Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F=600 N is applied to 10 kg block, then the tension in the string is

Options

A. 100N
B. 200N
C. 300N
D. 400N

Answer

D

Explanation

Here, `m_(1)=10kg, m=20kg, F=600N`
Let T be tension of the string and a be common acceleration of the system.
`a=(F)/(m_(1)+m_(2))=(600N)/(10kg+20kg)=(600)/(30)ms^(-2)=20ms^(-2)`

When force F is applied on 10kg block, then the tension in the string is
`T_(2)=m_(2)a=(20kg)(20ms^(-2))=400N`