IIT-JEE 2011 · Paper 1 · Physics Q46

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PhysicsInteger+4Question 46 of 69

Four solid spheres each of diameter $\sqrt5$ 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 $N\times10^{-4}\ \mathrm{kg\,m^2}$ , then $N$ is

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Answer and worked solution

Official answer: 9

Sphere data. $m=0.5$ kg, $r^2=\frac54$ cm²; each has $\frac25mr^2=0.25$ kg cm² about its own diameter.

About the diagonal. Two spheres lie on it; the other two are $\frac4{\sqrt2}=2\sqrt2$ cm away:
$I=4(0.25)+2(0.5)(8)=9\ \text{kg cm}^2=9\times10^{-4}\ \text{kg m}^2.$

Answer: 9

Topics:Parallel and perpendicular axis theorems Moment of inertia

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IIT-JEE 2011 · Paper 1, Physics question 46