JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Rotational Mechanics

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 \times 1 0^{- 4} kg - m^{2}

, then N is

Answer: 9

Solution

Radius of each sphere:

r = \frac{5}{2}

cm =

\frac{5}{200}

m

For a solid sphere:

I_{c m} = \frac{2}{5} m r^{2}

Two spheres lie on the diagonal axis, two are off-axis at distance

\frac{a}{2}

from it (where a = 0.04 m).

Moment of inertia:

I = 2 (\frac{2}{5} m r^{2}) + 2 (\frac{2}{5} m r^{2} + m (\frac{a}{2})^{2})

I = \frac{4}{5} m r^{2} + \frac{4}{5} m r^{2} + 2 m \frac{a ^{2}}{2} = \frac{8}{5} m r^{2} + m a^{2}

Substituting values:

I = \frac{8}{5} \times 0.5 \times \frac{5}{40000} + 0.5 \times 0.0016 = 1 0^{- 4} + 8 \times 1 0^{- 4} = 9 \times 1 0^{- 4} kg-m^{2}

Therefore, N = 9.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Rotational Mechanics

Solution

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