A) 10 m/s
B) 7 m/s
C) 12 m/s
D) None of these
Correct Answer: B
Solution :
\[mg-mg\cos \theta =\frac{m{{v}^{2}}}{l}\] or \[\frac{{{v}^{2}}}{l}=g(1-\cos \theta )\] \[{{v}^{2}}=gl(1-\cos \theta )\] ... (i) Applying conservation of energy \[\frac{1}{2}mgl=\frac{1}{2}m{{v}^{2}}+mgl(1-\cos \theta )\] \[{{v}^{2}}gl-2gl(1-\cos \theta )\] ... (ii) Solving Eqs. (i) and (ii), we get \[\theta ={{\cos }^{-1}}\frac{2}{3}\] From Eq. (i) \[{{v}^{2}}=10\times 15\left( 1-\frac{2}{3} \right)\] \[=150\left( \frac{1}{3} \right)=50\] \[\therefore \] \[v=\sqrt{150}\approx 7\,m/s\]
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