System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 8

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Question 1
1 / -0

The vector product of two vectors a and b is a vector c such c is perpendicular to the plane containing a and b and the direction is given by is given by

A
left handed screw rule

B
index finger rule

C
left hand rule

D
right handed screw rule

Solution

If vectors A and B lie in the plane of this page, the vector C will be perpendicular to this plane.
The sense (upward or downward) of the direction of the vector product is given by the direction of the advance of the tip of a right-handed screw when rotated from A to B through angle θ between them, the screw being placed with its axis perpendicular to the plane containing the two vectors.
Question 2
1 / -0

The moment of inertia of a body about any axis is

A
equal to the sum of the moment of inertia of the body about any axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

B
equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

C
equal to the difference of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

D
equal to the average of the moments of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Solution

according to parallel axis theorem the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its Centre of mass and the product of its mass and the square of the distance between the two parallel axes.

I = I_cm + Ma²
Question 3
1 / -0

A rigid body is in mechanical equilibrium if

A
∑ F ≠ 0 and ∑ τ ≠ 0

B
∑ F ≠ 0 and ∑ τ = 0

C
∑ F = 0 and ∑ τ ≠ 0

D
∑ F = 0 and ∑ τ = 0

Solution

for mechanical equilibrium of a rigid body, it must be in translational as well as in rotational equilibrium. Linear acceleration and angular acceleration must be zero. Hence net force and net torque will be zero that is

∑ F = 0 and ∑ τ = 0
Question 4
1 / -0

Centre of gravity can be defined

A
as the center of mass

B
as that point where the total gravitational force on the body is zero

C
as that point where the total gravitational torque on the body is zero

D
as that point where the total gravitational torque on the body is greater than zero

Solution

The point at which the entire weight of a body may be thought of as centered so that if supported at this point the body would balance perfectly, so it can also be defined as as that point where the total gravitational torque on the body is zero.