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A force $\vec{F}=\alpha \hat{i}+3 \hat{\jmath}+6 \hat{k}$ is acting at a point $\vec{r}=2 \hat{1}-6 \hat{j}-12 \hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: $(1)-$ (2) $-1$ (3) 2 (4) Zero
A force $\vec{F}=\alpha \hat{i}+3 \hat{\jmath}+6 \hat{k}$ is acting at a point $\vec{r}=2 \hat{1}-6 \hat{j}-12 \hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: $(1)-$ (2) $-1$ (3) 2 (4) Zero
Physics | Physics
A force $\vec{F}=\alpha \hat{i}+3 \hat{\jmath}+6 \hat{k}$ is acting at a point $\vec{r}=2 \hat{1}-6 \hat{j}-12 \hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: $(1)-$ (2) $-1$ (3) 2 (4) Zero

The Solution is (2).
If $\overrightarrow{\mathrm{L}}=$ constant then $\vec{\tau}=0$
So $\vec{r} \times \vec{F}=0 \Rightarrow \vec{F}$ should be parallel to $\vec{r}$ so coefficient should be in same ratio. so $\frac{\alpha}{2}=\frac{3}{-6}=\frac{6}{-12}$
So $\alpha=-1$

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