\(\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}\)
when the spring is cut into two equal halves, than the force constant of each part is 2k. When the mass is suspended from one of the halves, new time period is \(T^{\prime}=2 \pi \sqrt{\frac{m}{2 k}}=\frac{T}{\sqrt{2}}\)

As Temperature of room A is less than that of room B evidently Room A has more air in it.

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

Time period T = \(\frac{2\pi}{\sqrt GM_E} (R_E+ h)^\frac{3}{2}\)
\(\therefore\) It is independemt of mass

\(m=3 \mathrm{~kg} \quad r=40 \times 10^{-2} \mathrm{~m}=0.4 \mathrm{~m}\)
\(\text { Force }=30 N\)
Moment of inertia of a hollow-cylinder I= MR²
\(=3 \times\left(40 \times 10^{-2}\right)^{2} \)
\(=3 \times 0.4 \times 0.4=0.48 \mathrm{kgm}^{2} \)
\( F R =I \alpha \)
\(30 \times 40 \times 10^{-2}=0.48 d \)
\(d=\frac{12}{0.48}=\frac{1200}{48}=25 \mathrm{rad} \mathrm{s}^{-2} \)
\(\alpha=25 \mathrm{rad} \mathrm{s}^{-2} \)

Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

\(\Delta l =0.01 \)
\(l =3.51 \)
\(\% \text { error } =\frac{\Delta l}{l} \times 100=\frac{0.01}{3.51} \times 100 \)
\(=0.00284 \times 100 \)
\(=0.284=0.28 \% \)