The torque due to the force () about the origin, acting on a particle whose position vector is (), would be [2025]

Question

The torque due to the force ( $2 \hat{i} + \hat{j} + 2 \hat{k}$ ) about the origin, acting on a particle whose position vector is ( $\hat{i} + \hat{j} + \hat{k}$ ), would be [2025]

Smart Achievers · Accepted Answer

(3)

$\vec{τ} = \vec{r} \times \vec{F}$

$\hat{i} + \hat{j} + \hat{k}$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{matrix} |$

$\vec{τ} = \hat{i} (2 - 1) - \hat{j} [0] + \hat{k} (1 - 2) = \hat{i} - \hat{k}$

Solution

Q.
The torque due to the force ( $2 \hat{i} + \hat{j} + 2 \hat{k}$ ) about the origin, acting on a particle whose position vector is ( $\hat{i} + \hat{j} + \hat{k}$ ), would be [2025]

1 ( $\hat{i} - \hat{j} + \hat{k}$ )

2 ( $\hat{i} + \hat{k}$ )

3 ( $\hat{i} - \hat{k}$ )

4 ( $\hat{j} - \hat{k}$ )

Ans.
(3)

$\vec{τ} = \vec{r} \times \vec{F}$

$\vec{τ} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{matrix} |$

$\vec{τ} = \hat{i} (2 - 1) - \hat{j} [0] + \hat{k} (1 - 2) = \hat{i} - \hat{k}$

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Solution

Q. The torque due to the force (2i^+j^+2k^) about the origin, acting on a particle whose position vector is (i^+j^+k^), would be [2025]

1 (i^–j^+k^)

2 (i^+k^)

3 (i^–k^)

4 (j^–k^)

Ans. (3) τ→=r→×F→ τ→=|i^j^k^111212| τ→=i^(2–1)–j^[0]+k^(1–2)=i^–k^