Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio M/m will be

Question

Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio M/m will be (A) 2 (B) 3 (C) √2 (D) √3

Tardigrade Admin · Accepted Answer

Net torque about O should be zero. Hence textmg(( textl/2) textsin 60Â°)= textMg(( textl/2) textsin 30Â°) â´ ( textM/ textm)=( textsin 60Â° / textsin 30Â°)=√3

Q. Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio $M / m$ will be

$2$

$3$

$2$

$3$

Net torque about O should be zero. Hence
$mg (\frac{l}{2} sin 60°) = Mg (\frac{l}{2} sin 30°)$
∴ $\frac{M}{m} = \frac{sin 60°}{sin 30°} = 3$

Q. Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio M/m will be

2

3

2​

3​

Net torque about O should be zero. Hence mg(2l​sin 60°)=Mg(2l​sin 30°) ∴ mM​=sin 30°sin 60°​=3​

Q. Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio $M / m$ will be

$2$

$3$

$2$

$3$

Net torque about O should be zero. Hence
$mg (\frac{l}{2} sin 60°) = Mg (\frac{l}{2} sin 30°)$
∴ $\frac{M}{m} = \frac{sin 60°}{sin 30°} = 3$