Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Two solid spherical planets have equal radii and mass of 4M and 9M and distance between their centre is 6R. A body of mass m is projected toward heavy planet, then what should be the ditsance of body of mass m from lighter planet so that the gravitational force on it will be zero ?

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ANSWER : R = 2.4 R from PLANET of MASS 4M
2.

Two plates identical in size, one of black and rough surface (B_1) and the other smooth and polished (A_2) are in terconnected by a thin horizontal pipe with a mercury pellet at the centre. Two more plates A_1 (identical to A_2) and B_2 (identical B_1) are heated to the same temperature and placed closed to the plates B_1, and A_2 shown in Fig. The mercury pellet

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moves to the right
moves to the left
remains stationary
starts oscillating left and right

Solution :Smooth and polished PLATES are poor radiators of heat. Hence, heat coming out from A is small, even though B being a black and rough plate is a good absorber. Effectively the heat coming to the left of pellet P is small.
Black and rough plates are good radiators of heat. Hence plate `B_2` radiates heat to a satisfactory level , HOWEVER plate `A_2` being smooth and POLISHER, is a BAD absorber.Effectively the heat coming to the right of P is ALSO small.
3.

Figure shows two prpcesses A and B on a system. Let DeltaQ_(1) and DeltaQ_(2) be the heat given to the system in processes A and B respectively. Then

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`DeltaQ_(1) gtDeltaQ_(2)`
`DeltaQ_(1) =DeltaQ_(2)`
`DeltaQ_(1) ltDeltaQ_(2)`
`DeltaQ_(1) leDeltaQ_(2)`

ANSWER :A
4.

A student measured the length of a wire and wrote it as 2.50 can which instrument did he/she use to measure it?

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a screw gauge having 50 division in the circular scale and pitch as 1 mm.
a meter scale
a VERNIER caliper where the 10 division in vernier matches with a division in main scale.
a screw gauge having 100 division in the circular scale and pitch as 1 mm.

Solution :To get accuracy.
5.

A black body has maximum wavelength lamda_(m) at 2000 K. Its corresponding wavelength at 3000 K will be . . . . . ..

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`2/3lamda_(m)`
`3/2lamda_(m)`
`4/9lamda_(m)`
`9/4lamda_(m)`

Solution :According to Wien.s law,
`lamda_(m)T=b" (CONSTANT)"`
`:.(lamda_(m))_(1)T_(1)=(lamda_(m))_(2)T_(2)`
`:.(lamda_(m))_(2)=((lamda_(m))_(1)T_(1))/(T_(2))`
`=(lamda_(m)xx2000)/(3000)`
`=2/3lamda_(m)`
6.

what is hydrualicstress ?

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Solution :When the deforming FORCE acts everywhere on the SURFACE of the body normally so that CHANGE in volume of the body occures when a body immersed in a liquid . Hene it is called hydraulic stress .
7.

A body weights 500N on the surface of the earth. How much would it weight half way below the surface of the earth?

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`100N`
`500N`
`250N`
`125N`

ANSWER :C
8.

A particle executes simple harmonic with an amplitude of 5 cm. When the particle is at 4 cm from the meanposition the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

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`(8pi)/(3)`
`(4pi)/(3)`
`(3)/(8pi)`
`(7)/(3pi)`

SOLUTION :Velocity of particle PERFORMING SHM,
`V= pm omega sqrt(A^(2)-y^(2))`
`therefore v= pm omega sqrt(25-16)`
`therefore v= pm 3 omega"""…….."(1)`
and acceleration `a= pm omega^(2)y`
`therefore a= pm 4 omega^(2)"""…….."(2)`
From equation (1) and (2),
`a= v`
`therefore pm 4 omega^(2) = pm 3 omega`
`therefore omega = (3)/(4)`
`therefore (2pi)/(T) = (3)/(4)`
`therefore T= (8pi)/(3)s`.
9.

Four lenses are made of glass and the radius of curvature of each face is given below. Which will have the greatest focal power

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10CM CONVEX and 15CM convex
5cm convex and 10cm CONCAVE
15cm concave and plane
20cm convex 30CM concave

Answer :A
10.

The angle between twovectors 2 hat(i) + 3 hat(j) + hat(k) and - 3 hat(i) + 6 hat(k)is :

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`0^(@)`
`45^(@)`
`60^(@)`
`90^(@)`

Solution :`A. B = (2 hat(i) + 3 hat(j) + hat(k)) * (- 3 hat(i) + 6 hat(k))`
= - 6 + 6 = 0
Since A.B = 0 the angle between two vectors is `90^(@)`
11.

A steel wire is suspended vertically from a rigid supports when loaded in with a weight in air, it extends by L_(a) and when the weight is immersed completely in water, the extension is reduced to L_(w) Find the relative density of the material of the weight.

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Solution :`Y=(FL)/(Ac), Y=(MgL)/(AL_(a))....(1) Y=(MG(1-(d_(L))/(d_(s))L))/(AL_(W)) ....(2)`
`(MgL)/(AL_(a))=(MgL[1-(d_(L))/(d_(S))])/(AL_(w)), (1)/(L_(a))=(1-(d_(L))/(d_(S)))/(L_(w)) rArr (L_(w))/(L_(a))=1-(d_(L))/(d_(S)), (d_(L))/(d_(S))=1- (L_(w))/(L_(s))=(L_(a)-L_(w))/(L_(a)) rArr (d_(s))/(d_(L))=(L_(a))/(L_(a)-L_(w))`
12.

{:("Column-I","Column-II"),("A) Thermal expansion","P) Pendulum clock"),("B)" alpha","beta","gamma ,"Q) Depends, on dimensions Material, Temperature"),("C) Bimetallic strip","R) Depends on nature of the material only"),("D) Invar stell","S) Balance wheel of a watch"):}

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ANSWER :A-Q; B-R; C-S; D-P
13.

Condition that body can be parked in circular road with slope is …... .

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ANSWER :`tanthetalt=mu_S`
14.

A body having temperature 27^(0)C is kept in a room having temperature 27^(0)C. Does the body emit any radiation in this case when room temperature is same as the body temperature ?

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Yes
No
MAY emit, depends on the MATERIAL of body
Can.t say anything

ANSWER :A
15.

If the mass of the sun were ten times smaller and gravitational constant G were ten times larger in magnitude. Then,

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g' on the Earth will not CHANGE .
Reindrops willfall faster
Time period of a SIMPLE pendulum on the Earth would decrease
Walking on the ground would become more difficult.

Solution :`implies g_E = (GM_E)/(R_E^2)`
There will not be any any change in `g_E` due to decrease in mass of sun `g_E` doesn’t DEPEND on mas of sun .
And `g_E` depends on G.
`:. G_E prop G`
`:. (g.E)/(gE) = (G.)/G
`:.g._E = g_E xx (10G)/(G)`
`:. ` Gravitational ACCELERATION on earth increses by 10 times .
16.

Two identical stars of mass M orbit around their centre of mass. Each orbit is circular and has radius R, so that the two stars are always on opposite sides of the circle.The minimum energy required to separate the two stars to infinity is

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`(GM^(2))/(R)`
`(GM^(2))/(4R)`
`(GM^(2))/(2R)`
`2(GM^(2))/(R )`

Answer :C
17.

Two identical stars of mass M orbit around their centre of mass. Each orbit is circular and has radius R, so that the two stars are always on opposite sides of the circle. The orbital speed of the each star in the orbit is

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`SQRT((2GM)/(R ))`
`sqrt((GM)/(2R))`
`sqrt((GM)/(4R))`
`sqrt((GM)/(R ))`

ANSWER :C
18.

There is a soap bubble of radius 2.4 xx 10 ^(-4) min air in a cylinder which is originally at the pressure of 10^5 N//m^2. The air in the cylinder is now compressed isothermally until the radius of the bubble is halved. Calculate (approximately in multiple of 10^5N/m^2) the new pressure of air in the cylinder. The surface tension of the soap film is 0.08Nm^(-1).

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ANSWER :5
19.

The temperature of a black body increases from 7°Cto 287°C .The rate of energy radiation increases by

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`(287/4)^4`
16
4
2

Answer :C
20.

The acceleration 'a' in m//s^2of a particle is given by a=3t^2+2t+2where t is the time. If the particle starts out with a velocity u=2 m/s at t=0, then the velocity at the end of 2 second is

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12 m/s
18 m/s
27 m/s
36 m/s

Answer :B
21.

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Mass of oxygen molecules (m)=2.76xx10^(-26) kg Boltzmann's constant (k_(B))=1.38xx10^(-23)JK^(-1))

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Solution :`nu_("ESCAPE")=11200ms^(-1)`
Say at temperature T it attains `nu_("escape")`
`nu_("escape")=sqrt((3k_(B)T)/m_(O_(2)))`
`T=((nu_("escape"))^(2)xxm_(0_(2)))/(3k_(B))=((11200)^(2)xx2.76xx10^(-26))/(3xx1.38xx10^(-23))`
`(3.46xx10^(8)xx10^(-26))/(4.14xx10^(-23))=0.8357xx10^(5)`
`T=8.357xx10^(4)K`
22.

A circular disc of mass 100 g and radius 10 cm' is making 2 rps about an axis passing through its centre and perpendicular to its plane. Calculate its kinetic energy.

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SOLUTION :`mgh = (1//2)IOMEGA^(2) = MGR =(1//2) xx (3//2) MR^(2) xx (V^(2)//r^(2)) , v=2sqrt(rg//3)`, (ii) speed =2v = `4sqrt(rg//3)`
23.

For production of beats the two sources must have

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DIFFERENT FREQUENCIES and same AMPLITUDE
different frequencies
different frequencies, same amplitude and same phase
different frequencies and same phase

Answer :B
24.

The bridge of a sonometer is slightly displaced so that the length of wire is decreased by 0.5% and tension in the wire is increased by 1%. The fundamental frequency of wire

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INCREASES by `1%`
decreases by `1%`
increases by `1.5%`
decreases by `1.5%`

SOLUTION :`n prop (sqrt(T))/(l)implies(DELTAN)/(n)=(1)/(2)(DELTAT)/(T)-(Deltal)/(l)`
25.

The small particle is released from P. As it reaches Q its KE becomes K. The force exerted by it on hemisphere at Q is

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`(K)/(R )`
`(2K)/(R )`
`(3K)/(R )`
`(3K)/(2R )`

ANSWER :C
26.

A spherical planet far out in space has a mass M_(0) and diameter D_(0). A particle of mass 'm' falling freely near the surface of this planet will experience anacceleration due to gravity which is equal to

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`(GM_(0))/(D_(0)^(2))`
`4M(GM_(0))/(D_(0)^(2))`
`4(GM_(0))/(D_(0)^(2))`
`(GmM_(0))/(D_(0)^(2))`

Answer :C
27.

Centripetal acceleration is given by

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`(V^(2))/(R)`
`-(v^(2))/(r)`
`(r)/(v^(2))`
`-(r)/(v^(2))`

ANSWER :B
28.

The adjoining graph shows the distribution of kinetic energies E_(k) among the consultant molecules of a gas at a uniform temperature. N is the number of molecules each having energy in a small energy band around E-(k). Which of the following statements is true?

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Provided that the temperature does not change, the kinetic energy of each molecules is fixed.
The commonest value of kinetic energy is also the greatest kinetic energy of any of the molecules.
The total kinetic of the molecules is INDEPENDENT of the temperature of the gas.
The value `x` of `E_(k)` at which the peak of the curve occurs increasewhen the temperature rises.

Solution :For (a), remeber that molecules are continually in COLLISION with ONE another. For (b) look at the given graph carefully. For (d), remember that the root mean square speed increases with temperature INCREASE.
29.

The SI unit of velocity is

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`m//s`
`m sec^( 2)`
`mhr^(-2)`
`m//hr`

Solution :The SI UNIT of velocity is `m s^(-1)` or `m//s`.
30.

A glass plate of dimensions 30xx10cm is placed on a table on which there is a layer of machine oil of thicknes 0.01 mm. The co efficient of viscosity of machine oil is 0.5Ns//m^(2). The force requird to draw the plate horizontally with a velocity of 5ms^(-1)

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3000N
3750N
7500N
15000N

Answer :C
31.

A block of mass 3 kg which is on a smooth inclined plane making an angle of 30^(@) to the horizontal is connected by a cord passing over a light frictionless pulley to a second block of mass 2 kg hanging vertically. What is the acceleration of each block and what is the tension of the cord ?

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`0.98 m//s^(2), 17.6 N`
`1.98 m//s^(2), 19.6 N`
`0.49 m//s^(2), 9.8N`
`1.47 m//s^(2), 4.9 N`

ANSWER :A
32.

A locomotive accelerates a train of identical railway carts. The carts are numbered consecutively with the cart next to locomotive having the number 1. The tension in the connection between the carts with numbers 4 and 5 is three times begger than the tension in the connection between the carts with numbers 14 and 15. What is the number of the last cart? there is no resistance to the train's motion.

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SOLUTION :ACCELERATION of each CART will be same

Tension in CONNECTION between 4 and 5 cart.
`T_(4,5)=m(n-4)a`…(i)
Tension in connection between 14 and 15
`T_(14,15)=m(n-14)a` …(ii)
Given `T_(4,5)=3T_(14.15)`
`M(n-4)a=3m(n-14)aimplies n=19`.
33.

Two identical balls of mass M and mradius R each are placed on a frictionless plane. One of the balls is at rest while the second moves at a velocity v towards the first ball tangentially. The two balls stick together on collision. Then

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the speed of the centre of mass does not CHANGE because of the collision
the ANGULAR velocity of the system after the collision is `(5v)/(14R)`
the KINETIC ENERGY of the system after the collision is `3/7"Mv"^(2)`
someenergy is lost during the collision.

Answer :A::B::C::D
34.

A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10ms^(-1). A plumb bob is suspended from the roof of the car by a string of length 1m. The angle made by the string with vertical is (g = 10ms^(-2))

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`0^(@)`
`30^(@)`
`45^(@)`
`60^(@)`

SOLUTION :`TAN THETA=(v^(2))/(rg)`
35.

The bob A of a pendulum released from 30^(@) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.

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SOLUTION :It TRANSFERS its ENTIRE momentum to the ball on the table, and does not RISE at all.
36.

Obtain scalar product in terms of Cartesian component of vectors .

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Solution :`vec(A) and vec(B)` is written in Cartesian component as FOLLOW :
`vec(A) = A_(x) hat(i) +A_(y) hat(J) +A_(Z)hat(k)`
`vec(A) = A_(x) hat(i) +A_(y)hat(j) +A_(z)hat(k)`
`vec(B) = B_(x) hat(i) = B_(y) hat(j) +B_(z)hat(k)`
`:. vec(A) .vec(B) = (A_(x)hat(i) +A_(y)hat(j)+A_(z)hat(k)).(B_(x)hat(i) +B_(y)hat(j) +B_(z)hat(k))`
`A_(x)B_(x)(hat(i).hat(i))+A_(x) B_(y)(hat(i).hat(j))+A_(z)B_(z)(hat(i).hat(k))`
` + A_(y)B_(x) (hat(j).hat(i)) +A_(y)B_(z) (hat(j).hat(k))`
`+A_(z)B_(x)(hat(k).hat(i))+A_(z)B_(y)(hat(k).hat(j))+A_(z)B_(z)(hat(k).hat(k))`
In this equation`hat(i) . hat(i) = hat(j) .hat(j) = hat(k) .hat(k) = 1 ` and `hat(i).hat(j)= hat(j).hat(i) = 0 , hat(j) . hat(k) = hat(k) . hat(j) = 0 ` and `hat(k).hat(i) = hat(i) .hat(k) = 0 `
So , `vec(A) . vec(B) =A_(x)B_(x) +A_(y)B_(y) +A_(z)B_(z)`
37.

A long spring when stretched by 'x'cm has a potential energy 'V'. On increasing the stretching to 'nx' cm the potential energy stored in the spring will be

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`(V)/(n)`
nv
`n^(2)v`
`v//n^(2)`

Answer :C
38.

A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is 30.6 cal//kg^@C , the increase in temperature of the slab immediately after collision

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`6.72^@C`
`7.62^@C`
`5.62^@C`
`8.72^@C`

ANSWER :B
39.

(A) : Space rockets are usually launched in the equatorial line from west to east. (R) : The acceleration due to gravity is minimum at the equator.

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Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A.
Both .A. and .R. are true and .R. is not the correct explanation of .A.
.A. is true and .R. is FALSE
.A. is false and .R. is true

Answer :A
40.

A solid sphere of radius R floats in water to a depth of (R/2). Find the relative density of the material of the solid.

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Solution :Volume of submerged portion of the sphere to a DEPTH `(R/2)=5/24 piR^(3)`
We know that `(4/3piR^(3))RHO=((5)/(24) piR^(3))rho_(W)""i.e. rho/rho_(w)=(5 xx R^(3))/(24 piR^(3)) xx 3/4`
`rho/rho_(w)=(5)/(32)`
41.

The lower end of a capillary of radius r = 0.2 mm and length l = 8 cm is immersed in water whose temperature is constant and equal to T_(low) = 0^(@)C. Datermine the height h to which the water in the capillary rises, assuming that the thermal conductivity of the capillary is much higher than the termal conductivity of water in it. The heat exchange with the ambient should be neglected. Use the following temperature dependence of the surface tension of water : {:(T,.^(@)C,0,20,50,90),(sigma,mN//m,76,73,67,60):}

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SOLUTION :N//A
42.

Explain blood flow and heart attack with the help of Bernoulli's principle.

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Solution :Bernoulli.s principle helps in explaining BLOOD flow in artery.
The artery MAY get constricted DUE to the accumulation of plaque on its inner walls.
Thespeed of the flow of theblood in this CONSTRICTION region is raised which lowers the pressure inside and the artery may collapse due to the external pressure.
the heart exerts further pressure to open this artery and forces the blood through.
As the blood rushes through the opening ,the internal pressure once again drops.
Due to same reasons leading to a repaet collapse this may result in heart attack.
43.

Showthat are rateof flow of heat through a spherical shell , the inner and outerwalls of whichhave radiir_(1)and r_(2) and are maintained at differnetunifrom temperature theta_(1)andtheta_(2)respectievly , is givenbyH = 4Kr_(1)r_(2)(theta_(1) - theta_(2))/(r_(2) - r_(1)) where K is thermal conductivityof material of shell. He obtain the expression fortemperature distributions.

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Solution :LetO be the centreof sphericalshell , havinginnerand outerradii ` r_(1)`and `r_(2)`respectively , i.e.,`OA = r_(1), OB = r_(2) ` . Let`theta_(1)`and `theta_(2)`be the steady temperature of inner and outerall `(theta_(1) GT theta_(2))` respectively . Theshell may be supposed to be formedof largenumberof thinconcentricspherical shells . Consider one such shell of radius r andthickness dr .
Let `theta ` and `theta + d theta`therespectivelytemperature at distancer and r+dr from centreO.
Thetemperature gradient ` = (d theta)/(dr) therefore ` Rate of flowof heatacross the surfaceof the surfaceof the shell .
`H =- KA (delta theta)/(dr) (or) H =- K (4pir^(2) .(d theta)/(dr))..............(1)`
Kbeingthermalconductivityof the material of shelll.
Due tosymmeticnatureofshell H issamein all directionsis steady state.
Therefore from (1)
`d theta =- (H)/(4 pi K) (dr)/(r^(2))...........(2)int_(theta_(1))^(theta_(2)) d theta = (H)/(4 pi K) UNDERSET(r_(i))OVERSET(r_(2))int (dr)/(r^(2)) or [theta]_(theta_(1))^(theta_(2)) = - (H)/(4 pi K) [-(1)/(r)]_(r_(1))^(r_(2))`
(or) `theta_(2) - theta_(1) = (H)/(4 pi K) [ (1)/(r_(2)) - (1)/(r_(1))] (or) theta_(1) - theta_(2) = (H)/(4 pi K) [ (1)/(r_(1)) - (1)/(r_(2))]`
Thegives rateofflowof heat`H = (4 pi K(theta_(1) - theta_(2)))/(((1)/(r_(1)) - (1)/(r_(2)))) ...............(3)`
Temperaturedistribution :
Integratingequaiton(2) betweelimits ` theta_(1)`and `theta` we get`underset(theta_(1))overset(theta_(2))intd theta = - (H)/(4pi k)underset(ri)overset(r)int (dr)/(r^(2))`
(or) `theta - theta_(1) = - (H)/(4pi K) [-(1)/(r)]_(ri)^(r)............(4)`substitutingvalueof H from (3) in (4) .
we get `theta = theta_(1)+ (theta_(1) - theta_(2))(((1)/(r) - (1)/(r_(1))))/(((1)/(r_(1)) - (1)/(r_(2))))..............(5)`
THISIS therequiredexpression .
44.

A solid cylinder when dropped from a height of 2 m acquires a velocity while reaching the ground. If the same cylinder is rolled down from the top of an inclined plane to reach the ground with same velocity, what must be the height of the inclined plane? Also compute the velocity

Answer»

Solution : In the first case, ,
potential energy - kinetic energy
`mgh=(1)/(2) mv^(2)`
`mgxx2=(1)/(2) mv^(2)""...(1)`

In second case,
potential energy - translational kinetic energy + rotational kinetic energy
`mgh.=(1)/(2) mv^(2)+(1)/(2)1 OMEGA^(2)`
`mgh=(1)/(2)+(1)/(2) ((mr^(2))/(2))((v^(2))/(r^(2)))`
`:. mgh.=(3)/(4) mv^(2) ""....(2)`
Dividing. (2) by (1),
`(mgh.)/(mg xx2)=((3)/(4)mv^(2))/((1)/(2)mv^(2))=(3)/(4)xx(2)/(1)=(3)/(2)`
`k=3m`
From EQUATION (1) `2mg=(1)/(2) mv^(2)`
`v= sqrt(4)g=2sqrt(g)`
`v=2xx sqrt(9.81)`
`v=6.3 ms^(-1)`
45.

A railway coach of mass 8000 kg moving with speed of 54 ms^(-1)collide with rest coach of same like . Find the decrease in kinetic energy in this process.

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Solution :Hint use of the FORMULA of : DECREASES in KINETIC ENERGY = `(m_(1)m_(2)v_(li))/(2(m_(1)+m_(2))`
46.

A billiard ball moving at a speed 2m/s strikes an identical ball initially at rest, at a glancing blow. After the collision one ball is found to be moving at a speed of 1m/s at 60^(@) with the original line of motion. The velocity of the other ball shall be -

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`(3)^(1/2)m//s at 30^(@)` to the ORIGINAL DIRECTION.
`1m//s at 60^(@)` to the original direction
`(3)^(1//2)m//s at 60^(@)` to the original direction
`1m//s at 30^(@)` to the original direction

Answer :A
47.

The dimensions of mu_(0)I//MB where mu_(0) is the permeability of free space, I is the moment of inertia, M is the magnetic moment and B is the magnetic induction respectively, are those of

Answer»

time
`("time")^(2)`
`("time")^(3)`
`("time")^((1)/(2))`

Solution :`B= mu_(0)ni` (For a solenoid)
where n is the number of turns per unit length , I is the current and `mu_(0)` is the permeability of free space.
or `(mu_(0))/(B) = (1)/(ni) :. [(mu_(0))/(B)]= [(1)/(ni)]= (1)/([L^(-1)][A])= [LA^(-1)]`
`[M]= ["MAGNETIC moment"]= [ m xx 2L]`
[pole strenght `xx` magnetic lenght]
`=["weber xx L] = [ML^(2)T^(-2)A^(-1)xx L]= [ML^(3)T^(-2)A^(-1)]`
`= [LA^(-1)] xx ([ML^(2)])/([ML^(3)T^(-2)A^(-1)])= [T^(2)]= ("time")^(2)`
48.

A body falling vertically .downwards explodes into two equal fragments when it is at a height of 200 m and has a downward velocity of ,6just after the explosion one pf the fragments is seen to move .downward with a velocity of 8 ms^(-1) .Calculate the position of the centre pf mass of the system 5 s after ,the explosion.

Answer»


Solution :The EXPLOSION is not due to any EXTERNAL force. So after explosion ALSO the CM continues to move as if there was no explosion. Distance travelled after 5s is `y=y_(0) + (1//2) gt^(2) = 6 XX 5 +(1//2) 9.8 xx 25 y=152.5 m`. So C.M. is at a distance 200-152.5 = 47.5 m above the ground.
49.

In an adiabatic change, the pressure P and temperature T of a diatomic gas are related by the relation P prop T^C ,where C equals to :

Answer»

`5//3`
`2//5`
`3//5`
`7//2`

ANSWER :D
50.

For a body in S.H.M the velocity is given by the relation v= sqrt(144-16x^(2))m//"sec". The maximum acceleration is

Answer»

`12 m//"SEC"^(2)`
`16 m//"sec"^(2)`
`36 m//"sec"^(2)`
`48 m//"sec"^(2)`

Answer :D