A force F → = α i ^ + 3 j ^ + 6 k ^ is acting at a point r → = 2 i ^ - 6 j ^ - 12 k ^ . The value of α for which angular momentum about origin is conserved is:NEET 2015 (Phase 2)

If L → = constant then τ → = 0 So r → × F → = 0 ⇒ F → should be parallel to r → so coefficient should be in same ratio. So α 2 = 3 - 6 = 6 - 12 So α = - 1

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationAngular momentumEasy2 minPYQ_2015

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A force $\vec{F} = α \hat{i} + 3 \hat{j} + 6 \hat{k}$ is acting at a point $\vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k}$ . The value of $α$ for which angular momentum about origin is conserved is:NEET 2015 (Phase 2)

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Solution:

If $\vec{L} =$ constant then $\vec{τ} = 0$

So $\vec{r} \times \vec{F} = 0 \Rightarrow \vec{F}$ should be parallel to $\vec{r}$ so coefficient should be in same ratio. So $\frac{α}{2} = \frac{3}{- 6} = \frac{6}{- 12}$

So $α = - 1$

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Angular momentum

⏱ 2mℹ️ Source: PYQ_2015

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