A force F → = ( i ^ + 2 j ^ + 3 k ^ )   N acts at a point ( 4 i ^ + 3 j ^ - k ^ )   m . Then…

A force F=(i^+2j^+3k^) N acts at a point (4i^+3j^-k^) m. Then the magnitude of torque about the point (i^+2j^+k^) m will be x N-m.The value of x is..........

Solution

r=4-1i^+3-2j^+-1-1k^

=3i^+j^-2k^

τ=r×F=i^jk^31-2123

=i^7-j^11+k^5=7i^-11j^+5k^

=49+121+25=195

Asked in: JEE Main 2020 (05 Sep Shift 1)

Practice more Rotational Motion questions on Aicharya