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A stone of mass $1 \mathrm{~kg}$ is tied with a string and it is whirled in a vertical circle of radius $1 \mathrm{~m}$. If tension at the highest point is $14 \mathrm{~N}$, then velocity at lowest point will be closest to:
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Options:
  • A $3 \mathrm{~m} / \mathrm{s}$
  • B $4 \mathrm{~m} / \mathrm{s}$
  • C $6 \mathrm{~m} / \mathrm{s}$
  • D $8 \mathrm{~m} / \mathrm{s}$
Solution:
1190 Upvotes Verified Answer
The correct answer is: $6 \mathrm{~m} / \mathrm{s}$
The correct option is: $6 \mathrm{~m} / \mathrm{s}$
Mass of the stone, $\mathrm{m}=1 \mathrm{~kg}$
Radius, $\mathrm{r}=1 \mathrm{~m}$
Tension, $\mathrm{T}=14 \mathrm{~N}$
Let $\mathrm{v}_1$ is velocity at the top point and $\mathrm{v}_2$ is at the bottom point.
The tension at the top point is:
$\begin{aligned}& \mathrm{T}=\frac{\mathrm{mv_{1 } ^ { 2 }}}{\mathrm{r}}+\mathrm{mg} \\& 14=\frac{(1) \mathrm{v}_1^2}{1}+\mathrm{g} \Rightarrow 14=\mathrm{v}_1^2+10 \\& \mathrm{v}_1^2=4 \Rightarrow \mathrm{v}_1=2\end{aligned}$
So, total kinetic energy atthe bottom is:
$\begin{aligned}& \frac{1}{2} \mathrm{mv}_2^2=(\mathrm{mgh})_{\text {bottom }}+\frac{1}{2} \mathrm{mv}_1^2 \\& \frac{1}{2} \mathrm{mv}_2^2=20+2 \\& \mathrm{v}_2^2=44 \Rightarrow \mathrm{v}=\sqrt{44}=6.63 \mathrm{~ms}^{-1}\end{aligned}$

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