(d) $\tau_{x} = \ \frac{\pi\eta r^{4}}{2l}\ \theta_{x}\ \ \ and\ \ \ \tau_{y} = \ \ \frac{\pi\eta(2r)^{4}}{2l}\theta_{y}$ Since, $\tau_{x} = \ \ \tau_{y},$ $\therefore\ \ \theta_{x} = 16\theta_{y}\ \ \ \ or\ \ \frac{\theta_{x}}{\theta_{y}} = 16$
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Ph.D. in Physics with 19+ years of teaching experience.