Laws of Motion Test

Result

Laws of Motion Test - 52

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Question 1
1 / -0

If force F =(500 - 100)t, then impulse as a function of time will be:-

A
$$500t - 50 t^2$$

B
$$50 t - 10$$

C
$$50 - t ^2$$

D
$$100 t^2$$

Solution

We know that force is given as $$F=\dfrac{dP}{dt}$$
so impulse $$dP$$ is given as $$dP=Fdt$$
putting value of $$F$$ and integrating both sides we get $$\int dP=\int (500-100t)dt $$
or $$P=500t-50t^2+c$$ , $$c$$ is an integration constant.
as the $$P$$ is here the $$momentum$$ supplied or abstracted so $$c=0$$
Question 2
1 / -0

A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released, then:

The total acceleration of the sphere and the thread tension as a function of $$\theta$$ , the angle of deflection of the thread from the vertical will be

A
$$g\sqrt{1+3cos^2} \theta, T=3 mg\,cos\,\theta$$

B
$$g \, cos\,\theta, T=3 mg \, cos\,\theta$$

C
$$g\sqrt{1+3sin^2} \theta, T=5 mg\,cos\,\theta$$

D
$$g \, sin\,\theta, T=5 mg \, cos\,\theta$$

Solution
Question 3
1 / -0

The angle $$\theta$$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally will be

A
$$cos\, \theta =\dfrac{1}{\sqrt{3}}$$

B
$$cos\, \theta =\dfrac{1}{3}$$

C
$$sin\, \theta =\dfrac{1}{\sqrt{3}}$$

D
$$sin\, \theta =\dfrac{1}{\sqrt{2}}$$

Solution
Question 4
1 / -0

In the figure, a 4.0 kg ball is on the end of a 1.6 m rope that is fixed at O. The ball is held at point A, with the rope horizontal is given an initial downward velocity. The ball moves theough three qyarters of a circle with no friction and arrives at B, with the rope barely under tension. Thee initial velocity of the ball, at point A, is closest to

A
4.0 m/s.

B
5.6 m/s.

C
6.2 m/s

D
6.8 m/s

Solution

The motion of the ball is always perpendicular to the tension in the Rope. Therefore, the tension does no work on the system and the total mechanical energy of the ball is conserved.
$$E_A = E_B$$
$$mgR + \dfrac{1}{2} mv_A^2 = mg(2R) + \dfrac{1}{2} mv_B^2$$
$$v_A^2 = 2gR +v_B^2$$
Newton's second law for the ball at point B
$$mg + T = \dfrac{mv_B^2}{R}$$
$$T \approx 0$$
$$v_B = gR$$
$$v_A = 3gR$$
= 3(9.8)(1.6)
= 6.9 m/s
Since the velocity is near to 6.8m/s.
Hence (D) option is correct answer
Question 5
1 / -0

A car is moving in a circular path with a uniform speed $$v$$. When the car rotates through an angle $$\theta$$, the magnitude of change in its velocity is

A
$$\triangle v = v\sin \left (\dfrac {\theta}{2}\right )$$

B
$$\triangle v = v\sin \theta$$

C
$$\triangle v = 2v\sin \left (\dfrac {\theta}{2}\right )$$

D
$$\triangle v = 2v\sin \theta$$

Solution
Question 6
1 / -0

Which of the following groups of forces could be in equilibrium?

A
$$3 N, 4 N, 5 N$$

B
$$4 N, 5 N, 10 N$$

C
$$30 N, 40 N, 80 N$$

D
$$1 N, 3 N, 5 N$$

Solution

For the equilibrium of force, the resultant of two smaller is equal and opposite to third one.

$$C^2=A^2+B^2$$

Lets see,

A. $$5^2=3^2+4^2$$
$$25=25$$

B. $$10^2=5^2+4^2$$
$$100{\neq} 41$$

C. $$80^2=40^2+30^2$$
$$6400\neq2500$$

D. $$5^2=1^2+3^2$$
$$25\neq10$$

From the above discussion,
The correct option is A.
Question 7
1 / -0

One end of a string of length $$1.0m$$ is tied to a body of mass $$0.5kg$$. It is whirled in a vertical circle with angular velocity $$4 rad/s$$. The tension in the string when body is at the lower most point of its motion is equal to [Take $$g=10m/s^2$$].

A
$$3N$$

B
$$5N$$

C
$$8N$$

D
$$13N$$

Solution

$$ T - mg = \cfrac{mv^2}{r}$$
or, $$ T - mg = m \omega ^2 r $$
$$ \therefore T = mg + m\omega^2 r$$
$$ = (0.5 \times 10) + (0.5 \times 4^2 \times 1)$$
$$ = 5+8 = 13N$$
Question 8
1 / -0

The frictional force is?

A
Self adjustable

B
Not self adjustable

C
Scalar quantity

D
Equal to the limiting force

Solution

Friction force is self adjustable force. Because it tries to keep the object in rest until an unbalanced force is applied on the object. Therefore the magnitude of static friction is equal to external force till it remains as balanced force.
Question 9
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A ball of mas $$150 g$$, moving with an acceleration $$20 m/s^2$$, is hit by a force which acts on it for $$1$$sec. The impulsive force is:-

A
$$3 Ns$$

B
$$1 Ns$$

C
$$05 Ns$$

D
$$1.2 Ns$$

Solution

$$force = mass x acceleration$$
$$F = ma$$
$$F = 150/1000 x 20$$
$$F = 3N$$
$$Impulse = Force \times Time$$
$$Impulse = 3 \times 1 => 3 N s $$
Hence,
option $$A$$ is correct answer.
Question 10
1 / -0

An inclined track ends in a circular loop of diameter $$D$$. From what height on the track a particles should be released so that it completes that loop in the vertical plane?

A
$$\dfrac {5D}{2}$$

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

C
$$\dfrac {5D}{4}$$

D
$$\dfrac {4D}{5}$$

Solution