Rotational Motion — Page 7 | NEET Physics

Q61

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

A

\overrightarrow r .\overrightarrow \tau > 0

and

\overrightarrow F .\overrightarrow \tau < 0

B

\overrightarrow r .\overrightarrow \tau = 0

and

\overrightarrow F .\overrightarrow \tau = 0

C

\overrightarrow r .\overrightarrow \tau = 0

and

\overrightarrow F .\overrightarrow \tau \ne 0

D

\vec r.\vec \tau \ne 0

and

\overrightarrow F .\overrightarrow \tau = 0

Correct Answer

Option B

Solution

Torque is always perpendicular to

\overrightarrow F

as well as

\overrightarrow r

. $\therefore$

\overrightarrow r .\overrightarrow \tau = 0

as well as

\overrightarrow F .\overrightarrow \tau = 0

Q62

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

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

Total mass = M, total length = L Moment of inertia of OA about O = Moment of inertia of OB about O.

$\Rightarrow$ M.I total =

2 \times \left( {{M \over 2}} \right){\left( {{L \over 2}} \right)^2}.{1 \over 3} = {{M{L^2}} \over {12}}

Q63

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

A

\sqrt 2 :1

B

\sqrt 2 :\sqrt 3

C

\sqrt 3 :\sqrt 2

D

1:\sqrt 2

Correct Answer

Option D

Solution

M.I. of a circular disc,

M{k^2} = {{M.{R^2}} \over 2}

M.I. of a circular ring = MR 2 . $\therefore$ Ratio of their radius of gyration =

{1 \over {\sqrt 2 }}:1

$\Rightarrow$ 1 :

{\sqrt 2 }

Q64

A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is L A when it is at A and L B when it is at B, then

A L A = L B

B the relationship between L A and L B depends upon the slope of the line AB

C L A < L B

D L A > L B .

Correct Answer

Option A

Solution

Angular momentum = Linear momentum × perpendicular distance between line of action of liner momentum from origin Let d be the perpendicular distance. p A and p B be the linear momentum at A and B Angular momentum L A = p A × d L B = p B × d So, linear momentum will be equal, i.e., L A = L B .here, p A and p B are equal as have equal velocity.

L A = L B

Q65

A wheel has angular acceleration of 3.0 rad/sec 2 and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radian) of

A 10

B 12

C 4

D 6

Correct Answer

Option A

Solution

Given: Angular acceleration, $\alpha$ = 3 rad/sec 2 Initial angular velocity

{\omega _i}

= 2 rad/sec Time t = 2 sec Using,

\theta = {\omega _i}t + {1 \over 2}\alpha {t^2}

$\therefore$

\theta = 2 \times 2 + {1 \over 2} \times 3 \times 4 = 4 + 6 = 10\,

radian.

Q66

A uniform rod AB of length

l

and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml 2 /3, the initial angular acceleration of the rod will be

A

{{mgl} \over 2}

B

{3 \over 2}gl

C

{{3g} \over {2l}}

D

{{2g} \over {3l}}

Correct Answer

Option C

Solution

Torque about A,

\tau = mg \times {l \over 2} = {{mgl} \over 2}

\tau = I\alpha

$\therefore$ Angular acceleration,

\alpha = {\tau \over I} = {{mgl/2} \over {m{l^2}/3}} = {{3g} \over {2l}}

Q67

The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc

A

{1 \over 2}

MR 2

B MR 2

C

{2 \over 5}M{R^2}

D

{3 \over 2}M{R^2}

Correct Answer

Option D

Solution

Moment of inertia of a uniform circular disc about an axis through its centre and perpendicular to its plane is

{I_C} = {1 \over 2}M{R^2}

by the theorem of parallel axes.

$\therefore$ Moment of inertia of a uniform circular disc about an axis touching the disc at its diameter and normal to the disc is I.

I = {I_C} + M{h^2} = {1 \over 2}M{R^2} + M{R^2} = {3 \over 2}M{R^2}

Q68

A uniform rod AB of length

l

and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml 2 /3, the initial angular acceleration of the rod will be

A

{{mgl} \over 2}

B

{3 \over 2}gl

C

{{3g} \over {2l}}

D

{{2g} \over {3l}}

Correct Answer

Option C

Solution

Torque about A,

\tau = mg \times {l \over 2} = {{mgl} \over 2}

\tau = I\alpha

$\therefore$ Angular acceleration,

\alpha = {\tau \over I} = {{mgl/2} \over {m{l^2}/3}} = {{3g} \over {2l}}

Q69

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity

\omega

. The force exerted by the liquid at the other end is

A

{{M{L^2}{\omega ^2}} \over 2}

B

{{ML{\omega ^2}} \over 2}

C

{{M{L^2}\omega } \over 2}

D

ML{\omega ^2}

Correct Answer

Option B

Solution

The centre of the tube will be at length L/2.

So radius r = L/2.

The force exerted by the liquid at the other end = centrifugal force Centrifugal force =

Mr{\omega ^2} = M\left( {{L \over 2}} \right){\omega ^2} = {{ML{\omega ^2}} \over 2}

Q70

The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is

A MR 2

B

{1 \over 2}

MR 2

C

{3 \over 2}

MR 2

D

{7 \over 2}

MR 2

Correct Answer

Option C

Solution

Moment of Inertia of uniform circular disc with radius ‘R’ and mass ‘M’ about axis passing through C.M. and normal to disc will be : I CM = (1/2) MR 2 Moment of Inertia about axis tangential to disc : I T = I CM + MR 2 I T = (1/2) MR 2 + MR 2 = (3/2) MR 2