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Q.

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

6

b

9

c

12

d

3

answer is B.

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Detailed Solution

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=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