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…

A force F=αi^+3j^+6k^ is acting at a point r=2i^-6j^-12k^ . The value of α for which angular momentum about origin is conserved is:
  1. 2
  2. Zero
  3. 1
  4. 1

Solution

If L= constant then τ=0
So r×F=0F should be parallel to r so coefficient should be in same ratio. So α2=3-6=6-12
So α=-1 .

Asked in: NEET 2015 (Phase 2)

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