A) 1.5 m
B) 15 m
C) 18 m
D) 28 m
Correct Answer: C
Solution :
\[H=\frac{{{u}^{2}}}{2g}\,\Rightarrow \,H\propto \frac{1}{g}\]Þ \[\frac{{{H}_{B}}}{{{H}_{A}}}=\frac{{{g}_{A}}}{{{g}_{B}}}\] Now \[{{g}_{B}}=\frac{{{g}_{A}}}{12}\ \text{as}\,\,g\propto \rho R\] \\[\frac{{{H}_{B}}}{{{H}_{A}}}=\frac{{{g}_{A}}}{{{g}_{B}}}\]=12Þ\[{{H}_{B}}=12\times {{H}_{A}}=12\times 1.5=18m\]
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