Question

12. In the given diagram, reading of spring balance will
be (g = 10 m/s2)
[NCERT Pg. 100]
0000
3 kg
6 kg
(1) 30 N
(2) 40 N
(3) 60 N
(4) 80 N​

Answer 1

Answer:

Using FBD, 

10a=10g−T or T=100−10a

5a=T−5g or 5a=100−10a−5g

or, 15a=50

Thus, T=100−10(50/15)=200/3N

This is the reading of spring balance, which is none of the answers