Practice Start Test

Rotational Motion

NEET Physics · 99 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular velocity will be in the ratio
A 2 : 1
B 1 : 2
C 2:1\sqrt 2 :1
D 1:21:\sqrt 2
Correct Answer
Option C
Solution
K.E.=12Iω2K.E. = {1 \over 2}I{\omega ^2}
∴12I1ω12=12.2I1ω22\therefore {1 \over 2}{I_1}\omega _1^2 = {1 \over 2}.2{I_1}\omega _2^2
ω12ω22=21⇒ω1ω2=21{{\omega _1^2} \over {\omega _2^2}} = {2 \over 1} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {{\sqrt 2 } \over 1}
Q72
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ\theta . The frictional force
A dissipates energy as heat
B decreases the rotational motion
C decreases the rotational and translation motion
D converts translational energy to rotational energy.
Correct Answer
Option D
Solution

It is noted that net work done by frictional force when drum rolls down without slipping is zero, so W net = 0 W trans + W rotational = 0 ΔK trans + ΔK rotational = 0 ΔK trans = – ΔK rotational Hence, frictional force converts translational energy to rotational energy.

Q73
A round disc of moment of inertia II 2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia II 1 rotating with an angular velocity ω\omega about the same axis. The final angular velocity of the combination of discs is
A I2ωI1+I2{{{I_2}\omega } \over {{I_1} + {I_2}}}
B ω\omega
C I1ωI1+I2{{{I_1}\omega } \over {{I_1} + I{}_2}}
D (I1+I2)ωI1{{\left( {{I_1} + {I_2}} \right)\omega } \over {{I_1}}}
Correct Answer
Option C
Solution

The initial angular momentum of the system is

I1ω{I_1}\omega

and final angular momentum is

(I1+I2)ω′\left( {{I_1} + {I_2}} \right)\omega '

where

ω′\omega '

= final angular velocity of combination of discs On equating the initial and final angular momentum, we get

ω′=I1ω(I1+I2)\omega ' = {{{I_1}\omega } \over {\left( {{I_1} + {I_2}} \right)}}
Q74
Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side ll cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm 2 units will be
A 34{3 \over 4}mll 2
B 2mll 2
C 54{5 \over 4}mll 2
D 32{3 \over 2}mll 2
Correct Answer
Option C
Solution

I AX = m(AB) 2 + m(OC) 2 = m

ll

2 + m (

ll

cos 60º) 2 = m

ll

2 + m

ll

2 /4 = 5/4 m

ll

2

Q75
A wheel having moment of inertia 2 kg m 2 about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
A 2π15{{2\pi } \over {15}} N m
B π12{\pi \over {12}} N m
C π15{\pi \over {15}} N m
D π18{\pi \over {18}} N m
Correct Answer
Option C
Solution
ωf=ωi−αt⇒0=ωi−αt{\omega _f} = {\omega _i} - \alpha t \Rightarrow 0 = {\omega _i} - \alpha t

∴\therefore

α=ωit\alpha = {{{\omega _i}} \over t}

where α\alpha is retardation. The torque on the wheel is given by

τ=Iα=Iωt=I.2πυt=2×2×π×6060×60\tau = I\alpha = {{I\omega } \over t} = {{I.2\pi \upsilon } \over t} = {{2 \times 2 \times \pi \times 60} \over {60 \times 60}}
τ=π15\tau = {\pi \over {15}}

N m This is the torque required to stop the wheel in 1 min. (or 60 sec.).

Q76
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axes in the plane of the ring is
A 2 : 3
B 2 : 1
C 5:6\sqrt 5 :\sqrt 6
D 1:21:\sqrt 2
Correct Answer
Option C
Solution

Radius of gyration of disc about a tangential axis in the plane of disc is

52R=K1{{\sqrt 5 } \over 2}R = {K_1}

, radius of gyration of circular ring of same radius about a tangential axis in the plane of circular ring is

K2=32R{K_2} = \sqrt {{3 \over 2}} R
∴K1K2=56\therefore {{{K_1}} \over {{K_2}}} = {{\sqrt 5 } \over {\sqrt 6 }}
Q77
A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom ?
A 2gh\sqrt {2gh}
B 34gh\sqrt {{3 \over 4}gh}
C 43gh\sqrt {{4 \over 3}gh}
D 4gh\sqrt {4gh}
Correct Answer
Option C
Solution

K.E. =

12Iω2+12mv2{1 \over 2}I{\omega ^2} + {1 \over 2}m{v^2}

K.E. =

12(12mr2)ω2+12mv2{1 \over 2}\left( {{1 \over 2}m{r^2}} \right){\omega ^2} + {1 \over 2}m{v^2}
=14mv2+12mv2=34mv2= {1 \over 4}m{v^2} + {1 \over 2}m{v^2} = {3 \over 4}m{v^2}

Now, gain in K.E. = Loss in P.E.

34mv2=mgh⇒v=(43)gh{3 \over 4}m{v^2} = mgh \Rightarrow v = \sqrt {\left( {{4 \over 3}} \right)gh}
Q78
A stone is tied to a string of length ll and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is
A 2(μ2−gl)\sqrt {2\left( {{\mu ^2} - gl} \right)}
B u2−gl\sqrt {{u^2} - gl}
C u−u2−2glu - \sqrt {{u^2} - 2gl}
D 2gl\sqrt {2gl}
Correct Answer
Option A
Solution

The total energy at A = the total energy at B

⇒12mu2=12mv2+mgl\Rightarrow {1 \over 2}m{u^2} = {1 \over 2}m{v^2} + mgl
⇒v=u2−2gl\Rightarrow v = \sqrt {{u^2} - 2gl}

The change in magnitude of velocity =

u2+v2\sqrt {{u^2} + {v^2}}
=2(u2−gl)= \sqrt {2\left( {{u^2} - gl} \right)}
Q79
A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω\omega . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be :
A Mω4m{{M\omega } \over {4m}}
B MωM+4m{{M\omega } \over {M + 4m}}
C (M+4m)ωM{{\left( {M + 4m} \right)\omega } \over M}
D (M−4m)ωM+4m{{\left( {M - 4m} \right)\omega } \over {M + 4m}}
Correct Answer
Option B
Solution

Applying conservation law of angular momentum,

I1ω1=I2ω2{I_1}{\omega _1} = {I_2}{\omega _2}
I2=(Mr2)+4(m)(r2)=(M+4m)r2{I_2} = \left( {M{r^2}} \right) + 4\left( m \right)\left( {{r^2}} \right) = \left( {M + 4m} \right){r^2}

(Taking

ω1=ω{\omega _1} = \omega

and

ω1=ω1{\omega _1} = \omega _1

)

⇒Mr2ω=(M+4m)r2ω1\Rightarrow M{r^2}\omega = \left( {M + 4m} \right){r^2}{\omega _1}
⇒ω1=MωM+4m\Rightarrow {\omega _1} = {{M\omega } \over {M + 4m}}
Q80
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R. then the fraction of total energy associated with its rotational energy will be
A K2+R2R2{{{K^2} + {R^2}} \over {{R^2}}}
B K2R2{{{K^2}} \over {{R^2}}}
C K2K2+R2{{{K^2}} \over {{K^2} + {R^2}}}
D R2K2+R2{{{R^2}} \over {{K^2} + {R^2}}}
Correct Answer
Option C
Solution

Rotational energy =

12I(ω)2=12(mK2)ω2{1 \over 2}I{\left( \omega \right)^2} = {1 \over 2}\left( {m{K^2}} \right){\omega ^2}

Linear kinetic energy =

12mω2R2{1 \over 2}m{\omega ^2}{R^2}

∴\therefore Required fraction

=12(mK2)ω212(mK2)ω2+12mω2R2=K2K2+R2= {{{1 \over 2}\left( {m{K^2}} \right){\omega ^2}} \over {{1 \over 2}\left( {m{K^2}} \right){\omega ^2} + {1 \over 2}m{\omega ^2}{R^2}}} = {{{K^2}} \over {{K^2} + {R^2}}}
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