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(C) Let the beam have length $\ell$ and weight $W$ acting at its center $C$ (at distance $\ell/2$ from either end).Let the two persons be at positions

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$5 : 9$

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(C) Let the beam have length $\ell$ and weight $W$ acting at its center $C$ (at distance $\ell/2$ from either end).Let the two persons be at positions $P_1$ and $P_2$ from ends $A$ and $B$ respectively.Distance of $P_1$ from $A$ is $\ell/5$.Distance of $P_2$ from $B$ is $\ell/3$.Distance of $P_1$ from center $C$ is $\frac{\ell}{2} - \frac{\ell}{5} = \frac{3\ell}{10}$.Distance of $P_2$ from center $C$ is $\frac{\ell}{2} - \frac{\ell}{3} = \frac{\ell}{6}$.For rotational equilibrium,taking torque about the center $C$:$N_A 	imes \frac{3\ell}{10} = N_B 	imes \frac{\ell}{6}$.$\frac{N_A}{N_B} = \frac{\ell}{6} 	imes \frac{10}{3\ell} = \frac{10}{18} = \frac{5}{9}$.

Two persons of equal height are carrying a long uniform wooden beam of length $\ell$. They are at a distance of $\frac{\ell}{5}$ and $\frac{\ell}{3}$ from the nearest ends of the rod. The ratio of the normal reaction at their heads is

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