System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 23

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Question 1
1 / -0

The kinetic energy of a moving body is given by $$K = 2v^{2}$$, $$K$$ being in joules and $$v$$ in m/s. Its momentum when traveling with a velocity of $$2\ ms^{-1}$$ will be :

A
$$16\ kg m s^{-1}$$

B
$$4\ kg m s^{-1}$$

C
$$8\ kg m s^{-1}$$

D
$$2\ kg m s^{-1}$$

Solution

$$K=\dfrac{1}{2}mv^{2}=2v^{2}$$
$$\Rightarrow m=4kg$$
$$\vec{p}=m.\vec{v}$$
$$=4\times 2$$
$$=8\ kg m/s$$
Question 2
1 / -0

A torque of $$0.5 Nm$$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is $$50 N$$, the width of the screw driver is:

A
$$0.5 cm$$

B
$$0.75 cm$$

C
$$1 cm$$

D
$$1.5 cm$$

Solution

Let width be $$x$$. So the torque due to couple is $$2\times50\times x/2=50x$$
$$50x = 0.5\Rightarrow x=0.01m=1\ cm$$
Question 3
1 / -0

The position of axis of rotation of a body is changed so that its moment of inertia decreases by 36%. Find the % change in its radius of gyration.

A
It decreases by 18%

B
It increases by 18%

C
It decreases by 20%

D
It increases by 20%

Solution

$$\textbf{Step 1 - Finding relation between}$$ $$k_{1}\ and\ k_{2}$$
We know that $$I = mk^{2}$$
Initially, $$I_{1} = mk_{1}^{2}$$
and finally, $$I_{2} = mk_{2}^{2}$$
given that initial moment of inertia decrease by $$36\%$$
$$\therefore I_{2} = I_{1} - \dfrac {36}{100}I_{1}$$
$$\Rightarrow I_{2} = 0.64 I_{1}$$
$$\Rightarrow mk_{2}^{2} = 0.64\ mk_{1}^{2}$$
$$\Rightarrow k_{2} = 0.8 k_{1}$$, so $$k_{1} > k_{2}$$

$$\textbf{Step 2 - Calculating change in radius of gyration}$$
Change in radius of gyration $$= \dfrac {k_{1} - k_{2}}{k_{1}}\times 100$$
$$= \dfrac {k_{1} - 0.8k_{1}}{k_{1}} \times 100$$
$$= 20$$

Hence, the radius of gyration decreases by $$20\%$$
Question 4
1 / -0

The radius of gyration of a rotating metallic disc is independent of which of the following physical quantity.

A
Position of axis of rotation

B
Mass of disc

C
Radius of disc

D
Centre of mass of disc

Solution

$$I=mk^{2}$$
$$k=\sqrt{\dfrac{T}{m}}$$
$$I=\dfrac{mR^{2}}{2}$$
$$k=\dfrac{R}{\sqrt{2}}$$
$$\therefore r=R_{0} (1+ \alpha \Delta T)$$
$$\therefore$$ k is independent of mass.
Question 5
1 / -0

A wheel of radius $$R$$ is free to rotate about its natural axis and $$I$$ is its moment of inertia. If a tangential force $$F$$ is applied on the wheel along its rim, then angular acceleration of wheel is:

A
$$IFR$$

B
$$\dfrac{FR}{I}$$

C
$$\dfrac{IF}{R}$$

D
$$\dfrac{IR}{F}$$

Solution

The torque is given using the relation
$$\vec{\tau} = \vec{F} \times \vec{R}$$
or
$$\tau=I\alpha$$
or
$$\alpha=\dfrac{FR}{I}$$
Question 6
1 / -0

A flywheel making $$120$$ r.p.m is acted upon by a retarding torque producing angular retardation of $$\pi \ rad/s^{2}$$. Time taken by it to come to rest is:

A
$$1$$ s

B
$$2$$ s

C
$$3$$ s

D
$$4$$ s

Solution

$$T=I \alpha $$
$$\omega =\dfrac{120\times 2\pi }{60}$$
$$\omega =4\pi $$
$$\omega =\omega _{0}+\alpha t$$
$$0=4\pi -\pi t$$
$$t=4 s$$
Question 7
1 / -0

A brass disc is rotating about its axis. If temperature of the disc is increased then its:

A
radius of gyration increases, but moment of inertia remains the same

B
moment of inertia increases but radius of gyration remains the same

C
radius of gyration and moment of inertia both remain the same

D
radius of gyration and moment of inertia both increases

Solution

Radius of gyration and
moment of inertia will increase.
$$\because I=mk^{2}$$
$$k=\sqrt{\dfrac{I}{m}}$$
$$I=\dfrac{mR^{2}}{2}$$
$$k=\dfrac{R}{\sqrt{2}}$$
$$A=A_{0}(1+ 2\alpha \Delta T) (eq^{n} $$ of thermal expansion)
$$\pi r^{2}=\pi R_{0}^{2} (1+ 2\alpha \Delta T)$$
$$r^{2}=R_{0}^{2} (1+ 2\alpha \Delta T)$$
$$\therefore$$ r will increase on increasing the temperature
$$\therefore I\ and \ k$$ will increase on increasing the $$temperature$$
Question 8
1 / -0

Mass of thin long metal rod is 2 kg and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is $$0.5 kgm^{2}$$. Its radius of gyration is:

A
$$20 cm$$

B
$$40 cm$$

C
$$50 cm$$

D
$$1 m$$

Solution

$$I= 0.5kgm^2$$
m= 2kg
K= radius of gyration

$$2k^2= 0.5$$

$$k^2= \sqrt{0\cdot 25}$$

$$\therefore k =\sqrt{0.25} m = 50\: cm$$
Question 9
1 / -0

A wheel of moment of inertia $$5\times 10^{-3}kgm^{2}$$ is making $$20 \ rev/s$$. The torque required to stop it in $$10 \ s$$ is :

A
$$2\pi \times 10^{-2}Nm$$

B
$$2\pi \times 10^{2}Nm$$

C
$$4\pi \times 10^{-2}Nm$$

D
$$4\pi \times 10^{2}Nm$$

Solution

$$\omega =20\times 2\pi$$
$$\tau = I\omega$$
So, $$ \tau \Delta t=5\times 10^{-3}\omega$$
$$\tau =\dfrac{5\times 10^{-3}\times 20\times 2\pi}{10}=2\pi\times 10^{-2} Nm$$
Question 10
1 / -0

Four point size bodies each of mass $$m$$ are fixed at four corners of a light square frame of side length $$L$$. The radius of gyration of these four bodies about an axis perpendicular to the plane of frame passing through its centre is:

A
$$\sqrt{2}L$$

B
$$2 L$$

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

D
$$\dfrac{L}{2}$$

Solution

$$I= 4\left (m \left (\dfrac{L}{\sqrt2} \right )^2 \right )$$
$$= 2mL^2$$
So we have radius of gyration equivalence equation as,
$$4mk^2= 2mL^2$$
$$k= \dfrac{L}{\sqrt2} $$

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System of Particles and Rotational Motion Test - 23

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System of Particles and Rotational Motion Test - 23

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Question 1

$$16\ kg m s^{-1}$$

$$4\ kg m s^{-1}$$

$$8\ kg m s^{-1}$$

$$2\ kg m s^{-1}$$

Solution

Question 2

$$0.5 cm$$

$$0.75 cm$$

$$1 cm$$

$$1.5 cm$$

Solution

Question 3

It decreases by 18%

It increases by 18%

It decreases by 20%

It increases by 20%

Solution

Question 4

Position of axis of rotation

Mass of disc

Radius of disc

Centre of mass of disc

Solution

Question 5

$$IFR$$

$$\dfrac{FR}{I}$$

$$\dfrac{IF}{R}$$

$$\dfrac{IR}{F}$$

Solution

Question 6

$$1$$ s

$$2$$ s

$$3$$ s

$$4$$ s

Solution

Question 7

radius of gyration increases, but moment of inertia remains the same

moment of inertia increases but radius of gyration remains the same

radius of gyration and moment of inertia both remain the same

radius of gyration and moment of inertia both increases

Solution

Question 8

$$20 cm$$

$$40 cm$$

$$50 cm$$

$$1 m$$

Solution

Question 9

$$2\pi \times 10^{-2}Nm$$

$$2\pi \times 10^{2}Nm$$

$$4\pi \times 10^{-2}Nm$$

$$4\pi \times 10^{2}Nm$$

Solution

Question 10

$$\sqrt{2}L$$

$$2 L$$

$$\dfrac{L}{\sqrt{2}}$$

$$\dfrac{L}{2}$$

Solution

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