A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d-from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

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$\frac{w (d - x)}{x}$

$\frac{w (d - x)}{d}$

$\frac{wx}{d}$

$\frac{wd}{x}$

answer is B.

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Detailed Solution

Since the rod is in equilibrium

So the net torque about B is zero.

$τ_{1} = τ_{2}$

$N_{1} d = w (d - x) \Rightarrow N_{1} = \frac{w (d - x)}{d}$

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