A) \[\frac{4mg}{5}\]
B) \[\frac{5mg}{4}\]
C) \[\frac{3mg}{5}\]
D) \[\frac{5mg}{3}\]
Correct Answer: C
Solution :
Mostly student use T \[\cos {{53}^{o}}=mg\](balances forces) \[T\frac{3}{5}=mg\Rightarrow T=\frac{5mg}{3}\]and marked option [d]. Hence correct option is [c]. As velocity \[=0\Rightarrow \]no radial acceleration \[\Rightarrow T=mg\cos {{53}^{o}}=\frac{3mg}{5}\]
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