Rotational Motion — Page 9 | NEET Physics

Q81

A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with

A aluminium at interior and iron surround to it

B iron at interior and aluminium surround to it

C using iron and aluminium layers in alternate order

D sheet of iron is used at both external surface and aluminium sheet as internal layers.

Correct Answer

Option A

Solution

A circular disc may be divided into a large number of circular rings.

Moment of inertia of the disc will be the summation of the moments of inertia of these rings about the geometrical axis.

Now, moment of inertia of a circular ring about its geometrical axis is MR 2 , where M is the mass and R is the radius of the ring.

Since the density (mass per unit volume) for iron is more than that of aluminium, the proposed rings made of iron should be placed at a higher radius to get more value of MR 2 .

Hence to get maximum moment of inertia for the circular disc, aluminium should be placed at interior and iron at the outside.

Q82

A disc is rotating with angular speed

\omega

. If a child sits on it, what is conserved

A linear momentum

B angular momentum

C kinetic energy

D potential energy.

Correct Answer

Option B

Solution

If external torque is zero, angular momentum remains conserved.

[External torque is zero because the weight of child acts downward] L = I $\omega$ = constant

Q83

A solid sphere of radius R is placed on smooth horizontal surface. A horizontal force F is applied at height h from the lowest point. For the maximum acceleration of centre of mass, which is correct ?

A h = R

B h = 2R

C h = 0

D no relation between h and R.

Correct Answer

Option D

Solution

Since there is no friction at the contact surface (smooth horizontal surface) there will be no rolling.

Hence, the acceleration of the centre of mass of the sphere will be independent of the position of the applied force F.

Therefore, there is no relation between h and R.

Q84

A point P consider at contact point of a wheel on ground which rolls on ground without slipping then value of displacement of point P when wheel completes half of rotation (if radius of wheel is 1 m)

A 2 m

B

\sqrt {{\pi ^2} + 4} m

C

\pi \,m

D

\sqrt {{\pi ^2} + 2} \,m

Correct Answer

Option B

Solution

In half rotation point P has moved horizontally.

{{\pi d} \over 2} = \pi r = \pi \times 1m = \pi \,m

\left[ \because {radius = 1m} \right]

In the same time, it has moved vertically a distance which is equal to its diameter = 2 m.

$\therefore$ Displacement of P =

\sqrt {{\pi ^2} + {2^2}} = \sqrt {{\pi ^2} + 4}

m

Q85

For a hollow cylinder and a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier

A solid cylinder

B hollow cylinder

C both simultaneously

D can't say anything.

Correct Answer

Option A

Solution

Solid sphere reaches the bottom first because for solid cylinder

{{{K^2}} \over {{R^2}}} = {1 \over 2}

, and for hollow cylinder

{{{K^2}} \over {{R^2}}} = 1

Acceleration down the inclined plane

\propto {1 \over {{{{K^2}} \over {{R^2}}}}}

Solid cylinder has greater acceleration, so it reaches the bottom first.

Q86

As shown in the figure at point O a mass is performing vertical circular motion. The average velocity of the particle is increased, then at which point will the string break

A A

B B

C C

D D

Correct Answer

Option B

Solution

When a sphere is rotating in a vertical circle, it exerts the maximum outward pull when it is at the lowest point B.

Therefore, tension at B is maximum =

{{m{v^2}} \over R}

So, the string breaks at point B.

Q87

For the adjoining diagram, the correct relation between I 1 , I 2 , and I 3 is, (I-moment of inertia)

A I 1 > I 2

B I 2 > I 1

C I 3 > I 1

D I 3 > I 2

Correct Answer

Option B

Solution

As effective distance of mass from BC is greater than the effective distance of mass from AB, therefore I 2 > I 1 .

Q88

Mass per unit area of a circular disc of radius

a

depends on the distance r from its centre as

\sigma \left( r \right)

= A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

A

2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

B

\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

C

2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)

D

2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)

Correct Answer

Option A

Solution

dI = dm(r2) and dm = $\sigma$ 2 $\pi$ rdr $\therefore$ dI = $\sigma$ 2 $\pi$ rdr(r2) = $\sigma$ 2 $\pi$ r3dr Given

\sigma \left( r \right)

= A + Br $\therefore$ dI = 2 $\pi$ (A + Br)r3dr

\int {dI = \int\limits_0^a {2\pi {r^3}\left( {A + Br} \right)dr} }

$\Rightarrow$ I =

2\pi \left[ {{{A{r^4}} \over 4} + {{B{r^5}} \over 5}} \right]

=

2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

Q89

A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which on of the following will not be affected ?

A Angular velocity

B Angular momentum

C Moment of inertia

D Rotational kinetic energy

Correct Answer

Option B

Solution

Solid sphere is rotating in free space that means no external torque is operating on the sphere.

Angular momentum will remain the same since external torque is zero.

Q90

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is :

A 8.75 × 10–3 J

B 1.13 × 10–3 J

C 8.75 × 10–4 J

D 6.25 × 10–4 J

Correct Answer

Option C

Solution

K.E =

{1 \over 2}m{V^2} + {1 \over 2}{I_{cm}}{\omega ^2}

=

{1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}

=

{1 \over 2}m{V^2} + {1 \over 5}m{V^2}

=

{7 \over {10}}m{V^2}

=

{7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}

=

{{35} \over 4} \times {10^{ - 4}}

= 8.75 × 10–4 J