\(\mathrm{V}_{\mathrm{A}}: \mathrm{V}_{\mathrm{B}} =1: 2 \)
\(\mathrm{~A}_{\mathrm{A}}: \mathrm{A}_{\mathrm{B}} =\sqrt{k_{B}}: \sqrt{8 k_{A}} \)
\(=\sqrt{\frac{k_{B}}{8 k_{A}}} \)

\(v_{m s}=1.73 \sqrt{\frac{k T}{m}}\)
\(\text { AT Increased by } 10 \text { times }\)
\(v_{\mathrm{ms}} \propto \Delta T\)
\(\text { RMS speed increases by } 10 \text { times. }\)

\(\text { Rigidity modulus }=\frac{1}{3} \times \text { Young's modulus }\)
\(\text { Poisson ratio }=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)

\(\text { Error in radius }=2 \%\)
\(\Delta r=\frac{2}{100}=0.02\)
\(\text {Volume of the sphere }=\frac{4}{3} \pi r^{3}\)
\(V =\frac{4}{3} \pi r^{3} \)
\(\frac{d V}{V} =\frac{4}{3} \pi \times 3 r^{2} d r \)
\(=3 d r=3(2 \%)=6 \%\)