Q132 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Physics

The torque due to force $\mathbf{F}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ acting at a point $\mathbf{r}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ is

  • A
    $-4 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+22 \hat{\mathbf{k}}$
  • Correct answer
    B
    $4 \hat{\mathbf{i}}+17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}}$
  • C
    $8 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+22 \hat{\mathbf{k}}$
  • D
    $-4 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}}$

Solution

Given, $$ \begin{aligned} F & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{r}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \tau & =\mathbf{v} \times \mathbf{F}=(8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 8 & 2 & 3 \\ 3 & -2 & -1 \end{array}\right| \\ & =\hat{\mathbf{i}}(-2+6)-\hat{\mathbf{j}}(-8-9)+\hat{\mathbf{k}}(-16-6) \\ & =4 \hat{\mathbf{i}}+17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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