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Rotational Motion question

2022 · 29 Jul · Shift 2 · Q58
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  5. /2022 · 29 Jul · Shift 2 · Q58

Rotational Motion question

2022 · 29 Jul · Shift 2 · Q58

JEE MainPhysicsRotational MotionMCQ+4 / −1
The torque of a force 5i^+3j^−7k^5 \hat{i}+3 \hat{j}-7 \hat{k}5i^+3j^​−7k^ about the origin is τ\tauτ. If the force acts on a particle whose position vector is 2i+2j+k2 i+2 j+k2i+2j+k, then the value of τ\tauτ will be
  1. A
    11i^+19j^−4k^11 \hat{i}+19 \hat{j}-4 \hat{k}11i^+19j^​−4k^
  2. B
    −11i^+9j^−16k^-11 \hat{i}+9 \hat{j}-16 \hat{k}−11i^+9j^​−16k^
  3. C
    −17i^+19j^−4k^-17 \hat{i}+19 \hat{j}-4 \hat{k}−17i^+19j^​−4k^
  4. D
    17i^+9j^+16k^17 \hat{i}+9 \hat{j}+16 \hat{k}17i^+9j^​+16k^
View written solutionFree

Correct answer: C

  1. Given data

    Force vector: F⃗=5i^+3j^−7k^\vec F = 5\hat i + 3\hat j - 7\hat kF=5i^+3j^​−7k^

    Position vector: r⃗=2i^+2j^+k^\vec r = 2\hat i + 2\hat j + \hat kr=2i^+2j^​+k^

    Torque about origin is given by: τ⃗=r⃗×F⃗\vec \tau = \vec r \times \vec Fτ=r×F

  2. Compute the cross product

    \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 2 & 1 \\ 5 & 3 & -7 \end{vmatrix}$$

  3. Expand the determinant

    - \hat j\,(2\cdot(-7) - 1\cdot 5) + \hat k\,(2\cdot 3 - 2\cdot 5)$$

  4. Simplify each component

    • For i^\hat ii^: 2(−7)−3=−14−3=−172(-7) - 3 = -14 - 3 = -172(−7)−3=−14−3=−17

    • For j^\hat jj^​: −(2(−7)−5)=−(−14−5)=19-\big(2(-7) - 5\big) = -(-14 - 5) = 19−(2(−7)−5)=−(−14−5)=19

    • For k^\hat kk^: 2⋅3−2⋅5=6−10=−42\cdot 3 - 2\cdot 5 = 6 - 10 = -42⋅3−2⋅5=6−10=−4

    Hence, τ⃗=−17i^+19j^−4k^\vec \tau = -17\hat i + 19\hat j - 4\hat kτ=−17i^+19j^​−4k^

  5. Match with options

    This corresponds to: C: −17i^+19j^−4k^\boxed{\text{C: } -17\hat i + 19\hat j - 4\hat k}C: −17i^+19j^​−4k^​

  6. Comparison with stored answer

    Stored correct answer is C, which matches the derived answer.

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