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

 

 

Here, ${\mathrm{m}}_1=10 \mathrm{kg}, {\mathrm{m}}_2=20 \mathrm{kg}, \mathrm{F}=600 \mathrm{N}$

Let T be the tension of the string and a be common acceleration of the system.

$$\begin{gathered} \therefore a=\frac{F}{m_1+m_2} \\ =\frac{600 N}{10 kg+20 kg} \\ =\frac{600}{30} m s^{-2} \\ =20 m s^{-2} \end{gathered}$$

When a force F is applied on 10 kg block, then the tension in the string is

$$\begin{gathered} T=m_{2}a \\ =\left(20 kg\right)(20 m s^{-2}) \\ =400 N \end{gathered}$$