Four equal masses, $m$ each are placed at the corners of a square of length $(\ell)$ as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :

(C) $\sqrt{3} \mathrm{~m} \ell^{2}$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Rotational Mechanics

Four equal masses,

m

each are placed at the corners of a square of length

(ℓ)

as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :

Answer: (D)

Take

A

at the origin, with

B

(ℓ, 0)

C

(ℓ, ℓ)

and

D

(0, ℓ)

, as placed in the figure. The diagonal

D B

runs along the direction

(1, - 1)

, so the axis through

A

parallel to it is the line

x + y = 0

, and the distance of a corner

(x, y)

from that line is

\frac{∣ x + y ∣}{2}

.
This gives

r_{A} = 0

r_{B} = r_{D} = \frac{ℓ}{2}

and

r_{C} = 2 ℓ

I = m (\frac{ℓ ^{2}}{2}) + m (\frac{ℓ ^{2}}{2}) + m (2 ℓ^{2}) = 3 m ℓ^{2}

The moment of inertia about that axis is

3 m ℓ^{2}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option