\(\text { Force }=v \frac{d m}{d t}\)
\(=v \frac{d}{d t} \text { (volume } \times \text { density) }\)
\(=v \frac{d}{d t}(A x p) \)
\(=v A p \frac{d x}{d t} \)
\(=v \times A p \times v \)
\(=A p v^{2} \)
Power = Force x Velocity
\(=A p v^{2} \times v=A p v^{3}\)
\(\therefore \text { Power } \alpha v^{3}\)

\(\text { Kinetic energy } E_{k}=\frac{p^{2}}{2 m}\)
\(\frac{\Delta E_{k}}{E_{k}}=\frac{2 \Delta p}{p}\)
\(\text {Given that } \frac{\Delta p}{p}=0.1\)
∴ Increase in kinetic energy
\(\frac{\Delta E_{k}}{E_{k}}=2 \frac{\Delta p}{p} \)
\(\frac{\Delta E_{k}}{E_{k}}=2 \times 0.1=0.2 \% \)

Radius =R
\(v_{1}^{2}-v_{2}^{2}=4 g R\)
\(\text { Tension } T_{2}=\frac{m v_{2}^{2}}{R_{2}}-m g\)
\(\text {To find minimum speed, let } T_{2}=0\)
\(0 =\frac{m v_{2}^{2}}{R}-m g \)
\(\frac{m v^{2}}{R} =m g \)
\(v_{2}^{2}=R g \ v_{2} =\sqrt{g R} \)
\(\text { sub (2) in the eqn (1) we get }\)
\(v_{1}^{2}-(\sqrt{g R})^{2} =4 g R \)
\(v_{1}^{2}-g R =4 g R \)
\(v_{1}^{2} =4 g R+g R \)
\(=5 g R \)
\(v_{1} =\sqrt{5 g R} \)

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 = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

\(m_{1} =1, \quad m_{2}=2 \)
\(K . E . =m g(h-x) \)
\(\text { For both balls }(h-x)\)
\(=40 \text { i.e.) Same }\)
\(g=\text { constant }\)
\(\therefore K \cdot E_{1}=m_{1} g(h-x)=m_{1} g \times 40 \)
\(K \cdot E_{2}=m_{2} g(h-x)=m_{2} g \times 40 \)
\(\therefore \frac{K \cdot E_{1}}{K \cdot E_{2}}=\frac{m_{1} g \times 40}{m_{2} g \times 40}=\frac{m_{1}}{m_{2}} \)
\( \therefore K \cdot E_{1}: K \cdot E_{2}=1: 2 \)