An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is θ, then :

Question

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is θ, then : (A)  tan θ = (2/√3) (B)  tan θ = (1/3) (C)  tan θ = (1/2) (D)  tan θ = (1/2 √3)

Tardigrade Admin · Accepted Answer

Let mass of one rod is m. Balancing torque about hinge point. mg (C1P) = mg (C2N) mg ((L/2) sinθ) = mg ((L/2) cosθ - L sinθ) ⇒ (3/2) mg L sinθ = (mgL/2) cosθ ⇒ tan θ = (1/3)

Q. An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $A B = BC$ , and the angle made by $A B$ with downward vertical is $θ$ , then :

$tan θ = \frac{2}{3}$

$tan θ = \frac{1}{3}$

$tan θ = \frac{1}{2}$

$tan θ = \frac{1}{2 3}$

Let mass of one rod is $m$ .
Balancing torque about hinge point.
$m g (C_{1} P) = m g (C_{2} N)$
$m g (\frac{L}{2} sin θ) = m g (\frac{L}{2} cos θ - L sin θ)$
$\Rightarrow \frac{3}{2} m gL sin θ = \frac{m gL}{2} cos θ$
$\Rightarrow tan θ = \frac{1}{3}$

Q. An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB=BC, and the angle made by AB with downward vertical is θ, then :

tanθ=3​2​

tanθ=31​

tanθ=21​

tanθ=23​1​