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In a smooth hemispherical shell of radius R, a rod of mass $(\frac{\sqrt{3}}{2}) kg$ is placed horizontally and is in equilibrium. The length of rod is R. Find the normal reaction at any end of the rod (in N). [Take $g = 10 {m/s}^{2}$ ]

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\frac{5}{\sqrt{3}}

5 \sqrt{3}

answer is A.

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Detailed Solution

Given $L_{rod} = R_{1} m$

$M_{rod} = \frac{\sqrt{3}}{2} kg$

From figure $2Ncos30 = \frac{\sqrt{3}}{2} g$

$2N× \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} g$

$N = \frac{10}{2}$

$∴$ Normal reaction at any end

$N = 5 N$

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