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Q.

$\bar{F} = a \bar{i} + 3 \bar{j} + 6 \bar{k}$ and $\bar{r} = 2 \bar{i} - 6 \bar{j} - 12 \bar{k} .$ The value of ‘a’ for which the angular momentum is conserved is

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a

2

b

1

c

-1

d

0

answer is A.

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Detailed Solution

$\bar{τ} = \bar{r} \times \bar{F} = 0$
$r F \sin θ = 0$
$θ = 0$
$\bar{r}$ and $\bar{F}$ are parallel
$\frac{2}{a} = \frac{- 6}{3} = \frac{- 12}{6}$
$\frac{2}{a} = - 2 = - 2$
$a = - 1$

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F¯=ai¯+3j¯+6k¯ and r¯=2i¯−6j¯−12k¯.The value of ‘a’ for which the angular momentum is conserved is