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 \[\frac{{{v}^{2}}}{l}=g(1-cos\theta )\] \[{{v}^{2}}=gl(1-cos\theta )\] ?(1) Applying conservation of energy \[\frac{1}{2}mgl=\frac{1}{2}m{{v}^{2}}+mg(1-cos\theta )\] \[{{v}^{2}}=gl-2gl(1-cos\theta )\] ?(2) Solving Eqs. (1) and (2), we get \[\theta ={{\cos }^{-1}}\frac{2}{3}\] From Eq. (1) \[{{v}^{2}}=10\times 15\left( 1-\frac{2}{3} \right)=150\left( \frac{1}{3} \right)=50\] \[\therefore \] \[v=\sqrt{50}=7m/s\]
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