A force of (2i^+3j^+4k^)N acts on a particle whose position vector with respect to the origin of the coordinate system is (6i^+bj^+12k^)m. If the angular momentum of the body is constant, the value of 'b' is

The Angular momentum will be constant when the torque acting on the body is zero.So, τ=r→×F→=0(6i^+bj^+12k^)m×(2i^+3j^+4k^)N=0on solving we get b=9

A force of (2i^+3j^+4k^)N acts on a particle whose position vector with respect to the origin of the coordinate system is (6i^+bj^+12k^)m. If the angular momentum of the body is constant, the value of 'b' is

The Angular momentum will be constant when the torque acting on the body is zero.So, τ=r→×F→=0(6i^+bj^+12k^)m×(2i^+3j^+4k^)N=0on solving we get b=9

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A force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) N$ acts on a particle whose position vector with respect to the origin of the coordinate system is $(6 \hat{i} + b \hat{j} + 12 \hat{k}) m$ . If the angular momentum of the body is constant, the value of 'b' is

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The Angular momentum will be constant when the torque acting on the body is zero.

So, $τ = \vec{r} \times \vec{F} = 0$

$(6 \hat{i} + b \hat{j} + 12 \hat{k}) m \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) N = 0$

on solving we get b=9

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A force of (2i^+3j^+4k^)N acts on a particle whose position vector with respect to the origin of the coordinate system is (6i^+bj^+12k^)m. If the angular momentum of the body is constant, the value of 'b' is

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A force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) N$ acts on a particle whose position vector with respect to the origin of the coordinate system is $(6 \hat{i} + b \hat{j} + 12 \hat{k}) m$ . If the angular momentum of the body is constant, the value of 'b' is