AIPMT 2009 | Rotational Motion Question 56 | Physics

AIPMT 2009

MCQ (Single Correct Answer)

-1

If $$\overrightarrow F $$ is the force acting on a particle having position vector $$\overrightarrow r $$ and $$\overrightarrow \tau $$ be the torque of this force about the origin, then

$$\overrightarrow r .\overrightarrow \tau > 0$$ and $$\overrightarrow F .\overrightarrow \tau < 0$$

$$\overrightarrow r .\overrightarrow \tau = 0$$ and $$\overrightarrow F .\overrightarrow \tau = 0$$

$$\overrightarrow r .\overrightarrow \tau = 0$$ and $$\overrightarrow F .\overrightarrow \tau \ne 0$$

$$\vec r.\vec \tau \ne 0$$ and $$\overrightarrow F .\overrightarrow \tau = 0$$

AIPMT 2009

MCQ (Single Correct Answer)

-1

A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $$\omega $$. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

$${{\omega M} \over {M + 2m}}$$

$${{\omega \left( {M + 2m} \right)} \over M}$$

$${{\omega M} \over {M + m}}$$

$${{\omega \left( {M - 2m} \right)} \over {M + 2m}}$$

AIPMT 2008

MCQ (Single Correct Answer)

-1

The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is

$$\sqrt 2 :1$$

$$\sqrt 2 :\sqrt 3 $$

$$\sqrt 3 :\sqrt 2 $$

$$1:\sqrt 2 $$

AIPMT 2008

MCQ (Single Correct Answer)

-1

A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90^o. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

$${{M{L^2}} \over 6}$$

$${{\sqrt 2 M{L^2}} \over {24}}$$

$${{M{L^2}} \over {24}}$$

$${{M{L^2}} \over {12}}$$

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