Laws of Motion Test

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Laws of Motion Test - 32

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Question 1
1 / -0

A circular track of radius $$100$$ m is designed for an average speed $$54$$ km/h. Find the angle of banking.

A
$$ \tan ^{ -1 }{ \left( \dfrac { 3 }{ 20 } \right) } $$

B
$$ \tan ^{ -1 }{ \left( \dfrac { 9 }{ 40 } \right) } $$

C
$$ \tan ^{ -1 }{ \left( \dfrac { 3 }{ 10 } \right) } $$

D
None of these

Solution

$$ \tan { \theta } = \dfrac { { v }^{ 2 } }{ rg } $$

$$= \dfrac { 15\times 15 }{ 100\times 10 } =\dfrac { 9 }{ 40 } $$

$$ \theta =\tan ^{ -1 }{ (\dfrac { 9 }{ 40 } ) } $$
Question 2
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A force of 10 N acts on a body for 3 microsecond $$(\mu s)$$. Calculate the impulse. If mass of the body is 5 g, calculate the change of velocity.

A
$$6 \times 10^{-3} m s^{-1}$$

B
$$3 \times 10^{-3} m s^{-1}$$

C
$$8 \times 10^{-3} m s^{-1}$$

D
$$16 \times 10^{-3} m s^{-1}$$

Solution

Here, $$F=10 N, t=3 \mu s$$
$$=3 \times 10^{-6}s$$;
$$M =5 g = 5 \times 10^{-3} kg$$
Now, impulse $$= Ft = 10 \times 3 \times 10^{-6}$$
$$= 3 \times 10^{-5} Ns$$
Also, impulse $$=$$ change in momentum
$$=Mv_2 - Mv_1$$
$$= M(v_2-v_1)$$
$$\therefore (v_2 - v_1) = \displaystyle \frac{impulse}{M}=\frac{3 \times 10^{-5}}{5 \times 10^{-3}}$$
$$=6 \times 10^{-3} m s^{-1}$$
Question 3
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Which of the following must be true for the sum of the magnitude of the momenta of the individual particles in the system?

A
It must be zero

B
It could be non-zero, but it must be a constant

C
It could be non zero, and it might not be a constant

D
It could be zero, even if the magnitude of the total momentum is not zero

Solution

The sum of the magnitude of the momenta of the individual particles in the system is equal to the total mass times the velocity of centre of mass.
Velocity of COM depends upon the net external force, and as nothing is given about it we can't tell anything strictly about it.
Hence It could be non zero, and it might not be a constant
Question 4
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Water in a bucket is whirled in a vertical circle with a string to it. The water does not fallen even when the bucket is inverted at the top of its path . We conclude that in this position:

A
$$\displaystyle mg = \frac{mv^2}{r}$$

B
$$mg$$ is greater than $$\displaystyle \frac{mv^2}{r}$$

C
$$mg$$ is not greater than $$\displaystyle \frac{mv^2}{r}$$

D
$$mg$$ is not less than $$\displaystyle \frac{mv^2}{r}$$

Solution

Its because a centrifugal force $$\dfrac { m{ v }^{ 2 } }{ r } $$ acts on the water which is away from the center and opposite to force of gravity $$mg$$. If the $$mg$$ is greater then $$\dfrac { m{ v }^{ 2 } }{ r } $$ water will spill out. but if $$\dfrac { m{ v }^{ 2 } }{ r } $$ is greater then $$mg$$ water will not spill out of the bucket.

Note: for keeping the water in bucket while rotating it in a vertical circle you will have to keep the velocity above a certain value to make $$\dfrac { m{ v }^{ 2 } }{ r } $$ greater then $$mg$$.
Question 5
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A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $$\overrightarrow{a}$$ is correctly shown in

A

B

C

D

Solution

The bob has both radial as well as tangential acceleration when it is at a displacement less than its maximum displacement.
From figure $$a_t =g sin\theta$$and $$a_r = \dfrac{T}{m} - g cos\theta$$
Thus resultant acceleration $$\vec{a} = \vec{a_r} + \vec{a_t}$$ points in the direction as shown in the figure.
Question 6
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The magnitude of force (in $$N$$) acting on a body varies with time $$t$$ (in $$\mu s)$$ as shown. $$AB, BC$$ and $$CD$$ are straight line segments. The magnitude of total impulse of force on the body from $$t=4\mu s$$ to $$t=16\mu s$$ is :

A
$$ { 10 }^{ -3 }Ns$$

B
$$ { 10 }^{ -2 }Ns$$

C
$$ { 10 }^{ -4 }Ns$$

D
$$5\times { 10 }^{ -4 }Ns$$

Solution

The magnitude of impulse is area under the curve of $$F-t$$ diagram

$$I= $$ area from $$4 $$ to $$6 \mu s +$$ area from $$6 $$ to $$16 \mu s $$

$$I= \dfrac{1}{2}(200+800) \times (6-4)+\dfrac{1}{2} \times 800 \times (16-6)$$

$$I=1000+4000 N\mu s$$

$$I=5 \times 10^{-3} Ns$$
Question 7
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A stone tied to string of length $$l$$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed $$u$$. The magnitude of the change in velocity as it reaches a position, where the string is horizontal is

A
$$\sqrt{u^2-2gl}$$

B
$$\sqrt{2gl}$$

C
$$\sqrt{u^2-gl}$$

D
$$\sqrt{2(u^2-gl)}$$
Question 8
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A stone of mass $$1000g$$ tied to a light string of length $$10/3m$$ is whirling in a vertical circle. If the ratio of the maximum tension to minimum tension is $$4$$ and $$g=10{ ms }^{ -2 }$$, then the speed of stone at the highest point of circle is :

A
$$20{ ms }^{ -1 }$$

B
$$10\sqrt { 3 } { ms }^{ -1 }$$

C
$$5\sqrt { 3 } { ms }^{ -1 }$$

D
$$10{ ms }^{ -1 }$$

Solution

Minimum tension is at topmost point of the circle
Maximum tension is bottom point of the circle.
$$ \dfrac{Max. Tension}{Min. Tension} = \dfrac { \dfrac { m{ { v }_{ b }^{ 2 } } }{ r } +mg }{ \dfrac { m{ v }_{ t }^{ 2 } }{ r } -mg } =4 $$
and
Conservation of energy is :
$$ \dfrac { { v }_{ b }^{ 2 }-{ v }_{ t }^{ 2 } }{ 2 } =2gr $$
Using both equations,
$$ { v }_{ t }^{ 2 }=3gr$$
So
$$ {v}_{t} = \sqrt { 3gr } = 10\ m/s$$
Question 9
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A particle of mass, $$m$$, is tied to a light string and rotated with a speed, $$v$$, along a circular path of radius, $$r$$. If $$T=$$ tension in the string and $$mg =$$ gravitational force on the particle, then the actual forces acting on the particle are

A
$$mg$$ and $$T$$ only.

B
$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed inwards.

C
$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed outwards.

D
Only a force $$mv^2/r$$ directed outwards.

Solution

The force $$mv^2/r$$ directed outwards, called centrifugal force, is not a real force.
At A, $$mv_1^2/l = T_1 +mg$$
The resultant of $$mg$$ and tension ($$T$$) will provide the centripetal force.
Question 10
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A stone of mass $$1\ kg$$ tied to a light inextensible string of length. $$L = \dfrac{10}{3}$$ metre is whirling in a circular path of radius, $$L$$, in a vertical plane. If the ratio of maximum tension to the minimum tension is $$4$$ and if $$g$$ is taken to be $$10\ m/s^{2}$$, the speed of the stone at the highest point of circle is :

A
$$20\ m/s$$

B
$$10 \sqrt{3}\ m/s$$

C
$$10\ m/s$$

D
$$5\sqrt{2}\ m/s$$

Solution

When the stone is revolved in a vertical circle, tension is maximum at bottom and minimum at top. So we have, Tension at bottom as $$T_B$$ and Tension at top as $$T_T$$.

We know, $$\dfrac { { T }_{ B } }{ { T }_{ T } } =4$$

$$\Rightarrow \dfrac { \dfrac { mv_{ B }^{ 2 } }{ L } +mg }{ \dfrac { mv_{ T }^{ 2 } }{ L } -mg } =4$$

$$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } +mg=4\left( \dfrac { mv_{ T }^{ 2 } }{ L } -mg \right)$$

$$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } =\dfrac { 4mv_{ T }^{ 2 } }{ L } -5mg$$

$$\Rightarrow mv_{ B }^{ 2 } = 4mv_{ T }^{ 2 }-5mgL$$ ......$$(I)$$

Now using the law of conservation of energy, at top and bottom of the vertical circle, we have:

$$\dfrac { 1 }{ 2 } mv_{ T }^{ 2 }\quad +2mgL=\frac { 1 }{ 2 } mv_{ B }^{ 2 }$$

$$\Rightarrow m{ { v }_{ B } }^{ 2 }=m{ { v }_{ T } }^{ 2 }+4mgL$$ ......$$(II)$$

from $$(I)$$ and $$(II)$$, we have

$$4mv_{ T }^{ 2 }-5mgL=m{ { v }_{ T } }^{ 2 }+4mgL\\ \Rightarrow 3mv_{ T }^{ 2 }=9mgL\\ \Rightarrow { v }_{ T }=\sqrt { 3gL } \\ \Rightarrow { v }_{ T }=\sqrt { 3\times 10\times \left( \dfrac { 10 }{ 3 } \right) } =10\ m/s$$