System of Particles and Rotational Motion - Result Question 24

24. 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

[2015]

(a) $\frac{Wd}{x}$

(b) $\frac{W(d-x)}{x}$

(c) $\frac{W(d-x)}{d}$

(d) $\frac{Wx}{d}$

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Answer:

Correct Answer: 24. (c)

Solution:

(c) By torque balancing about $B$

$N_A(d)=W(d-x)$

$N_A=\frac{W(d-x)}{d}$