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(50)\begin{aligned}\frac{1}{4} \text { turn } & =\frac{\pi}{2} \mathrm{rad} \\\text { weight }=m g & =25(9.81) \\& =245.25\end{aligned}Using Belt Friction\begin{aligned}\frac{T_{1}}{T_{2}} & =e^{-\mu \theta} \\\Rightarrow \frac{P}{245.25} & =e^{-(0.5 \times \pi / 2)} \\\Rightarrow & =111.86 \mathrm{~N} \\& \simeq 112 \mathrm{~N} \rightarrow \text { option (5) is }\end{aligned}correct. ...