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Rotational Motion question

2003 · Shift 0 · Q147
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Rotational Motion question

2003 · Shift 0 · Q147

JEE MainPhysicsRotational MotionMCQ+4 / −1
Let F→\overrightarrow FF be the force acting on a particle having position vector r→,\overrightarrow r ,r, and τ→\overrightarrow \tauτ be the torque of this force about the origin. Then
  1. A
    r.→τ→=0  \overrightarrow {r.} \overrightarrow \tau = 0\,\,r.τ=0 and F.→τ→e0  \overrightarrow {F.} \overrightarrow \tau e 0\,\,F.τe0
  2. B
    r.→τ⃗e0  \overrightarrow {r.} \vec \tau e 0{\mkern 1mu} {\mkern 1mu}r.τe0 and F.→τ→=0  \overrightarrow {F.} \overrightarrow \tau = 0\,\,F.τ=0
  3. C
    r.→τ⃗e0 \overrightarrow {r.} \vec \tau e 0{\mkern 1mu}r.τe0 and F.→τ→e0\overrightarrow {F.} \overrightarrow \tau e 0F.τe0
  4. D
    r.→τ⃗=0 \overrightarrow {r.} \vec \tau = 0{\mkern 1mu}r.τ=0 and F.→τ→=0  \overrightarrow {F.} \overrightarrow \tau = 0\,\,F.τ=0
View written solutionFree

Correct answer: D

  1. Write the torque in vector form

    The torque of force F⃗\vec FF about the origin for a particle at position r⃗\vec rr is

    τ⃗=r⃗×F⃗.\vec \tau = \vec r \times \vec F.τ=r×F.

  2. Check r⃗⋅τ⃗\vec r \cdot \vec \taur⋅τ

    We need

    r⃗⋅τ⃗=r⃗⋅(r⃗×F⃗).\vec r \cdot \vec \tau = \vec r \cdot (\vec r \times \vec F).r⋅τ=r⋅(r×F).

    Now, r⃗×F⃗\vec r \times \vec Fr×F is perpendicular to r⃗\vec rr. Hence their dot product is zero:

    r⃗⋅τ⃗=0.\vec r \cdot \vec \tau = 0.r⋅τ=0.

  3. Check F⃗⋅τ⃗\vec F \cdot \vec \tauF⋅τ

    Similarly,

    F⃗⋅τ⃗=F⃗⋅(r⃗×F⃗).\vec F \cdot \vec \tau = \vec F \cdot (\vec r \times \vec F).F⋅τ=F⋅(r×F).

    Since r⃗×F⃗\vec r \times \vec Fr×F is also perpendicular to F⃗\vec FF, we get

    F⃗⋅τ⃗=0.\vec F \cdot \vec \tau = 0.F⋅τ=0.

  4. Conclusion

    Both quantities are zero:

    r⃗⋅τ⃗=0,F⃗⋅τ⃗=0.\vec r \cdot \vec \tau = 0, \qquad \vec F \cdot \vec \tau = 0.r⋅τ=0,F⋅τ=0.

    Therefore, the correct option is

    D.\boxed{\text{D}}.D​.
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