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Rotational Motion question

2019 · 11 Jan · Shift 2 · Q51
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Rotational Motion question

2019 · 11 Jan · Shift 2 · Q51

JEE MainPhysicsRotational MotionMCQ+4 / −1
The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :
  1. A
    π8{\pi \over 8}8π​
  2. B
    π6{\pi \over 6}6π​
  3. C
    π4{\pi \over 4}4π​
  4. D
    π3{\pi \over 3}3π​
View written solutionFree

Correct answer: B

  1. Use the formula for torque magnitude

    The magnitude of torque on a particle is τ=rFsin⁡θ\tau = rF\sin\thetaτ=rFsinθ where:

    • τ=2.5 N m\tau = 2.5\,\text{N m}τ=2.5N m
    • r=5 mr = 5\,\text{m}r=5m
    • F=1 NF = 1\,\text{N}F=1N
    • θ\thetaθ is the angle between r⃗\vec rr and F⃗\vec FF

  2. Substitute the given values

    2.5=5⋅1⋅sin⁡θ2.5 = 5 \cdot 1 \cdot \sin\theta2.5=5⋅1⋅sinθ

    2.5=5sin⁡θ2.5 = 5\sin\theta2.5=5sinθ

  3. Solve for sin⁡θ\sin\thetasinθ

    sin⁡θ=2.55=0.5\sin\theta = \frac{2.5}{5} = 0.5sinθ=52.5​=0.5

  4. Find the angle

    θ=sin⁡−1(0.5)=π6\theta = \sin^{-1}(0.5) = \frac{\pi}{6}θ=sin−1(0.5)=6π​

    Since the options contain the acute angle, we take θ=π6\boxed{\theta = \frac{\pi}{6}}θ=6π​​

  5. Check options

    • A: π8\frac{\pi}{8}8π​ ❌
    • B: π6\frac{\pi}{6}6π​ ✅
    • C: π4\frac{\pi}{4}4π​ ❌
    • D: π3\frac{\pi}{3}3π​ ❌

Therefore, the correct option is B.

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