+91-9566306857  (or)  +91-9176170648

Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

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

1 Answer

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


answered Jul 16, 2013 by meena.p
edited Feb 10, 2014 by meena.p
How did u get a=F/3a

Related questions