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A small body slides down a smooth uneven surface from a height $H$, which eventually emerges into a circular loop of radius $R( < H)$. What should be the value of $H$, so that the force on the body at $A$ is $\sqrt{2}$ times its weight?
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Verified Answer
The correct answer is:
$H=\frac{3 R}{2}$
Given, height through which body slides $=H$
Radius of circular loop $=R$
Force on body $A, F_A=\sqrt{2}$ weight $(w)$
Let, velocity of body at $A=v_A$
Acceleration due to gravity $=g$
$$
\frac{m v_A^2}{R}=\sqrt{2} w=\sqrt{2} m g \Rightarrow v_A^2=\sqrt{2} \mathrm{Rg}
$$
By using law of conservation of energy
Energy at position $(P)=$ Energy at position $(A)$
$$
\begin{array}{rlrl}
& & m g H & =\frac{1}{2} m v_A^2+m g R \\
\Rightarrow & 2 g H & =v_A^2+2 g R=\sqrt{2} g R+2 g R=(\sqrt{2}+2) g R \\
\Rightarrow & & H & =\left(\frac{1}{\sqrt{2}}+1\right) R=1.7 R=1.5 R=\frac{3}{2} R
\end{array}
$$
Radius of circular loop $=R$
Force on body $A, F_A=\sqrt{2}$ weight $(w)$
Let, velocity of body at $A=v_A$
Acceleration due to gravity $=g$
$$
\frac{m v_A^2}{R}=\sqrt{2} w=\sqrt{2} m g \Rightarrow v_A^2=\sqrt{2} \mathrm{Rg}
$$
By using law of conservation of energy
Energy at position $(P)=$ Energy at position $(A)$
$$
\begin{array}{rlrl}
& & m g H & =\frac{1}{2} m v_A^2+m g R \\
\Rightarrow & 2 g H & =v_A^2+2 g R=\sqrt{2} g R+2 g R=(\sqrt{2}+2) g R \\
\Rightarrow & & H & =\left(\frac{1}{\sqrt{2}}+1\right) R=1.7 R=1.5 R=\frac{3}{2} R
\end{array}
$$
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