System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 55

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Question 1
1 / -0

Due to slipping, points $A$ and $B$ on the rim of the disc have the velocities shown. Distance between centre of disc and point on disc which is having zero velocity.

A
0.04$$\mathrm { m }$$

B
0.05$$\mathrm { m }$$

C
0.06$$\mathrm { m }$$

D
None

Solution

Angular velocity with respect to $$\frac{{{v_1}}}{r}$$
$$ = \frac{{4.5}}{{0.4}}$$
If $$A$$ be at rest let a point at distance $$x$$ from
$$A$$ has zero velocity . Then
$$\begin{array}{l} { V_{ A } }+rq=0 \\ 1.5=-k\left( { \frac { { 4.5 } }{ { 0.4 } } } \right) \\ -\frac { { 0.4 } }{ 3 } =k \\ k=-0.133 \end{array}$$
Distance of this point from center $$=0.2+0.133$$m
$$=0.333$$m
Question 2
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Find ratio of radius of gyration of a disc and ring of same radii at their tangential axis in plane.

A
$$\sqrt{\dfrac{5}{3}}$$

B
$$\sqrt{\dfrac{5}{6}}$$

C
$$1$$

D
$$\dfrac{2}{3}$$

Solution

For disc
$$MK^2_1 = \dfrac{5}{4} MR^2 \Longrightarrow K_1 = \dfrac{\sqrt{5}}{2}R$$ (Ref. Image [i])
For Ring

$$MK^2_2 = \dfrac{3}{2}MR^2 \Longrightarrow K_2 = \sqrt{\dfrac{3}{2}}R$$ (Ref. Image [ii])

So $$\dfrac{k_1}{k_2} = \dfrac{\sqrt{5} \times \sqrt{2}}{2 \times \sqrt{3}} = \sqrt{\dfrac{5}{6}}$$
Question 3
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The moment of inertia of a thin scale of length L and mass M about an axis passing through the center of mass and perpendlcular to its lenqth would be

A
$$\frac { 7M L ^ { 2 } } { 48 }$$

B
$$\frac { M L ^ { 2 } } { 4 }$$

C
$$\frac { M L ^ { 2 } } { 3 }$$

D
$$\mathrm { ML } ^ { 2 }$$

Solution

$$I = {I_{cm}} + m{\left( {\dfrac{1}{4}} \right)^2}$$
$$ = \dfrac{{M{l^2}}}{{12}} + \dfrac{{M{l^2}}}{{16}}$$
$$ = \dfrac{{4M{l^2} + 3M{l^2}}}{{48}}$$
$$ = \dfrac{{7M{l^2}}}{{48}}$$
Hence,
option $$(A)$$ is correct answer.
Question 4
1 / -0

A bullet of mass 0.01$$\mathrm { kg }$$ and traveling at a speed of 500$$\mathrm { m } / \mathrm { sec }$$ strikes a block of which suspended by a string of length 5$$\mathrm { m }$$ . The centre of gravity of the block is found to vertical distance of 0.1$$\mathrm { m }$$ . What is the speed of the bullet after it emerges from the block?

A
359$$\mathrm { m } / \mathrm { s }$$

B
220$$\mathrm { m } / \mathrm { s }$$

C
204$$\mathrm { m } / \mathrm { s }$$

D
284$$\mathrm { m } / \mathrm { s }$$

Solution

By energy conservation of the suspended block we can say
initial $$KE=$$ final potential energy
$$\dfrac{1}{2}m{v^2} = mgH$$
$$\dfrac{1}{2}m{v^2} = gH$$
$$v = \sqrt {2gH} $$
$$v = \sqrt {2 \times 10 \times 0.1} = 1.41\,m/s$$
now$$,$$ by momentum conservation
$${P_{bullet}} = {P_{block}} + {P_{bullet}}$$
$$0.01 \times 500 = 1 \times 1.41 + 0.01 \times {v_f}$$
$$5 - 1.41 = 0.01 \times {v_f}$$
$$0.01{v_f} = 3.59$$
$${v_f} = 359\,m/s$$
Hence,
option $$(A)$$ is correct answer.
Question 5
1 / -0

Find radius of gyration of a solid circular disc of mass M, radius R, about an axis parallel to its diameter and passing through its edge

A
$$R \dfrac { \sqrt { 5 } } { 2 }$$

B
$$R \dfrac { \sqrt { 3 } } { 2 }$$

C
$$\dfrac { R } { \sqrt { 2 } }$$

D
$$\dfrac { 3 R } { 2 }$$

Solution

$$\begin{array}{l}\text { Moment of Inertia about an axis passing } \\\text { through centre in the plane of the } \\\text { disc is } \dfrac{M R^{2}}{4} \\\text { about a tangent }=\dfrac{M R^{2}}{4}+M R^{2} \\ I=\dfrac{5 M R^{2}}{4}\end{array} $$
$$\begin{array}{l}\text { Let } k \text { be radius of Gyration } \\\qquad \begin{aligned}\dfrac{5 M R^{2}}{4} &=M k^{2} \\\Rightarrow & k=\dfrac{\sqrt{5}}{2} R\end{aligned} \\\text { Hence option } A \text { is correct }\end{array}$$
Question 6
1 / -0

Two thin rods each of mass $$m$$ and length 1 are joined to form L shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod end and perpendicular to both the ends is

A
$$\dfrac{{5m{l^2}}}{3}$$

B
$$\dfrac { m l ^ { 2 } } { 6 }$$

C
$$\dfrac { 5 } { 7 } M R ^ { 2 }$$

D
$$\dfrac { 7 } { 12 } M R ^ { 2 }$$

Solution

$$\begin{array}{l} I=\dfrac { { m{ L^{ 2 } } } }{ 3 } +\dfrac { { m{ L^{ 2 } } } }{ { 12 } } M\left( { \dfrac { { 5{ l^{ 2 } } } }{ 4 } } \right) \\ =\dfrac { { m{ L^{ 2 } } } }{ 3 } +\dfrac { { m{ L^{ 2 } } } }{ { 12 } } +\dfrac { { 5m{ l^{ 2 } } } }{ 4 } \\ =\dfrac { { 20m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 5m{ l^{ 2 } } } }{ 3 } \end{array}$$
$$\therefore $$ Option $$A$$ is correct .
Question 7
1 / -0

Which of the following in the measure of inertia

A
momentum

B
mass

C
acceleration

D
None of these

Solution
Question 8
1 / -0

The moment of inertia of a thin uniform rod about a transverse axis passing through centre is given to be I. The moment of inertia of the same rod about a transverse axis passing through a point midway between the centre and the end of the rod is:

A
$$4I$$

B
$$9I/4$$

C
$$7I/4$$

D
$$5I/2$$

Solution

$$\begin{array}{l}\text { Let } L \text { be Length of rod, } m \text { be mas of rod } \\\text { we know that } I=\dfrac{m L^{2}}{12} \\\text { By using parallel axis theorem } \\\qquad I^{\prime}=I+\dfrac{m L^{2}}{16}\end{array}$$
$$\begin{aligned} &=\dfrac{m L^{2}}{12}+\dfrac{m L^{2}}{16} \\ &=\dfrac{7 m L^{2}}{48}=\dfrac{7 \times m L^{2}}{4 \times 12}=\dfrac{7I}{4} \\\text { Hence option } C \text { is correct }\end{aligned}$$
Question 9
1 / -0

Radius of gyration of the system of two rods each of mass ' $$m ^ { \prime }$$ and length $$l ^ { \prime }$$ about an axis shown is:

A
$$l \sqrt { \frac { 1 } { 72 } }$$

B
$$l \sqrt { 7 }$$

C
$$l \sqrt { \frac { 7 } { 72 } }$$

D
None

Solution

$$\begin{array}{l} 2m{ K^{ 2 } }=\frac { { m{ l^{ 2 } } } }{ { 12 } } +\frac { { m{ l^{ 2 } } } }{ { 12 } } +\frac { { m{ l^{ 2 } } } }{ { 36 } } \\ K=l\sqrt { \frac { 7 }{ { 72 } } } \\ Ans.\, \, (C) \end{array}$$
Question 10
1 / -0

Two small kids weighing 10kg and 15kg are trying to balance a seesaw of total length 5m with the fulcrum at the centre. If one of the kids is sitting at an end, the other should sit at

A
1.67 m from the centre

B
2.50 m from the centre

C
2 m from the centre

D
1 m from the centre

Solution

$$\begin{array}{l}\text { Let, } B \text { of mass } 15\mathrm{~kg}\text { is sitting } \\\text { at distance } x \text { from centre. }\end{array}$$

$$\begin{array}{r}\text { By rotational equilibrium, }\\\qquad\begin{aligned}\left(\tau_{A}\right) &=\left(m_{A} g \times 2\cdot 5\right) \\\text { and }\left(\tau_{B}\right) &=\left(m_{B} g\times x\right)\end{aligned}\end{array}$$

$$\begin{array}{l}\text { Note that, } \tau_{A} \text { and }\tau_{B}\text { are in opposite directions. } \\\text { Thus, } m_{A} g \times 2 \cdot 5=m_{B} g \times x . \\\therefore x=\frac{10}{15}\times 2.5 \mathrm{~m}=1.67 \mathrm{~m}\end{array}$$