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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2015 Exam Questions

Multiple Choice Questions

1.

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 )


C.

The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.

Consider the stones thrown up simultaneously as shown in the diagram below.

Considering the motion of the second particle with respect to the first we have relative acceleration
|a21| = |a2-a1| = g-g = 0



Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
negative 240 space equals space 10 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
or space straight t squared minus 2 straight t minus 48 space equals space 0
or space straight t squared space minus 8 straight t space plus space 6 straight t space minus 48 space equals space 0
or space space straight t squared space equals space 8 comma negative 6 space left parenthesis not space possible right parenthesis
Thus, distance covered by the second particle with respect to the first particle in 8s is
S12 = (v21) t = (40-10)(8s)
 = 30 x 8 = 240m
Similarly, time taken by the second particle to strike the ground is given by
negative 240 space equals space 40 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
minus 240 space equals space 40 space straight t space minus 5 straight t squared
5 straight t squared minus 40 straight t minus 240 space equals 0
straight t squared minus 8 straight t minus 48 space equals space 0
straight t squared minus 12 straight t space plus space 4 straight t minus 48 space equals 0
straight t left parenthesis straight t minus 12 right parenthesis plus 4 left parenthesis straight t minus 12 right parenthesis space equals 0
straight t space equals space 12 comma negative 4 space left parenthesis not space possible right parenthesis
Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.

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2.

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :

  • h2/4R

  • 3h/4

  • 5h/8

  • 3h2/8R


B.

3h/4

We know that centre of mass of uniform solid cone of height h is at height h/4 from base therefore,




h-zo = h/4
zo = h-h/4 = 3h/4

740 Views

3.

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = , the potential at the centre of the cavity thus formed is (G = gravitational constant)

  • -GM/2R

  • -GM/R

  • -2GM/3R

  • -2GM/R


B.

-GM/R

Consider cavity as negative mass and apply superposition of gravitational potential.
Consider the cavity formed in a solid sphere as shown in figure
V (∞) = 0


According to the question, we can write potential at an internal point P due to the complete solid sphere.
straight V subscript straight s space equals space fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets 3 straight R squared minus open parentheses straight R over 2 close parentheses squared close square brackets
equals negative fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets 3 straight R squared minus straight R squared over 4 close square brackets
equals negative fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets fraction numerator 11 straight R squared over denominator 4 end fraction close square brackets space equals space fraction numerator negative 11 GM over denominator 8 straight R end fraction
Mass of removed part,
=fraction numerator straight M over denominator begin display style 4 over 3 xπR cubed end style end fraction space straight x 4 over 3 straight pi open parentheses straight R over 2 close parentheses cubed space equals space straight M over 8
Potential at point P due to removed part

straight V subscript straight c space equals space fraction numerator negative 3 over denominator 2 end fraction space straight x space fraction numerator GM divided by 8 over denominator straight R end fraction space equals space fraction numerator negative 11 GM over denominator 8 straight R end fraction space minus open parentheses fraction numerator negative 3 GM over denominator 8 straight R end fraction close parentheses
Thus, potential due to remaining part at point P,

straight V subscript straight p space equals space straight V subscript straight s minus straight V subscript straight c space equals space fraction numerator negative 11 space GM over denominator 8 straight R end fraction space minus open parentheses negative fraction numerator 3 GM over denominator 8 straight R end fraction close parentheses
fraction numerator left parenthesis negative 11 space plus 3 right parenthesis GM over denominator 8 straight R end fraction space equals space minus GM over straight R

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4.

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

  • 44%

  • 50%

  • 56%

  • 62%


C.

56%

354 Views

5.

From a solid sphere of mass M and radius R, a cube of the maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is

  • fraction numerator MR squared over denominator 32 square root of 2 straight pi end root end fraction
  • fraction numerator MR squared over denominator 16 square root of 2 straight pi end root end fraction
  • fraction numerator 4 MR squared over denominator 9 square root of 3 straight pi end root end fraction
  • fraction numerator 4 MR squared over denominator 3 square root of 3 straight pi end root end fraction

C.

fraction numerator 4 MR squared over denominator 9 square root of 3 straight pi end root end fraction

Consider the cross-sectional view of a diametric plane as given the figure.
Using geometry of the cube


PQ space equals space 2 straight R space equals space left parenthesis square root of 3 right parenthesis space straight a space or space space straight a space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction
Volume density of the solid sphere

straight rho space equals space fraction numerator straight M over denominator begin display style 4 over 3 end style πR cubed end fraction space equals space fraction numerator 3 over denominator 4 straight pi end fraction open parentheses straight M over straight R cubed close parentheses
mass space of space cube space left parenthesis straight m right parenthesis space space equals space left parenthesis straight rho right parenthesis left parenthesis straight a cubed right parenthesis
space equals space open parentheses fraction numerator 3 over denominator 4 straight pi end fraction straight x space straight M over straight R cubed close parentheses open square brackets fraction numerator 2 straight R over denominator square root of 3 end fraction close square brackets cubed
space equals space fraction numerator 3 straight M over denominator 4 πR cubed end fraction space straight x space fraction numerator 8 straight R cubed over denominator 3 square root of 3 end fraction space equals fraction numerator 2 straight M over denominator square root of 3 straight pi end fraction
Moment of inertia of the cube about the given axis is 
straight I subscript straight y space equals space ma squared over 12 space left parenthesis straight a squared plus straight a squared right parenthesis space space equals ma squared over 6
rightwards double arrow space straight I subscript straight Y space equals space ma squared over 6 space equals space fraction numerator 2 straight M over denominator square root of 3 straight pi end root end fraction space straight x 1 over 6 space straight x space fraction numerator 4 straight R squared over denominator 3 end fraction space equals space fraction numerator 4 MR squared over denominator 9 square root of 3 straight pi end root end fraction

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6.

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both the cases body is brought from an initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:

  • ln2,4ln2

  • ln2,ln2

  • ln2,2ln2

  • 2ln2,8ln2


B.

ln2,ln2

Since entropy is a state function, therefore a change in entropy in both the processes must be same .

448 Views

7.

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its Bob, the time period changes TM. If Young's modulus of the material of the wire is Y, then 1/Y is equal to (g = gravitational acceleration)

  • open square brackets open parentheses straight T subscript straight M over straight T close parentheses squared minus 1 close square brackets straight A over Mg
  • open square brackets open parentheses straight T subscript straight M over straight T close parentheses squared minus 1 close square brackets Mg over straight A
  • open square brackets 1 minus space open parentheses straight T subscript straight M over straight T close parentheses squared close square brackets straight A over Mg
  • open square brackets 1 minus space open parentheses straight T subscript straight M over straight T close parentheses squared close square brackets Mg over straight A

A.

open square brackets open parentheses straight T subscript straight M over straight T close parentheses squared minus 1 close square brackets straight A over Mg

Time period, straight T space equals space 2 straight pi square root of straight L over straight g end root
When additional mass M is added to its Bob

straight T subscript straight M space equals space 2 straight pi square root of fraction numerator straight L plus increment straight L over denominator straight g end fraction end root comma where Δ is increase in length
As we know that,

straight Y space equals space fraction numerator Mg divided by straight A over denominator increment straight L divided by straight L end fraction space equals space fraction numerator MgL over denominator straight A increment straight L end fraction
rightwards double arrow increment straight L space equals space MgL over AY
straight T subscript straight M space equals space MgL over AY
straight T subscript straight M space equals space 2 straight pi square root of fraction numerator straight L plus begin display style MgL over AY end style over denominator straight g end fraction end root
rightwards double arrow open parentheses straight T subscript straight M over straight T close parentheses squared space equals space 1 plus Mg over AY
or space Mg over AY equals space 1 minus open parentheses straight T subscript straight M over straight T close parentheses squared
1 over straight Y space equals space straight A over Mg open square brackets open parentheses straight T subscript straight M over straight T close parentheses squared minus 1 close square brackets

692 Views

8.

The period of oscillation of a simple pendulum is straight T space equals space 2 straight pi space square root of straight L over straight g end root . The measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

  • 2%

  • 3%

  • 1%

  • 5%


B.

3%

Time period,

straight T space equals space 2 straight pi space square root of straight L divided by straight g end root
thus, changes can be expressed as

plus-or-minus fraction numerator 2 increment straight T over denominator straight T end fraction space equals space plus-or-minus fraction numerator increment straight L over denominator straight L end fraction space plus-or-minus fraction numerator increment straight g over denominator straight g end fraction
According to the question, we can write

fraction numerator increment straight L over denominator straight L end fraction space equals space fraction numerator 0.1 over denominator 20.0 end fraction cm space equals space 1 over 200
Again time period

straight T space equals space 90 over 100 straight s
and space increment straight T space equals space 1 over 100 straight s
rightwards double arrow space fraction numerator increment straight T over denominator straight T end fraction space equals space 1 over 90
Now comma space
because space straight T space equals space 2 straight pi space square root of straight L over straight g end root
straight g space equals 4 straight pi squared straight L over straight T squared
fraction numerator increment straight g over denominator straight g end fraction space equals space fraction numerator increment straight L over denominator straight L end fraction space plus space fraction numerator 2 increment straight T over denominator straight T end fraction
or space fraction numerator increment straight g over denominator straight g end fraction straight x space 100 percent sign space equals space open parentheses fraction numerator increment straight L over denominator straight L end fraction close parentheses straight x 100 percent sign space plus space open parentheses fraction numerator 2 increment straight T over denominator straight T end fraction close parentheses straight x space 100 percent sign
space equals space open parentheses 1 over 200 straight x 100 close parentheses percent sign space plus space 2 space straight x space 1 over 90 space straight x space 100 percent sign
space equals space 2.72 percent sign space almost equal to space 3

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9.

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

  • 100N

  • 80 N

  • 120 N

  • 150 N


C.

120 N

In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, fA = 20 N and for block B.
fB = fA +100 = 120 N

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10.

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume straight u equals space straight U over straight V space proportional to space straight T to the power of 4 and pressure.straight p space equals space 1 third open parentheses straight U over straight V close parentheses If the shell now undergoes an adiabatic expansion the relation between T and R is

  • T ∝ e-R

  • T ∝ e-3R

  • T ∝ (1/R)

  • T ∝(1/R3)


C.

T ∝ (1/R)

According to given equation,

straight p space equals space 1 third open parentheses straight U over straight V close parentheses
rightwards double arrow nRT over straight V space equals space 1 third open parentheses straight U over straight V close parentheses
left square bracket because space pV space equals space nRT right square bracket
or space nRT over straight V space proportional to 1 third straight T to the power of 4
or space VT cubed space equals constant
or space 4 over 3 πR cubed straight T cubed space equals space constant
or space TR space equals constant
straight T proportional to 1 over straight R

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