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A particle of mass $\mathrm{m}$ is attached to a thin uniform rod of length $\mathrm{a}$ and mass $\mathrm{4 \mathrm{~m}}$ . The distance of the particle from the centre of the mass of the rod is $\mathrm{\frac{a}{4}}$ .. The moment of inertia of the combination about an axes passing through 0 normal to the rod is -

Option: 1
$\frac{64}{48} \mathrm{ma}^2$

Option: 2
$\frac{91}{48} \mathrm{ma}^2$

Option: 3
$\mathrm{\frac{27}{48} m a^2}$

Option: 4
$\frac{51}{48} \mathrm{ma}^2$

Answers (1)

best_answer

Momernot of inertia $\mathrm{=m\left(\frac{3 a}{4}\right)^2+m, \times \frac{a^2}{3}}$

For the centre of rod -

$\mathrm{\left(\frac{m_1 a^2}{12}+\frac{m_1 a^2}{4}\right) =\frac{m_1 a^2}{3} }$

$\mathrm{m_1 =4 m }$

$\mathrm{\therefore \text { Total (I) } =m\left(\frac{3 a}{4}\right)^2+\frac{4 m a^2}{3} }$

$\mathrm{ =\frac{9 m a^2}{16}+\frac{4 m a^2}{3} }$

$\mathrm{ =\frac{(27+64)}{48} m a^2 }$

$\mathrm{ I =\frac{91}{48} ma^2 }$

Hence option 2 is correct.

Posted by

shivangi.bhatnagar

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