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QJEEM19Pha51e: JEE Main 2019 Physics - Class 11 JEEM19Pha51e Solution: An L-shaped object, made of thin rods of uniform mass density,…

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\theta , thrown :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Rotational Motion Question 177 English

Multiple Choice Options

  • A tan θ\theta = 123{1 \over {2\sqrt 3 }}
  • B tan θ\theta = 12{1 \over 2}
  • C tan θ\theta = 23{2 \over {\sqrt 3 }}
  • D tan θ\theta = 13{1 \over 3}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Morning Slot Physics - Rotational Motion Question 177 English Explanation
Assume mass of each part = m

Here, C1RC1A=sin⁡θ{{{C_1}R} \over {{C_1}A}} = \sin \theta

∴\therefore    C1R = C1A(sin θ\theta )

= 12{1 \over 2} sin θ\theta

BYAB=sin⁡θ{{BY} \over {AB}} = \sin \theta

⇒\Rightarrow    BY = Lsin θ\theta

As By = MP

∴\therefore    MP = Lsin θ\theta

C2MBC2{{{C_2}M} \over {B{C_2}}} = cos θ\theta

⇒\Rightarrow    C2M = BC2cos θ\theta

⇒\Rightarrow    C2M = L2cos⁡θ{L \over 2}\cos \theta

∴\therefore    C2P = C2M −- MP = L2cos⁡θ{L \over 2}\cos \theta −- Lsin θ\theta

Now balancing torque about hinge paint A,

mg(C1R) = mg(C2P)

⇒\Rightarrow    mg (L2sin⁡θ)\left( {{L \over 2}\sin \theta } \right) = mg (L2cos⁡θ−Lsin⁡θ)\left( {{L \over 2}\cos \theta - L\sin \theta } \right)

⇒\Rightarrow     sin⁡θ2{{\sin \theta } \over 2} = cos⁡θ2−{{\cos \theta } \over 2} - sin θ\theta

⇒\Rightarrow     3sin⁡θ2{{3\sin \theta } \over 2} = cos⁡θ2{{\cos \theta } \over 2}

⇒\Rightarrow    tan θ\theta = 13{1 \over 3}

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