Four holes of radius 5\,{cm} are cut from a thin square plate of 20\,{cm} and mass 1\,{kg}. The moment of inertia of the remaining portion about Z-axis is

Correct Answer is : 0.0017 kg−m20.0017\,{kg}-{m}²0.0017kg−m2

Correct Answer is : 0.0017 kg−m20.0017\,{kg}-{m}²0.0017kg−m2

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BITSAT 2022 Paper

Section: Physics

Question : 11 of 130

Marks: +1, -0

Four holes of radius

5\,\text{cm}

20\,\text{cm}

1\,\text{kg}

Z

Solution: 👈: Video Solution

Area mass density,

\sigma = \frac{M}{16R^2} (\because \text{Area} = 4R \times 4R = 16R^2)

Mass of each hole

m_1 = \sigma \pi R^2 = \frac{M}{16R^2} \pi R^2 = \frac{\pi M}{16}

Distance between centre of plate and centre of hole

x = \frac{\sqrt{(2R)^2+(2R)^2}}{2} = \frac{2\sqrt{2}R}{2}

Moment of inertia of one hole at about Z-axis

I_1 = \frac{1}{2} m_1 R^2 + m_1 x^2 = \frac{5\pi}{32} M R^2

Moment of inertia of whole plate about Z-axis,

I = \frac{M(4R)^2}{6} = \frac{8}{3} M R^2

Required moment of Inertia

I_0 = I - 4I_1, = \left[\frac{8}{3} - 4\left(\frac{5\pi}{32}\right)\right] M R^2 = \left[\frac{8}{3} - \frac{5\pi}{8}\right] M R^2

Given, R=5cm and M=1kg

So,

I_0 \left[\frac{8}{3} - \frac{5\pi}{8}\right] 1 \times 25 \times 10^{-4} = 0.0017\,\text{kg}-\text{m}^2

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