Laws of Motion Test

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Laws of Motion Test - 33

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Question 1
1 / -0

A body of mass $$2kg$$ is lying on a floor. The coefficient of static friction is $$0.54$$. What will be the value of frictional force if the force is $$2.8N$$ and $$g=10{ ms }^{ -2 }$$?

A
$$zero$$

B
$$2N$$

C
$$2.8N$$

D
$$8N$$

Solution

Now the maximum possible value of friction is $$N\mu$$.

$$N=2\ g=20$$ Newtons

so friction $$=20\times 0.54$$

$$=10.8$$ Newtons

Since this force is less than the applied force of $$2.8$$ Newtons
and as friction is a dynamic force it will adjust itself and hence the
friction force will be equal to $$f=2.8$$ Newtons
Question 2
1 / -0

A balloon has $$5g$$ of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of $$4cm/s$$. If the balloon shrinks completely in $$2.5$$ seconds, then the mean force acting on the balloon is:

A
$$2dyne$$

B
$$50dyne$$

C
$$8dyne$$

D
$$8N$$

Solution

$$Mean force = \dfrac {change\ in\ momentum}{time}$$
Units are given in CGS unit, Hence force will be in $$dyne$$
hence $$Mean\ force = \dfrac{5\times 4}{2.5} = 8\ dyne$$
Question 3
1 / -0

A body of mass $$1\ kg$$ is rotating in a vertical circle of radius $$1\ m$$. What will be the difference in its kinetic energy at the top and bottom of the circle?
Take $$g = 10\ m/s^{2}$$

A
$$50\ J$$

B
$$30\ J$$

C
$$20\ J$$

D
$$10\ J$$

Solution

According to work energy theorem, $$\Delta K.E.=W$$ and here work is done by the gravitational force.
$$\Rightarrow \Delta K.E.=W = mg\ (2r) = 1 \times 10 \times 2(1)=20\ J$$
Question 4
1 / -0

Let $$\theta$$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, then tension in the string is mg $$\cos\theta$$

A
always

B
never

C
at the extreme position

D
at the mean position

Solution

For simple pendulum

$$\dfrac { m{ v }^{ 2 } }{ r } =T-mg\cos { \phi } $$

But when $$\phi =\theta $$, i.e., the bob is at extreme position. its velocity is zero, hence the equation becomes: $$T-mg\cos { \theta }=0$$

$$\Rightarrow \quad T = mg\cos { \theta } $$
Question 5
1 / -0

The velocity of a particle at highest point of the vertical circle is $$\sqrt{3rg}$$. The tension at the lowest point, if mass of the particle is $$m$$, is

A
$$2\ mg$$

B
$$4\ mg$$

C
$$6\ mg$$

D
$$8\ mg$$

Solution

Let the velocity at lowest point be $$v$$. then according to law of conservation of energy.
energy at lowest point= energy at highest point

$$\Rightarrow \dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}m(\sqrt{3rg})^2 \\ \Rightarrow v=\sqrt{7rg}$$

At lowest point,

$$\dfrac { m{ v }^{ 2 } }{ r } =T-mg \\ \Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{7rg} )^{ 2 } }{ r }$$

$$\Rightarrow T=mg+7mg=8mg $$
Question 6
1 / -0

A stone of mass $$1\ kg$$ is tied of the end of a string $$1\ m$$ long. It is whirled in a vertical circle. If the velocity of stone at the top is $$4\ m/s$$. What is the tension in the string at the lowest point?
Take $$g = 10\ m/s^{2}$$

A
$$6\ N$$

B
$$66\ N$$

C
$$5.2\ N$$

D
$$76\ N$$

Solution

Let the velocity at lowest point is $$v$$ and at highest point be $$u$$, then according to law of conservation of energy.
energy at lowest point= energy at highest point.

$$\Rightarrow\dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}mu^2 \\ \Rightarrow

v=\sqrt{u^2+4rg}=\sqrt{4^2+(4\times 1 \times 10)}=\sqrt{56}$$
At lowest point,

$$\dfrac { m{ v }^{ 2 } }{ r } =T-mg$$

$$\Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{56} )^{ 2 } }{ r }$$ $$\quad$$ $$(r=1)$$

$$\Rightarrow T=mg+56\ m=66 \times m =66 \times 1 = 66\ N $$.
Question 7
1 / -0

For traffic moving at $$60 km/hr$$ along a smooth circular track of radius $$0.1 km$$, the correct angle of banking should be:

A
$$\displaystyle tan^{-1} \left [ \frac{(100 \times 9.8)}{\left ( \frac{50}{3} \right )^2} \right ]$$

B
$$\displaystyle tan^{-1} \left [ \frac{\left ( \frac{50}{3} \right )^2}{(100 \times 9.8)} \right ]$$

C
$$tan^{-1} \displaystyle \left ( \frac{60^2}{0.1} \right )$$

D
$$tan^{-1} \sqrt{(60 \times 0.1 \times 9.8)}$$

Solution

Relation between banking angle, speed and radius of circular track is given by
$$\theta =\tan ^{ -1 }{ \left[ \dfrac { v^{ 2 } }{ rg } \right] } $$
here
$$v=60 \mathrm { km/hr}=60 \times \dfrac{5}{18} \mathrm{ m/s}=\dfrac{50}{3} \mathrm{ m/s}$$
and $$r=0.1 \mathrm{km}=100\mathrm{m}$$
$$\Rightarrow \theta =\tan ^{ -1 }{ \left[ \dfrac { \left( \dfrac { 50 }{ 3 } \right) ^{ 2 } }{ (100\times 9.8) } \right] } $$
Question 8
1 / -0

If a particle of mass m moving with velocity $$v_1$$ is subject to an impulse $$I$$ which produces a final velocity $$v_2$$, then $$I$$ is given by :

A
$$m(v_1 - v_2)$$

B
$$m(v_1 + v_2)$$

C
$$m(v_2 - v_1)$$

D
$$\dfrac{m\left(v_2^2 - v_1^2\right)}{2}$$

Solution

Impulse is equal to the change in momentum caused by the force.
$$I=m\times v_{2} - m\times v_{1}$$
Question 9
1 / -0

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle.
There is a change in:

A
momentum of the rocket

B
mass of the rocket

C
both A & B

D
None of the above

Solution

As the rocket is moving upward by ejecting gases, its mass is getting decreased gradually. As momentum is the product of the mass and velocity that is $$p = m\times v$$. Hence momentum decreases as mass is decreasing.
Question 10
1 / -0

A cricket ball of mass 500 gm strikes a bat normally with a velocity 30 m/s and rebounds with a velocity 20 m/s in the opposite direction. The impulse of the force exerted by the ball on the bat is:

A
0.5 N.s

B
1.0 N.s

C
25 N.s

D
50 N.s

Solution

Impulse is given by:
$$I=m\Delta v=0.5\times(20-(-30))=0.5\times50$$
$$I=25 Ns$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 33

Result

Laws of Motion Test - 33

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SHARING IS CARING

Question 1

$$zero$$

$$2N$$

$$2.8N$$

$$8N$$

Solution

Question 2

$$2dyne$$

$$50dyne$$

$$8dyne$$

$$8N$$

Solution

Question 3

$$50\ J$$

$$30\ J$$

$$20\ J$$

$$10\ J$$

Solution

Question 4

always

never

at the extreme position

at the mean position

Solution

Question 5

$$2\ mg$$

$$4\ mg$$

$$6\ mg$$

$$8\ mg$$

Solution

Question 6

$$6\ N$$

$$66\ N$$

$$5.2\ N$$

$$76\ N$$

Solution

Question 7

$$\displaystyle tan^{-1} \left [ \frac{(100 \times 9.8)}{\left ( \frac{50}{3} \right )^2} \right ]$$

$$\displaystyle tan^{-1} \left [ \frac{\left ( \frac{50}{3} \right )^2}{(100 \times 9.8)} \right ]$$

$$tan^{-1} \displaystyle \left ( \frac{60^2}{0.1} \right )$$

$$tan^{-1} \sqrt{(60 \times 0.1 \times 9.8)}$$

Solution

Question 8

$$m(v_1 - v_2)$$

$$m(v_1 + v_2)$$

$$m(v_2 - v_1)$$

$$\dfrac{m\left(v_2^2 - v_1^2\right)}{2}$$

Solution

Question 9

momentum of the rocket

mass of the rocket

both A & B

None of the above

Solution

Question 10

0.5 N.s

1.0 N.s

25 N.s

50 N.s

Solution

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