A force F- - - - 3- - 6k- is acting at a point r- - 2- - 6- - 12k- The value of - for which angular momentum about origin is conserved is-

Here, F⃗ = αî + 3ĵ + 6 k̂ or, F⃗ = F_xî + F_yĵ + F_zk̂ and r⃗ = 2 î 6 ĵ 12 k̂ or r⃗ = x î y ĵ z k̂ ∴F_x/x = F_y/y = F_z/z or, α/2 = 3/ 6 = 6/ 12 or, 6α ...

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Byju's Answer

Standard XIII

Physics

The Principle

A force F⃗ = ...

Question

A force $\to F = α^i + 3^j + 6^k$ is acting at a point $\to r = 2^i - 6^j - 12^k .$ The value of $α$ for which angular momentum about origin is conserved is:

-1

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zero

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Solution

The correct option is A -1
Here, $\to F = α^i + 3^j + 6^k$
or, $\to F = F_{x}^i + F_{y}^j + F_{z}^k$
and $\to r = 2^i - 6^j - 12^k$
or $\to r = x^i - y^j - z^k$
$∴ \frac{F_{x}}{x} = \frac{F_{y}}{y} = \frac{F_{z}}{z}$
or, $\frac{α}{2} = \frac{3}{- 6} = \frac{6}{- 12}$
or, $- 6 α = 6 \Rightarrow α = - 1$

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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:

The correct option is A -1Here, →F=α^i+3^j+6^k or, →F=Fx^i+Fy^j+Fz^k and →r=2^i−6^j−12^k or →r=x^i−y^j−z^k ∴Fxx=Fyy=Fzz or, α2=3−6=6−12 or, −6α=6⇒α=−1

A force $\to F = α^i + 3^j + 6^k$ is acting at a point $\to r = 2^i - 6^j - 12^k .$ The value of $α$ for which angular momentum about origin is conserved is: