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Rotational Motion question

2021 · Q171
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Rotational Motion question

2021 · Q171

NEETPhysicsRotational MotionMCQ+4 / −1
A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g = 10 m/s2)

NEET 2021 Physics - Rotational Motion Question 15 English
  1. A
    112{1 \over {12}}121​kg
  2. B
    12{1 \over {2}}21​kg
  3. C
    13{1 \over {3}}31​kg
  4. D
    16{1 \over {6}}61​kg
View written solutionFree

Correct answer: A

By balancing torque

2g ×\times× 20 = 0.5g ×\times× 60 + mg ×\times× 120

⇒\Rightarrow⇒ m = 0.56{{0.5} \over 6}60.5​ kg = 112{1 \over {12}}121​ kg

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