Laws of Motion Test

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Laws of Motion Test - 47

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Question 1
1 / -0

A simple pendulum with bob of mass $$m$$ and length $$x$$ is held in position at an angle $$1$$ and then angle $$2$$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $${v}_{1}$$ and $${v}_{2}$$ respectively. Then, $$\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } $$ is

A
$$\cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } $$

B
$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } } $$

C
$$\sqrt { \cfrac { 2gx(1-\cos { { \theta }_{ 1 } } ) }{ 1-\cos { { \theta }_{ 2 } } } } $$

D
$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 2gx(1-\cos { { \theta }_{ 2 } } ) } } $$

Solution

A simple pendulum with bob of mass m and length x is held in position at an angle 1 and then angle 2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are $$v_1$$ and $$v_2$$ respectively.
Kinetic Energy = loss in PE
$$m v_1^2 /2 = m g L (1 - cos θ_1)$$
So,
Similarly $$m v_2^2 /2 = m g L (1 - cos θ_2)$$

Ratio $$\dfrac{v_1}{ v_2} =\sqrt{\dfrac{ (1 - cos θ_1)}{ (1 - cos θ_1)}}$$ .
Question 2
1 / -0

A curved road of diameter $$1.8\ km$$ is banked so that no friction is required at a speed of $$30\ m/s$$. What is the banking angle?

A
$$6^{\circ}$$

B
$$16^{\circ}$$

C
$$26^{\circ}$$

D
$$0.6^{\circ}$$

Solution

Radius of curved road $$r = \dfrac{D}{2} = \dfrac{1.8 \ km}{2} = 900 \ m$$
Speed $$v = 30 \ m/s$$
So, banking angle $$\tan \theta =\dfrac{v^2}{rg} $$
$$\therefore$$ $$\tan\theta =\dfrac{(30)^2}{900\times 9.8} = 0.1$$
$$\implies \ \theta = tan^{-1}(0.1) = 6^o$$
Question 3
1 / -0

A particle of mass m strikes a wall with speed v at an angle $$30^0$$ with the wall elastically as shown in the figure. The magnitude of impulse imparted to the ball by the wall is :

A
mv

B
$$\dfrac{mv}{2}$$

C
2mv

D
$$\sqrt{3}$$ mv

Solution

Momentum changing to normal = impulse.
Impulse imparted $$= 2mv \sin 30°$$
=mv
Question 4
1 / -0

A bullet is fired from a gun. The force on the bullet is given by $$F = 600 - 2\times 10^{5}t$$, where $$F$$ is in newtons and $$t$$ in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

A
$$9\ Ns$$

B
Zero

C
$$0.9\ Ns$$

D
$$1.8\ Ns$$

Solution

Force at which time zero u
$$F=0=600-2\times { 10 }^{ 5 }t\\ t=\dfrac { 600 }{ 2\times { 10 }^{ 5 } } =3\times { 10 }^{ -3 }s$$
Impulse imparted $$=\int _{ 0 }^{ t }{ Fdt } $$
$$=\int _{ 0 }^{ t }{ (600-2\times { 10 }^{ 5 } } t)dt\\ =\left[ 600t-1\times { 10 }^{ 5 }{ t }^{ 2 } \right] _{ 0 }^{ 3\times { 10 }^{ -3 } }\\ =1.8-0.9\\ =0.9Ns$$
Question 5
1 / -0

If n balls hit elastically and normally on a surface per unit area and all units has mass m and are moving with same velocity u, then force on surface is:

A
$$mun$$

B
$$2mun$$

C
$$\dfrac{1}{2}mu^2n$$

D
$$mu^2n$$

Solution

Impulse acting per second = 2 n m u
F.T = 2 n m u
F = 2 n m u
Question 6
1 / -0

A motorcyclist rides in vertical circle in a hollow sphere of radius 5 m. Find the minimum speed required, so that he does not have any contact with the sphere at the highest point .

A
$$4.6 \ m/s$$

B
$$7 \ m/s$$

C
$$7.5 \ m/s$$

D
$$8 \ m/s$$

Solution

Given,

$$r=5m$$

$$g=9.8m/s^2$$

The minimum speed , $$v=\sqrt{rg}$$

$$v=\sqrt{5\times 9.8}$$

$$v=7m/s$$
The correct option is B.
Question 7
1 / -0

A car sometimes overturns while taking a turn. When it overturns, it is

A
The inner wheel which leaves the ground first

B
The outer wheel which leaves the ground first

C
Both the wheels leaves the ground simultaneously

D
Either wheel which leaves the ground first

Solution

While turning the maximum allowed speed is:
$$V_{max}=\sqrt{\mu rg}$$
where,
$$\mu$$ is coefficient of friction,
$$r$$ is radius of the turn and
$$g$$ acceleration due to gravity.
For the inner wheel the radius of the turn is less than the outer wheel and therefore its maximum speed is also less than for the outer wheel. Thus the inner wheel leaves the ground first.
Question 8
1 / -0

A particle is projected so as to just move along a vertical circle of radius $$r$$ with the help of the massless string. The ratio of the tension in the string when the particle is at the lowest and highest point on the circle is?

A
$$1$$

B
Finite but large

C
Zero

D
Infinite

Solution

$$T-mg\cos\theta=\cfrac{mV^2}{r}$$
At lowest point $$A$$:
$$T_A=\cfrac{mV_A^2}{r}+mg\cos (0^o)$$
$$\implies T_A=\cfrac{mV_A^2}{r}+mg$$
At highest point $$B$$:
$$T_B=0$$
$$\therefore \cfrac{T_A}{T_B}=\cfrac{({mV_A^2/r})+mg}{0}$$
$$\implies \cfrac{T_A}{T_B}=\infty$$
Question 9
1 / -0

A solid sphere is placed on a smooth horizontal plane. A horizontal impluse $$I$$ is applied at a distance $$h$$ above the central line as shown in the figure. Soon after giving the impulse the sphere starts rolling.
The ratio $$h/R$$ would be

A
$$\dfrac {1}{2}$$

B
$$\dfrac {2}{5}$$

C
$$\dfrac {1}{4}$$

D
$$\dfrac {1}{5}$$

Solution

After application of impulse let velocity $$=v$$
$$I=mv\Rightarrow v=I/m$$
Angular impulse about C.M
$$gmv=I\omega\\ \Rightarrow hmv=\cfrac{2}{5}mR^2\omega\Rightarrow\omega=\cfrac{5}{2}\cfrac{hv}{R^2}$$
For pure rolling
$$v=r\omega\Rightarrow \omega=v/r\Rightarrow \cfrac{v}{R}=\cfrac{5}{2}\cfrac{hv}{R^2}\Rightarrow\cfrac{h}{R}=\cfrac{2}{5}$$
Question 10
1 / -0

A $$1$$m long uniform beam is being balanced as shown in the given figure. What are the forces X and Y?

A
Force X-$$1$$N, Force Y-$$0$$N

B
Force X-$$2$$N, Force Y-$$1$$N

C
Force X-$$3$$N, Force Y-$$3$$N

D
Force X-$$4$$N, Force Y-$$7$$N

Solution