M.Phil | Rabindra Bharati University - [RBU], Kolkata | 2010 |
M.A | Rabindra Bharati University - [RBU], Kolkata | 2005 |
B.A | BB College, Asansol | 2003 |
12th | BB College, Asansol | 2000 |
10th | St. Patrick's HS School, Asansol | 1998 |
16 Years
Almost 16 years of teaching English, Maths and Science as a Tutor (face-to-face) and 8 years as a Professor of English (both Regular and Distance). I am also an author for English and Science books.
I Can Manage Both
My approach is simple. First, I explain the context and then provide relevant and necessary information. Also, I guide my students to write to-the-point and brief answers.
200.00
Class 11 - 12 | English, All Boards | INR 150.00 /hour |
College Level | English, BA Tuition | INR 200.00 /hour |
Test Preparation | NET, TET, SET, UGC Net | INR 300.00 /hour |
Competitive Exams | SSC Exams, SSB Exam, NDA, LIC | INR 250.00 /hour |
Given, AB is the diameter of the larger circle.
P and Q are the Centres of the smaller circles.
M is the point where the circles with centres P and Q touch each other.
So, AB = AM + MB ----------------------------- (i)
Thus, AM is the diameter of the circle with centre P
So, AM = AP + PM
= PM + PM (AP = PM, both being radii of the circle with centre P)
= 2PM ----------------------------- (ii)
And MB is the diameter of the circle with centre Q
So, MB = MQ + QB
= MQ + MQ (MQ = QB, both being radii of the circle with centre Q)
= 2MQ ----------------------------- (iii)
From (i), AB = AM + MB
= 2PM + 2MQ (from ii and iii)
AB = 2(PM + MQ)
or, PM + MQ = AB/2
or, MQ = AB/2 - PM
= (AB - 2PM)/2
Total weight of mixture = 25 kg Total weight of water = 20% of 25 kg = 5 kg. Now, 20% weight = 5 kg Let the extra amount of water be 'x' kg. Thus, (5+x)/(25+x)Ã—100 = 33.33 Or, 500 + 100x = 33.33x + 833.35 Or, (100 - 33.33)x = 833.35 - 500 Or, 66.67x = 333.35 Or x = 5 kg. So, the weight of water required to be added = 5 kg.
Let the total population be 'x'. Now, males:females = 3:2 Sum of ratios = 3+2 = 5 So, males = 3x/5 And females = 2x/5. No. Of male graduates = 20% of 3x/5 = (20/100)(3x/5) = 3x/25 No. Of female graduates = 25% of 2x/5 = (25/100)(2x/5) = x/10 Thus, no. Of graduates = 3x/25 + x/10 = 11x/50. Thus, no. Of non-graduates = x - 11Ã—/50 = 39x/50 Hence, percentage of non-graduates = (39x/50)/(x)Ã—100 = 39Ã—2 = 78% So, the correct answer is Option D.
The No. of marbles kept originally = 15
no. of white marbles = 5
Thus, fraction of white marbles = 5/15 = 1/3
Let the number of white marbles added be 'x'.
Thus, (5+x)/(15+x) = 3/5
or, 5(5 + x) = 3(15 + x)
or, 25 + 5x = 45 + 3x
or, 5x - 3x = 45 - 25
or, 2x = 20
or, x = 10.
So, the correct option is (D).
There are mistakes and missing informations in the question. Where is M? Are you sure that it is AM and not AD? Also, isn't <CAB = 90 degrees?
Surface Area of a Sphere = 4πr^2
Curved Surface Area of Cylinder = 2πrh
Thus, ratio of Surface Area of a Sphere : Curved Surface Area of Cylinder = 4πr2 : 2πrh
= 2r : h
= d : h
So, the correct option is (A).
Option B. OM = 2 cm and OQ = 7 cm
Given, Mode = 2
So, 3x - 10 = 2
or, 3x = 10 + 2
or 3x = 12
or, x = 4
And 4y + 6 = 2
or, 4y = 2 - 6
or, 4y = - 4
or, y = - 1
Thus, x + y = 4 - 1 = 3
So, the correct option is B.
Given, (a + b)/2 = 9
Thus, a = 18 - b
And, b = 18 - a
Also, c = 5a - 4
Thus, c = 5(18 - b) - 4
= 90 - 5b - 4
= 86 - 5b
So, (b + c)/2 = (b + 86 - 5b)/2
= (86 - 4b)/2
= 43 - 2b
= 43 - 2(18 - a)
= 43 - 36 + 2a
= 7 + 2a
So, Option C is correct.
Even numbers between 21 and 43 are:
22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42
Thus, Median = 6th number
= 32
So the correct option is B.
Re-arrenging the series:
1, 2, 4, 5, 7, 8, 10, 10, 12, 14, 15
Since, the series has 11 numbers, so the Median will be the 6th number = 8
And as 10 is the most repeated number so the Mode = 10
Hence, Option A is the correct answer.
Correct option is (B).
Working shown in the attached file.
The question is erroneous. Instead of <QAB = 90, it should be <PQB = <PQA = 90
The solution is provided in the attachment.
Please check the attachment for the solution.
Option C
Option D
Let the monthly salary be R 100.
Money given to mother = R 25
Money remaining = R 75
Money paid for rent = R 11.25
Money kept aside for monthly expenses = R 18.75
Thus, money kept in bank = R [100 – (25 + 11.25 + 18.75)]
= R [100 – 55]
= R 45
Sum of amount given to mother and amount kept in bank = R [25 + 45] = R 70
When assumed sum is R 70, then actual sum is R 42000
So, when assumed sum is R 100, then actual sum is R 60000
Speed of car in the 1st part of the journey = 39(45/60) km/hr = 52 km/hr
Speed of car in the 2nd part of the journey = 25 (35/60) km/hr = 300/7 km/hr = 42.85 km/hr
Average speed = (52 + 42.85)/2 km/hr = (94.85)/2 km/hr = 47.43 km/hr
So, the correct answer is E (None of the above).
Let the duration of the investment be ‘t’ and let the annual profits be divided proportionately by each.
Also, let the total profits be ‘x’.
So, A = R 64000; B = R 52000; C = R 36000
Hence, A:B:C = 64:52:36 = 16:12:9
Hence, the profit shares = 16x : 12x : 9x
If 16x = 35584
And 9x = (9 x 35584)/16 = R 20016
So, C’s share of the annual profits = R 20016
So, the correct option is E.
The solution is worked out in the attached document.
Hence, from the derivation in the attachment,
h = 4a
or h = 4(8) where a = 8
or h = 32 cm
and r = 8root2
Let the price of the bike be R 100.
Thus, increased price = 25% of R100 = R125
Now, decrease in % of price with respect to the new price = (25/125) x 100 = 20%
Let the number be ‘x’.
Therefore the new number = 2x/3
And the actual new number = 3x/2
Difference = (3x/2 – 2x/3) = (9x – 4x)/6 = 5x/6
Thus, percentage error in calculation = (5x/6)/(3x/2) x 100 = (10x/18x) x 100 = 1000/18 = 55.56%
Let the original fraction be x/y
Now, x + 25% of x = 5
Or, x + 25x/100 = 5
Or, 5x/4 = 5
Or, x = 4
And, y – 10% of y = 9
Or, y – 10y/100 = 9
Or, 9y/10 = 9
Or, y = 10
Thus, x/y = 4/10 = 2/5
Let the distance to be covered be x km
Ratio of time taken:
A:B:C = 12:15:10
Sum of ratios = 12 + 15 + 10 = 37
Speed of A = 60 km/hr
Thus, 12x/37 = 60
Or, x/37 = 5
Or, x = 5(37) = 185
Hence, Speed of B = 15(185)/37 = 75 km/hr
Let the charges per hour be R 100
Thus, increase in charges = 25% of 100
So, new charges = R 125
Let the time taken to play a game be ‘T’ hr
Thus, the decreased time = t hr
Now, actual expenditure = R 110
So, 125t = 100T + 10T
Or, 125t = 110T
Or, t = 22T/25
Now, decrease in time = T – t = T – 22T/25 = 3T/25
And % decrease = [(3T/25)/T]100 = 12%
[(4356)/2]/(11/4) = 792
792/6 = 132
So, 132 x 132 = √17424
so, ? = 17424
centre = (5, 7)
x2 + y2 + 4x + 8y – 45 = 0
x2 + 4x + y2 + 8y – 45 = 0
or, x2 + 4x + 4 + y2 + 8y + 16 = 25
or, (x + 2)2 + (y + 4)2 = 52
so, centre = (– 2, – 4)
Let the Time taken be T hr.
So, for T/2, Distance = 80 x T/2
And, for T/2, Distance = 40 x T/2
total distance = 60 = 80T/2 + 40T/2
= 60 = 120 T/2
= 60 = 60T
= 1 hr = T
So, Avg. Speed = 60/1 = 60 km/hr
A
I
I
C ------------------------------------I B
The question is wrong as C cannot be to the West of A. There is some missing data in this case.
The Relative Velocity of the ground with respect to Train B is 90 km/hr.
C/5 = (F-32)/9 where C = Celsius and F = Farenheiht
1. Archemedes Number
2. Chandrasekhar Number
3. Decibel
R = [–(–10) ± √(100 – 1160)]/2x29 = [100 ± √(– 1060)]/58 = [50 ± √(– 265)]/29 = [50 ± 16.28i]/29
A = - 14
T5 = 2
Now, t5 = a + (n-1)d
2 = – 14 + (5 – 1)d
16 = 4d
D = 4
Sn = n/2[2a – (n – 1)d]
40 = n/2[– 28 – (n – 1)4]
80 = n[- 28 – 4n + 4]
80 = n[- 24 – 4n]
20 = n[6 – n]
n2 – 6n + 20 = 0
n = [–(–6) ± √(36 – 80)]/2 = [6 ± √(– 44)]/2 = [6 ± 2√(– 11)]/2 = 3 ± 1(3.32) = 6.32 or 0.32
A = - 14
T5 = 2
Now, t5 = a + (n-1)d
2 = – 14 + (5 – 1)d
16 = 4d
D = 4
Sn = n/2[2a + (n – 1)d]
40 = n/2[– 28 + (n – 1)4]
80 = n[- 28 + 4n - 4]
80 = n[- 32 + 4n]
20 = n[-8 + n]
n2 – 8n - 20 = 0
n2 - 10 n + 2n - 20 = 0
n(n - 10) + 2(n - 10) = 0
(n - 10)(n + 2) = 0
n = 10 or - 2
As n cannot be negetive, na = 10 isthe correct answer.
Let the age of the son be ‘x’ years.
Thus, Mr. Mohan’s age = 4x years
After 10 years,
Son’s age = x + 10 years
Mr. Mohan’s age = 4x + 10
Now, 2(x + 10) = 4x + 10
Or 2x + 20 = 4x + 10
Or 2x = 10
Or x = 5 years
X-ray
Newton's third law of motion states that 'for every action there is an equal and opposite reaction'. When a gun is being shot, there are forces acting within it to make the bullet move forward and out. These forces are in the form of the kinetic energy of the bullet and the potential energy that is tored in it. When the bullet moves out with the help of the combination of force, acceleration and energy, there is also an equal and opposite combination of foce, energy and acceleration working on it, known as the 'recoil'. This recoil leads us to fall back. This is the same principle that provides 'thrust' to rockets before they can lift off to space.
The Archemedes' principle is used in designing ships, submarines, boats and hydrometers.
Archemedes himself used it to find out the purity of gold.
Fishes use the Archemedes priciple to rise up to the surface of waterbodies or to sink doen to their bottom.
Power (D) = 1/Focal Length (F)
or, 0.2 = 1/F
or, F = 1/0.2
= 10/2
= 5 m.
Hence, the options given are wrong.
1 Newton is the measurement of the force exerted when a mass of 1 kg moves with an acceleration of 1 m per second sq.
Hence, 1 N = 1 kg x 1 m/s2
= 1 kgm/s2
The vector product of two vectors a and b is given by a vector whose magnitude is given by |a||b|sinθ (where0âˆ˜≤θ≤180âˆ˜), which stands for the angle between the two vectors. Note that the direction of the resultant vector is denoted by a unit vector ^n whose direction is perpendicular to both the vectors a and b in a way that a, b and ^n are oriented in right-handed system.
Right-handed orientation happens when vector a is twisted in the direction of vector b, then the direction of the unit vector ^n goes in the direction in which a right-handed screw would spin if moved in a similar manner. Also, these given vectors a and b cannot be called null vectors or non-parallel in nature. The right-hand thumb rule gives a clear picture of the direction of the resultant vector.
Therefore, we can conclude that, a×b = |a||b|sinθ ^n, where a×b stands for the cross product of two vectors. In any given situation, if the vectors are null or both the vectors are parallel to each other, then the cross product cannot be defined.
In this case, we can conclude that a×b = 0
Physical Representation
Let’s assume that a and b are the adjacent sides of the parallelogram OACB and the angle between the vectors a and b is θ. Then, we can say that the area of the parallelogram is denoted by |a×b| = |a||b|sinθ.
Properties of Cross Product of Two Vectors
i) The vector product never has a Commutative Property. It is denoted by,
a×b = – (b×a)
ii) The property given below is true in the case of vector multiplication:
(ka)×b= k(a×b) =a×(kb)
iii) If the vectors mentioned are collinear then
a×b= 0
(Since the angle between both the vectors would be 0, then sin 0 = 0)
iv) As per the above property
We can conclude that the vector multiplication of a vector with itself would be a×a= |a||a|sin0 ^n = 0 Also, when it comes to unit vector notation
^j×^j=^j×^j=^k×^k =0
As per the above discussion
^i×^j=^k=−^j×^i
^j×^k=^i=−^k×^j
^k×^i=^j=−^i×^k
This example can be explained better with the help of the following diagram. When moving in clockwise direction and considering the cross product of any two pair of the unit vectors, we can derive the third one and we get the negativeresultant in anticlockwise direction.
v) a × b in terms of unit vectors can be represented as
a =a1^i+a2^j+a3^k
b =b1^i+b2^j+b3^k
Then →a×→b =(a1^i+a2^j+a3^k)(b1^i+b2^j+b3^k)
When expanded, we would get
|a||b|sinθ ^n = (a2b3–a3b2)^i+(a3b1–a1b3)^j+(a1b2–a2b1)^k
vi) Distributive Law: a×(b+c) = a×b+a×c
Moment of inertia of a rod of length r = 1/3 (mr2)
Given, mass of rod = m, length = l and < of rotation = theta
Let, the radius of the circle of raotation be r.
Thus, r = l cos theta
and I = 1/3 (l cos theta)2
A ray of light bends when it travels from a optically denser medium to an optically lighter one or vice versa. This phenomenon is known as Refraction of Light.
This change of direction of light is because of the change in the speed of light in different media. Also, the angle at which the ray of light moves from one medium to another, influences the angle of refraction of light. The greater the angle, the greater is the refraction.
Let the age of the son be x years.
Thus, Mohan's age is 4x years.
After 10 years,
Son's age is (x + 10) years
ans Mohan's age is (4x + 10) years
Now, 4x + 10 = 2(x + 10)
Or, 4x + 10 = 2x + 20
Or, 2x = 10
Or, x = 5
So, present age of Mohan's son is 5 years.
Your question is incomprehensible. Please rewrite the question for us to answer it properly.
Given a = - 15; d = 9; n = 30
Thus, S30 = 30/2 [2a + (30 – 1)d]
= 15 [2(–15) + 29(9)]
= 15 [–30 + 261]
= 15[231]
= 3465
Let the CP = â‚¹ 100
Thus, MP = â‚¹ 140
Discount = 10%
So, SP = â‚¹ 140 - â‚¹ 14 = â‚¹ 126
So, Profit = â‚¹ 26
When Profit = â‚¹ 26, CP Profit = â‚¹ 100
When Profit = â‚¹ 325, CP Profit = â‚¹ (325 x 100)/26 = â‚¹ 1250
Let the CP = â‚¹ 100
Thus, MP = â‚¹ 140
Discount = 10%
So, SP = â‚¹ 140 - â‚¹ 14 = â‚¹ 126
So, Profit = â‚¹ 26
When Profit = â‚¹ 26, CP = â‚¹ 100
When Profit = â‚¹ 325, CP = â‚¹ (325 x 100)/26 = â‚¹ 1250
Physical quantities having units but no dimensions are: plane angle, angular displacement and solid angle.
Let mass of the body be ‘m’ and velocity be ‘v’.
So, KE = ½ (mv2)
Now, increase in mass = 10% of m
Thus, new mass = m + m/10 = 11m/10
Increase in velocity = 20% of v
Thus, new velocity = v + v/5 = 6v/5
Hence, new KE = ½[(11m/10)(6v/5)2]
= 198mv2/250
So, increase in KE = 198mv2/250 – mv2/2 = 73mv2/250
Hence, % increase in KE = [(73mv2/250)/(mv2/2)]100 = 58.4%
2. (i) Speed may be described as the distance traversed by a body with rerspect to time.
(ii) Instanteneous velocity may be described as the velocity of a body at a certain point of time.
(iii) Velocity may be defined as the displacement of a body or a the ditance covered by it in a particular direction with respect to time.
(iv) Acceleratio is the rate of change of velocity with rspect to time.
Given focal length = 10 mm = 10/1000 m = 1/100 m = 0.01 m
Thus, P = 1/f = 1/0.01 D = 100D
Pressure = Force/Area = mg/cm2
Density = mass/volume = m/cm3
Thus, m x mg/cm2 x cm3/m = mg cm = mgh = Potential Energy.
Thus, it is the formula for finding Potential Energy.
Since the bodies are of equal dimension and masses, then the centre of mass will be at the centre of the sphere B where the spheres are arranged as A, B and C respectively. Hence, distance of the Centre of Mass from the centre of A is equal to its diametre as its radius is the same as the radius of sphere B.
Given, v = v0; u = 0; height = S and acceleration = g
So, v02 = u2 + 2gS
As u = 0,
v02 = 2gS
S = v02/2g
While, tripling the height, the acceleration due to gravity remains the same.
So, 3S = 3 v02/2g
Or, H = 3S
= (√3 v0) 2/2g
Hence, the velocity in case the height is tripled is √3 v0
Relative density of a substance means the density of an object at a particulart temperature with respect to the density of water at the same temperature.
A pentavalent dopant such as Antimony are known as donor impurities since they donate an extra electron in the crystal structure which is not required for covalent bonding purposes and is readily available to be shifted to the conduction band. This electron does not give rise to a corresponding hole in the valence band because it is already excess, therefore upon doping with such a material, the base material such as Germanium contains more electrons than holes, hence the nomenclature N-type intrinsic semiconductors.
On the other hand when a trivalent dopant such as Boron is added to Germanium additional or extra holes get formed due to the exactly reverse process of what was described in the upper section. Hence this dopant which is also known as acceptor creates a P-type semiconductor.
The portion of the energy band structure of a semiconductor or insulator between the valence andconduction bands. This portion does not contain a continuum of states as do the valence and conduction bands but may contain discrete states introduced by dopant or deep level impurities in the material.
The speed of a film (or film speed), refers to the measure of a film's sensitivity to light. Each film speed is best suited for a different type of photography. The lower the speed, the longer an exposure to light is necessary to produce image density. If the film speed is higher, it requires less exposure but generally has reduced quality in the form of grain and noise. Noise and grain are the abnormalities in brightness and colour in images; they look similar to a layer of "snow" on a television set. They're measured using the ISO system from the International Organization for Standardization, and are the giant numbers typically seen on a box of film. The abbreviation ASA (American Standard Association) is also used in conjunction with film speed. ASA and ISO are interchangeable.
Total time taken = 1 hr and 30 minute = 90 minutes
Total stoppage time = 12 x 2 = 24 minutes
Thus, total travelling time = 90 – 24 = 66 minutes = 1 1/10 hr = 11/10 hr
Speed = 40 km/hr
Thus, distance = Speed x Time = 40 x 11/10 = 44 km
Let the Principal be P and Rate be R
Thus, A = 2P
Hence, SI = A – P = 2P – P = P
T = 10 years
Thus, R = (100 x P)/(P x 10) = 10%
CP = â‚¹ 12500; Transport = â‚¹ 300; Installation = â‚¹ 800
Thus, total CP = â‚¹ 13600
P% = 15% = (P/CP)100
Hence, P = (15 x 13600)/100 = â‚¹ 2040
Thus, SP = CP + P = â‚¹ 13600 + â‚¹ 2040 = â‚¹ 15640
Investment of A:B:C = â‚¹ 12000 : â‚¹ 15000 : â‚¹ 18000 = 4 : 5 : 6
Sum of ratios = 4 + 5 + 6 = 15
A’s share = 4x/15
B’s share = 5x/15
C’s share = 6x/15
A’s share of profit = â‚¹ 3000
Thus, 4x/15 = â‚¹ 3000
Or, x = â‚¹ (750 x 15) = â‚¹ 11250 = Total Profit
Let the CP be â‚¹ 100.
Therefore, the MP is 40% of CP = â‚¹ 140.
Discount is 20%of MP = â‚¹ (20 x 140)/100 = â‚¹ 28.
Thus, actual SP = â‚¹ (140 – 28) = â‚¹ 112.
Thus, Profit = â‚¹ (112 – 100) = â‚¹ 12.
Profit % = (P/CP) x 100 = (12/100) x 100 = 12%
Given A = {2, 4, 5} and B = {7, 8, 9}
Thus, A x B = {(2, 7), (2, 8), (2, 9), (4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9)}
Hence, n(A x B) = 9
So, the correct option is B.