Laws of Motion Test

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Laws of Motion Test - 55

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Question 1
1 / -0

The length of simple pendulum is 1 m and mass of its bob is 50 g. The bob is given sufficient velocity so that the bob describe vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is.

A
0.98 J

B
1.96 J

C
4.9 J

D
9.8 J

Solution

A bob of mass $$m = 0.05\ Kg$$ is performing vertical circle of radius $$(r) = 1m$$.
In the boundary case.
a minimum velocity of $$\mu = \sqrt{5gr}$$ has to provided to it
so it completes a vertical circle.
So, kinetic energy $$(K_1)$$ at the lowest point,
$$K_1 = \dfrac{1}{2}m\mu^2 = \dfrac{5}{2}mgr$$
and $$PE = 0$$, raking it as reference at highest point
$$PE : \mu_2 =mgh = 2mgr$$ $$(as \, h = 2r)$$
From conservation of energy.
$$K_1 + \mu_1= K_2+\mu_2 \Rightarrow K_2 = K_1 - \mu_2 = \dfrac{1}{2}mgr$$
Thus, loss in $$KE$$:
$$\Delta K = K_1-K_2 = 2mgr = 2\times 0.05 \times 9.8 \times 1 = 0.98\ J$$ (Ans)
Question 2
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A car moves around a curved road of radius $$R_1$$ at constant speed v without sliding. If we double the car's speed, what is the least radius that would now keep the car safe from sliding?

A
$$2R_1$$

B
$$4R_1$$

C
$$6R_1$$

D
None of these
Question 3
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A force of $$25$$N acts on a body at rest for $$0.2$$s and a force of $$70$$N acts for the next $$0.1$$s in opposite direction. If the final velocity of the body is $$5ms^{-1}$$, the mass of the body is:

A
$$1$$ kg

B
$$2$$ kg

C
$$0.8$$ kg

D
$$0.4$$ kg

Solution

Resulatant impule = Change in momentum
$$7N\cdot s-5N\cdot s=2N\cdot s$$
$$mv=2N\cdot s$$
$$m=\dfrac { 2 }{ 5 } kg$$
Question 4
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Block $$A$$ of mass $$4kg$$ is to be kept at rest against a smooth vertical wall by applying a force $$F$$ as shown in fig. The force required is $$\left( g = 10 \mathrm { m } / \mathrm { s } ^ { 2 } \right)$$

A
$$40 \sqrt { 2 } N$$

B
$$20 \sqrt { 2 } N$$

C
$$10 \sqrt { 2 } N$$

D
$$15 \sqrt { 2 } N$$

Solution

Hence (A) is correct option.
Question 5
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What is the impulse of force shown in the following fingre?

A
205 Ns

B
450 Ns

C
800 Ns

D
1000 Ns

Solution

Impulse, $$I=\int{F(t).dt}=$$ Area of F and t graph.

Where,

$$ F=\,force $$

$$ t=\,time $$

$$I=\dfrac{1}{2}Ft=\dfrac{1}{2}\times 100\times \left( 10-1 \right)=450\,Ns$$

Hence, impulse is $$450\,Ns$$
Question 6
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A body of mass tied at the end of a strong of length $$l$$ is projected with velocity $$\sqrt{4lg}$$, at what height will it leave the circular path:

A
$$\dfrac{5}{3}l$$

B
$$\dfrac{3}{5}l$$

C
$$\dfrac{1}{3}l$$

D
$$\dfrac{2}{3}l$$

Solution

REF.Image
In a vertical circle the tension at the string is given by :-
$$T= \dfrac{mu^{2}}{l}-mg(2-3 cos \theta )$$
Here $$u^{2}= 4gl$$, and particle stop circular motion
when $$T=0$$
$$\Rightarrow T=0\Rightarrow \dfrac{mu^{2}}{L}= mg(2-3 cos\theta )$$
$$\Rightarrow \dfrac{m.4gL}{L}= mg(2-3 cos \theta )\Rightarrow cos\theta = -2/3$$
$$\therefore $$ Height, $$h=l\left | cos\theta \right |=\frac{2}{3}l$$
Hence it attains maximum height,
$$H= h+L= \dfrac{5}{3}L$$ (Ans)
Question 7
1 / -0

A small block slides with velocity $$v _ { 0 } = 0.5 \sqrt { g r }$$ on the horizontal friction less surface as shown in the figure. The block leaves the surface at point C. The angle $$\theta$$ in the figure is:

A
$$\cos ^ { - 1 } \dfrac { 4 } { 9 }$$

B
$$\cos ^ { - 1 } \dfrac { 3 } { 4 }$$

C
$$\cos ^ { - 1 } \dfrac { 1 } { 4 }$$

D
$$\cos ^ { - 1 } \dfrac { 4 } { 5 }$$

Solution

Refer this concept for the solution when a body sides on the spherical path of mass$$(m)$$ and velocity $$(v)$$.
The required centripetal force is provided by the net component of weight of the body along $$BO$$, i.e.
$$\dfrac{mv^2}{r} = mg \cos \theta -R$$
When body detached from the surface $$R=0$$
then, $$\dfrac{mv^2}{r} = mg \cos \theta$$
$$\dfrac{v^2}{r} = g \cos \theta$$ ...$$(1)$$ use this for the solution
according to question given:
$$v_o = 0.5 \sqrt{gr}$$ put in $$(1)$$
$$\dfrac{(0.5\sqrt{gr})^2}{2} = g \cos \theta$$
$$\dfrac{0.25 gr}{r} =g \cos \theta$$
$$\cos \theta = 0.25$$
$$\cos \theta = \dfrac{1}{4}$$
$$\theta = \cos^{-1}\dfrac{1}{4}$$
Question 8
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A ball of mass $$150g$$ moving with an acceleration $$20m/s^2$$ is hit by force , which acts on it for $$0.1sec.$$ The impulsive force is

A
$$0.5$$N-s

B
$$0.1$$N-s

C
$$0.3$$N-s

D
$$30$$N-s

Solution
Question 9
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In which of the following cases , the net force is zero?
I) A ball freely falling from a certain height
II) A cork floating on the surface of water
III) An object floating in air

A
I and II only

B
II and III only

C
III and I only

D
I, II and III

Solution

$$ \begin{array}{l} \text { In first case gravity is there so net fore is } \\ \text { not zero? while in others it is balanced by } \\ \text { buoyancy of air and liquid. } \\ \text { so net force is zero in II, } III \text { case } \\ \text { so Ans is (B). } \end{array} $$
Question 10
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A body is standing on a merry-go-round which is rotating uniformly in a horizontal plane. His frame of reference is

A
an inertial frame

B
a non- inertial frame

C
either inertial or non- inertial

D
neither inertial nor non- inertial

Solution

When a body is standing on a merry-go-around which is rotating uniformly in a horizontal plane. His frame of reference is an inertial frame because a system of co-ordinates whose axes can be suitably chosen is said to be a frame of reference. In general we have to solve the problem of mechanics using inertial frame, which was said that the same time it is possible to solve the same problem using a non-inertial frame, this example, the motion. This example, the motion, centripetal acceleration is present therefore strictly speaking any frame or reference fixed on the merry-go-round can not be taken as an inertial frame where the acceleration will be zero and speed will be uniform.