JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 12)
A body of $$m \mathrm{~kg}$$ slides from rest along the curve of vertical circle from point $$A$$ to $$B$$ in friction less path. The velocity of the body at $$B$$ is:
_4th_April_Evening_Shift_en_12_1.png)
(given, $$R=14 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$$ and $$\sqrt{2}=1.4$$)
10.6 m/s
19.8 m/s
16.7 m/s
21.9 m/s
Explanation
From energy conservation $$\rightarrow$$
$$\begin{aligned} & m g(R+R \sin 45)=\frac{1}{2} m v^2 \\ & \Rightarrow 10\left(1+\frac{1}{\sqrt{2}}\right) \times 14=\frac{1}{2} v^2 \\ & \Rightarrow 10\left(1+\frac{\sqrt{2}}{2}\right) \times 28=v^2 \\ & \Rightarrow 10(1+0.7) \times 28=v^2 \\ & \Rightarrow v=21.81 \end{aligned}$$
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