c
By momentum conservation along \(y\)

\(m _{1} u _{1} \sin \theta_{1}= m _{2} u _{2} \sin \theta_{2}\)

i.e. \(mu _{1} \sin \theta_{1}=10 mu _{2} \sin \theta_{2}\)

\(\Rightarrow u _{1} \sin \theta_{1}=10 u _{2} \sin \theta_{2}\) \(...(i)\)

\(kf _{ m _{ 1 }}=\frac{1}{2} ki _{ m _{ 1 }}\) i.e. \(\frac{1}{2} mu _{1}^{2}=\frac{1}{2} \times \frac{1}{2} mu ^{2}\)

i.e. \(\sqrt{ u _{1}=\frac{ u }{\sqrt{2}}}\) \(...(ii)\)

Also collision is elastic : \(k _{ i }= k _{ f }\)

\(\frac{1}{2} m u^{2}=\frac{1}{2} m u_{1}^{2}+\frac{1}{2} \cdot 10 m \cdot u_{2}^{2}\)

\(\frac{1}{2} mu ^{2}=\frac{1}{2} \times \frac{1}{2} mu ^{2}+\frac{1}{2} \times 10 m \cdot u _{2}^{2}\)

\(\frac{1}{4} mu ^{2}=\frac{1}{2} \times 10 \times mu _{2}^{2}\)

\(u _{2}=\frac{ u }{\sqrt{20}}\)\(...(iii)\)

Putting \((ii)\) \(\&\) \((iii)\) in \((i)\)

\(\frac{ u }{\sqrt{2}} \sin \theta_{1}=10 \cdot \frac{ u }{\sqrt{20}} \sin \theta_{2}\)

\(\sin \theta_{1}=\sqrt{10} \sin \theta_{2} \quad \rightarrow\) Hence \(n =10\)