System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 10

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Question 1
4 / -1

Arrange the comparative stabilities of the three objects shown in the figure in terms of work and potential energy.

A
2nd object < 3rd object < 1st object

B
3rd object < 1st object < 2nd object

C
1st object < 2nd object < 3rd object

D
1st object < 3rd object < 2nd object

Solution

A object’s stability depends upon its centre of gravity.the larger the centre of gravity the more stable,By looking at the figures we can easily figure out which is more stable.
Question 2
4 / -1

The figure shows a dish kept at an orientation with a marble placed on it. If the marble in figure is displaced slightly, it would rattle about for a while and come to rest in its original place. This is an example of :

A
Centre of gravity principle

B
Stable Equilibrium

C
metastable equilibrium

D
unstable equilibrium

Solution

If a dish is kept at an orientation with a marble placed on it and if the marble in figure is displaced slightly, it would rattle about for a while and come to rest in its original place. This is the example of Stable Equilibrium.
When a body tries to regain its equilibrium position after being slightly displaced and released, it is said to be in stable equilibrium.
Question 3
4 / -1

The rock has a mass of 1 kg. What is the mass of the measuring stick if it is balanced by a support force at the one-quarter mark? The centre of mass of the measuring stick is at the one-half mark, one-quarter of the stick’s length beyond the pivot.

A
0.5 kg

B
4 kg

C
2 kg

D
1 kg

Solution

The stick is balanced by the 1 kg rock that is one-quarter of its length to the left and by its own weigh which is at the Centre of Mass, one-quarter of its length to the right. Therefore, the mass of the stick must be one kilogram, 1 kg.
Question 4
4 / -1

A pair of equal and opposite forces with different line of action are said to form a

A
inertia

B
moment

C
couple

D
momentum

Solution

When two forces of equal magnitude opposite in direction and acting along parallel straight lines, then they are said to form a couple. The perpendicular distance between the two force forming a couple is called the arm of the couple.
Question 5
4 / -1

A meter rule is supported at its centre. It is balanced by two weights, A and B. If A and B are placed at a distance 20 cm, and 40 cm from the centre of scale, find the weight of B. The weight of A is 50N.

A
30 N

B
40 N

C
25 N

D
20 N

Solution

In balanced condition torque of the both sides must be equal so Fa × Da = Fb × Db
50 × 20 = Fb × 40
Fb = 25 N
Question 6
4 / -1

In order to rotate a body, we need to apply

A
two forces at two ends but in the same direction

B
a large force at any point on the body

C
Two equal forces at two ends but in the opposite direction

D
a small force but exactly at its centre of gravity

Solution

In order to rotate a body, we need to apply two equal forces at two ends but in the opposite direction.
Question 7
4 / -1

Determine the force that must be applied to end of the class 1 lever shown below to lift the 500 lb load.

A
500 lb

B
2500 lb

C
5 lb

D
100 lb

Solution

Balance the 2 opposite moments of inertia on the 2 sides
1 × 500 = F × 5
F = 100
Question 8
4 / -1

Five forces are separately applied to a flat object lying on a table of negligible friction as shown in figure. The force which will not cause the object to rotate about the centre of the object is

A
F₂

B
F₃

C
F₄

D
F₁

Solution

The line of action of F2 passes through the centre of rotation, and thus cannot produce a torque about the centre (r = 0).
Question 9
4 / -1

Which of the following appliance works on the principle of moments.

A
pendulum clock

B
Spring balance

C
Balance wheel

D
Beam balance

Solution

In beam balance we consider the torque at centre of beam which is moment of force at its each end.
Question 10
4 / -1

There are two blocks of mass 3 kg and 5 kg hanging from the ends of a rod of negligible mass. The rod is marked in eight equal parts as shown. At which of the points indicated should a string be attached if a rod is to remain horizontal when suspended from the string?

A
D

B
F

C
B

D
E

Solution

This is based on torque balance, That is F.d=const.
Therefore 3×d=5×(8-d)
by solving we get d=5.
and at point 5=E