Laws of Motion Test

Result

Laws of Motion Test - 71

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A uniform rod of mass $$M$$ and length $$L$$ is pivoted at one end such that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end of the rod is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $$\theta$$ with the vertical is

A
$$g\sin { \theta } $$

B
$$\cfrac { g }{ L } \sin { \theta } $$

C
$$\cfrac { 3g }{ 2L } \sin { \theta } $$

D
$$6gL\sin { \theta } $$

Solution

$$\tau_o= I\alpha$$
Moment of inertia of rod about O, $$I= \dfrac{ML^2}{3}$$
$$\tau_o= Mg\dfrac{L}{2}sin\theta$$
thus
$$Mg\dfrac{L}{2}sin\theta= \dfrac{ML^2}{3} \alpha$$

$$\implies \alpha= \dfrac{3g}{2L}sin\theta$$
Question 2
1 / -0

A small block of mass m is pushed on a smooth track from position A with a velocity $$2\sqrt5$$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
When the block reaches point B, what is the direction (in terms of angle with horizontal) of acceleration of the block?

A
$$tan^{-1}(\frac{1}{2})$$

B
$$tan^{-1}(2)$$

C
$$sin^{-1}(\frac{2}{3})$$

D
The block never reaches point B.

Solution
Question 3
1 / -0

A cyclist going around a circular road of radius $$10m$$ is observed to be bending inward 30$$^\circ $$ with vertical. Frictional force acting on the cyclist is (Given: $$g = 10 m/s^2$$, mass of the cyclist is $$90$$ Kg)

A
$$532 N$$

B
$$800N$$

C
$$1559 N$$

D
$$520 N$$

Solution
Question 4
1 / -0

A solid sphere is placed on a horizontal plane. A horizontal impulse $$I$$ is applied at a distance $$h$$ above the central line as shown in the figure. Soon after giving impulse the sphere starts rolling

A
$$\cfrac{1}{2}$$

B
$$\cfrac{2}{5}$$

C
$$\cfrac{1}{4}$$

D
$$\cfrac{1}{5}$$

Solution
Question 5
1 / -0

A mass $$m$$ is revolving in a vertical circle at the end of a string of length $$20\ cm$$. By how much times does the tension of the string at the lowest point exceed the tension at the topmost point-

A
$$2\ mg$$

B
$$4\ mg$$

C
$$6\ mg$$

D
$$8\ mg$$

Solution

Let tension $$T_1$$ at top most point is given by
$$T_1 + mg = \dfrac{mV^2_1}{l}$$ ...(1)

similarly, the tension $$T_2$$ at the lowest point is given by
$$T_2 -mg = \dfrac{mV_2^2}{l}$$ ...(1)

here, tension acting in words while balancing centrifugal force and weight of the body acting outwards
$$T_2 - T_1 = \dfrac{mv^2_2}{l} +mg - \dfrac{mv_1^2}{l} +mg$$

$$\dfrac{m}{l} (v^2_2 -V^2_1) + 2mg$$

we know from kilometres
$$v^2_1 = v^2_2 - 2g(2l)$$
$$T_2 T_1 = \dfrac{m}{l} [2g(2l)] +2mg$$

$$= 6mg$$
Hence (C) is correct answer
Question 6
1 / -0

A rod of mass $$m$$ and length $$l$$ in placed on a smooth table. An another particle of same mass $$m$$ strikes the rod with velocity $${v}_{0}$$ in a distance perpendicular to the rod at distance $$x(</2)$$ from its centre. Particle sticks to the rod. Let $$\omega$$ be the angular speed of system after collision, then find maximum possible value of impulses (by varying $$x$$) that can be imparted to the particle during collision. Particle still sticks to the rod.

A
$$\dfrac { m{ v }_{ 0 } }{ 2 }$$

B
$$\dfrac { 2m{ v }_{ 0 } }{ 3 }$$

C
$$\dfrac { 3m{ v }_{ 0 } }{ 4 }$$

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

Solution
Question 7
1 / -0

As shown in the given figure the ball is given sufficient velocity at the lowest point to complete the circle. Length of string is $$1m$$. Find the tension in the string, when it is at $$60^{\circ}$$ with vertical position.
(Mass of ball $$= 5\ kg$$).

A
$$160\ N$$

B
$$180\ N$$

C
$$200\ N$$

D
$$225\ N$$

Solution
Question 8
1 / -0

A mass of $$6 \ kg$$ is suspended by a rope of length $$2 \ m$$ from a ceiling. A force of $$50 \ N$$ is applied in horizontal direction at the mid point of the rope. What is the angle of the rope with the vertical in equilibrium?

A
$${\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right)$$

B
$${\tan ^{ - 1}}\left( {\dfrac{5}{4}} \right)$$

C
$${\tan ^{ - 1}}\left( {\dfrac{5}{6}} \right)$$

D
None of these

Solution

Conditions of total equilibrium quint the sum of all
the external forces acting on the body is zero, and
the sum of all the torques from external forces $$c_{0}$$
zero - These two Conditions must be simultaneously satisfied in equilibrium
for horizontal equilibrium

$$T_{3}=T_{1} \sin \theta$$

$$for\ vertical\ equilibrium,$$

$$T_{2}=T_{1} \cos \theta$$

$$Taking\ ratios$$

$$\dfrac{T_{3}}{T_{2}}=\tan \theta$$

for the equilibrium body
$$T_{2}=m g$$

$$T_{2}=6 \times 10N \\$$

$$\therefore \quad \tan \theta=\dfrac{5}{6}$$

$$\theta_{2}$$

$$\theta=\tan ^{-1}\left(\dfrac{5}{6}\right)$$

$$\operatorname{option\ c}= \tan ^{-1}\left(\dfrac{5}{6}\right)$$
Question 9
1 / -0

The position time plot for a $$400 gm$$ object are shown. The impulses at $$0$$ sec, $$1$$ sec, $$3$$ sec are :-

A
$$ 1.6 N-s, 1.6 N-s, 1.6 N-s $$

B
$$ 1.6 N-s, 1.6 N-s, -1.6 N-s $$

C
$$ 1.6 N-s, 0, 1.6 N-s $$

D
$$ 1.6 N-s, 0, -1.6 N-s $$

Solution
Question 10
1 / -0

A ball of mass 0.2 Kg is dropped from a certain height above the ground.It bounces back after an elastic collision with the floor.If the speed with which the ball strikes the ground is 10m/s, then the impluse imparted by the ball on the floor is:

A
12 Kg m/s

B
8 Kg m/s

C
4 Kg m/s

D
2 Kg m/s

Solution