System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

A metallic sphere having mass $$2 \,kg$$ is moving with a velocity of $$10 \,m /s.$$ The momentum of the sphere in kg metre / sec. will be-

A
$$1 / 5$$

B
$$5$$

C
$$12$$

D
$$20$$

Solution

$$P=m×v$$
$$P=2×10$$
$$P=20m/s$$
Question 2
1 / -0

In kinetic theory of gases , a molecule of mass $$m$$ of an ideal gas collides with a wall of vessel with velocity $$V$$. The change in the linear momentum of the molecule is

A
$$2 \,mV$$

B
$$mV$$

C
$$-mV$$

D
Zero

Solution
Question 3
1 / -0

The total momentum of the molecules of $$1\,gm\, mol $$of a gas in a container at rest of $$300\,K $$ is

A
$$2 \times \sqrt{3 R \times 300} \, gm \times cm / sec $$

B
$$2 \times 3 \times R \times 300 \, gm \times cm / sec $$

C
$$1 \times \sqrt{3 \times R \times 300} \, gm \times cm / sec $$

D
Zero

Solution
Question 4
1 / -0

A $$57.0-g$$ tennis ball is traveling straight at a player at $$21.0\ m/s$$. The player volleys the ball straight back at $$25.0\ m/s$$. If the ball remains in contact with the racket for $$0.060\ 0\ s$$, what average force acts on the ball?

A
$$22.6\ N$$

B
$$32.5\ N$$

C
$$43.7\ N$$

D
$$72.1\ N$$

E
$$102\ N$$

Solution
Question 5
1 / -0

Three identical thin rods each of mass $$m$$ and length $$L$$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

A
$$\dfrac{2mL^{2}}{3}$$

B
$$\dfrac{3mL^{2}}{2}$$

C
$$\dfrac{4mL^{2}}{3}$$

D
$$\dfrac{3mL^{2}}{4}$$

Solution

Sum of MI of the two rods, at vertex of which the axis passes, is $$2\times mL^2/3=2mL^2/3$$
For the third rod, the distance between center of the rod and the axis is $$d=\sqrt3L/2$$
Using parallel axis theorem, its MI is $$mL^2/12+3mL^2/4=5mL^2/6$$
So total MI of the system is $$2mL^2/3+5mL^2/6=3mL^2/2$$
Question 6
1 / -0

A uniform sheet each of thickness 10 units is
cut into the shape as shown. Compute then x
and y-coordinates of the centre of mass of the
piece from point A.

A
17.5, 22.5

B
22.5, 17.5

C
10.5, 22.5

D
190/7, 125/7
Question 7
1 / -0

In the system shown, a force of $$100$$ N is applied at the end shown. What is the magnitude $$\tau $$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is $$0.1$$)

A
$$10$$

B
$$5$$

C
$$20$$

D
$$25$$

Solution

For the motion to start, the 'moment-couple' on the drum must be equal to the product of frictional force of the rod against the drum and the drum radius. Since the rod is in equilibrium, we can take torque about the fulcrum to find the normal force of the drum on the rod (Normal force N acts at centre of rod)
$$\sum \tau _{fulcrum}=0$$
or
$$ -100(2+2)+N(2)=0$$
or
$$N=200\ N$$
Friction force$$f=\mu N$$ $$= (0.1)(200) = 20 N $$
$$\tau =$$ Torque produced by couple$$=$$$$(20)1$$
or $$\tau =20\ Nm$$
Question 8
1 / -0

A car (open at the top) of mass 9.75 kg is coasting along a level track at 1.36 m/s, when it begins to rain hard. The raindrops fall vertically with respect to the ground. When it has collected 0.5 kg of rain, the speed of the car is

A
0.68 m/s

B
1.29 m/s

C
2.48 m/s

D
9.8 m/s

Solution

Initial $$m_{1} = 9.75$$ kg
final $$m =9.75 + 0.5 = 10.25$$
$$m_{1}v_{1} = m_{2}v_{2}$$
$$(9.75)(1.36) = (10.25)v_{2}$$
$$v_{2} = 1.29$$ m/s
Question 9
1 / -0

Two particles A and B are revolving with constant angular velocity on two concentric circles of radius $$1 m$$ and $$2 m$$ respectively as shown in figure. The positions of the particles at $$t = 0$$ are shown in figure. If $$m_{A} = 2 kg$$, $$m_{B} = 1 kg$$ and $$P_{A}$$ and $$P_{B}$$ are linear momentum of the particles then what is the maximum value of $$\left |P_{A} + P_{B} \right |$$ in $$kgm/s$$ in subsequent motion of the two particles.

A
$$1$$

B
$$7$$

C
$$8$$

D
$$5$$

Solution

Since the angular velocities of both the masses are constant, their linear velocities would also remain constant using the relation $$v=r \omega$$.
Thus at each point their velocities change only in direction. Thus the momentum of particle A is $$P_{A}= m_A v_A $$ i.e. $$2\times2 = 4 kgm/s$$ and that of B is $$P_{B}= m_B v_B $$ i.e. $$1\times3 = 3 kgm/s$$.
We will get their momentum as a maximum when the particles travel parallel to each other in the same direction.
Thus we get the maximum value of their momentum simply as the sum of their individual momentums i.e. $$ |P_A +P_B| = 4-3 = 1 kgm/s$$ at all points.
Question 10
1 / -0

An automobile is started from rest with one of its doors initially at right angles. lf the hinges of the door are toward the front of the car, the door will shut as the automobile picks up speed. The acceleration a is constant and the centre of mass is at a distance d from the hinges. The time $$T$$ needed for the doors to close is given by:
$$\left [ Given:\displaystyle\int_{0}^{\frac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=N \right ]$$(Consider the door as a square plate)

A
$$\mathrm{T}=(\sqrt{\dfrac{3\mathrm{d}}{2\mathrm{a}}}) \mathrm{N}$$

B
$$\mathrm{T}=(\sqrt{\dfrac{\mathrm{d}}{3\mathrm{a}}}) \mathrm{N}$$

C
$$\mathrm{T}=(\sqrt{\dfrac{2\mathrm{d}}{\mathrm{a}}}) \mathrm{N}$$

D
$$\mathrm{T}=(\sqrt{\dfrac{2\mathrm{d}}{3\mathrm{a}}}) \mathrm{N}$$

Solution

Equation of motion is
$$I\alpha =Ma(dcos\theta )$$
Now moment of inertia of the door about the hinges is given as
$$I=\dfrac{Md^{2}}{3}+Md^{2}$$
$$=M\left ( \dfrac{4}{3}d^{2} \right )$$
Hence,
$$\dfrac{4}{3}Md^{2}\alpha =Ma dcos\theta $$

$$\Rightarrow \dfrac{4}{3}d\times\omega \dfrac{d\omega }{d\theta }=acos\theta \left ( \because \alpha =\omega \dfrac{d\omega }{d\theta } \right )$$

$$\Rightarrow \dfrac{4}{3}d\int \omega d\omega =\int acos\theta d\theta $$

$$\Rightarrow \dfrac{4}{3}d\dfrac{\omega ^{2}}{2}=asin\theta $$

$$\omega ^{2}=\dfrac{3a}{2d}sin\theta +C$$

where, C$$=$$constant of integration

At $$\theta =0$$ (initially), we have $$\theta =0$$
$$\Rightarrow C=0$$
$$\Rightarrow \omega^{2}=\dfrac{3a}{2d}sin\theta $$

$$\Rightarrow \omega =\sqrt{\dfrac{3a}{2d}sin\theta }$$

$$\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}dt$$ and integrating between $$t=0$$ and $$t=T$$
(corresponding to $$\theta =0$$ and $$\theta =\dfrac{\pi }{2}$$)

$$\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}.T$$

$$\Rightarrow T=\left ( \sqrt{\dfrac{2d}{3a}} \right )N$$

$$\left ( since,given:\displaystyle N=\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }} \right )$$