KCET · Physics · Oscillations
A force \( \vec{F}=5 \hat{i}+2 \hat{j}-5 \hat{k} \) acts on a particle whose position vector is \( \vec{r}=\hat{i}-2 \hat{j}+\hat{k} \).
What is the torque about the origin ?
- A \( 8 \hat{i}+10 \hat{j}+12 \hat{k} \)
- B \( 8 \hat{i}+10 \hat{j}-12 \hat{k} \)
- C \( 8 \hat{i}-10 \hat{j}-8 \hat{k} \)
- D \( 10 \hat{i}-10 \hat{j}-\hat{k} \)
Answer & Solution
Correct Answer
(A) \( 8 \hat{i}+10 \hat{j}+12 \hat{k} \)
Step-by-step Solution
Detailed explanation
Given \( \vec{F}=5 \hat{i}+2 \hat{j}-5 \hat{k} ; \vec{r}=\hat{i}-2 \hat{j}+\hat{k} \)
We know torque
\(\tau=\vec{r} \times \vec{F}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 1 \\5 & 2 & -5\end{array}\right|\)
Therefore,
\(\tau=\hat{i}(10-2)-\hat{j}(-5-5)+\hat{k}(2+10) \)
\(\Rightarrow \tau=8 \hat{i}+10 \hat{j}+12 \hat{k}\)
Thus, torque about origin is \( 8 \hat{i}+10 \hat{j}+12 \hat{k} \)
We know torque
\(\tau=\vec{r} \times \vec{F}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 1 \\5 & 2 & -5\end{array}\right|\)
Therefore,
\(\tau=\hat{i}(10-2)-\hat{j}(-5-5)+\hat{k}(2+10) \)
\(\Rightarrow \tau=8 \hat{i}+10 \hat{j}+12 \hat{k}\)
Thus, torque about origin is \( 8 \hat{i}+10 \hat{j}+12 \hat{k} \)
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