Laws of Motion Test

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Laws of Motion Test - 48

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Question 1
1 / -0

A railway track is banked for a speed v, by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. The radius of curvature of the track is r. Then,

A
$$\dfrac{h}{d}=\dfrac{v^2}{rg}$$

B
$$tan(sin^{-1}\dfrac{h}{d})=\dfrac{v^2}{rg}$$

C
$$tan^{-1}(\dfrac{h}{d})=\dfrac{v^2}{rg}$$

D
$$\dfrac{h}{r}=\dfrac{v^2}{dg}$$

Solution

$$ Ncos\theta = mg $$
$$ N sin \theta = mv^2 / r $$
$$ tan \theta = v^2 / rg $$
for train, AB= d and BB = h
$$ sin \theta = h/d $$
$$ \theta = sin ^{-1}(h/d) $$
$$ tan \theta = tan (sin ^{-1 }h/d)= v^2 / rg$$
So, option B is correct.
Question 2
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A particle is tied to one end of a light inextensible string and is moved in a vertical circle, the other end of the string is fixed at the centre. Then for a complete motion in a circle, which is correct.
(air resistance is negligible).

A
Acceleration of the particle is directed towards the centre

B
Total mechanical energy of the particle and earth remains constant

C
Tension in the string remains constant

D
Acceleration of the particle remains constant

Solution

At any time force acting on particle vary and hence acceleration (net) will have different direction at different times. Tension also changes and its minimum at top point. Magnitude of acceleration also varies.
Considering earth and particle as a system and no external force on system is acting, total mechanical energy will be conserved.
Question 3
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Two identical balls $$A$$ and $$B$$ are connected with massless rod $$AB$$ and moves in a smooth circular track of radius $$R. OB$$ is horizontal and $$PQ = R\left (1 - \dfrac {\sqrt {3}}{2}\right )$$. Find the equation by which maximum value of $$\theta$$ can be obtained, where $$\theta$$ is the angle that new position of ball $$B$$ makes at $$O$$ with horizontal.

A
$$\sin \theta \left (\dfrac {2 + \sqrt {3}}{2}\right ) - \dfrac {1}{2} \cos \theta = \dfrac {\sqrt {3}}{2}$$

B
$$\cos \theta \left (\dfrac {2 + \sqrt {3}}{2}\right ) - \dfrac {1}{2} \sin \theta = \dfrac {\sqrt {3}}{2}$$

C
$$\sin \theta \left (\dfrac {2 - \sqrt {3}}{2}\right ) - \dfrac {1}{2} \cos \theta = \dfrac {\sqrt {3}}{2}$$

D
$$\sin \theta \left (\dfrac {2 - \sqrt {3}}{2}\right ) - \dfrac {1}{2} \cos \theta = \dfrac {1}{2}$$
Question 4
1 / -0

"To every action, there is equal and opposite reaction". It is Newton's

A
First Law

B
Second Law

C
Third Law of Motion

D
None of these

Solution
Question 5
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A simple pendulum of length $$L$$ carries a bob of mass $$m$$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal the net force on the bob is:

A
$$\sqrt { 10 } mg$$

B
$$\sqrt { 5 } mg$$

C
$$4mg$$

D
$$1mg$$

Solution

V=minimum horizontal speed= $$\sqrt { 5Lg } $$
using energy conservation,
$$\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL\\ \therefore \quad \frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ v }^{ 2 }-mgL\\ =\frac { 1 }{ 2 } m{ 5gL }-mgL\\ \therefore \quad { v }^{ 2 }=\frac { 3gL }{ 2 } \times 2=3gL\\ T=\frac { m{ v }^{ 2 } }{ L } =3mg\\ \therefore \quad force=\quad \sqrt { { \left( 3mg \right) }^{ 2 }+{ \left( mg \right) }^{ 2 } } \quad \\ =\sqrt { 10 } mg$$
Question 6
1 / -0

A ball of mass m strikes a rigid wall with speed u and rebounds with the same speed. The impulse imparted to the ball by the wall is:

A
2mu

B
mu

C
Zero

D
-2mu

Solution

The situation is as shown in the figure.
$$p_{initial} = mu, p_{final}=-mu$$
Impulse imparted to the ball = change in momentum
= $$p_{initial} = mu, p_{final}=-mu-mu=-2mu$$
Question 7
1 / -0

A car is moving on a frictionless (in the radial direction) circular banked road, with a banking angle of $$15^o$$, find the velocity of the car.
Mass of car is $$600\ kg$$, the radius of the circular road $$=10m$$
Assume $$g=9.81\ m/s^2$$

A
4.26 m/s

B
5.12 m/s

C
51.2 m/s

D
42.6 m/s

Solution

We know that the force acting on the car = mg$$tan(\theta)$$
$$\Rightarrow \dfrac{mv^2}{r}=mg\ tan(\theta)$$
$$\Rightarrow v^2=rg\ tan(\theta)$$
$$\Rightarrow v^2=10\times 9.81 \times 0.268$$
$$\Rightarrow v=5.12\ m/s$$
Question 8
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A bullet of mass 40 g moving with a speed of $$90 ms^{-1}$$ enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistance force exerted by the block on the bullet is

A
180 N

B
220 N

C
270 N

D
320 N

Solution

Here, u=90 $$ms^{-1}$$, v= 0

m=40gm=$$\dfrac{40}{1000} kg$$= 0.04 kg, s=60 cm= 0.6 m

Using $$v^2-u^2 = 2as$$

$$\therefore (0)^2-(90)^2=2a\times 0.6$$

$$a=\dfrac{(90)^2}{2\times0.6}= -6750ms^{-2}$$

-ve sign shows the retradation.$$\therefore$$ The average resistive force exerted by block on the bullet is
$$F=m\times a= (0.04 kg)(6750 ms^{-2})= 270 N$$
Question 9
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If a car is moving at a velocity $$v$$ in a circular road with coefficient of friction $$\mu$$ and banking angle $$\theta$$. Find the radius of the road such that friction force doesn't act on the car in the radial direction.

A
$$r=\dfrac{v^2}{g\tan \theta}$$

B
$$r = \dfrac{\sqrt{v}}{g\tan \theta}$$

C
$$r = \sqrt{\dfrac{v^2}{g\tan \theta}}$$

D
$$r=\dfrac{v^2}{\sqrt{g}}$$

Solution

For the force of friction to be zero, the centripetal force has to be provided by the normal reaction force only. Which means $$N\ sin(\theta)=\dfrac{mv^2}{r}$$ .....................$$(1)$$
now the other component of the normal reaction $$N\ cos(\theta)$$ balances the weight of the car.
$$\Rightarrow N\ cos(\theta)=mg$$ ....................$$(2)$$

Using this in $$(1)$$ we get
$$mg\ tan(\theta)=\dfrac{mv^2}{r}$$
$$\Rightarrow r=\dfrac{v^2}{g\ tan(\theta)}$$
Question 10
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The velocity of a body moving in a vertical circle of radius r is $$\sqrt{7gr}$$ at the lowest point of the circle. What is the ratio of maximum and minimum tension?

A
$$4: 1$$

B
$$\sqrt 7:1$$

C
$$3: 1$$

D
$$2: 1$$

Solution

Since $$V_{b} >\sqrt{5 gr}\rightarrow$$ It will complete vertical circular
$$\downarrow$$
velocity at bottom
Thus $$T_{bottom}-T_{top}=6\ mg$$
$$T_{max}-T_{min}=6\ mg ...... (1)$$
Now,
$$\Rightarrow T_{max}-mg=m(a_{c})$$
$$T_{max}-mg =\dfrac{mv^{2}}{r}$$
$$T_{max}=\dfrac{m}{r}(7 gr)+mg$$
$$T_{max}=8\ mg$$
From $$(1)$$
$$T_{min}=2\ mg$$
Thus $$\dfrac{T_{max}}{T_{min}}=4:1$$