A force F = + 3j + 6k is acting at a point r = 2i - 6j - 12k. The value of α for which angular momentum about origin is conserved is (a) zero (b) 1 (c) - 1 (d) 2

Answer: (c). L is conserved when torque τ= r× F=0, i.e. F∥ r. r = 2 i-6 j-12 k = -2(- i+3 j+6 k) So F=α i+3 j+6 k is parallel to r only if α=-1. So the answer is (c) -1.

A force F = + 3j + 6k is acting at a point r = 2i - 6j - 12k. The… | NEET 2015 (AIPMT)

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Angular Momentum & its Conservation

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

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Answer: (c)

$\vec L$ is conserved when torque $\vec\tau=\vec r\times\vec F=0$ , i.e. $\vec F\parallel\vec r$ .
$\vec r = 2\hat i-6\hat j-12\hat k = -2(-\hat i+3\hat j+6\hat k)$
So $\vec F=\alpha\hat i+3\hat j+6\hat k$ is parallel to $\vec r$ only if $\alpha=-1$ .
So the answer is (c) $-1$ .

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