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A small disc of mass $m$ slides down with initial velocity zero from the top $(A)$ of a smooth hill of height $H$ having a horizontal portion $(B C)$ as shown in the figure. If the height of the horizontal portion of the hill is $h$, then the maximum horizontal distance covered by the disc from the point $D$ is
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The correct answer is:
$H$
According to law of conservation of energy at point $B$ shown in the figure given in question part,
Loss in $\mathrm{PE}=$ Gain in $\mathrm{KE}$
$\Rightarrow \quad m g(H-h)=\frac{1}{2} m v^2$
$\Rightarrow \quad v=\sqrt{2 g(H-h)}$
Now, $\quad h=\frac{1}{2} g t^2 \Rightarrow t=\sqrt{\frac{2 h}{g}}$
$\therefore$ Distance covered in horizontal portion,
$s=v \times t$
$=\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}$
$\Rightarrow \quad s=\sqrt{4 h(H-h)}$ ...(i)
For maximum value of $s$
$\frac{d s}{d h}=0$
$\Rightarrow \quad 2(H-2 h)=0 \Rightarrow H=2 h$
$\Rightarrow \quad h=\frac{H}{2}$
Substituting the value of $h$ in Eq. (i), we get,
$\begin{aligned} s & =\sqrt{4 \frac{H}{2}\left(H-\frac{H}{2}\right)} \\ & =\sqrt{2 H \cdot \frac{H}{2}}=H\end{aligned}$
Loss in $\mathrm{PE}=$ Gain in $\mathrm{KE}$
$\Rightarrow \quad m g(H-h)=\frac{1}{2} m v^2$
$\Rightarrow \quad v=\sqrt{2 g(H-h)}$
Now, $\quad h=\frac{1}{2} g t^2 \Rightarrow t=\sqrt{\frac{2 h}{g}}$
$\therefore$ Distance covered in horizontal portion,
$s=v \times t$
$=\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}$
$\Rightarrow \quad s=\sqrt{4 h(H-h)}$ ...(i)
For maximum value of $s$
$\frac{d s}{d h}=0$
$\Rightarrow \quad 2(H-2 h)=0 \Rightarrow H=2 h$
$\Rightarrow \quad h=\frac{H}{2}$
Substituting the value of $h$ in Eq. (i), we get,
$\begin{aligned} s & =\sqrt{4 \frac{H}{2}\left(H-\frac{H}{2}\right)} \\ & =\sqrt{2 H \cdot \frac{H}{2}}=H\end{aligned}$
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