A body weighs 8 gm, when placed in one pan and 18gm, when placed in the other pan of a false balance. If the beam is horizontal (when both the pans are empty), the true weight of the body is

The correct option is 12 gm.

To find the true weight of the body, let us analyze the principle of moments for a false balance (a balance with arms of unequal lengths).
Let the true weight of the body be $W$.
Let the lengths of the left and right arms of the balance be $L_1$ and $L_2$ respectively.

Since the beam remains horizontal when both pans are empty, the moments of the empty pans balance each other. Therefore, we can focus on the torque due to the added weights.

Case 1: When the body is placed in the left pan, it is balanced by a weight of 8 gm in the right pan. By the principle of moments:
$W\times L_1=8\times L_2$

From this, we can write:
$\frac{L_1}{L_2}=\frac{8}{W}$ (Equation 1)

Case 2: When the body is placed in the right pan, it is balanced by a weight of 18 gm in the left pan. By the principle of moments:
$18\times L_1=W\times L_2$

From this, we can write:
$\frac{L_1}{L_2}=\frac{W}{18}$ (Equation 2)

Equating Equation 1 and Equation 2:
$\frac{8}{W}=\frac{W}{18}$

Cross-multiplying to solve for $W$:
$W^2=8\times 18$

$W^2=144$

Taking the square root on both sides:
$W=\sqrt{144}=12$ gm

Thus, the true weight of the body is 12 gm.