AIPMT July 2015 physics questions with solutions

Showing 1–10 of 11 questions

• MCQmediumRotational Motionquestion based on pure rotation about fixed axis

  An automobile moves at 54 km/h. A wheel has radius 0.45 m and moment of inertia 3 kg·m². If the vehicle stops uniformly in 15 s, find average braking torque on the wheel.

  Options

  • Option A

    6.66 N·m

  • Option B

    8.58 N·m

  • Option C

    10.86 N·m

  • Option D

    2.86 N·m

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• MCQmediumCenter of maas , Momentum and Collisioncollision and types of collision

  A block of mass M moving with speed v hits an identical stationary block elastically on a frictionless plane. The first block emerges at speed v/3 and an angle θ to its initial direction. Find the second block's speed.

  Options

  • Option A

    (2√2/3)v

  • Option B

    3v/4

  • Option C

    (3/√2)v

  • Option D

    (√3/2)v

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• MCQmediumRotational Motiontorque and rotational equilibrium

  A force $\vec{F}$=α$\hat{i}+3\hat{j}+6\hat{k}$ acts at r=2$\hat{i}-6\hat{j}-12\hat{k}$. Find α for conservation of angular momentum about the origin.

  Options

  • Option A

    -1

  • Option B

    2

  • Option C

    zero

  • Option D

    1

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• MCQmediumCenter of maas , Momentum and Collisioncollision and types of collision

  A ball is thrown vertically downward from height 20 m with initial speed v₀. On impact it loses 50% of its energy and rebounds to the original height. Find v₀; take g=10 m s⁻².

  Options

  • Option A

    14 m s⁻¹

  • Option B

    20 m s⁻¹

  • Option C

    28 m s⁻¹

  • Option D

    10 m s⁻¹

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• MCQmediumGravitationSatellites, Orbital Motion & Kepler’s Laws

  A remote-sensing satellite is in a circular orbit 0.25×10⁶ m above Earth’s surface. Given Earth radius 6.38×10⁶ m and g=9.8 m s⁻², find its orbital speed.

  Options

  • Option A

    7.76 km s⁻¹

  • Option B

    8.56 km s⁻¹

  • Option C

    9.13 km s⁻¹

  • Option D

    6.67 km s⁻¹

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• MCQmediumGravitationSatellites, Orbital Motion & Kepler’s Laws

  A light satellite moves in an elliptical orbit around Earth. Which statement is correct?

  Options

  • Option A

    Its angular momentum changes direction while keeping constant magnitude.

  • Option B

    Its total mechanical energy varies periodically.

  • Option C

    Its linear momentum magnitude stays constant.

  • Option D

    Its acceleration is always directed towards Earth’s centre.

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• MCQmediumCircular Motionkinematics of circular motion

  The position vector of a particle is R = 4 sin(2πt)i + 4 cos(2πt)j, where R is in metres and t is in seconds. Which statement is WRONG?

  Options

  • Option A

    Acceleration is along -R.

  • Option B

    Magnitude of acceleration is v²/R, where v is the speed.

  • Option C

    Magnitude of velocity is 8 m s⁻¹.

  • Option D

    The path is a circle of radius 4 m.

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• MCQmediumRotational Motionmoment of inertia and kinetic energy in pure rotation

  Masses m₁ and m₂ are at the ends of a light rod of length L. Where must a transverse rotation axis pass (at a distance x from mass m₁) to minimize the work needed to achieve a prescribed angular speed ω₀?

  

  Options

  • Option A

    x=m₁L/(m₁+m₂)

  • Option B

    x=m₁L/m₂

  • Option C

    x=m₂L/m₁

  • Option D

    x=m₂L/(m₁+m₂)

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• MCQmediumCircular Motioncentripital force and centrifugal force

  Stones of masses m and 2m move in horizontal circles. The heavier stone travels radius r/2 and the lighter travels radius r. If their required centripetal forces are equal, the lighter stone's tangential speed is n times the heavier stone's speed. Find n.

  Options

  • Option A

    4

  • Option B

    1

  • Option C

    2

  • Option D

    3

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• MCQmediumKinematics part-3 Relative motionRelative motion in 2-D and shortest distance problem

  Two particles A and B move with constant velocities $\overrightarrow{v}_1$and $\overrightarrow{v}_2$. At the initial moment their position vectors are $\overrightarrow{r_1}$ and $\overrightarrow{r}_2$ respectively. The condition for particles A and B to collide is:

  Options

  • Option A

    ($\vec{r}_1-\vec{r}_2$)/|$\vec{r}_1-\vec{r}_2$| = ($\vec{v}_2-\vec{v}_1$)/|$\vec{v}_2-\vec{v}_1$|

  • Option B

    $\vec{r}_1.\vec{v}_1$= $\vec{r}_2.\vec{v}_2$

  • Option C

    $\vec{r}_1$×$\vec{v}_1$=$\vec{r}_2$×$\vec{v}_2$

  • Option D

    $\vec{r}_1-\vec{v}_1$= $\vec{r}_2-\vec{v}_2$

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