A) \[\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\]
B) \[\text{CHC}{{\text{l}}_{\text{3}}}\]
C) \[\text{CC}{{\text{l}}_{\text{4}}}\]
D) \[\text{C}\text{N}\]
Correct Answer: B
Solution :
\[{{I}_{AD}}=\frac{m{{d}^{2}}}{3}=4{{I}_{EF}}\] \[mg{{h}_{1}}=\frac{1}{2}m{{u}^{2}}\] \[{{h}_{1}}=\frac{{{u}^{2}}}{2g}\] Differentiating the equation, we get \[mg{{h}_{2}}=mg\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\] \[=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta \] \[=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}60\] \[=\frac{1}{2}m{{v}^{2}}{{\left[ \frac{1}{2} \right]}^{2}}=\frac{1}{2}m{{v}^{2}}\left[ \frac{1}{4} \right]\] \[{{h}_{2}}=\frac{{{u}^{2}}}{8g}\]
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