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MCQMediumJEE 2023Equilibrium of Rigid Bodies

JEE Physics 2023 Question with Solution

An object of mass 8 kg8 \, \text{kg} is hanging from one end of a uniform rod CD of mass 2 kg2 \, \text{kg} and length 1 m1 \, \text{m} pivoted at its end C on a vertical wall. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is:

A horizontal rod CD is pivoted at C on a vertical wall, with cable AB making 30 degree angle at point B, and an 8 kg mass hanging from end D. Distances shown are 60 cm from C to B and 40 cm from B to D.
  • A

    240 N240 \, \text{N}

  • B

    90 N90 \, \text{N}

  • C

    300 N300 \, \text{N}

  • D

    30 N30 \, \text{N}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A uniform rod of mass 2 kg2 \, \text{kg} and length 1 m1 \, \text{m} is hinged at CC. A mass of 8 kg8 \, \text{kg} hangs at the end. The cable is attached at point BB, 60 cm60 \, \text{cm} from CC, and makes an angle of 30∘30^\circ with the rod.

Find: The tension TT in the cable.

Free body diagram of the rod showing hinge at C, upward component T sin 30 at 60 cm, rod weight 20 N acting at 50 cm, and hanging load 80 N acting at 100 cm from C.

Taking torque about point CC:

(T2)×60=20×50+80×100\left(\frac{T}{2}\right) \times 60 = 20 \times 50 + 80 \times 100

Therefore,

3T=100+8003T = 100 + 800 T=300 NT = 300 \, \text{N}

Therefore, the tension in the cable is 300 N300 \, \text{N}. The correct option is C.

The solution states the answer as C. The torque balance uses g=10 m/s2g = 10 \, \text{m/s}^2 and distances in cm, which is valid since all lever arms are in the same unit.

Torque Balance with Components

Given: Weight of the rod acts at its centre, 0.5 m0.5 \, \text{m} from CC. Weight of the hanging object acts at 1 m1 \, \text{m} from CC. Only the vertical component of tension produces torque about CC.

Find: Tension TT.

The vertical component of tension is:

Tsin⁡30∘=T2T \sin 30^\circ = \frac{T}{2}

Now balance clockwise and anticlockwise torques about CC:

(Tsin⁡30∘)(0.6)=(2g)(0.5)+(8g)(1)\left(T \sin 30^\circ\right)(0.6) = (2g)(0.5) + (8g)(1)

Using g=10 m/s2g = 10 \, \text{m/s}^2,

(T2)(0.6)=20(0.5)+80(1)\left(\frac{T}{2}\right)(0.6) = 20(0.5) + 80(1) 0.3T=10+80=900.3T = 10 + 80 = 90 T=900.3=300 NT = \frac{90}{0.3} = 300 \, \text{N}

Therefore, the correct option is C.

Common mistakes

  • Using the full tension TT as the torque-producing force is incorrect because the cable is inclined. Only the perpendicular component Tsin⁡30∘T \sin 30^\circ produces torque about the hinge. Resolve the tension first.

  • Taking the rod's weight to act at the end DD is wrong. For a uniform rod, its weight acts at the centre of mass, which is 0.5 m0.5 \, \text{m} from CC. Always place the rod's weight at its midpoint.

  • Ignoring the given position of point BB leads to an incorrect lever arm. The tension acts at 60 cm60 \, \text{cm} from CC, not at the full 1 m1 \, \text{m} length. Use the actual point of application.

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