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(B) Let $m$ be the mass of the particle and $V$ be its speed. The particle moves in a horizontal circle of radius $r = \sqrt{R^2 - d^2}$.The forces ac

Vedclass · Accepted Answer

$\sqrt {\frac{{g({R^2} - {d^2})}}{d}} $

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(B) Let $m$ be the mass of the particle and $V$ be its speed. The particle moves in a horizontal circle of radius $r = \sqrt{R^2 - d^2}$.The forces acting on the particle are the normal reaction $N$ (directed towards the centre of the sphere) and the weight $mg$ (acting downwards).Resolving the normal reaction $N$ into horizontal and vertical components:Horizontal component: $N \sin 	heta = \frac{mV^2}{r}$ (providing the centripetal force)Vertical component: $N \cos 	heta = mg$ (balancing the weight)Dividing the two equations: $	an 	heta = \frac{V^2}{rg}$.From the geometry of the bowl,$	an 	heta = \frac{r}{d}$.Equating the two expressions for $	an 	heta$: $\frac{V^2}{rg} = \frac{r}{d} \Rightarrow V^2 = \frac{r^2 g}{d}$.Substituting $r^2 = R^2 - d^2$,we get $V = \sqrt{\frac{g(R^2 - d^2)}{d}}$.

$A$ particle moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $R$. The plane of motion is at a depth $d$ below the centre of the hemisphere. The speed of the particle is:

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