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Q)

Two blocks are in contact on a frictionless table. One has mass m and another mass '2m'. A force F is applied on '2m' as shown. Now the force F is applied on 'm' from right. In the two cases respectively the ratio of force of contact between the blocks is

a) same b) 2 :1 c) 1:2 d) 1:3

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A)
case I
$a=\large\frac{F}{3a}$
$N_1=ma$
$\qquad=m \times \large\frac{F}{3a}=\large\frac{F}{3}$
case II
$a=\large\frac{F}{3a}$
$N_2=2m \times a$
$\qquad=2m \times \large\frac{F}{3a}=\large\frac{2}{3}$$F$
$\large\frac{N_1}{N_2}=\frac{1}{2}$
Hence c is the correct answer.

How did u get a=F/3a

Comment
A)
Case 1 F=ma. (1) Total m=2m+m=3m According to first F=3ma Therefore, a=F/3m N1=ma=m*F/3m=F/3 (2) Case 2 a=F/3m N2=2ma=2m*F/3m=2F/3. (3) Now, divide equation (2) and (3) N1/N2=1/2
I think a=F/3m hona chahia.