(4)

[IMAGE 62]

As we know that the moment of inertia of a solid sphere about XY is,

$I_{XY}=\frac{7}{5}M{R}^2$    $\text{Also,}$ $I_{XY}=M{K}^2$

$\text{So, }M{K}^2=\frac{7}{5}M{R}^2$

$\Rightarrow R^2=\frac{5}{7}{K}^2$

$\text{or }R=K\sqrt{\frac{5}{7}} \text{or }R=\frac{5\sqrt{5}}{\sqrt{7}} (\because K=5 \text{m})$

$\text{On comparing with }R=\frac{5x}{\sqrt{7}},\text{ the value of }x\text{ is }\sqrt{5}.$

(1)

Moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod

            $I_{\text{cm}}=\frac{M{l}^2}{12}$                                                     ...(i)

where, M = Mass of the rod, $l$ = Length of the rod

Given, $M=400\text{g},$ $l=?, I_{\text{cm}}=2400$$\text{g}$ ${\mathrm{cm}}^2$

From eq (i), $l=\sqrt{\frac{I_{\text{cm}}\times 12}{M}}=\sqrt{\frac{2400\times 12}{400}}=8.484\approx 8.5\text{cm}$

(2)

$\frac{E_{\text{Sphere}}}{E_{\text{Cylinder}}}=\frac{\frac{1}{2}{I}_s{\omega }_s^2}{\frac{1}{2}{I}_c{\omega }_c^2}=\frac{I_s{\omega }_s^2}{I_c{\omega }_c^2}$

Here, $I_s=\frac{2}{5}m{R}^2, I_c=\frac{1}{2}m{R}^2, {\omega }_c=2{\omega }_s$

$\frac{E_{\text{Sphere}}}{E_{\text{Cylinder}}}=\frac{\frac{2}{5}m{R}^2\times {\omega }_s^2}{\frac{1}{2}m{R}^2\times {\left(2{\omega }_s\right)}^2}=\frac{4}{5}\times \frac{1}{4}=\frac{1}{5}$

(1)

Here, $l_1+l_2=l$

[IMAGE 63]

Centre of mass of the system,

$l_1=\frac{m_1\times 0+m_2\times l}{m_1+m_2}=\frac{m_{2}l}{m_1+m_2}$

          $l_2=l-l_1=\frac{m_{1}l}{m_1+m_2}$

Required moment of inertia of the system,

     $I=m_1{l}_1^2+m_2{l}_2^2$

       $=(m_1{m}_2^2+m_2{m}_1^2)\frac{l^2}{{{\displaystyle (}{\displaystyle m}{\displaystyle {}_1}{\displaystyle +}{\displaystyle m}{\displaystyle {}_2}{\displaystyle )}}^2}$

       $=\frac{m_1{m}_2(m_1+m_2)l^2}{{{\displaystyle (}{\displaystyle m}{\displaystyle {}_1}{\displaystyle +}{\displaystyle m}{\displaystyle {}_2}{\displaystyle )}}^2}=\frac{m_1{m}_2}{m_1+m_2}{l}^2$