Moment of Inertia - Free MCQ Practice Test with solutions, NEET Physics

We know that MI of a ring is mr²
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r² = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2

If K is radius of gyration, then moment of inertia for the rod is given as -

I = MK²                        (1)

But moment of inertia of uniform rod of length L and Mass M about an axis passing through its center the perpendicular to its length  is given as - 

comparing equation (1) and (2), we get

Given,
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR²/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×1²/2/2×(1/2)²/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.

Moment of inertia of solid sphere I_s= 2/5MR²
moment of inertia of hollow sphere I_h =2/3MR²
given mass of solid sphere =√45 kg.
I_s=I_h
2MR²/5=2MR²/3
given their masses are equal 2 (√45)²/5= 2 R²/3
45/5=R²/3
9=R²/3
9×3=R²
27=R²
√27=R
√3×9=R
3√3 m=R.

The moment of inertia (M.I.) of a sphere about its diameter = 2/5 MR²
According to the theorem of parallel axes, 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.
The M.I. about a tangent of the sphere = 2/5 MR²+ MR²=7/5 MR²

As we have learned Moment of inertia for continuous body -

r is perpedicular distance of a particle of mass dm of rigid body from axis of rotation taking a strip of radius x  and thickness dx
MI of ring dI = dm x² 

The problem involves a thin circular disk in the xy plane, and we need to find the ratio of its moment of inertia about the z and z' axes.

• The moment of inertia about the z-axis is calculated using the formula for a circular disk, which is (1/2)MR².
• For the z' axis, which is the axis through the centre parallel to the plane, the formula is (1/4)MR² + (1/12)MR².
• Simplifying, the moment of inertia about the z' axis becomes (1/3)MR².
• The ratio of the moment of inertia about the z and z' axes is therefore 1 : 3.

So, Option (d) is the correct answer.

For a uniform rod with negligible thickness, the moment of inertia about its centre of mass is:

Where M = mass of the rod and L = length of the rod
∴ The moment of inertia about the end of the rod is

Option D is correct: π r⁴/16.

I_x is defined as \int y² dA. For the quarter circle in the first quadrant:

I_x = \int_x=0^r \int_y=0^{√(r2 − x2)} y² dy dx.

Evaluate the inner integral: \int₀^{√(r2 − x2)} y² dy = (1/3)(r² − x²)^3/2.

Thus I_x = (1/3) \int₀^r (r² − x²)^3/2 dx.

Use the substitution x = r θsin, with dx = r cosθ dθ and limits 0 → π/2. Then (r² − x²)^3/2 = r³ cos³θ and the integrand becomes r⁴ cos⁴θ.

So I_x = (1/3) r⁴ \int₀^π/2 cos⁴θ dθ.

Use the known value \int₀^π/2 cos⁴θ dθ = 3π/16. Therefore I_x = (1/3) r⁴ × 3π/16 = π r⁴/16.

By symmetry I_y = π r⁴/16, and the polar moment about the corner is π r⁴/8.

Moment of inertia of circular plate,

Moment of inertia of Square plate,

 A moment of inertia is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element’s distance from the axis.

Moment of inertia of a thin spherical shell of mass M and radius R about its diameter.
I = 2/3 MR² 
For a rigid body system, the moment of inertia is the sum of the moments of inertia of all its particles taken about the same axis.


where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.

Moment of inertia of a disc,
I=1/2MR²
Disc is melted and recasted into a solid sphere.

The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.

The moment of inertia of a body is given by I =  m R²
where m = Mass and R = Distance from the axis
The moment of inertia of a solid cone 
m is the mass of the cone and r is the base radius of the cone and h is the height of the cone.

About the vertical axis 

Moment of inertia of a body depends on the mass of the body, distribution of mass in the body, position of axis of rotation of the body and also depends on the distance of body from the axis of rotation.

The moment of inertia of the removed part about the axis passing through the centre of mass and perpendicular to the plane of the disc = I_cm + md²
= [m × (R/3)²]/2 + m × [4R²/9] = mR²/2
Therefore, the moment of inertia of the remaining portion = moment of inertia of the complete disc – moment of inertia of the removed portion
= 9mR²/2 – mR²/2 = 8mR²/2
Therefore, the moment of inertia of the remaining portion (I _remaining) = 4mR².

From the perpendicular axis theorem, we have
I_z = I_x + I_y 
where, I_z is moment of inertia about Z-axis, I_x is moment of inertia about X-axis and I_y is moment of inertia about Y-axis.

Moment of inertia of a disc, I_z = 1/2MR²  
Now, I_x = I_y, because two perpendicular diameters are equivalent.

Also every diameter passes through centre of mass.

Using parallel axis theorem,

I = m_X r_X² + m_Y r_Y²
I = (0.7)× (0.1)² + (0.5)× (0.4)²
I = (0.7) x (0.01) + (0.5) x (0.16)
I = 0.007 + o.08
I = 0.087 kg m²
Therefore, the moment of inertia of the system is 0.087 kg m².

Moment of inertia of a hollow cylinder of mass M and radius r about its own axis is Mr²

The correct option is (1)
A ring about its axis, perpendicular to the plane of the ring.
(1) Moment of inertia of a ring about its axis and perpendicular to its plane = Mr²
(2) Moment of inertia of a disc about its axis and perpendicular to its plane = 1/2Mr² = 0.5Mr
(3) Moment of inertia of a solid sphere about one of its diameter =2/5Mr²=0.4Mr²
(4) Moment of inertia of a spherical shell about one of its diameter = 2/3Mr² = 0.66Mr²
Therefore, the moment of inertia of the ring is highest.

I = m₁ r₁² + m₂ r₂² + m₃ r₃² + m₄ r₄²
I = (0.2) × (0.4)² + (0.2) × (0.4 )² + (0.2) × (0.4)² + (0.2) × (0.4)²
I = 0.032 + 0.032 + 0.032 + 0.032
I = 0.128 kg m²
Moment of inertia of the balls about the axis 0.128 kg m²

Moment of inertia

For the centre of rod