System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 73

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 system consists of 3 particles each of mass m located at points (1, 1), (2, 2) and (3, 3). The coordinates of the centre of mass are

A
(6, 6)

B
(3, 3)

C
(1, 1)

D
(2, 2)

Solution
Question 2
1 / -0

A man weighing $$60kg$$ is standing on a trolley weighing $$240 kg$$. The trolley is resting on friction less horizontal rails. If the man starts walking on the trolley with a velocity of $$1ms^{-1}$$, then after $$4s$$, his dispalcement relative to the ground is

A
$$1m$$

B
$$48m$$

C
$$3.2m$$

D
$$2.4m$$

Solution
Question 3
1 / -0

A body of mass 'm' falls from a height 'h' and rebounds. If 'e' is the coefficient of restitution between the ground and the body, the change in linear momentum of the body is

A
$$m(1+e) \sqrt{2gh}$$

B
$$(me) \sqrt{2gh}$$

C
$$3(1 + e) \sqrt{2gh}$$

D
$$m(1 - e) \sqrt{2gh}$$

Solution
Question 4
1 / -0

Figure shows a thin metallic triangular sheet ABC. The mass of the sheet is M. The moment of inertia of the sheet about side AC is :

A
$$\frac{Ml^2 }{18}$$

B
$$\frac{Ml^2 }{12}$$

C
$$\frac{Ml^2 }{6}$$

D
$$\frac{Ml^2 }{4}$$

Solution
Question 5
1 / -0

Two particles of masses 4 kg and 8 kg are separated by a distance of 12 m. If they are moving toward each other under the influence of a mutual force of attraction, then the two particles will meet each other at a distance of

A
6 m from 8 kg mass

B
2 m from 8 kg mass

C
4 m from 8 kg mass

D
8 m from 8 kg mass

Solution

Mutual force is nothing but an $$internal $$ force, so the position of the $$center $$ of mass $$shouldn't $$ change.
Position of the center of mass from mass $$4kg$$ will be $$\dfrac{4}{4+8}12meter=4meter$$ so at a distance $$d=12m-4m=8m$$ from $$8Kg$$ mass.
They need to move to the position of center of mass.
Question 6
1 / -0

A scalene triangular lamina of uniform mass density and negligible thickness has one of its vertices at the origin. The position vectors of its other two vertices are $$\vec a$$ and $$\vec b$$. The location of its centre of mass will be

A
$$\frac{\vec a + \vec b}{2}$$

B
$$\frac{\vec a + \vec b}{3}$$

C
$$\frac{2(\vec a + \vec b)}{3}$$

D
$$\frac{3(\vec a + \vec b)}{4}$$

Solution
Question 7
1 / -0

The moment of inertia of a straight thin rod of mass M and length L about an axis perpendicular to its length and passing through its one end is -

A
$$\frac { { ML }^{ 2 } }{ 12 } $$

B
$$\frac { { ML }^{ 2 } }{ 3 } $$

C
$$\frac { { ML }^{ 2 } }{ 2 } $$

D
$${ ML }^{ 2 }$$

Solution
Question 8
1 / -0

Two carbon atoms have a non-covalent interaction whose potential energy varies with distance as $$E(r) = 32000000/{r}^{12} 729/{r}^{6},$$ where r is measured in Angstroms. The equilibrium inter-atomic distance will be

A
r=0

B
r=0.667

C
r=6.67

D
r=66.7

Solution
Question 9
1 / -0

A rod of steel is fixed at $$(1,0)$$ and toy is fixed at $$(2,1)$$ on it. The rod is rotated through an angle $$15^{o}$$ in clockwise direction, the new position of the toy will be

A
$$\left(1+\sqrt {6},\dfrac {1}{\sqrt {2}}\right)$$

B
$$\left(1-\sqrt {6},\sqrt {2}\right)$$

C
$$\left(1+\sqrt {6},\sqrt {2}\right)$$

D
$$none\ of\ these$$

Solution
Question 10
1 / -0

The angular velocity of a body changes from $${\omega _1}to\;{\omega _2}$$ without applying torque.The ratio of initial radius of gyration to the final of gyration is-

A
$$\sqrt {{\omega _2}} :\sqrt {{\omega _1}} $$

B
$$\sqrt {{\omega _1}} :\sqrt {{\omega _2}} $$

C
$${\omega _2}:{\omega _1}$$

D
$${\omega _1}:{\omega _2}$$

Solution