Four solid spheres each of diameter and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is , then is [2011]

Question

Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]

Smart Achievers · Accepted Answer

(9)

Let the four spheres be A, B, C & D

$m_{A} = m_{B} = m_{C} = m_{D} = 0.5 kg$

Moment of inertia of the system about the diagonal of the square

$I_{X Y} = I_{A} + I_{B} + I_{C} + I_{D} = 2 I_{A} + 2 I_{B}$

$= 2 [\frac{2}{5} M R^{2} + M a^{2}] + 2 [\frac{2}{5} M R^{2}]$

$= 4 \times \frac{2}{5} M R^{2} + 2 M a^{2} = M [\frac{8}{5} R^{2} + 2 a^{2}]$

$= 0.5 [\frac{8}{5} \times {(\frac{\sqrt{5}}{2})}^{2} + 2 \times 8] \times 10^{- 4}$

$= 0.5 [2 + 16] \times 10^{- 4}$

$= 9 \times 10^{- 4} = N \times 10^{- 4} {kg m}^{2}$ $∴$ $N = 9$

Solution

Q.
Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]

Want Us to Email you About Special Offers & Updates?

Solution

Q. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N×10-4 kg m2, then N is [2011]

Ans. (9) Let the four spheres be A, B, C & D mA=mB=mC=mD=0.5 kg Moment of inertia of the system about the diagonal of the square IXY=IA+IB+IC+ID=2IA+2IB =2[25MR2+Ma2]+2[25MR2] =4×25MR2+2Ma2=M[85R2+2a2] =0.5[85×(52)2+2×8]×10-4 =0.5[2+16]×10-4 =9×10-4=N×10-4 kg m2 ∴ N=9

Want Us to Email you About Special Offers & Updates?

Q.
Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]