Q.A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. Ratio of the tension in the string close to the boy to that in the other string is (A) 2 (B) 1.5 (C) 4/3 (D) 5/3

[!TLDR] Taking torques about the far end and solving for both string tensions from force- and torque-balance gives the near-string tension 3mg again …

Let the rod have length L, weight 2mg acting at its centre (L/2), and the boy of weight 3mg hang at one-third of the length, i.e. at L/3 from one end (call it end A, with the near string at A and the far string at end B, a distance L away). Force balance (vertical): $T_A+T_B=2mg+3mg=5mg$. Torque balance about end A (so the unknown $T_A$ drops out): the rod's weight at L/2 and the boy's weight at L/3 both produce clockwise moments about A, balanced by $T_B$ acting at L: $T_B\cdot L=2mg\left(\frac{L}{2}\right)+3mg\left(\frac{L}{3}\right)=mgL+mgL=2mgL$, so $T_B=2mg$. Then $T_A=5mg-2mg=3mg$. The string CLOSE to the boy (at end A, since the boy is at L/3 from A) has tension $T_A=3mg$; the str …

Use vertical force balance (sum of the two tensions = total weight) plus torque balance about one end of the rod to solve for …

Taking torques about the wrong point (e.g. about the rod's centre instead of an end), which needlessly complicates solving for the two unknown …