BC ⇒ BC = 60 √ 3 = 60 × √ 3 √ 3 × √ 3 = 60 √ 3 3 = 20 √ 3 = 20 × 1.73 = 34.6 m ⋯ ( 2 ) tan60BDBC 360BC BC603603336033 203201.7334.6 m2 Distance between the two points A and C = AC = BA + BC = 60 + 34.6 [∵ Substituted value of BA and BC from (1) and (2)] = 94.6 m

Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30° ABC = DAB = 30° (because DA || BC) tan 30 ° = AC BC => tan 30 ° = 80 BC => BC = 80 tan 30 ° = 80 ( 1 √ 3 ) = 80 × 1.73 = 138.4 m tan30ACBC tan3080BC BC = 80tan308013 801.73138.4 m i.e., Distance of the bus from the foot of the tower = 138.4 m

Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC Given that BA = 10 m , BAC = 60° sin 60° = BC BA √ 3 2 = BC 10 BC = 10 × √ 3 2 = 5 × 1.73 = 8.65 m

D- 254m. Let DC be the tower and A and B be the objects as shown above. Given that DC = 600 m , DAC = 45°, DBC = 60° tan 60° = DC BC √ 3 = 600 BC BC= 600 √ 3 ⋯ ( 1 ) tan 60°DCBC 3600BC BC= 60031 tan 45° = DC AC 1 = 600 AC AC= 600 ⋯ ( 2 ) tan 45°DCAC 1600AC AC= 6002 Distance between the objects = AB = (AC - BC) = 600 − 600 √ 3 6006003 [∵ from (1) and (2)] = 600 ( 1 − 1 √ 3 ) = 600 ( √ 3 − 1 √ 3 ) = 600 ( √ 3 − 1 √ 3 ) × √ 3 √ 3 = 600 √ 3 ( √ 3 − 1 ) 3 = 200 √ 3 ( √ 3 − 1 ) = 200 ( 3 − √ 3 ) = 200 ( 3 − 1.73 ) = 254 m 600113 600313 60031333 6003313 200331 20033 20031.73 254 m

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m We have DAC = 30°, DBC = 45° Let DC = h tan 30 ° = DC AC => 1 √ 3 = h AC => AC = h √ 3 ⋯ ( 1 ) tan30DCAC 13hAC AC = h31 tan 45 ° = DC BC => 1 = h BC => BC = h ⋯ ( 2 ) tan45DCBC 1hBC BCh2 We know that, AB = (AC - BC) => 40 = (AC - BC) => 40 = ( h √ 3 − h ) 40h3h[∵ from (1) & (2)] => 40 = h ( √ 3 − 1 ) => h = 40 ( √ 3 − 1 ) = 40 ( √ 3 − 1 ) × ( √ 3 + 1 ) ( √ 3 + 1 ) = 40 ( √ 3 + 1 ) ( 3 − 1 ) = 40 ( √ 3 + 1 ) 2 = 20 ( √ 3 + 1 ) = 20 ( 1.73 + 1 ) = 20 × 2.73 = 54.6 m 40h31 h4031 40313131 403131 40312 2031 201.731202.7354.6 m

C-10 m. Given that AC = 15 m, ADB = 30°, AEC = 60° Let DE = h Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE tan 60 ° = AC CE => √ 3 = 15 CE => CE = 15 √ 3 ⋯ ( 1 ) tan60ACCE 315CE CE = 1531 tan 30 ° = AB BD => 1 √ 3 = 15 − h BD tan30ABBD 1315hBD => 1 √ 3 = 15 − h ( 15 √ 3 ) 1315h153(∵ BD = CE and substituted the value of CE from equation 1 ) => ( 15 − h ) = 1 √ 3 × 15 √ 3 = 15 3 = 5 => h = 15 − 5 = 10 m 15h131531535 h15510 m i.e., height of the electric pole = 10 m

Let AB be the tower. Let C and D be the positions of the boat Then, ACB = 45° , ADC = 30°, BC = 100 m tan 45 ° = AB BC => 1 = AB 100 => AB = 100 ⋯ ( 1 ) tan45ABBC 1AB100 AB = 100 1 tan 30 ° = AB BD tan30ABBD => 1 √ 3 = 100 BD 13100BD(∵ Substituted the value of AB from equation 1) => BD = 100 √ 3 1003 CD = (BD - BC) = ( 100 √ 3 − 100 ) = 100 ( √ 3 − 1 ) 100310010031 It is given that the distance CD is covered in 10 seconds. i.e., the distance 100 ( √ 3 − 1 ) 10031 is covered in 10 seconds. Required speed = Distance Time DistanceTime = 100 ( √ 3 − 1 ) 10 = 10 ( 1.73 − 1 ) 1003110101.731 = 7.3 meter/seconds = 7.3 × 18 5 185 km/hr = 26.28 km/hr

Let AB be the tower. Let D and C be the positions of the car. Then, ADC = 30° , ACB = 45° Let AB = h, BC = x, CD = y tan 45 ° = AB BC = h x => 1 = h x => h = x ⋯ ( 1 ) tan45ABBChx 1hx hx1 tan 30 ° = AB BD = AB (BC + CD) = h x+y => 1 √ 3 = h x+y => x + y = √ 3 h => y = √ 3 h - x tan30ABBDAB(BC + CD)hx+y 13hx+y x + y = 3h y = 3h - x => y = √ 3 h − h y = 3hh (∵ Substituted the value of x from equation 1 ) => y = h ( √ 3 − 1 ) y = h31 Given that distance y is covered in 8 minutes. i.e, distance h ( √ 3 − 1 ) h31 is covered in 8 minutes. Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1). Let distance h is covered in t minutes. since distance is proportional to the time when the speed is constant, we have h ( √ 3 − 1 ) ∝ 8 ⋯ ( A ) h ∝ t ⋯ ( B ) (A) (B) => h ( √ 3 − 1 ) h = 8 t => ( √ 3 − 1 ) = 8 t => t = 8 ( √ 3 − 1 ) = 8 ( 1.73 − 1 ) = 8 .73 = 800 73 minutes = 10 70 73 minutes ≈ 10 minutes 57 seconds

D-24.8 m. PR represents the ladder and RQ represents the wall. cos 60° = PQ PR 1 2 = 12.4 PR PR = 2 × 12.4 = 24.8 m

C-46th

B-Both the arms are of same length

B-12m. Basic: tan of angle = opposite side / adjecent side ; tan 30 = 1/ ( square root of 3) = 1 / (1.73) Given man is 2 meter tall so actual height will be 2m addition to obtained opposite side X. Given adjecent side is 10 * (square root of 3) 10 * 1.73 Now tan angle is ===> X / ( 10 * 1.73) = 1 / 1.73 ===> X * 1.73 / 10 * 1.73 = 1 ===> X = 10 m ( obtained opposite side ) therefore Actual height is X + 2 = 10 + 2 = 12 meter.

A-346m. given height of the tower = 200m angle of elevation = 30deg => tan (30deg) = high of tower/distance from p to foot of the tower => tan (30deg)= 200/d => d= 200/tan (30deg) : tan( 30) = 1/sqrt(3) => d= 346m

C-Data inadequate. One of AB, AD and CD must have given. So, the data is inadequate.

C- 273 m. Let AB be the lighthouse and C and D be the positions of the ships. AB = 100m,∠ACB=30∘, ∠ADB=45∘ AB AC =tan 30∘= 1 √ 3 =>AC=AB∗ √ 3 =100 √ 3 m AB AD =tan 45∘=1 =>AB=AD=100m CD=AC+AD =(100 √ 3 +100) m =100( √ 3 +1) m =100∗2.73 m=273m

A- [M]=[T−11 C−22 h]M∝TXVYhZMTXVYhZ M′L0T0=(T′)X(L1T−1)Y(M1L2T−1)ZML0T0TXL1T1YM1L2T1Z M1L0T0=MZLY+2Z+TX−Y−ZM1L0T0MZLY2ZTXYZ Z=1Z1 Y+2Z=0Y2Z0 Y=−2Y2 X−Y−Z=0XYZ0 X+2−1=0X210 X=1X1 M⟹T−1C−2h1MT1C2h1

Ans. Option C - greater than 1 Area of the square is more than that of the rhombus, for having greater height.

The total cost of paving = Rs. 5.5*3.75*800 = Rs. 16500

Exclusions from the Cost of Inventories In determining the cost of inventories , it is appropriate to exclude certain costs and recognise them as expenses in the period in which they are incurred. Examples of such costs are: (a) abnormal amounts of wasted materials, labour, or other production costs; (b) storage costs, unless those costs are necessary in the production process prior to a further production stage; (c) administrative overheads that do not contribute to bringing the inventories to their present location and condition; and (d) selling and distribution costs.

The mainly issues which deal with Accounting Standard are Recogntion, Measurement/Valuation, Presentation, Disclosure.

On 28 December 1885, the Indian National Congress was founded at Gokuldas Tejpal Sanskrit College in Bombay, with 72 delegates in attendance.

Standard equation for quadratic polynomial : ax2+bx+c =0, can be taken as x2+b/ax+c/a=0 (divided both sides by a). here, sum of zeros = -3/2=-b/a ,which gives b/a=3/2 product of zeros = 1= c/a putting the values of b/a and c/a in general equation, we get; x2+3/2x+1=0 , solving further..we get 2x2+3x+1=0.

The Principle of Moments, also known as Varignon's Theorem, states that the moment of any force is equal to the algebraic sum of the moments of the components of that force. It is a very important principle that is often used in conjunction with the Principle of Transmissibility in order to solve systems of forces that are acting upon and/or within a structure.

a = 1.25m/s^2 Now , v = u + at = 1.25 x 8 = 10 m/s Hence s = 1/2 at^2 = (1.25)(64)/2 = 40 m Now, H= -40m ( due to downward direction.) and 40 = 10t - 5t^2 ( s= ut +1/2 at^2)(placing above obtained and given values) therefore , t^2 - 2t + 8 = 0 **t = 4**

Actually the bigger molecules like oxygen occupy the lattice points.In ccp unit cell the effective number of atoms occupying lattice points is 8(1/8)+6(1/2)=4 and in ccp the number of tetrahedral voids is 8 so divalents ions will be 8/8 and number of octahedral voids are 4 so one half would be 2 finally,XY2O4.

A- −11 (i) We know that μ=sinrsinrμsinrsinr and i+r=90∘ir90 or r=90∘−ir90i μ=sinisin(90−i)μsinisin90i =tanitani or i=tan−1(μ)itan1μ =tan−1

Ans:- A If charges were kept at all corners resultant electric field would have been zero So, removal of one charge would result in the same electric field as due to one charge alone. So, E = kQ/r^2

B- (n-3) :2. The total number of circular permutations of n people taken all at a time, N = (n - 1)! Let event A = Two specified people sit together. A group of two people, and remaining (n - 2) people can sit on a round table in [{1 + (n - 2)} - 1]! i.e. (n - 2)! ways. The two specified people can interchange their places in 2! ways. Hence favourable ways, N(A) = (n - 2)! 2! P(A)PA = (n−2)!2!(n−1)!n22n1 = 2n−12n1 P(A¯¯¯¯)PA = 1−2n−112n1 = n−3n−1n3n1 Therefore the odds against two specified people sitting next to each other are P(A¯¯¯¯)PA to P(A)PA i.e. (n−3):2

B-15. Let the no. of participants be nn. If no participant fell ill, then the no. of games played = nC2nC2 Each player will play n−1n1 games in the tournament. These n-1 games include 1 game which both the player played with each other. This one game is counted with both the players. The two participants played 3 games each and fell ill. ∴ The no. of games played = nC2−[(n−1)+(n−1)−1]+6=84nC2n1n11684 (given) ⇒n(n−1)2nn12−2n+9=84 ⇒n2−5n−150=0 ⇒n=15or−10(notpossible) ⇒n=15

C-60°.

C. Malonic acid

Ans:- A. after 20s the the distance covered by the train is s = 1/2( a t^2) s = 1/2 *1/2 *20*20 s = 100 m the velocity of the train after 20s is v2 = u2 +2as (u = 0 , since starting from rest ) v2 = 2*0.5*100 v =10 m/s so after 20 secs the head light of train covers 100 m distance and then the back light is switched on,in order to see both the actions in same time the observer should cover the 300 m distance in 20 secs relative to train and in opposite direction. V of observer with respect to train should be = 300/20 = - 15 m/s (- sign for opposite direction) - 15 = Vframe - Vtrain - 15 = Vframe - 10 Vframe = - 5 m/s Hence The correct answer is 5 m/s in the opposite direction. (A)

Given that the velocity at the 2n second interval = v0 = a x 2n. This gives a = v0/2n Displacement after n seconds is 0.5 a nxn Displacement after 2n seconds = 0.5 a . (2n) x (2n) Subtracting, the displacement in the last second is 0.5 x a x (3n^2) Replacing the value of a, we get the distance covered in last n second = 0.5 x v0 / 2n x 3 n^2 = (3/4) x v0 x n

C- mgR. K.E. = ½ mv2e ­where ve = escape velocity = v2gR ? K.E = ½ m x 2gR = mgR 3

C- 4/3 atm

A- The distance between the objective and the eyepiece is 16.02 m

Answer: Option A 381 m Explanation: Let C and D be the position of the aeroplanes. Given that CB = 900 m, CAB = 60°, DAB = 45° From the right ABC, tan 60 ° = CB AB √ 3 = 900 AB AB = 900 √ 3 = 900 × √ 3 √ 3 × √ 3 = 900 √ 3 3 = 300 √ 3 tan60CBAB 3900AB AB9003 900333900333003 From the right ABD, tan 45 ° = DB AB 1 = DB AB DB = AB = 300 √ 3 tan45DBAB 1DBAB DBAB3003 Required height = CD = (CB-DB) = ( 900 − 300 √ 3 ) = ( 900 − 300 × 1.73 ) = ( 900 − 519 ) = 381 m

Let CB be the pole and point D divides it such that BD : DC = 1 : 9 Given that AB = 15 m Let the the two parts subtend equal angles at point A such that CAD = BAD = θ From "Angle Bisector Theorem", we have BD DC = AB AC BDDCABAC ⇒ 1 9 = 15 AC 1915AC [∵ BD : DC = 1 : 9 and AB = 15(given)] => AC = 15 × 9 m ...(eq: 1) From the right ABC, CB = √ AC 2 − AB 2 CBAC2AB2(∵ Pythagorean theorem) = √ ( 15 × 9 ) 2 − 15 2 1592152(∵AC=15×9(eq:1) and AB=15 m(given)] = √ 15 2 × 9 2 − 15 2 = √ 15 2 ( 9 2 − 1 ) = √ 15 2 × 80 = √ 15 2 × 16 × 5 = 15 × 4 × √ 5 = 60 √ 5 m

Let ED be the building and AC be the tower. Given that ED = 230 m, ADC = b, AEB = a Also given that tan a = 5/12 and tan b = 4/5 Let AC = h Required Distance = Distance between the top of these buildings = AE From the right ABE, tan ( a ) = AB BE tan(aABBE => 5 12 = (h-230) BE 512(h-230)BE [∵ tan(a)=5/12(given), AB = (AC-BC) = (AC-ED) = (h-230)] => BE = 12 (h-230) 5 ⋯ ( e q : 1 ) 12(h-230)5eq1 From the right ACD, tan ( b ) = AC CD tan(bACCD => 4 5 = h CD 45hCD [∵ tan(b) = 4/5(given), AC=h] => CD = 5 h 4 ⋯ ( e q : 2 ) 5h4eq2 From the diagram, BE = CD ⇒ 12 (h-230) 5 = 5 h 4 12(h-230)55h4 (from eq:1 & eq:2) ⇒ 48 h − ( 4 × 12 × 230 ) = 25 h ⇒ 23 h = ( 4 × 12 × 230 ) 48h41223025h 23h412230 ⇒ h = ( 4 × 12 × 230 ) 23 = 480 m h41223023480 m ⋯ ( e q : 3 ) eq3 AB = (AC - BC) = (480 - 230) [∵ Since AC=h=480(from eq:3) and BC=ED=230 m(given)] = 250 m In the triangle ABE, tan(a) = 5/12. Let's figure out the value of sin(a) now. Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = 5/12 hypotenuse = √ 5 2 + 12 2 = 13 5212213 i.e., sin(a) = opposite side hypotenuse = 5 13 opposite sidehypotenuse513 We have seen that sin(a) = 5 13 513 ⇒ AB AE = 5 13 ⇒ AE = AB × 13 5 = 250 × 13 5 = 650 m ABAE513 AE = AB135 250135650 m i.e., Distance between the top of the buildings = 650 m

Let ED be the taller tower and AB be the shorter tower. Let C be the point of observation. Given that ACB = 22.5° and DCE = 67.5° Given that C is the midpoint of BD. Hence, BC = CD From the right ABC, tan 22.5 ° = AB BC ⋯ ( e q : 1 ) tan22.5ABBCeq1 From the right CDE, tan 67.5 ° = ED CD ⋯ ( e q : 2 ) tan67.5EDCDeq2 ( e q : 2 ) ( e q : 1 ) ⇒ tan 67.5 ° tan 22.5 ° = ( ED CD ) ( AB BC ) eq2eq1tan67.5tan22.5EDCDABBC = ED AB EDAB(∵ CD=BC) ⇒ tan ( 90 ° − 22.5 ° ) tan 22.5 ° = ED AB tan9022.5tan22.5EDAB ⇒ cot 22.5 ° tan 22.5 ° = ED AB cot22.5tan22.5EDAB [∵ tan(90-θ)=cot θ] ⇒ ( 1 tan 22.5 ° ) tan 22.5 ° = ED AB 1tan22.5tan22.5EDAB [ ∵ cot θ = 1 tan θ ] cotθ1tanθ ⇒ ED AB = 1 ( tan 22.5 ° ) 2 = 1 ( √ 2 − 1 ) 2 = ( 1 √ 2 − 1 ) 2 = [ ( √ 2 + 1 ) ( √ 2 − 1 ) ( √ 2 + 1 ) ] 2 = [ ( √ 2 + 1 ) ( 2 − 1 ) ] 2 = [ ( √ 2 + 1 ) 1 ] 2 = ( √ 2 + 1 ) 2 = ( 2 + 2 √ 2 + 1 ) = ( 3 + 2 √ 2 ) EDAB1tan22.52 12121212 2121212 212122112 2122221 322 Required ratio = ED : AB = ( 3 + 2 √ 2 ) : 1 3221

D-15 m. Let P be the point of observation and QR be the tower. Given that θ = sin − 1 ( 3 5 ) θsin135 and PQ = 20 m Let the height of the tower, QR = h and PR = x From the right PQR, sin θ = QR PR ⇒ sin [ sin − 1 ( 3 5 ) ] = h x ⇒ 3 5 = h x ⇒ x = 5 h 3 ⋯ ( e q : 1 ) sinθQRPR sinsin135hx 35hx x5h3eq1 From Pythagorean theorem, we have PQ 2 + QR 2 = PR 2 20 2 + h 2 = x 2 PQ2QR2PR2 202h2x2 20 2 + h 2 = ( 5 h 3 ) 2 202h25h32(∵ Substituted the value of x from eq:1) 20 2 + h 2 = 25 h 2 9 16 h 2 9 = 20 2 4 h 3 = 20 h = 3 × 20 4 = 15 m 202h225h29 16h29202 4h320 h320415 m i.e., height of the tower = 15 m

Let RQ be the pole and PQ be the shadow Given that RQ = 18 m and PQ = 6 √ 3 63 m Let the angle of elevation, RPQ = θ From the right PQR, tan θ = RQ PQ = 18 6 √ 3 = 3 √ 3 = √ 3 tanθRQPQ1863333 ⇒ θ = tan − 1 ( √ 3 ) = 60°

Let BD be the tower and A and C be the positions of the persons. Given that BD = 50 m, BAD = 30°, BCD = 60° From the right ABD, tan 30° = BD BA ⇒ 1 √ 3 = 50 BA ⇒ BA = 50 √ 3 tan 30° = BDBA 1350BA BA503 From the right CBD, tan 60° = BD BC ⇒ √ 3 = 50 BC ⇒ BC = 50 √ 3 = 50 × √ 3 √ 3 × √ 3 = 50 √ 3 3 tan 60° = BDBC 350BC BC503503335033 Distance between the two persons = AC = BA + BC = 50 √ 3 + 50 √ 3 3 = √ 3 ( 50 + 50 3 ) = 200 √ 3 3 m 5035033 350503 20033 m i.e., the distance travelled by the car in 10 seconds = 200 √ 3 3 20033 m Speed of the car = Distance Time DistanceTime = ( 200 √ 3 3 ) 10 = 20 √ 3 3 meter/second = 20 √ 3 3 × 18 5 km/hr = 24 √ 3 km/hr

D- MgL/18. unit length of rope is M/L given L/3 part of rope is hanging we have to solve using integration methods, lets say 'x' part of rope is consided with an element 'dx' which is pulled up with limits from 0 to L/3 Potential energy = mgh here dE= integral of ( Mx/L*g*dx) E = integral of ( M/L*g*x.dx) limits from 0 to L/3 P.E = MgL/18

Let A be the foot and C be the summit of a mountain. Given that CAB = 45° From the diagram, CB is the height of the mountain. Let CB = x Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that DAY = 30° It is also given that from the point D, the elevation is 60° i.e., CDE = 60° From the right ABC, tan 45 ° = CB AB tan45CBAB => 1 = x AB 1xAB [∵ CB = x(the height of the mountain)] => AB = x ... (eq:1) From the right AYD, sin 30 ° = DY AD sin30DYAD => 1 2 = DY 2 12DY2 (∵ Given that AD = 2) => DY = 1 ... (eq:2) cos 30 ° = AY AD cos30AYAD => √ 3 2 = AY 2 32AY2(∵ Given that AD = 2) => AY = √ 3 3 ... (eq:3) From the right CED, tan 60 ° = CE DE tan60CEDE ⇒ tan 60 ° = (CB - EB) YB tan60(CB - EB)YB[∵ CE=(CB-EB) and DE=YB)] ⇒ tan 60 ° = (CB - DY) (AB - AY) tan60(CB - DY)(AB - AY)[∵ EB=DY and YB=(AB-AY)] ⇒ tan 60 ° = (x - 1) ( x − √ 3 ) tan60(x - 1)x3 [∵ CB=x, DY=1(eq:2), AB=x(eq:1) and AY = √ 3 3(eq:3)] ⇒ √ 3 = (x - 1) ( x − √ 3 ) ⇒ x √ 3 − 3 = x − 1 ⇒ x ( √ 3 − 1 ) = 2 ⇒ 0.73 x = 2 ⇒ x = 2 0.73 = 2.7 3(x - 1)x3 x33x1 x312 0.73x2 x20.732.7 i.e., the height of the mountain = 2.7 km

C-9:4.

B- 5.00 ± 11%. Weight in air =(5.00±0.05)N Weight in water=(4.00±0.05)N Loss of weight in water =(1.00±0.1)N Now relative density =weightinair weight loss in water i.e. R . D=5.00±0.051.00±0.1 Now relative density with max permissible error =5.001.00±(0.055.00+0.11.00)×100=5.0±(1+10) =5.0±11

D-3.65 m. Let AB be the man and CD be the window Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m, DAF = 45° , CAF = 60° From the diagram, AF = BE = 5 m From the right AFD, tan 45 ° = DF AF 1 = DF 5 DF = 5 ⋯ ( 1 ) tan45DFAF 1DF5 DF = 51 From the right AFC, tan 60 ° = CF AF √ 3 = CF 5 CF = 5 √ 3 ⋯ ( 2 ) tan60CFAF 3CF5 CF532 Length of the window = CD = (CF - DF) = 5 √ 3 − 5 535[∵ Substituted the value of CF and DF from (1) and (2)] = 5 ( √ 3 − 1 ) = 5 ( 1.73 − 1 ) = 5 × 0.73 = 3.65 m

A-216. We have two set of numbers: 1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets: 1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120. 0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96 120+96=216

Let AB and CD be the poles with heights h and 2h respectively. Given that distance between the poles, BD = 200 m Let E be the middle point of BD, AEB = θ CED = (90-θ) (∵ given that angular elevations are complementary) Since E is the middle point of BD, we have BE = ED = 100 m From the right ABE, tan θ = AB BE tan θ = h 100 h = 100 tan θ ⋯ ( 1 ) tanθABBE tanθh100 h = 100tanθ1 From the right EDC, tan ( 90 − θ ) = CD ED cot θ = 2h 100 [ ∵ tan ( 90 − θ ) = cot θ ] 2h = 100 cot θ ⋯ ( 2 ) tan90θCDED cotθ2h100tan90θcotθ 2h 100cotθ2 ( 1 ) × ( 2 ) => 2 h 2 = 100 2 122h21002 [ ∵ tan θ × cot θ = tan θ × 1 tan θ = 1 ] tanθcotθtanθ1tanθ1 => √ 2 h = 100 2h100 => h = 100 √ 2 = 100 × √ 2 √ 2 × √ 2 = 50 √ 2 = 50 × 1.41 = 70.5 h1002100222 502501.4170.5 => 2h = 2 × 70.5 = 141 2h270.5141 i.e., the height of the poles are 70.5 m and 141 m.

Let DC be the wall, AB be the tree. Given that DBC = 30°, DAE = 60°, DC = 11 m tan 30° = DC BC 1 √ 3 = 11 BC BC = 11 √ 3 m AE = BC = 11 √ 3 m ⋯ ( 1 ) tan 30°DCBC 1311BC BC = 113 m AE = BC =113 m1 tan 60 ° = ED AE tan60EDAE √ 3 = ED 11 √ 3 3ED113 [∵ Substituted value of AE from (1)] ED = 11 √ 3 × √ 3 = 11 × 3 = 33 113311333 Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

it can be arranged as a chessboard with m-1 steps up down and n-1 steps left right. shortest possible distance will be m-1 steps down and n-1 steps right. no. of ways is….(m+n-2)! / (m-1)!(n-1)!

B-3. y = 4 cos2 (1/2 t) sin (1000 t) = 2 x 2 cos2(t/2)sin (1000 t) = 2 x (1 + cos t) sin (1000 t) = 2 sin (1000 t) + 2 cos t sin (1000 t) = 2 sin (1000 t) + sin [(1000+1)t] +sin [(1000−1)t] = 2 sin (1000 t) + sin (1001)t +sin (999)t = y1 + y2 + y3 As the above expression involves three different waves the displacement given is a result of superposition of three different waves.

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes. Given that CA = 150 m, BCA = 60° tan 60 ° = BA CA √ 3 = B A 150 BA = 150 √ 3 tan60BACA 3BA150 BA1503 i.e, the distance travelled by the balloon = 150 √ 3 1503 meters time taken = 2 min = 2 × 60 = 120 seconds Speed = Distance Time DistanceTime = 150 √ 3 120 = 1.25 √ 3 = 1.25 × 1.73 = 2.16 meter/second

A-9 m. Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation. Given that AD = 18 m, ABC = 60°, DBC = 30° Let DC be h. tan 30 ° = DC BC 1 √ 3 = h BC h = BC √ 3 ⋯ ( 1 ) tan30DCBC 13hBC hBC31 tan 60 ° = AC BC √ 3 = 18 + h BC 18+h = BC × √ 3 ⋯ ( 2 ) tan60ACBC 318hBC 18+hBC32 ( 1 ) ( 2 ) => h 18 + h = ( BC √ 3 ) ( BC × √ 3 ) = 1 3 => 3 h = 18 + h => 2 h = 18 => h = 9 m 12h18hBC3BC313 3h18h 2h18 h9 m i.e., the height of the tower = 9 m

Answer: Option C Explanation: Consider the diagram shown above. AC represents the hill and DE represents the pole Given that AC = 100 m XAD = ADB = 30° (∵ AX || BD ) XAE = AEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60 ° = AC CE => √ 3 = 100 CE => CE = 100 √ 3 ⋯ ( 1 ) tan60ACCE 3100CE CE = 10031 tan 30 ° = AB BD => 1 √ 3 = 100 − h BD tan30ABBD 13100hBD => 1 √ 3 = 100 − h ( 100 √ 3 ) 13100h1003(∵ BD = CE and substituted the value of CE from equation 1) => ( 100 − h ) = 1 √ 3 × 100 √ 3 = 100 3 = 33.33 => h = 100 − 33.33 = 66.67 m 100h131003 100333.33 h10033.3366.67 m i.e., the height of the pole = 66.67 m

Let BD be the lighthouse and A and C be the two points on ground. Then, BD, the height of the lighthouse = 60 m BAD = 45° , BCD = 60° tan 45 ° = BD BA ⇒ 1 = 60 BA ⇒ BA = 60 m ⋯ ( 1 ) tan45BDBA 160BA BA60 m1 tan 60 ° = BD BC ⇒ √ 3 = 60 BC ⇒ BC = 60 √ 3 = 60 × √ 3 √ 3 × √ 3 = 60 √ 3 3 = 20 √ 3 = 20 × 1.73 = 34.6 m ⋯ ( 2 ) tan60BDBC 360BC BC603603336033 203201.7334.6 m2 Distance between the two points A and C = AC = BA + BC = 60 + 34.6 [∵ Substituted value of BA and BC from (1) and (2)] = 94.6 m

Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30° ABC = DAB = 30° (because DA || BC) tan 30 ° = AC BC => tan 30 ° = 80 BC => BC = 80 tan 30 ° = 80 ( 1 √ 3 ) = 80 × 1.73 = 138.4 m tan30ACBC tan3080BC BC = 80tan308013 801.73138.4 m i.e., Distance of the bus from the foot of the tower = 138.4 m

Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC Given that BA = 10 m , BAC = 60° sin 60° = BC BA √ 3 2 = BC 10 BC = 10 × √ 3 2 = 5 × 1.73 = 8.65 m

Man per unit length =MlMl Consider an element of width dx at a distance x from 0 mass of element =dm=M2dmM2dxdx M.I= man × perpendicular distance from axis of rotation Hence Ml2sinθ12Ml2sinθ12 is a correct answer.

D-6mg. This question solve by method of conservation of energy. at the highest position tension in string is zero .now we proceed mg=mv^2/l at tje highest position v^2=gl now at the lowest position apply conservation of energy 1/2mv^2+mg.2l=1/2mv'^2 v'2=5gl where v'is velocity of bob at lowest position now, T=6mg

Time taken to go uptoU and then come back to the ground required 12 sec. So 6 s up then 6 s in coming down. the time taken between climbing from T to U and then coming to T again(crossing T twice) =6 s so time taken to go upto T from the ground=(12-6)/2=3s now if we consider the journey from ground to U: 0=u-6x10 ....... putting in 1st eq. u=60 m/s now tower height:H= H=60x3-1/2 x10x3x3.....putting in second eq H=135 m

D-4. Since density is mass÷volume, For a cylinder it becomes- mass÷(pi×radius^2×length) Relative percent errors for- Mass-0.003÷0.3×100=1% Radius^2-2×0.005÷0.5×100=2% Length-0.06÷0.6×100=1% THE UNITS CANCEL THEMSELVES FROM THE NUMERATOR AND DENOMINATOR. therefore, maximum percentage error in density-1%+2%+1%=4% so, the answer is 4%

Component of velocity of ball relative to lift are ux=4cos30∘=23–√m/sux4cos3023ms uy=4sin30∘=2m/suy4sin302ms and acceleration of ball relative to lift is a=10+2=12m/s2a10212ms2 in negative y direction acting downwards. Therefore Time of flight T=2uyaT2uya =2×2122212 =1/3s

Answer: Option D Explanation: Let DC be the tower and A and B be the objects as shown above. Given that DC = 600 m , DAC = 45°, DBC = 60° tan 60° = DC BC √ 3 = 600 BC BC= 600 √ 3 ⋯ ( 1 ) tan 60°DCBC 3600BC BC= 60031 tan 45° = DC AC 1 = 600 AC AC= 600 ⋯ ( 2 ) tan 45°DCAC 1600AC AC= 6002 Distance between the objects = AB = (AC - BC) = 600 − 600 √ 3 6006003 [∵ from (1) and (2)] = 600 ( 1 − 1 √ 3 ) = 600 ( √ 3 − 1 √ 3 ) = 600 ( √ 3 − 1 √ 3 ) × √ 3 √ 3 = 600 √ 3 ( √ 3 − 1 ) 3 = 200 √ 3 ( √ 3 − 1 ) = 200 ( 3 − √ 3 ) = 200 ( 3 − 1.73 ) = 254 m

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m We have DAC = 30°, DBC = 45° Let DC = h tan 30 ° = DC AC => 1 √ 3 = h AC => AC = h √ 3 ⋯ ( 1 ) tan30DCAC 13hAC AC = h31 tan 45 ° = DC BC => 1 = h BC => BC = h ⋯ ( 2 ) tan45DCBC 1hBC BCh2 We know that, AB = (AC - BC) => 40 = (AC - BC) => 40 = ( h √ 3 − h ) 40h3h[∵ from (1) & (2)] => 40 = h ( √ 3 − 1 ) => h = 40 ( √ 3 − 1 ) = 40 ( √ 3 − 1 ) × ( √ 3 + 1 ) ( √ 3 + 1 ) = 40 ( √ 3 + 1 ) ( 3 − 1 ) = 40 ( √ 3 + 1 ) 2 = 20 ( √ 3 + 1 ) = 20 ( 1.73 + 1 ) = 20 × 2.73 = 54.6 m

Consider the diagram shown above. AC represents the tower and DE represents the pole. Given that AC = 15 m, ADB = 30°, AEC = 60° Let DE = h Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE tan 60 ° = AC CE => √ 3 = 15 CE => CE = 15 √ 3 ⋯ ( 1 ) tan60ACCE 315CE CE = 1531 tan 30 ° = AB BD => 1 √ 3 = 15 − h BD tan30ABBD 1315hBD => 1 √ 3 = 15 − h ( 15 √ 3 ) 1315h153(∵ BD = CE and substituted the value of CE from equation 1 ) => ( 15 − h ) = 1 √ 3 × 15 √ 3 = 15 3 = 5 => h = 15 − 5 = 10 m 15h131531535 h15510 m i.e., height of the electric pole = 10 m

Answer: Option A Explanation: Consider the diagram shown above. Let AB be the tower. Let C and D be the positions of the boat Then, ACB = 45° , ADC = 30°, BC = 100 m tan 45 ° = AB BC => 1 = AB 100 => AB = 100 ⋯ ( 1 ) tan45ABBC 1AB100 AB = 100 1 tan 30 ° = AB BD tan30ABBD => 1 √ 3 = 100 BD 13100BD(∵ Substituted the value of AB from equation 1) => BD = 100 √ 3 1003 CD = (BD - BC) = ( 100 √ 3 − 100 ) = 100 ( √ 3 − 1 ) 100310010031 It is given that the distance CD is covered in 10 seconds. i.e., the distance 100 ( √ 3 − 1 ) 10031 is covered in 10 seconds. Required speed = Distance Time DistanceTime = 100 ( √ 3 − 1 ) 10 = 10 ( 1.73 − 1 ) 1003110101.731 = 7.3 meter/seconds = 7.3 × 18 5 185 km/hr = 26.28 km/hr

Answer: Option B Explanation: Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car. Then, ADC = 30° , ACB = 45° Let AB = h, BC = x, CD = y tan 45 ° = AB BC = h x => 1 = h x => h = x ⋯ ( 1 ) tan45ABBChx 1hx hx1 tan 30 ° = AB BD = AB (BC + CD) = h x+y => 1 √ 3 = h x+y => x + y = √ 3 h => y = √ 3 h - x tan30ABBDAB(BC + CD)hx+y 13hx+y x + y = 3h y = 3h - x => y = √ 3 h − h y = 3hh (∵ Substituted the value of x from equation 1 ) => y = h ( √ 3 − 1 ) y = h31 Given that distance y is covered in 8 minutes. i.e, distance h ( √ 3 − 1 ) h31 is covered in 8 minutes. Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1). Let distance h is covered in t minutes. since distance is proportional to the time when the speed is constant, we have h ( √ 3 − 1 ) ∝ 8 ⋯ ( A ) h ∝ t ⋯ ( B ) (A) (B) => h ( √ 3 − 1 ) h = 8 t => ( √ 3 − 1 ) = 8 t => t = 8 ( √ 3 − 1 ) = 8 ( 1.73 − 1 ) = 8 .73 = 800 73 minutes = 10 70 73 minutes ≈ 10 minutes 57 seconds

D-24.8 m. Angle = 60º and distance between wall and ladder= 12.4 m. Length of ladder= 2 x distance between wall and ladder = (2 x 12.4) m = 24.8 m.

An organization whose membership consists of workers and union leaders, united to protect and promote their common interests. The principal purposes of a labor union are to (1) negotiate wages and working condition terms, (2) regulate relations between workers (its members) and the employer, (3) take collective action to enforce the terms of collective bargaining, (4) raise new demands on behalf of its members, and (5) help settle their grievances. A trade union may be: (a) A company union that represents interests of only one company and may not have any connection with other unions.

C- pineal gland secretes melatonin in animals and this hormone regulates sleep.

Actually the bigger molecules like oxygen occupy the lattice points.In ccp unit cell the effective number of atoms occupying lattice points is 8(1/8)+6(1/2)=4 and in ccp the number of tetrahedral voids is 8 so divalents ions will be 8/8 and number of octahedral voids are 4 so one half would be 2 finally, AB2O4.

Ans- 0.98 m from mass of 0.3 kg.

At equilibrium ΣF=0ΣF0 kx0+(AL2σg)−0AL2σg Mg =0 x00 = Mg [1−LAσ2M]

It’s a question that is almost impossible to answer because it depends on so many different factors. According to English UK, it takes about 120 hours to progress from one level to the next.... Cambridge ESOL suggests slightly longer: 200 hours.

To be afraid of something means the fear is inside you whereas to be scared of someone denotes that is it external fear.

Read dictionary and when you dont know the meaning of any word google it immediately ...

Solution : λ=589nmλ589nm Δλ=589.6−589.0Δλ589.6589.0 =0.6nm0.6nm V=ΔλλVΔλλ =0.65890.6589×3×108m/s3108ms =3.06×105m/s3.06105ms =30.6km/s

When something is moving away from something else, the light that is emitted is predominantly at the red end of the spectrum. Red light wavelengths tend to be more stretched out the futher you are from the object. Blue light wavelengths tend to be closer together. Blueshift light tends to be emitted from objects that are moving closer. By looking at the blue and red shift, we can determine whether the object is coming closer to us such as in the case of the Andromeda galaxy or moving away from us.

In electrodynamics, linear polarization or plane polarization of electromagnetic radiation is a confinement of the electric field vector or magnetic field vector to a given plane along the direction of propagation. The orientation of a linearly polarized electromagnetic wave is defined by the direction of the electric field vector. For example, if the electric field vector is vertical (alternately up and down as the wave travels) the radiation is said to be vertically polarized.

Sound waves are longitudinal. Only transverse waves can be polarised. Longitudinal waves cannot be polarised. Transverse: vibrations perpendicular to the direction of travel of the wave . Longitudinal: vibrations parallel to the direction of travel of the wave. Sound waves are longitudinal so they cannot be polarised. Polarised: the wave vibrations occur in one plane only Unpolarised: the wave vibrations occur in all planes perpendicular to the direction of travel. The reason that only transverse waves can be polarised is that their vibrations can potentially occur in all directions perpendicular to the direction of travel. It is therefore possible to confine the vibrations to a single plane. Or put another way an unpolarised transverse wave is a 3D shape, it is therefore possible to confine it to a 2D shape (a plane). The vibrations of a longitudinal wave occur along a single line, it is therefore not possible to confine that to a plane, i.e. a 1D shape cannot be confined within a 2D shape.

C- Endorphin

As a general guide, if we consider diffraction of a wave with wavelength λλ from an object of size dd then the characteristic angle of the diffraction is given by: sinθ=O(λd) sinθOλd where the O()O symbol means of order i.e. roughly the same as. So for example in a Young's slits experiment, where dd is the slit spacing, the angle of the first maximum is given by: sinθ=λd sinθλd Alternatively, the diffraction pattern from a pinhole of diameter dd has the first minimum at: sinθ≈1.22λd sinθ1.22λd It should now be obvious why diffraction of sound is so commonly observed in everyday life. The wavelength of sound is of order one metre, so objects around a metre in size will diffract it strongly. By contrast the wavelength of light is around half a micron, so you need to get the size down to the micron scale before light starts scattering strongly.

Both diffraction and interference occur in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs. This is what give the diffraction pattern. I think what you are confused about is the single slit experiment. In the single slit experiment, light passes through a single slit which produces an interference pattern. This not do to diffraction of the light. In the single slit experiment, the slit is large enough that diffraction does not occur. Instead every point along the width of the slit acts like it's own point light source, which all interfere with one another. In fact, if you make the slit narrow enough in the single slit experiment, diffraction occurs and the interference pattern goes away.

Fringe width is inversely proportional to the distance (d) between the sources. When coherent sources are placed very close, fringe width is very large and a single fringe may occupy the whole field of view and the pattern would not be detected. Similarly, if the coherent sources are far apart, fringe with will be very small and the pattern cannot be detected.

A low flying aircraft reflects the T.V screen can be due to interference between the direct signal and the reflected signal. So May be due to interference between the direct and reflected signal...

usually generate light by black body radiation, that is by heating something until it glows. There are many sources for black body radiation, but the dominant one is usually random thermal motion causing random transient electric dipoles within the black body. The changes in these dipoles generates eletromagnetic radiation, and because the dipole changes are random so is the EM radiation they generate. If you take two points in the black body that are close compared to the wavelength of the lattice vibrations then their motion, and hence the EM generated, will be closely correlated. However as you increase the separation between the two points the correlation will decrease and at macroscopic distance will be essentially zero.

Intensity of a wave(carrier of energy) is defined as the energy crossing per unit area per second(Power/Area). Intensity of a particle(Photon,quantum of energy) is defined the flux of no. of particles per second(means no. of photons passing per area per time). So in the photon picture of light if no. of photons increases then intensity of light increases. Albert Einstein used this concept to describe photo-electric effect. He proposed that a photon collides with a single electron and transfers its all energy to the electron and electron leaves the metal surface. So if intensity of light is increased then more photon will incident on the surface and so more electrons will emerge.

That is because of a term called the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation. That is the amplitude is not uniform in all directions. The amplitude of the secondary wavelets goes as 1+cosθ1cosθ, where θθ is the angle made from the direction of motion. So, when θ=0θ0, that is along the direction of propagation of the wave, then the amplitude contribution of the secondary wavelets is maximum. However, if θ=πθπ, then the amplitude contribution is zero.

An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionised gas (plasma). The SI unit for measuring an electric current is the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second. Electric current is measured using a device called an ammeter. Electric currents cause Joule heating, which creates light in incandescent light bulbs. They also create magnetic fields, which are used in motors, inductors and generators. The moving charged particles in an electric current are called charge carriers. In metals, one or more electrons from each atom are loosely bound to the atom, and can move freely about within the metal. These conduction electrons are the charge carriers in metal conductors.

It depends on the direction magnetic field if the field is in direction of electric field then the resistance will decrease because all domains of conductor are arranged in direction of electric field and resistance decreases.

D - Cochlea The organ of Corti or spiral organ, is the receptor organ for hearing and is located in the mammalian cochlea. This highly varied strip of epithelial cells allows for transduction of auditory signals into nerve impulses' action potential.

The parasympathetic system is responsible for stimulation of "rest-and-digest" or "feed and breed" activities that occur when the body is at rest, especially after eating, including sexual arousal, salivation, lacrimation (tears), urination, digestion and defecation. Its action is described as being complementary to that of the sympathetic nervous system, which is responsible for stimulating activities associated with the fight-or-flight response.

Depends on the medium. Pure glass, sure the photons just pass on through with little resistance and coupling.

D- limbic system. Emotions like pleasure, fear and brain are carried out by the limbic system, located in the temporal lobe.

In diffraction grating, the spectrum of the light obtained is very sharp and clear than that of two slit. This helps for precise measurement of wavelength of the monochromatic light. So, diffraction grating is better than two slits set up for measuring the wavelength of monochromatic light.

In optics, the corpuscular theory of light, arguably set forward by Descartes (1637) states that light is made up of small discrete particles called "corpuscles" (little particles) which travel in a straight line with a finite velocity and possess impetus. This was based on an alternate description of atomism of the time period.This theory cannot explain refraction diffraction,interference and polarization.Newton was the pioneer of this theory.

C-33N. Mass =70 %*1.54 =1.078 g/ml Mass in 1000 ml solution=1.078*1000=1078 g Molar mass of orthophosphoric acid=98 g Normality of orthophosphoric acid=(Mass of solute in 1L)/(Molar mass/basicity) =(1.078)/(98/3) =33 N So,the normality of orthophosphoric acid having purity of 70 % by weight and specific gravity 1.54 is 33N.

A-40ml. Oxalic acid dehydrate H2C2O4 . 2H2O : mw = 126 It is a dibasic acid, hence equivalent weight = 63 ⇒ Normality = 6.3/63 × 1000/250 = 0.4 N ⇒ N1V1 = N2V2 0.1 ×V1 = 0.4 × 10 Hence, V1 = 40 mL.

C- The reflected ray rotates through an angle 2θ. A ray of light is reflected by an angle (angle of incidence or angle of reflection; since both are equal) about the normal after reflection. Now, when the mirror itself is rotated by theta, the normal also rotates by theta. So the ray of light is reflected at an angle theta about the Normal. Hence the total angle becomes: theta(angle between the incident ray and normal) + theta ( angle between the normal and the reflected ray)=2*theta.

When we stretch a wire, its length increases but its area of cross section also decreases. However the volume of the wire remans constant. If L1 is the initial length L2 is the final length, A1 is the original area of cross section and A2 is the final area of cross section we have L1A1 = L2A2 = V. We know that the resistance R1 = rho L1/A1 = rho L1 X L1 /L1 X A1 = rho L1^2/V Similarly R2 = rho L2^2/V Dividing we get R2/R1 = (L2) ^2 /(L1) ^2 = (1.1 )^2 = 1.21 since L2 is 10 percent longer than L1 So R2 = 1.21 R1 = 1.21 x 10 = 12.1 ohms

B-0.1 mm/sec. Density of Cu=9×103kg/m3 (mass of 1 m3 of Cu) 6.0×1023 atoms has a mass = 63×103kg \ Number of electrons per m3are =6.0×102363×10−3×9×103=8.5×1028 Now drift velocity =vd=ineA =1.18.5×1028×1.6×10−19×π×(0.5×10−3)2 =0.1×10−3m/sec...

Consider the two identical sphere A and B each having charge q are placed at a distance r apart as shown below. Now, ... (1) [5×0.00005 N = 5×10-5 N] When an identical uncharged sphere C is touched to A, both will share the charge on A equally i.e., each of the spheres A and C will now possess charge q/2. The sphere C placed at the mid point between A and B experiences forces FCA (due to sphere A) and FCB (due to sphere B) in the directions as shown in the figure below. Now, and Since FCB>FCA, the resultant force on the sphere C is given by, Using eqn (1), we have, F' = 5×10-5 N (along CA)

when 2 identicle metalic surface bought in contact then charge onn them are equalised due to the flow of free elctron let the initial charge on B and C is q when third sphere come in contact with B then new charge on B =q/2 and charge on the third conductor =q/2 and third conductor come in contact with C so total charge =q/2 + q =3q/2 charge will equally distribute ,so charge on C = 3q/4 now force = k (q/2)(3q/4) /d2 = 3F/8

C-10/3R. From the First law of Thermodynamics we have :- Q=w+ΔU(1) (1) QwΔU where, QQ = Heat supplied to the gas ww = Work done by the gas ΔUΔU = Change in the internal energy of the gas Given :- w=Q4(2) (2) wQ4 From (1)1 & (2)2 we have :- ⟹ΔU=3Q4(3) (3) ΔU3Q4 Now, ΔU=nRΔTγ−1(4) (4) ΔUnRΔTγ1 where, nn = No. of moles of gas ΔTΔT = Change in temperature RR = Universal Gas Constant γγ = Specific heat ratio For a diatomic gas :- γ=75(5) (5) γ75 Using (3)3, (4)4 & (5)5 we get :- ⟹nΔT=3Q10R(6) (6) nΔT3Q10R Also note that :- Q=nCΔT(7) (7) QnCΔT where, CC = Molar specific heat for the process So now replace nΔTnΔT in (7)7 by (6)6 :- ⟹3QC10R=Q3QC10RQ ⟹C=10/3R

A-2/3 Q. For monoatomic gas at constant pressure: ΔUQ=13⇒ΔU=Q3ΔUQ13ΔUQ3 From 1st law of thermodynamics: W=ΔQ−ΔU=Q−Q/3= 2Q/3

B-1.047 s. Length of the line = Distance between extreme positions of oscillation = 4 cm So, Amplitude a=2cm. also vmax=12cm/s. ∵vmax=ωa=2πTa ⇒T=2πavmax=2×3.14×212=1.047sec

Since card is replaced everytime, Therefore the probability of picking a card of hearts is 13/52 = 1/4 each time a card is drawn. Now, Probability that atleast 3 trials are needed = 1 - (P[1 trial is needed] + P[2 trials are needed]) P[2 trials] = 1st trial gives non-heart card and 2nd trial gives a heart card = 1 - (1/4 + (3/4 * 1/4)) = 1 - (1/4 + 3/16) = 1 - 7/16 = 9/16 = 0.5625 Therefore, Required Probability = 0.5625

D--+ ve ions and – ve ions will start moving randomly. In the absence of electric field the ions in the solution move randomly due to thermal energy.

D- none of above. During adsorption, there is always decrease in surface energy which appears as heat. Therefore adsorption always takes place with evolution of heat, i.e. it is an exothermic process and since the adsorption process is exothermic, the physical adsorption occurs readily at low temperature and decreases with increasing temperature. (Le Chatelier's principle).

B-2.5 cm. SOLUTION HINT : For concave lens 1/v−1/4=1/-20 v = 5 cm h2/h1=|v|/|u|⇒h2=2×5/4=2.5 cm

1/v1 + 1/u1 = 1/f1 ….(1) When an image is virtual 1/(-v2) + 1/u2 = 1/f ….(2) From equation (1) V1 = u1f/ u1 – f Or, v1/u1 = f/u1 –f = magnification (m) From equation (2) V2 = u2f/ f –u2 Or, v2/u2 = f/f-u2 = magnification (m1) If m = m1 Or, f/u1 –f = f/f-u2 Or, f = u1 + u2/2 = Focal length

B- W1=W2 . At 70 degrees the liquid converts to a different liquid having some density other than that of alochol. Now the law of buoyancy doesn't change on changing the density of liquid. So, W1=W2.

B-2R. Since the spheres are in contact at the centre of each cube face, fd = 2 R.

A- phosphorus. Solution: P4O10 + 6Fe + 3O2 2Fe3(PO4)2 Fe3(PO4)2 + 2Fe3C + 3Fe 2Fe3P + 6FeO + 2CO FeO + SiO2 FeSiO3 (furnace lining)

Whenever the current flowing through a coil changes, an emf is generated across it. This induced emf is proportional to the change in current. ie EMF α dI/dt The constant of proportionality is called Self Indusctance (denoted by L) and measured in Henry. EMF =- L (dI/dt) The negative sign indicates the EMF generated is opposite to the flow of current

Suppose a current i is passed through the inner solenoid . A magnetic field is produce inside whereas field outside it is zero. is permeability of vaccum. Let and be the no of turns it each solenoid of equal length l. Therefore, flux through each turn of solenoid is The total flux through all the turns in a length l of is And we know that where M is the mutual inductance. Therefore mutual inductance of two long coaxial solenoids of same lenth...

The relative permeability of iron compared to air is about 200. Thus, introduction of iron core will magnify the inductance by 200 times compared to air core. In other words, inductance will increase 200 times.

The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons. (a) Decreased death rate (b) Increased birth rate and longevity The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has also resulted in an increase in the longevity of an individual.

Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenarios are (1)Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc. (2)Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society

Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies. Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Apomixis is the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also, by sowing hybrid seeds, it is difficult to maintain hybrid characters as characters segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also, it is a cost-effective method for producing seeds.

Tapetum is the innermost layer of the microsporangium. It provides nourishment to the developing pollen grains. During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids, and other nutritious material required for the development of pollen grains. It also produces the exine layer of the pollen grains, which is composed of the sporopollenin.

Parthenocarpy is the process of developing fruits without involving the process of fertilization or seed formation. Therefore, the seedless varieties of economically important fruits such as orange, lemon, water melon etc. are produced using this technique. This technique involves inducing fruit formation by the application of plant growth hormones such as auxins.

Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil), which is used in various plant hybridization techniques. Emasculation is performed by plant breeders in bisexual flowers to obtain the desired variety of a plant by crossing a particular plant with the desired pollen grain. To remove the anthers, the flowers are covered with a bag before they open. This ensures that the flower is pollinated by pollen grains obtained from desirable varieties only. Later, the mature, viable, and stored pollen grains are dusted on the bagged stigma by breeders to allow artificial pollination to take place and obtain the desired plant variety.

Fruits derived from the ovary and other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits which develop from the ovary, but do not consist of the thalamus or any other floral part. In an apple, the fleshy receptacle forms the main edible part. Hence, it is a false fruit.

The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for some time and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the growing embryo and after the formation of the endosperm, further development of the embryo from the zygote starts.

Triple fusion is the fusion of the male gamete with two polar nuclei inside the embryo sac of the angiosperm. This process of fusion takes place inside the embryo sac. When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of synergids and releases two male gametes there. Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the zygote (syngamy). The other male gamete fuses with the two polar nuclei present in the central cell to form a triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion. It results in the formation of the endosperm. One male gamete nucleus and two polar nuclei are involved in this process.

Various artificial hybridization techniques (under various crop improvement programmes) involve the removal of the anther from bisexual flowers without affecting the female reproductive part (pistil) through the process of emasculation. Then, these emasculated flowers are wrapped in bags to prevent pollination by unwanted pollen grains. This process is called bagging. This technique is an important part of the plant breeding programme as it ensures that pollen grains of only desirable plants are used for fertilization of the stigma to develop the desired plant variety.

Self-incompatibility is a genetic mechanism in angiosperms that prevents self-pollination. It develops genetic incompatibility between individuals of the same species or between individuals of different species. The plants which exhibit this phenomenon have the ability to prevent germination of pollen grains and thus, prevent the growth of the pollen tube on the stigma of the flower. This prevents the fusion of the gametes along with the development of the embryo. As a result, no seed formation takes place.

There are two types of flowers present in plants namely Oxalis and Viola − chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species. Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers.

The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs. The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re-divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus. Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8 nucleate.

The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspore develops into the female gametophyte, while the remaining three degenerate.

A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower. Examples of plants bearing bisexual flowers are: (1) Water lily ( Nymphaea odorata) (2) Rose (Rosa multiflora ) (3) Hibiscus (Hibiscus Rosa-sinensis ) (4) Mustard ( Brassica nigra) (5) Petunia (Petunia hybrida)

Zoospore...zygote 1. A zoospore is a motile asexual spore that utilizes the flagella for movement. A zygote is a non-motile diploid cell formed as a result of fertilization. 2. It is an asexual reproductive structure. It is formed as a result of sexual reproduction.

Fringe width is inversely proportional to the distance (d) between the sources. When coherent sources are placed very close, fringe width is very large and a single fringe may occupy the whole field of view and the pattern would not be detected. Similarly, if the coherent sources are far apart, fringe with will be very small and the pattern cannot be detected.

because the polaroids reduce the intensity of sunlight and makes u feel comfortable.. Thus this concept is used

At polarising angle, ip+rp=90°... rp=90−ip=90−55.3 =34.7

The refractive index of a material depends on the wavelength of the light. As the polarising angle depends on the refractive index, it also depends on the wavelength of the light.

The polarization of an electromagnetic wave follows the direction of the electric field. For example, if the electric component is oscillating along the x-axis and the magnetic field is oscillating in the y-axis, the polarization will be along the x-axis.

Plane of vibration : The vibrations of EM wave occuring in a plane. This plane is always perpendicular to the plane of polarisation.

For longitudinal waves the direction of propagation is parallel to the direction of travel and they cannot be polarized, for transverse the direction of travel is perpendicular to the direction of propagation and they can be polarized..

Solution : Wave length of light in water (λ′=λ/μ) decreases.... So width of central maximum (β0αλ) also decreases...

waves diffracted from the edge of circular obstacle interfere constructive at the center of the shadow resulting in the formation of a bright spot .

Angular separation between diffraction fringe θ=β/D When the distance D is doubled the angular separation θθ remains unchanged

The size of the slit should be of the order of the wavelength of the light used.

The central fringe obtained with a Lloyd's mirror is dark because the reflection of light from the denser mirror surface introduces a path difference ofλ/2. In case of biprism, the waves from both the coherent sources reach the central point in the same phase so the central fringe is bright.

This is because the wavelength of light waves is much smaller than the wavelength of water waves. Consequently, the interference fringes have much smaller width in case of light waves than in water waves.

The wavelength of light in water (λ′=λ/μ)is less than that in air. When the apparatus is immersed in water, fringe width(β∝λ′) decreases.

Energy gets transferred from the regions of destructive interference to the regions of constructive interference.

Intensity of maxima decreases and that of minima increases.

Ifμ is-the refractive index and t the thickness of the thin film, then the entire interference pattern gets displaced by distance, Δx=D/d(μ−1)t As μ depends on λ (increases with decreasing λ), the violet fringe is shifted farther than the red fringe. So there is a kind of dispersion in the central white fringe.

The given path difference satisfies condition for the minimum of intensity for yellow Hence if yellow light is used, a dark fringe will be formed at the given point. If white light is used, all components white light except the yellow one would be present a point.

When the distance d between the coherent sources is large, the fringe width (β∝1/d) becomes too small to be detected. The interference pattern cannot be observed.

Fringe width, β=λD/d i.e., β∝1/d, when d→0,β→∞. Fringe width is very large. Even a single fringe occupy the entire screen. The interference pattern cannot be observed.

Original fringe width, β=λD/d New fringe width, β′=λ.2D/d/2=4×D/d=4β i.e., fringe width increases four times.

It is because a broad source is equivalent to a large number of narrow sources lying close to each other. Different pairs of narrow sources will produce their own interference patterns which will overlap each other. So the fringe system is lost.

Fringe width, β=Dλ/d When the distance D between the slits and the screen is doubled, the fringe width also gets doubled.

Fringe width, β=Dλ/d As the separation 'd' between the two slits decreases, fringe width β increases.

Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Coherent sources have a constant phase difference. This ensures that the positions of maxima and minima do not change with time i.e., a sustained interference pattern is obtained.

The positions of bright and dark fringes will change rapidly. Such rapid changes cannot be detected by our eyes. A uniform illumination is seen on the screen i.e., interference pattern disappears.

The conditions for obtaining sustained interference of light are (i) The two light sources should be coherent, (ii) The two light sources should be narrow and placed close to each other.

Two light waves interfering with each other are said to show coherence if the initial phase difference between them remains constant with time.

Frequency is the characteristic of the source while wavelength is the characteristic of the medium. When monochromatic light travels from one medium to another, its speed changes, so its wavelength (λ=c/v) changes but frequency v remains unchanged.

Zero. No phase change occurs during refraction.

When a wave is reflected into rarer medium from the surface of a denser medium, it undergoes a phase change of π radian.

(i) Energy of the wave decreases because a part of the light wave is reflected back into air. (ii) Frequency of the wave remains unchanged.

No. If they intersect, then there will be two rays or two directions of propagation of energy at the point of intersection which is not possible.

Yes. In a homogeneous medium, the amplitude is constant on a given wavefront.

2π radian.

Light undergoes interference, diffraction and polarisation. These phenomena establish that light is a wave motion.

B.Tech course is somewhat more practically intensive than the BE course which is more theoretical.The difference (which is quite minor) between the two degrees can be understood from their names itself, Engineering Vs. Technology. Engineering is a "science", dealing more with theoretical concepts & principles; technology deals more with the application of these theories for practical purposes. But, both B.E. & B.Tech courses offered by various engineering colleges/ institutes, have similar curriculum more or less. Both degrees are valued equally by prospective employers. What matters more is the reputation of the engineering college from where you obtain your B.E/ B.Tech degree.

generally in DNA adenine binds with thymine and guanine binds with cytosine. so if cytosine is 20% then automatically guanine will be 20%. so remaining 60% will be equally distributed in adenine and thymine.. 60/2=30% will be adenine

Avoid sexual intercourse with unknown or multiple partners. Use of condoms during sexual intercourse. Consulting a qualified doctor immediately for treatment in case of doubt of infection.

It's isolated trapezium 1st find area of rectangular base=13 height=x to find height AEC IS RIGHT ANGLE TRIANGLE 6^2+X^2=15^2 X^2=189 X=13.74 area of rectangular=13*13.74=178.72=179 area of 2 triangle=1/2*2*6*13.74=82.44 total area=261.44

In Young's double slit experiment, the fringe width is given by β=Dλ/d=Dc/dv Clearly, (i) When light of smaller frequency (v) is used, fringe width increases. (ii) Wavelength of light in water decreases, so fringe width (β∝λ) also decreases.

This is because sound waves get easily diffracted round the edge of the wall while light waves do not.

The wavelength of light is much smaller than the size of the wall. So diffraction of light around the edge of the wall is very poor. We cannot see around the house.

For diffraction to take place, the wavelength should be of the order of the size of the obstacle. The radio waves (particularly short radio waves) have wavelengths of the order of the size of the building and other obstacles coming in their way and hence they easily get diffracted. Since wavelength of the light waves is very small, they are not diffracted by buildings.

Diffraction effect is more pronounced if the size of obstacle or aperture is of the order of the wavelength of the waves. As the wavelength of light( ~10−6m) is much smaller than the size of the objects around us, so diffraction of light is not easily seen. But sound waves have large wavelength. They get easily diffracted by the objects around us.

Sex education in the school must be encouraged as it provides right information’s to the students and in removing certain myths and misconceptions about sex-related aspects existing in their minds. It will also inform them about the causes of occurrence of STDs, AIDS, etc. and also about the safe and hygienic sexual practices.

Plants part are divided into two parts. vegetative and reproductive parts. Vegetative reproduction is a type of reproduction in which their is no involvement of reproductive parts. Here any part of the plant is cut and with appropriate nutrient and growth hormone condition is able to develop into a fully grown plant. Their are various techniques which are used for vegetative reproduction. Plant should be perennial to be get vegetatively propagated...

For differences between asexual reproduction and sexual reproduction refer image..Vegetative reproduction is considered as a type of asexual reproduction because here also characters of parent plants are preserved. Here propagation takes place where new individual arises from any vegetative part of the parent (root, stem, leaf etc.), and possesses exactly the same characteristics of the parent plant from which it was sampled. Moreover the formation of new individuals does not involve two parents, process involved is asexual.

In asexual reproduction, offsprings are produced by a single parent without the involvement of gamete formation and fusion. Thus offsprings resemble the parent genetically and morphologically. In sexual reproduction, fusion of gametes occur and fertilization and meiosis are essential events. Here, offsprings differ from parents due to appearance of variations.

B-3. Fig, pine apple and mulberry are composite fruits.

C-Outer Phloem - Outer Cambium - Middle Xylem - Inner Cambium - Inner Phloem... A vascular bundle having the phloem strands on both outer and inner sides of xylem is called bicollatoral.

C- Epidermis, Hypodermis, Cortex, Endodermis. Sequence of cellular layers from the periphery towards the cortex in an old dicot stem is epidermis, hypodermis, cortex, endodermis, pericycle and vascular bundles.

D- 40 cm. Solution : xCM=m1x1+m2x2=m3x3/m1+m2+m3 =300(0)+500(40)+400(70)/300+500+400 =20000+28000/1200 =48000/1200 =40cm

A-1 MeV.

D-1.78.

C-10-37C.

Answer : 11RT Explanation : U=n1f1/2RT+n2f2/2RT, f - No.of degrees of freedom f1 for a diatomic gas = 5 f2 for a monoatomic gas = Therefore , U=2×5/2RT+4×3/2RT =11RT

C- A condition of no current flow through the galvanometer. Because In zero deflection condition, potentiometer draws no current

900 J. Let the stiffness constant of the spring be k Therefore, the elastic potential energy stored by the spring when it is stretched through a distance of 'a' from its natural length = (1/2) X k X a X a = 25 J (given). So, k X a X a = 50 J Now, the total elastic potential energy stored by the spring when it is stretched through an additional distance of '5a' that is the total distance of '6a' from its natural length = (1/2) X k X 6a X 6a = 18 X k X a X a = 18 X 50 J = 900 J

Answer: 4% Explanation: We know, Time period (T) of a simple pendulum is related with acceleration due to gravity (g)and length of the pendulum(l) as follows T = 2 π √ l g Squaring and rearranging we get g = 4 π 2 l T 2 Taking log we get log g = log 4 + 2 log π + log l − 2 log T Differentiating we have Δ g g = Δ l l − 2 Δ T T Maximum error ( Δ g g ) max % = Δ l l % + 2 Δ T T % ( Δ g g ) max % = 2 + 2 ⋅ 1 = 4

KE¹ = mgh. KE² = mg × 0.8h. KE lost = KE¹- KE² = 0.2mgh . Fraction of KE lost : 0.2mgh/mgh = 0.2 .

As ucosθ0=u/2 ∴θo=60° So R=u2sin(2×60°)/g =√3/2 u2/g

A molecule of diatomic gas has five degrees of freedom (3 tanslational and 2 rotational).

Reason is Viscosity. Explanation : When water in a container or glass tumbler is set into rotation by glass rod , the rotation starts from co-axial cylindrical layers. On stopping the rotation by glass rod, the speed of different layers are seen decreasing and liquid finally comes to rest.Which shows that the liquid experiences a internal friction. Hence Viscosity is the property of fluids by virtue of which an internal resistance or drag[fluid friction ] comes into play when liquid is in motion . The internal friction tries to stop the motion of liquid.

Given : mass of an electron=9.11 x 10⁻³¹kg =9.11 x 10⁻²⁸g 1kg=1000g Number of electrons present in 1kg will be : =1000/9.11 x 10⁻²⁸ =109.76 x 10 ²⁸ ∴ Number of electrons present in 1kg would be 109. 76x 10²⁸ electrons.

Mass of the body is m Acceleration of the body is a Using Newton’s second law of motion, the force experienced by the body is given by the equation: F = ma Both m and a are constants. Hence, force will also be a constant. F = ma = Constant The velocity for uniformly accelerated motion is v = u + at = 0 + at = at Power is given by the relation: = Fv P = ma ∙ at P = ma^ 2 t As acceleration is constant, hence P∝t

0.5 N, in vertically downward direction, in all cases Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as: F = m × a Where, F = Net force m = Mass of the pebble = 0.05 kg a = g = 10 m/s2 ∴F = 0.05 × 10 = 0.5 N The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If his/her maths base is good, then time required will lessen by a month or so. Overall, 6-7 months is required for a student to complete the syllabus and cover up the various portions of each subject. After that, continuous practice is required to adapt to the qstns asked, and devising a strategy to excel in the exam.

A.C. voltage, V= V0 sin ωt As, t=π/ω=1/2.2π/ω=1/2T therefore, first half cycle (T/2). Hence, average value of AC voltage, Eav2V0π

A.C. (Alternating Current) Circuits rely on an A.C. power source. Now, should you wish to reduce the current flowing in such a circuit, use an inductor or a capacitor since these devices would reduce the current flowing through the circuit while dissipating no power within themselves, thus causing no power loss!

Transformers are used in AC

cosϕ=0.5=1/2 => ϕ=60° Hence the phase difference is 60°

No emf is induced in the metallic case. When l and v are parallel, no emf is induced.

Intensity of the bulb decreases because there will be self induction effect when soft iron core is inserted.

Here, XC is the capacitive-reactance and is the frequency in an a.c circuit.

Tier 1 cutoff is around 130 for general. CBI SI post vacancies will be around 200 approx according to previous data. Expect CBI SI general vacancies (70 -80 ) to be filled under top 1000 ranks . (Again it depends on other top posts’ vacancies). Everything depend on Tier 2 exam level as it is the MAKE or BREAK stage . Tier 2 in SSC CGL 2016 was of fairly difficult level so we can take Tier 2 this will be of same level. So, as an approximation, you need to score around 490 out of 600 to be safe . Then around 65/100 in Tier 3 . Total around 555/700 to be safe for CBI as competition is increasing .

Ans- 100...

Suppose that a shop has an equal number of LED bulbs of two different types. ==> Therefore Probability of Taking Type 1 Bulb => 0.5 Probability of Taking Type 2 Bulb => 0.5 The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. Prob(100+ | Type1) => 0.7 Prob(100+| Type2) => 0.4 Prob(100+) => Prob(100+ | Type1) * Prob(Type1) + Prob(100+ | Type2) * Prob(Type2) = 0.7 * .5 + .4 * .5 = 0.55

Ans : 13 Hz. When 33 Hz and 46 Hz are multiplied with each other, we get signals of frequency 13 Hz and 79 Hz so on. Since we are passing the output through low pass filter, it allows only low frequency signals till it’s cut off frequency(23 Hz). So only 13 Hz is passed.

The second moment of Poisson Distribution is: Here we have, λ2+λ=2λ2λ2. Solving we get, λ=−2,1λ21. The average number of events in an interval is designated λλ (lambda). Lambda is the event rate, also called the rate parameter. Since the number of events cannot be −ve... So the mean of random variable is 1

C-20:21. Q can finish the task in 25 days, working alone for 12 hours a day. What can we learn from this line?? That Q, in 25*12 hours can complete a work alone. That is, in his work in 1 hour is 1/25*12=1/300. R can finish the task in 50 days, working alone for 12 hours per day. Now, similarly, R's one hour work is 1/600. Now, Q has worked for 5 days of 12 hours(60 hours ) and R for 7 days of 18 hours( 126 hours). We finally know, how many hours both worked, and their capacity for a hour. ratio of work done by Q and R after 77 days from the start of the project= Work done by Q/Work done by R=(60* 1/300)/(126* 1/600)=20/21.

C-6.02. michal ------------------10km----------------------------I LIVE AHMED---------5KM----------I LIVE SUSAN---------------7KM--------------------I LIVE SUSAN--ARUN--AHMED--------------------ILIVE SO ARUN IS BETWEEN 7 KM AND MORE THEN 5 KM SO HERE ANS IS LESS THEN 7 KM

Total fruits = 5692000 ={ Unripe(15%)→853800 Ripe(85%)→4838200 } ={ Unripe(15%){ Apples(45%)→384210 Orange→Don′t care } Ripe(85%){ Apples(34%)→1644988 Orange→Don′t care } } So, Total Apples = 384210 + 1644988 = 2029198-A

The distance after which the police catches the bburglar is 's' and time is 't'after the burglar car stated. Then, the time taken by police van to catch the burglar is 't−5'Now, applying second equation of motion to the burgar's car,s= ut 12at2s= 0×t 12 ×2×t2 (u=0 since car starts from rest)s= t2 −−−−(1)Applying the second equation of motin to the police car,s = ut 12at2s= 20×(t−5) 12 ×0×(t−5)2 (since it moves with uniform velocity a=0)s= 20t−100 −−−−− (2)since eq 1= eq 2t2 −20t 100 = 0(t−10)2 = 0 t −10 = 0 t=10 st−5 = 5 s....Therefore the police car over takes the burglar car after 5s it has reached the secene

B-3 km. Explanation: Let the distance be 3x km, then, x/3+x/4+x/5=47/60....47x/60=4760....x=1. So total distance = 3*1 = 3 Km

C- increase by 8%. Let length and width be xx and yy units respectively. Area=xyAreaxy After the length and width changes, the new area is Area=(x+x5)(y−y10)Areaxx5yy10 ⟹Area=6x5×9y10=54xy50=27xy25Area6x59y1054xy5027xy25 Percentage change in area=|27xy25−xy|xy×100=225×100=8

Ans-32 km. Suppose he travels on foot for x hours. Then, he travels on bicycle for ( 7 − x ) hours 8 x + 16 ( 7 − x ) = 80 ⇒ 8 x = 32 ⇒ x = 4 Distance travelled in foot = 8 × 4 = 32 km

A-6600. Solution: Let initial amount that the man invests be A Let x be the capital => Profit on 1st investment + Profit on 2nd investment + Profit on 3rd investment => S.I. = P X R X T => Remainder of capital = 1 - 1/3 - 1/4 = 5/12x (x being the capital) => a*1/3 *7/100 + 1/4 *a * 8/100 + 5x/12*a* 10/100 = 561 Therefore, a = 6600

Ans- 40 days... Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y. Then, 4x + 6y = 1 and 3x + 7y = 1 . 8 10 Solving the two equations, we get: x = 11 , y = 1 400 400 1 woman's 1 day's work = 1 . 400 10 women's 1 day's work = 1 x 10 = 1 . 400 40 Hence, 10 women will complete the work in 40 days.

E- none of these.... North-west direction is 1350 clockwise to the south direction. Similarly, North-east direction is 1350 clockwise to the south direction.

C- Grand daughter... As Amar is son of Mayank then Beena is grand daughter of Mayank....

Ans-- South-east... The man initially faces in the direction OP. On moving 90° clockwise, he faces in the direction OX. On further moving 180° anticlockwise, he faces in the direction OY. Finally, on moving 90°anticlockwise, he faces in the direction OZ, which is South-east.

New Average weight = (60 * 55 - 40)/60 = 54.33 kg

Let ‘y’ be the invested amount A.T.Q. 1.3y(1.12)2=81536 On Solving y=Rs.50,000/-

A=7x, C=9x, B=7x+29, D=9x+33 A.T.Q. 29 + 7x =15+9x ⇒ x = 7 ∴ D received 96 chocolates

CP of 10 calculators , 16 watches =Rs.56100/- CP of 15 calculators, 24 watches=1.5×56100=84150/- Profit=20% ∴S.P. of 15 calculators, 24 watches =1.2×84150 = Rs. 100,980/-

Ans-- A- 96 .... Area of square= (side)2. => 729=(side)2. => Side=√729= 27... Now l=27. Diff.= 27-6 = 21= b. Now perimeter of rec. = 2(L+B). => 2(27+21). => 2(48). =>96....

Price of 36 kg sugar @ Rs. 45 = 36 * 45 = Rs. 1620. Price of 24 kg sugar @ Rs. 42 = 24 * 42 = Rs. 1008. Total Cost Price = 1620 + 1008 = Rs. 2628. Total SP = 2628 + 20% of 2628 = Rs. 3153.6. Total Sugar = 36 = 24 = 60 kg. SP per kg = 3153.6 / 60 = Rs. 52.56.

His Per day pay = Rs. 26. Total pay employee got = Rs. 829. Total pay he gets if he did not remain idle a single day, = 26 *56 = Rs.1456. He Forfeits or fined = 1456 - 829 = Rs. 627. Per day he Forfeits Rs. 7. Means per idle day he looses = 26 +7 = Rs. 33. So, Total idle days = 627/33 = 19 days.

Ratio of students in 2012 in institutes A and B, = 7 : 15. Let number of students in institute A in 2012 = 700. And Number of students in institutes B in 2012 = 1500. 25% increase in the number of students in 2013, Now, number of students in Institute A, = 700 + 25% of 700. = 875. Number of students in B in 2013 as 26% students increased in B, = 1500 + 26% of 1500 = 1890. Current Ratio of the students, = 875 /1890 = 25 : 54.

Home___________D Km____________Station. Let Distance of the station = D km. And actual time taken = t hours. Given, S1 = 6 kmph. Time taken to reach station with speed 6 kmph, = t + 8 min = [t +(8/60)] = t + (2/15)= (15t +2)/15 D = S1 *[t +(2/15)] = 6 *[t +(2/15)] -------- (1)[As, Distance = Speed *time] S2 = 10 kmph. Time taken to reach station with speed 10 kmph, = t - 7 min = t - (7/60) = (60t -7)/60 D = 10* [(60t - 7)/60] ----------------- (2) Comparing (1) and (2) 10* [(60t - 7)/60] = 6 *[t +(2/15)] (60t -7)/6 = 6 * [(15t +2)/15] (60t -7)/6 = (30t +4)/5 300t - 35 = 180t +24 t =15/4...

E-20016. Required ratio of profit of A : B : C = 64 : 52 : 36 = 16 : 13 : 9 Now, let the share of profit of A be 16x, that of B be 13x and that of C be 9x The share of profit of A = 35584 16x = 35584 x = 35584/16 = Rs 2224 Hence the share of profit of C = 9 x 2224 = Rs 20016

C-48. Let the units digit be x and the tens digit be y Then, the number = 10y + x According to the question, x + y =12... (1) And 10x + y = 10y + x+36 9x - 9y =36 x - y = 4....(2) By solving (1) and (2), we get x = 8 Putting the value of x in (1), we get y = 4 Hence, the number = 10y + x = 10 x 4 + 8 = 48

D-48 km/hr. Average speed = Total distance/Total time (39 + 25)/(45 +35) minutes = 64/80 x 60 = 48 km/hr

Answer : 26 metric tonnes Description : As 9 engines consumes 24 metric tonnes of coal, when each working for 8 hours. Therefore, 1 engines consume in 1 hour = 24/(8*9) = 3 3 engines of former type = 4 engines of latter type. 6 engine of former type = 8 engine of latter type. Then, Coal required for 8 engines each running for 13 hours a day = 24*6*13/(8*9) = 26 metric tonne

A-40. Explanation: Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years. Sum of the present ages of wife and child (20 x 2 + 5 x 2) years = 50 years. Husband's present age = (90 - 50) years = 40 years.

Explanation: Total weight increased = (8 x 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg.

Let the average after 7th inning = x Then average after 16th inning = x - 3 16(x-3)+87 = 17x x = 87 - 48 = 39

If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, average speed of the whole journey = 2xy/x+y kmph. By using the same formula, we can find out the average speed quickly. Average speed = 2×84×56/84+56=2×84×56/140=2×21×56/35 =2×3×56/5=33/65=67.2

B-174. Explanation : Total runs scored by the player in 40 innings = 40 × 50 Total runs scored by the player in 38 innings after excluding two innings = 38 × 48 Sum of the scores of the excluded innings = 40 × 50 - 38 × 48 = 2000 - 1824 = 176 Given that the scores of the excluded innings differ by 172. Hence let's take the highest score as x + 172 and lowest score as x Now x + 172 + x = 176 => 2x = 4 => x =42 = 2 Highest score = x + 172 = 2 + 172 = 174

for SSC CHSL first of all go to the syllabus and read it carefully. after that you have to select books that are essential for preparation.. there is a wide range of books in market .

A-1800. 6 machines produce 270 bottles per minute; 1 machine produces 270/6=45 bottles per minute; 10 machines produce 45*10=450 bottles per minute; In 4 minutes 10 machines produce 450*4=1,800 bottles.

D-13. Let the required number of days be x. Less persons, More days (Indirect Proportion) More working hours per day, Less days (Indirect Proportion) Persons 30 : 39 :: 12 : x Working hours/day 6 : 5 30 x 6 x x = 39 x 5 x 12 x = (39 x 5 x 12) (30 x 6) x = 13.

D-195. Le the required time be x seconds. More metres, More time (Direct Proportion) 0.128 : 25 :: 1 : x 0.128x = 25 x 1 x = 25 = 25 x 1000 0.128 128 x = 195.31. Required time = 195 sec (approximately).

Let the height of the building x metres. Less lengthy shadow, Less in the height (Direct Proportion) 40.25 : 28.75 :: 17.5 : x 40.25 x x = 28.75 x 17.5 x = 28.75 x 17.5 40.25 x = 12.5

30*8*104*3/2 = 26*9*x => x = 160 additional workmen = 160-104=56 men

Ans-. Amount of work completed by 1 man in 1 day, working 1 hours a day = xx3=1x2Amount of work y men in y days, working y hours a day = y3×1x2=y3x2

B-12. 3 pumps take 16 hrs total (8 Hrs a day) If 1 pump will be working then, it will need 16*3=48 hrs 1 pump need 48 Hrs If I contribute 4 pumps then 48/4=12 hrs.

A-9. M1 = 9 D1 = 12 H1 = 5 W1 = 1 M2 = 4 D2 = 30 H2 = x W2 = 2 M1D1H1W1=M2D2H2W2⇒9×12×51=4×30×x2⇒9×12×5=2×30×x⇒3×12×5=2×10×x⇒3×6×5=10×x⇒3×3×5=5×x⇒x=3×3=9

A-49 rounds. Let the required number of rounds be x More radius, less rounds(Indirect proportion) Hence we can write as radius14:20}::x:70 ⇒14×70=20x⇒14×7=2x⇒x=7×7=49

C-110. Explanation: Originally let there be x men. Less men, More days (Indirect Proportion) (x-10) : x :: 100 :110 10x= 1100 x= 110

A-885.5 km. Let the required actual distance be x km More scale distance, More actual distance(direct proportion) Hence we can write as scale distance.6:80.5}::6.6:x ⇒.6x=80.5×6.6⇒.1x=80.5×1.1⇒x=80.5×11=885.5

price of 357 apples = Rs.1517.25price of 1 apple = Rs. 1517.25357price of 49 dozens apples = Rs.(49×12×1517.25357)≈Rs. 2500

Persons worked = 104 Number of hours each person worked per day = 8 Number of days they worked = 30 Work completed = 2/5 Remaining days = 56 - 30 = 26 Remaining Work to be completed = 1 - 2/5 = 3/5 Let the total number of persons who do the remaining work = x Number of hours each person needs to be work per day = 9 More days, less persons(indirect proportion) More hours, less persons(indirect proportion) More work, more persons(direct proportion) Hence we can write as DaysHoursWork30:268:935:25⎫⎭⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪::x:104 ⇒30×8×35×104=26×9×25×x⇒x=30×8×35×10426×9×25=30×8×3×10426×9×2=30×8×104/26×3×2=30×8×4/3×2=5×8×4=160Number of additional persons required = 160 - 104 = 56

Let the required time be x seconds. More metres, More time (Direct Proportion) 0.128 : 25 :: 1 : x 0.128x = 25 x 1 x = 25 = 25 x 1000 0.128 128 x = 195.31. Required time = 195 sec (approximately).

Let the required number of working hours per day be x. More pumps, Less working hours per day (Indirect Proportion) Less days, More working hours per day (Indirect Proportion) Pumps 4 : 3 :: 8 : x Days 1 : 2 4 x 1 x x = 3 x 2 x 8 x = (3 x 2 x 8) (4) x = 12.

Let the required number of revolutions made by larger wheel be x. Then, More cogs, Less revolutions (Indirect Proportion) 14 : 6 :: 21 : x 14 x x = 6 x 21 x = 6 x 21/ 14 x = 9.

B-30. There is a meal for 200 children. 150 children have taken the meal. Remaining meal is to be catered to 50 children. Now, 200 children 120 men. 50 children 120/200 x 50 = 30 men.

Let the man's rate upstream be x kmph and that downstream be y kmph. Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs. x *8 4/5 = (y x 4) 44/5 x =4y y = 11/5x. Required ratio = (y + x)/2 : (y - x)/2 = (16x/5* 1/2) : (6x /5* 1/2) = 8/5: 3/5 = 8 : 3.

Speed downstream = (22 + 4) = 26 kmph Time = 24 minutes = 24 60 hour = 2 5 hour Distance travelled = Time × speed = 2 5 ×26 = 10.4 km

Speed downstream = (14 + 1.2) = 15.2 kmph Speed upstream = (14 - 1.2) = 12.8 kmph Total time taken = 4864/ 15.2 + 4864/ 12.8 = 320 + 380 = 700 hours

Let the number of subjects = x Then, total marks he scored for all subjects = 63x If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65 => Total marks he would have scored for all subjects = 65x Now we can form the equation as 65x - 63x = additional marks of the student = 20 + 2 = 22 => 2x = 22 => x = 22/2 = 11

There are 2 grandparents of 67yrs (67*2) 2 parents of 35yrs (35*2) 3 grandchildren of 6yrs (6*3) average = sum of total ages/no.of persons 67 x 2 + 35 x 2 + 6 x 3 divided by 2 + 2 + 3 =31 5/7years

In a 30 days month you will get 5 sundays if the 1st day is sunday. So 5 sundays x 510 = 2550 Remaining 25 days X 240 days =6000. So 2550 +6000= 8550/30 = 285.

D-67 kg. Let Kiran's weight = x. Then According to Kiran, 65 < x < 72 ----(equation 1) According to brother, 60 < x < 70 ----(equation 2) According to mother, x ≤ 68 ----(equation 3) Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as 65

Let P, Q and R represent their respective monthly incomes. Then, we have: P + Q = (5050 x 2) = 10100 .... (i) Q + R = (6250 x 2) = 12500 .... (ii) P + R = (5200 x 2) = 10400 .... (iii) Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv) Subtracting (ii) from (iv), we get P = 4000. P's monthly income = Rs. 4000.

A-23 years. Let the average age of the whole team by x years. 11x - (26 + 29) = 9(x -1) 11x - 9x = 46 2x = 46 x = 23. So, average age of the team is 23 years.

D-19. Average of 20 numbers = 0. Sum of 20 numbers (0 x 20) = 0. It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009. Required sale = Rs. [ (6500 x 6) - 34009 ] = Rs. (39000 - 34009) = Rs. 4991.

Ans-27ft.

Perimeter = Distance covered in 8 min. = (12000/60 x 8) m = 1600 m. Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2.

Ans-60 m.

A-176. Explanation: area of the room = 544 * 374 sq cm size of largest square tile = H.C.F of 544cm and 374cm= 34cm area of 1 tile = 34*34 sq cm no. of tiles required = (544*374) / (34 * 34) = 176

C-40 cm. Explanation: Let breadth = x x cm Then, length = 2 x 2x cm Area = x × 2 x = 2 x 2 x2x2x2 sq.cm. New length = ( 2 x − 5 ) 2x5 cm New breadth = ( x + 5 ) x5 cm New area = ( 2 x − 5 ) ( x + 5 ) 2x5x5 sq.cm. Given that, new area = initial area + 75 75 sq.cm. ⇒ ( 2 x − 5 ) ( x + 5 ) = 2 x 2 + 75 ⇒ 2 x 2 + 10 x − 5 x − 25 = 2 x 2 + 75 ⇒ 5 x − 25 = 75 ⇒ 5 x = 75 + 25 = 100 ⇒ x = 100/ 5 = 20 cm Length = 2 x = 2 × 20 = 40

Average of the two areas = 700 2 = 350 one-fifth of the average of the two areas = 350 5 = 70 350570 ⇒ Difference of the two areas = 70 70 Let area of the smaller part = x x hectares. Then, area of the larger part = x + 70 x70 hectares. x + ( x + 70 ) = 700 ⇒ 2 x = 630 ⇒ x = 315

length = 9 feet breadth = 37 − 9/ 2 = 14 feet Area = 9 × 14 = 126 square feet

When a bar magnet is brought near a magnetic material like iron, steel it gets attracted towards and sometimes gets lifted by the magnet if the material density is less. This is called Single touch method.

D-60. The interval between red lights is 20 seconds and the interval between green lights is 12 seconds. Their LCM is 69 seconds. So the lights will flash together in 60 seconds

D-1590. Let the number that Anita wanted to multiply be 'X'. She was expected to find the value of 35X. Instead, she found the value of 53X. The difference between the value that she got (53X) and what she was expected to get (35X) is 540. i.e., 53X - 35X = 540 or (53 - 35) * X = 540 X = 30 Therefore, the correct product = 53 * 30 = 1590

Explaination

Ans--- √a1a2

The coefficient of restitution (COR) is the ratio of the final to initial relative velocity between two objects after they collide. ... A perfectly inelastic collision has a coefficient of 0, but a 0 value does not have to be perfectly inelastic.

Ans-36.87°.

Ans-4cms-2.

Percentage= camera users/ total phone users*100. => 35/500*100. =>7%. So7% of users have camera phones..

B-700 . 1 min= 4 pages. So 1400 pages = 350 min.... Now there are 2 copies so 350*2= 700 min....

A-160. Julie has 250 rs.. she purchased sweater for RS. 75. So 250-75= 175. But at the end shw had 15 rs left with her so,. 175-15= 160... So, she spend 160 on jacket..

Please refer given image....

Ans-36.87°.

As we know that Sound is a mechanical wave and the speed of sound in air is given by : v = √(γRT/M) , which depends on temperature . But , light is an electromagnetic wave and the speed of light wave is given by : c = √(1/εoμo) , which is independent of temperature .

Ans-4 cms.

Please refer image for answer

If a ball is thrown up, the direction of motion of the body is the same as the direction of its velocity whereas the acceleration due to gravity acts on it in the downward direction.Thus, the direction in which an object moves is given by the direction of velocity and not by the direction of acceleration.

Please refer image for this answer...

Ans. r=√3/4 a

Ams-137.04.

The strong electrolyte is B. Since strong electrolyte are completely or near completely dissociated with high concentrations. So dilution has only a small effect on small electrolytes molar conductivity. It is opposite for a which is aweal electrolytes increasing strongly molar conductivity when diluting.

The structural formula of the other compound is [Cr(H2O)4Cl2]Cl⋅2H2O.

If anaerobic condition are not maintained then enzymes cannot act and hence it cannot be mix properly...

A can do work in 6 days = 1/6. B can do work in 12 days= 1/12. They can complete work together= 1/6+1/12. => 3/12 => 1/4... So they can complete the work in 4 days....

D-44. If age of the teacher was 21, average would not have changed. Since average increased by 1 1, Age of the teacher = 21 + 23 × 1 = 44

B-22 years. Let the son's present age be x years. Then, man's present age = (x + 24) years. (x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 x = 22.

Ans- 50 years. Let the age before 10 years = x x. Then, 125 x 100 = x + 10 ⇒ 125 x = 100 x + 1000 ⇒ x = 1000 25 = 40 Present age = x + 10 = 40 + 10 = 50

Ans--- 1) G is sitting second to the right of E... 2) K is professor.... 3) Third to the right...

Ans---1) 36 N > J > K > M > I > L K=31 L=12 From the question , J Possibly handles only 36. Because K handled 31 projects so that J must be greater than 31. So answer is 36. Ans---2) (I+K) = 33+L. => 33+12. => (I+K)= 45. Now K =31 so I = 45-31 = 14....

B-4:7. 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

D-36 days.

C-23/35. 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

D-156.

A- The inductance of a coil will increase under all the following conditions except when more length for the same number of turns is provided....

A- inductance. An inductor stores the kinectic energy of moving electron in form of magnetic field.

C-As per Faraday's laws of electromagnetic induction, an e.m.f. is induced in a conductor whenever it cuts magnetic flux...

A-The property of coil by which a counter e.m.f. is induced in it when the current through the coil changes is known as self-inductance...

A- Rs. 450/-. Let CPa and CPb be the cost prices of the items A and B Given that CPb = CPa + 150 Also given that item A was sold at 1.1 CPa . Item B was sold at 0.8 CPb or 0.8 (CPa + 150) By the question, 1.1 CPa / (0.8 CPa +120) = 11/12 This gives 13.2 CPa = 8.8 CPa + 1320, Solving we get CPa = 300 Therefore CPb = 300 + 150 = 450

Let CPa and CPb be the cost prices of the items A and B Given that CPb = CPa + 150 Also given that item A was sold at 1.1 CPa . Item B was sold at 0.8 CPb or 0.8 (CPa + 150) By the question, 1.1 CPa / (0.8 CPa +120) = 11/12 This gives 13.2 CPa = 8.8 CPa + 1320, Solving we get CPa = 300 Therefore CPb = 300 + 150 = 450

A- 5th to the right.

A-GCD

C-73.17%. Required percentage = (70 + 80) x 100 % (95 + 110) = 150 x 100 % 205 = 73.17%.

D-7:9. Required ratio = (75 + 65)/(85+65) = 140 /180 = 7/9.

C-25:21. Solutions: Ratio of speed of camel and elephant = 5/3 : 7/5 = 5/3 X 15 : 7/5 X 15 = 25:21

B- (18,25)

B-7. (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12. or 137 + x = 144 or x = 144 - 137 = 7.

B-(47+43). As we know an−bn=(a−b)(an−1+an−2b+…+abn−2+bn−1) anbnaban1an2babn2bn1 . Now, if n n is odd then bn=−(−b)n bnbn , so using the above an+bn=an−(−b)n=(a−(−b))(an−1+an−2(−b)+…+a(−b)n−2+(−b)n−1)= anbnanbnaban1an2babn2bn1 =(a+b)(an−1−an−2b+…−abn−2+bn−1) aban1an2babn2bn1 (4743+4343)=(47+43)(4742−474143+…−47∗4341+4342) 47434343474347424741434743414342 (4747+4347)=(47+43)(4746−474543+…−47∗4345+4346) Common is (47+43)

Ans-75 cm

Focal length of concave lens = -20 cm = -20/100 m power = P₁ = 1/-20/ 100 = -5 D focal length of convex lens = +25 cm = 25/100 m power = P₂ = 1/25/100 = 4 D So net power = P = P₁ + P₂ = -5D + 4D = -1 D (implies combination behave as concave lens).

Coarse grains generally refers to cereal grains other than wheat and rice — in the OECD countries, those used primarily for animal feed or brewing. The climatic condition for Jute Cultivation is mentioned below : Temperature: 25°C is ideal for jute cultivation. Rainfall: 150 cm to 200 cm rainfall is essential for jute cultivation. Moisture in the wind is necessary. Soil: New alluvial or loamy soil or river basin is ideal for jute cultivation. Land: Plain land or gentle slope or low land is ideal for jute cultivation. But huge number of cheap labor is required along with irrigation facilities. HYV Seeds like Basudev, Sobajsora etc., plenty of fertilizer are required. The ideal conditions for the production of cotton are: Climate: Cotton grows well in warm and moist climate where summer is long and where there is salinity in the soil. Temperature: Cotton grown well in a temperature of 24°C. But cotton bursts out, high temperature is injurious. Rainfall: 60-100 cm rainfall is essential for the cultivation of cotton. But rainfall after the cotton comes out of the cocoon in injurious. Soil: Loamy soil mixed with lime and potash is good for cotton. Cotton grows well in Black Cotton Soil which is sticky in nature and has water retentive capacity. Land: Plain lands or gentle slope lands are good for cotton production. Drainage is necessary for the cultivation of cotton.

Area = 1/2 * diagonal(square) Area = 1/2 *12^2 Area = 1/2 * 144 Area = 72 cm^2

Ans-150. As 2 = 1x1x1 + 1x1 12 = 2x2x2 + 2x2 36 = 3x3x3 + 3x3 so on ... so, ans = 5x5x5 + 5x5 = 125 + 25 = 150

Ans-12 πF, 4πF...

Potential of any point a with respect to any other point b,..

Ans- -6.8 eV. The energy of an electron in a Bohr atom is expressed as where, k = constant, Z = atomic number, N = orbit number E_n = - 〖kZ〗^2/n^2 = - 27.6eV for H (n = 1) When n = 2, E2 = E_2 = (- 27.6)/2^2 eV = - 6.80 eV (n can have only integral value 1, 2, 3,……. ∞)

Ans-- 1.08°. Refractive index of air w.r.to glass, gηa = 1.50 Refractive index of air w.r.to water, wηa = 133 Therefore, aηg x gηw x wηa= 1 1/1.50 xgηw 1.33 = 1 gηw = 1.50/ 1.33 = 1.13 gηw = 1/sin C Sin C = l/gηw = 1/1.13 = 0.88 C = 1.08° , is the critical angle.

Ans-- 2.52V. V = BvL = 0.9 x 7 x 0.4 = 2.52V

Peltier’s Effect 1. It takes place only at the junction. 2. Heat is evolved or absorbed. 3. It is a reversible process. 4. It depends on the direction of current 5. Heat is evolved or absorbed is proportional to the current. 6. Amount of heat evolved or absorbed depends on the nature of the metals and temperature of the conductor. Joule’s Effect 1. It takes place throughout the conductor. 2. Heat is always produced. 3. It is an irreversible process. 4. It is independent of the direction of the current. 5. Heat produced is proportional to the square of the current. 6. Amount of heat produced depends on the resistance of the conductor.

The energy of a parallel plate capacitor increase when the area of plates is doubled...

Ans- 1.7 kgs. Flower-nectar contains 50% of non-water part. In honey this non-water part constitutes 85% (100-15). Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg Therefore amount of flower-nectar needed $= ({0.85/0.5} × 1)kg = $1.7 kgs

Ans-36,000. Let the percentage of the total votes secured by Party D be x% Therefore, the percentage of total votes secured by Party R = (x - 12)% As there are only two parties contesting in the election, the sum of the votes secured by the two parties should add up to 100% i.e., x + x - 12 = 100 2x - 12 = 100 or 2x = 112 or x = 56%. If Party D got 56% of the total votes, then Party R would have got (56 - 12) = 44% of the total votes. 44% of the total votes = 132,000 i.e., 4410044100 * T = 132,000 => T = 132000∗1004413200010044 = 300,000 votes. The margin by which Party R lost the election = 12% of the total votes = 12% of 300,000 = 36,000.

Ans-- 8%. let 1 lt cost 1 rupees now again 1 lt cost is 125/115 increase=125/115-1; %reduce=inc*100/inc cost=(2*23*100)/(23*25) =8%

Ans-6.02 km. ---------------------- 5 km ahmed i live --------------------------------------------------------------------------------10 km michael ----------------------------------------7km susan question say that Arun is father who away from Ahmed and closer to susan means value lies greater than 5 and less than 7......only one value which is 6.02

P(hitting one shot)=4C1*(1/4)*(3/4)^3=108/256 P(hitting two shots)=4C2*(1/4)^2*(3/4)^2=54/256 P(hitting three shots)=4C3*(1/4)^3*(3/4)=12/256 P(hitting all 4 shots)=4C4*(1/4)^4=1/256 So, P(hitting)= 108/256+54/256+12/256+1/256= 175/256

a weighted average question. Since the given average of 50% and 60% is 55% (right in the middle), it means the number of people surveyed in Eros (n) is same as the number of people surveyed in Angie. So n = 100 Total = 100 + 100 = 200

female = x Male = 750-x x/2 + 750-x/4 = 1/3* (750)= 250 x= 250 x/2 = 125

by double matrix chart... A not A B not B Total for A - 100 For A and B both - 50 For A alone 50 For neither is 0. we then have that 150 people not voted for A, and voted for B only.

59% of the employees E are workers. Since the factory consists of only workers, executives, and clerks, the remaining 100 – 59 = 41% of the employees must include only executives and clerks. Since we are given that the number of executives is 460 and the number of clerks is 360, which sum to 460 + 360 = 820, we have the equation (41/100)E = 820, or E = 100/41× 820 = 2000.

C-9. Let kk be an integer such that p9+q10=kp9q10k. Since all three numbers pp, qq, and kk are integers, among the given options, the only case could be: integer (p9)p9 + integer (q10)q10 = integer (k)k The only value that fits the condition is (C). So, Ans - C.

42.42 = 14K + Km/50...we can rewrite the number as follows: 42+ 0.42= 14K + Km/50........Since K is integer, then 42=14K..........K=3 0.42=Km/50......42/100=3m/50......m=7 k+m=3+7=10

Plug in some numebrs and check - m = 5 n = 3 r = 2 What is the remainder when 2m is divided by 2n ? 2m = 10 2n = 6 So, Remainder when 2m is divided by 2n is = 4 2r = 4 Thus, answer must be 2r

Using number properties: the average value of an even list of consecutive numbers will be the middle number. list of numbers: m m+1 m+2 m+3 m+4 average given in question stem: n average as per the number properties rule: m+2 Therefore m+2 = n New list: " 9 consecutive integers that start with (m + 2)" ... since m+2 = n: n n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8 Average (using the same number properties rule): n n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8

smallest of five consecutive even integer = x greatest of five consecutive even integers = x+8 greatest of five consecutive even integers - smallest of five consecutive even integer = x + 8 - x = 8

LCM (5, 6, 7, 8, 9) - 1 = 2519.

Number of Multiples of 3 from 1 through 30 = 30/3 = 10 Number of Multiples of 13 from 1 through 30 = 30/13 = 2 Number of Multiples of 3 and 13 both from 1 through 30 = Number of Multiples of 13*3(=39) = 0 Total favourable cases = 10 + 2 - 0 = 12 Probability = 12 / 30 = 2/5

P(zero shots hit plane) = P(1st shot misses AND 2nd shot misses AND 3rd shot misses AND 4th shot misses) = P(1st shot misses) x P(2nd shot misses) x P(3rd shot misses) x P(4th shot misses) = 0.3 x 0.4 x 0.5 x 0.6 = 3/10 x 4/10 x 5/10 x 6/10 = 360/10000 = 0.036 So....... P(at least 1 shot hits plane) = 1 - P(zero shots hit plane) = 1 - 0.036 = 0.964

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is: = 6*6*6*6=64 n(S) = 64 Let X be the event that all dice show the same face. X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} n(X) = 6 Hence required probability = n(X)/ n(S) = 6/ 64*4 = 1/ 216

11/12 36 possible outcomes P(11, that is either [5,6] or [6,5]) = 2/36. P(12, that is [6,6]) = 1/12 hence P(11 or 12) = P(sum>= 11) = 3/36 = 1/12 P(sum=11) = 1 - 1/12 = 11/12

In a chessboard the 4 squares at corners share a side only with 2 squares. If the first square is at corner, then the other square must be from those 2. out of four one square can be chosen in 4c1 ways; hence total no. of ways = 4*2=8 the remaining 24 squares of sidemost rows share a side with 3 squares; hence total no. of ways = 24*3=72 other 36 squares share a side with 4 squares; thats y total no. of ways= 36*4=144 144+72+8=224 are the no. of ways in which 2 squares sharin a side can be selected. total no of ways of selecting 2 squares out of 64 squares are= 64c2 the required probability= 224/(64c2)= 1/9

Each of the span is an independent event and the outcome of the 15 th span will not depend on the outcome of the earlier spans. So, Either 35:1

There are a total of 90 two digit numbers. Every third number will be divisible by '3'. Therefore, there are 30 of those numbers that are divisible by '3'. Of these 30 numbers, the numbers that are divisible by '5' are those that are multiples of '15'. i.e. numbers that are divisible by both '3' and '5'. There are 6 such numbers -- $15, 30, 45, 60, 75$ and $90$. We need to find out numbers that are divisible by '3' and not by '5', which will be: $30 - 6 = 24$. 24 out of the 90 numbers are divisible by '3' and not by '5'. The required probability is therefore, $24/90$ = 4/15

P(getting a 6) = 1/6 P(not getting a 6 on one throw) = 5/6 P (not getting a 6 on three throws) = 5/6*5/6*5/6 = 125/216 P(getting at least a 6) = 1-125/216= 91/216

35X=y ; 53X=y 540; 53X=35X 540; 18X=540; X=540/18=30; 53x30=1590

Galvanizing means coating Zinc on iron surface as in steel pipes for use domestic in plumbing before the advent of plastic water pipes...

80 minutes = 4 half-lives of A = 2 half-lives of B Let the initial number of nuclei in each sample be N NANA after 80 minutes =N/2^4 => Number of A nuclides decayed =15/16N NBNB after 80 minutes =N/2^4 => Number of B nuclides decayed =3/4N Required ratio =15/16/3/4 =5/4

Telescope creates a focused image of a distant object, so that the object looks nearer. Magnifying power of telescope, M = 20, implies that the tree appears 20 times nearer.

Equation of line parallel to x = y = z through (1, –5, 9) is x -1/1= y- 5/1 = z -9/ 1 = a Now if If P(a+ 1, a– 5, a+ 9) be point of intesection of line and plane then a+1 + a-5 + a+9 =5. Solving this a=-10 which means Coordinates point are (–9, –15, –1) and hence distance is 10 * root(3)

Let length of two parts be ‘a’ and ‘2 - a’ As per condition given, we write a=4xand2−a=2πra4xand2a2πr x=a4xa4 and r=2−a2πr2a2π ∴A(square)=(a4)2=a216Asquarea42a216 and A(circle)=π[(2−a)2π]2=π(4+a2−4a)4π2Acircleπ2a2π2π4a24a4π2 =(a2−4a+4)4πa24a44π f(a)=a216+a2−4a+44πfaa216a24a44π ∴f(a)=a2π+4a2−16a+1616πfaa2π4a216a1616π f′(a)=116πfa116π[2aπ+8a−16]2aπ8a16 f′(a)=0=>2aπ+8a−16=0fa02aπ8a160 => 2aπ+8a=162aπ8a16 x=a4=2π+4xa42π4 and r=2−a2πr2a2π =2−8π+42π28π42π =2π+8−82π(π+4)2π882ππ4 =1π+41π4 x=2π+4x2π4 and r=1π+4r1π4 x=2r

By Coulomb’s Law, Force is inversely proportional to square of distance between two charged particles. So if distance increases 2 times, force reduces 4 times.

On the basis of Huygen's wave theory, the laws of refraction Given : AEB is the incident wave front and DGC be the refracted wavefront. BC = V1t V1 = velocity of light in medium 1. AD = V2t V2 = velocity of light in medium 2.

Ans-- The reading on Ammeter is 0.5 A

234Th90 → 234Pa91 + 0e−1 + energy

Energy of photon = kinetic energy of photon electron + threshold frequency

Condition for achromatic doublet of two lens:. w/f + w/f' =0 w+w' are the dispersive power of two lens, f + f' are the respective focal length.

When the magnet are grounded concave, the fields lines then pass radially by across the pole pieces such a field is known as radial field.

Magnetic flux density at the axial position::: |B axial| = myu/ 4π =2|M|r/(r^2-l^2)^2

C-Mega electron volt(MeV)

C-Band spectrum

B-7.5 m. Length of the carpet = (total cost)/(Rate /m) = (8100)/(46) = 180 m Area of the room = Area of the carpet = (180 * 75/100) = 135 sq m Hence, Breadth of the room = (area)/(length) = 135/18 = 7.5 m

B- Rs. 500. Let the total sale be Rs. x. Then, 2.5%. of x = 12.50 (25/100 * 1/100 * x) = 125/10 x = 500.

D-10%. Let the rate be R% p.a. Then, 5000 x R x 2 + 3000 x R x 4 = 2200. 100 100 100R + 120R = 2200 R = 2200 = 10. 220 Rate = 10%.

Part filled by A in 1 min = 1 . 20 Part filled by B in 1 min = 1 . 30 Part filled by (A + B) in 1 min = 1 + 1 = 1 . 20 30 12 Both pipes can fill the tank in 12 minutes.

A- Sunday. The year 2004 is a leap year. So, it has 2 odd days. But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only. The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004. Given that, 6th March, 2005 is Monday. 6th March, 2004 is Sunday (1 day before to 6th March, 2005).

B-10:13. Required Ratio will be (83+108+74+88+98/98+112+101+133+142)=(451/586)=1/1.3=10:13

A-544.44 lakhs. Total expenditure of the Company during 2000 = Rs. (324 + 101 + 3.84 + 41.6 + 74) lakhs = Rs. 544.44 lakhs.

C-69%. Required percentage = (288 + 98 + 3.00 + 23.4 + 83) x 100 % (420 + 142 + 3.96 + 49.4 + 98) = 495.4 x 100 % 713.36 69.45%.

C-1%. Required percentage = (3.00 + 2.52 + 3.84 + 3.68 + 3.96) x 100 % (288 + 342 + 324 + 336 + 420) = 17 x 100 % 1710 1%.

D-36.66 lakhs.. Average amount of interest paid by the Company during the given period = Rs. 23.4 + 32.5 + 41.6 + 36.4 + 49.4 lakhs 5 = Rs. 183.3 lakhs 5 = Rs. 36.66 lakhs.

C-1.5, 1.6 km/hr. Rate in still water = 1 / 2 (10 + 7) km/hr = 8.5 km/hr. Rate of current = 1 / 2(10 - 7) km/hr = 1.5 km/hr.

C-9. Any side of triangle is less than the sum of other two sides so in this case x < 22. Also any side of triangle is larger than positive difference of other two sides so x > 2. 2 < x < 22 Total of 19 possible values. As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values

D--2π/3. sin B = b / 2R = AC / 2 = R / 2R [Given AB = AC = R] = 1/2 B = p / 6 or 5p / 6 When B = 5p / 6, C = 5p / 6 [AB = AC ? B = C] ?B + C > p So, B = 5p / 6 not possible Hence, B should bep / 6,C = p / 6 A = p - (p / 6 + p / 6) A = 2p / 3

E-- 15kHz.

C- constant... AM frequency of the carrier wave remains unchanged

C--+Variation in amplitude or frequency of carrier wave is called modulation...

D- signal.

A--With a 30 m long coaxial cable, bandwidth can exceed 100MHz

B-CD data are stored as a series of tiny indentations known as “pits”, means digital signal...

D-all of above. express the samples of the analog signal , we use the notation (with square brackets), where integer values of index the samples. .... Attempting to recover by ideal low pass filtering instead produces since the latter has a lower frequency.

A--- Each digit in a binary number is known as a bit...

B- more than 3 MHz. The sky waves are the radiowaves of frequency between 2 MHz to 30 MHz.

A-A digital quantity has only 2 values...

C-Bit on left hand side has highest value...

B- same as Earth....A geosynchronous satellite is a satellite in geosynchronous orbit, with an orbital period the same as the Earth's rotation period.

A-First communication satellites used frequencies of 6 GHz for uplink..

B--Unwanted signal that distorts a transmitted signal is called noise...

A- amplitude...the alteration of the amplitude of the carrier wave, whereas FM involves changes in the frequency of the carrier wave.

D-2. In mathematics and digital electronics, a binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically 0 (zero) and 1 (one). The base-2 numeral system is a positional notation with a radix of 2.

D--Phenomena in which signal transmitted in one circuit creates undesired effect in other circuit is known as crosslinking..

B-- Voltage signal generated by a microphone is analogue in nature...

A--At end of communication system, signal is converted from radio to sound..

A---If frequency of modulated wave is less than frequency of carrier wave, then input signal is negative...

C--Decrease in strength of signal is known as attenuation...

B- As compared to sound waves frequency of radio waves is  higher...

A-64 μs...The frequency of H pulses is 15625 Hz. Hence . T = 1/f = 1/15625 = 64 μs.

D--When the channel is noisy, producing a conditional probability of error p = 0.5; the channel capacity and entropy function would be, respectively,0&1...

B--- push pull amplifier is current amplifier...A push–pull amplifier is a type of electronic circuit that uses a pair of active devices that alternately supply current..

C-Isotropic antenna is one which radiates equally in all directions. It is a standard reference antenna.

D-0.05... T=1/f = 1/20 = 0.05..

B- Frequency modulation gives noise free reception and is used for radio broadcasts.

D- both (b) and (c). Both the case, i.e., ZL > Z0 and ZL < Z0 give rise to standing waves.

C-- 0.1 Pa (= 0.5 x 10-4 W/m2) is taken as standard sound pressure for defining S/N ratio of microphones.

B- 1,2,3. FM equipments are complex as compared to AM equipments.

A- 1.92 ms. (2 ms) (1 - 0.04) = 1.92 ms.

C---CC connection has very high input impedance and low output impedance.

C--When matching over a range of frequencies is desired, double stub is necessary.

A-Frequency shift keying (FSK) is a system of frequency modulation used in telegraphy.

C--4. Doubling of frequency increase SNR in frequency modulation by 4. It has no effect in AM.

B--1000 kHz. Frequency present in the sidebands is equal to = fc ± fm, fc ± 2fm , fc ± 3fm .

D--0.5. m1 = 0.4, m2 = 0.3.

D-- Some types of traps used in video IF amplifier of colour TV are shunt traps, series traps, absorption traps...

B--Rate of information depends on bandwidth.

1. Equilibrium can be attained from either side. 2. Equilibrium is dynamic in nature, i.e., at equilibrium reaction does not stop. 3. At equilibrium, there is no change in the concentration of various species. 4. The equilibrium state remains unaffected by the presence of catalyst. Catalyst helps to attain the equilibrium state rapidly. 6. The observable physical properties of the process become constant.

A--Photography is no longer an art form. Nowadays everyone has access to digital cameras that only need to be pointed at the subject in order to generate a perfect image. The writer of the argument apparently assumes that the selection of the subject is not an important artistic factor in photography...

D---1/6...Probability of getting an even number in one on the first dice is (when you get (2,4 or 6) that is 3 outcomes out of 6 possible outcomes That is 3/6. Probability of NOT getting an even number id 1-3/6 = 3/6 or 1/2.........................(1) The pribability of getting a multiple of 3 is when you get 1 or 2 or 4 or 5 on the second dice Or 4/6 = 2/3 So probability of NOT getting the above is 1-2/3 = 1/3.........................................................(2) Hence the required probability is 1/2 x 1/3 or 1/6..

A--360 sq. metre....l+b =42 and l = b+6 Solving b =18 and l = 24 diagonal of the rectangle = sq root of 18^2+24^2 = 30 m. Area of triangle = 1/2 (24)(30) = 360 sq m.

E--2.4 km. speed of train = length of train / time taken to cross pole = 480/30 = 16m/s In three minute it will cover = 16 * 3 * 60 = 2880 m Of this, train length = 480 m Therefore platform = 2880 - 480 = 2400m = 2.4km

C---2.1 years...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

E--823.08 ltr...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.

D-69. The total number of marks = 20 × 68 = 1360 Add difference of marks misread against the actual marks i.e. 20 = 1360 + 20 = 1380 Now the correct average = 1380/20 = 69

Delayed Marriages. ... Medical Facilities. ... Legislative Actions. ... Providing Incentives. ... Spread Awareness. ... Women Empowerment. ... Eradicate Poverty. ... Education.

(a) Birds have feathers which help them to fly. (b)Forelimbs are modified into wings for flight. (c)The avian flight muscles are used for flying. (d)The long bones are hollow and connected to air cavities to reduce weight. (e)The bones are light. (f)Feathers decrease water loss from body. (g)High metabolic rate to lift body from ground.

Gas bladder or air bladder is a gas filled sac present in fishes. It helps in maintaining buoyancy. Thus, it helps fishes to ascend or descend and stay in the water current.

If we consider specific characteristics instead of fundamental features then each organism will be placed in a separate group and the entire objective of classification would not be achieved. Classification of animals is also important in comparing different organisms and judging their individual evolutionary significance. If only a single characteristic is considered, then this objective would not be achieved.

The liver has two large sections, called the right and the left lobes. ... The liver also detoxifies chemicals and metabolizes drugs. As it does so, the liver secretes bile that ends up back in the intestines. The liver also makes proteins important for blood clotting and other functions.

A--Rs. 480. SP = CP + Profit and CP = SP + loss If the profit and loss is same with two different SP, then CP = (SP1 + SP2) / 2 = (524+436)/2 = 960/2 = 480

B-30 min... Part filled by (A + B) in 1 minute = 1 + 1 = 1 . 60 40 24 Suppose the tank is filled in x minutes. Then, x 1 + 1 = 1 2 24 40 x x 1 = 1 2 15 x = 30 min.

A-102102...Total amount after 4 years = 29282 + 26620 + 24200 + 22000 = Rs. 102102

D-32 years...Father's age was 28 28 years when my sister was born. My sister's age was 4 4 years when my brother was born. Therefore, father's age was 28 + 4 = 32 28432 years when my brother was born.

Before a fruit can be formed, the flowers must bloom so the male and female parts can develop and produce pollen and receptive ovules. Within the flower, pollen is produced in the stamens, while the female ovules form inside a pistil.

Let u=x2ux2. Then du=(x2)′dx=2xdxdux2′dx2xdx (steps can be seen here) and we have that xdx=du2xdxdu2. The integral can be rewritten as ∫xcos(x2)dx=∫12cos(u)du xcosx2dx12cosudu Apply constant multiple rule ∫cf(u)du=c∫f(u)ducfuducfudu with c=12c12 and f(u)=cos(u)fucosu: ∫12cos(u)du=(12∫cos(u)du) 12cosudu12cosudu The integral of the cosine is ∫cos(u)du=sin(u)cosudusinu: 12∫cos(u)du=12sin(u) 12cosudu12sinu Recall, that u=x2ux2: 12sin(u)=12sin(x2) 12sinu12sinx2 Therefore, ∫xcos(x2)dx=12sin(x2) xcosx2dx12sinx2 Add the constant of integration: ∫xcos(x2)dx=12sin(x2)+C xcosx2dx12sinx2C Answer: ∫xcos(x2)dx=12sin(x2)+Cxcosx2dx12sinx2C.

For a ball dropped from a building, speed in the downward direction after time t is given by υy1=0+gt=−10t For a ball thrown up with υ=40ms−1, Speed after time t is given by υy2=40−10t \ Relative speed of balls =υy2−υy1=40ms−1

Yes, the man will be able to land on the next building...

Given, Speed of first car = 18 km/h Speed of second car = 27 km/h Relative speed of each car w.r.t. each other = 18 + 27 = 45km/h

If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity. According to the problem, acceleration of object is given by the relation a=g-bv When speed becomes constant acceleration a = dv/dt = 0 (uniform motion). where, g = gravitational acceleration Clearly, from above equation as speed increases acceleration will decrease. At a certain speed say v0, acceleration will be zero and speed will remain constant. Hence, a = g- bv0 = 0 => v0 = g/b

A--- The direction of ball would change after 10 secs. Then ball and the train are in relative motion and their relative velocity is zero.

Ans. The height = u^2/2g = 28.2*28.2/(2*9.8) m = 40.57 m = 40.6 m (approx)

45kmph=12.5m/s total length discovered by train=850m+150m=1000m velocity=Dispacement/time t=s/v t=1000m / (12.5m/s) t=80s time taken to cross the bridge is 80s..

S=-40m, u=10m/s, a=-10m/s2 By formulae s=ut+1/2 at2 -40=10×t +1/2 ×-10 ×t2 -40=10t-5t2 5t2-10t+40 On simplifying We get t2-2t+8 On simplifying We get Time t=4s

A scientific attitude is a disposition to act in a certain way or a demonstration of feelings and/or thoughts. Studies of the actions of scientists have led to lists of scientific attitudes such as displayed below.

The scientific method has ancient precedents and Galileo exemplifies a mathematical understanding of nature which is the hallmark of modern natural scientists. Galileo proposed that objects falling regardless of their mass would fall at the same rate, as long as the medium they fall in is identical.

Either way (so-called "pure" or "applied" research), science aims to increase our understanding of how the natural world works. The knowledge that is built by science is always open to question and revision.

Physics is called exact science because it is based on measurement of fundamental quantities. The main aim of science is to find the truth behind the various processes taking place in the universe..

Any object that rotates has angular momentum while it's spinning. This can be anything from obvious things like a top spinning on a table to things we might not think about like a doorknob turning. Angular momentum gives us a measurement of an object's ability to keep spinning. The more angular momentum something has, the more it will want to keep rotating.

B- decreases....As a ball rises the vertical component of its velocity decreases...

Tapi river rises near Multai in the Betul district of Madhya Pradesh at an elevation of about 752 m and flows for about 724 km before outfalling into the Arabian Sea through the Gulf of Cambay.

Bond-dissociation energy (BDE or D0) is one measure of the strength of a chemical bond. It can be defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at 0 K (absolute zero).

The center of the nervous system is the brain. The brain takes in what your eyes see and ears hear, and if you decide that you want to move around, your brain tells your muscles to do it. Your brain makes your muscles move by sending tiny electrical signals to them through your nerves. Remember how neurons can be really long? Well, nerves are just a lot of those really long neurons all bunched together. Those really long neurons each send a small electrical shock to your muscles, which makes them move, moving your body. The nervous system is really complicated, but it can be divided into two really general parts. One is the Central Nervous System (or CNS). The CNS consists of your brain and spinal cord. The brain and spinal cord are inside your skull and vertebrae (the vertebrae make up your backbone). These bones protect the CNS when you get into accidents.

Legal services at Judiciary. Lok Adalat got initiated with a strong objective to take the justice system much closer to the people. It also aim at lessening the pending litigation . Its significance lies in actual justice delivery at door step of people and actual reduction in the numbers of real litigation.

Devasthan Rahasya Prema Roothi Rani Vardaan Seva Sadan Premashram Nirmala (novel) Kaayakalp Gaban Karmabhoomi Godaan

Dry Ice is the common name for solid carbon dioxide (CO2). It gets this name because it does not melt into a liquid when heated; instead, it changes directly into a gas (This process is known as sublimation).

Yes,,With the right information, and preparation, you can certainly clear the CLAT.

Robert Browning's “My Last Duchess”...

Boric acid is formed when brax is acidified. Na2B4O7 +2HCl +5H2O ---->2NaCl +4H3BO3