JEE physics

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Question
CBSEENPH11020260

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 )

  • WiredFaculty
  • WiredFaculty
  • WiredFaculty
  • WiredFaculty

Solution

C.

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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
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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
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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
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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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Question
CBSEENPH11020262

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

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  • 100N

  • 80 N

  • 120 N

  • 150 N

Solution

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

Question
CBSEENPH11020263

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%

Solution

C.

56%

Question
CBSEENPH12039446

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

Solution

B.

3h/4

We know that centre of mass of uniform solid cone of height h is at height h/4 from base therefore,
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h-zo = h/4
zo = h-h/4 = 3h/4

Question
CBSEENPH12039447

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

Solution

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
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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
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Moment of inertia of the cube about the given axis is 
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