Question

Three blocks of masses $m_{1}, \, m_{2}$ and $m_{3}$ are connected to a massless string on a frictionless table as shown in the figure. They are pulled with a force of $40 N.$ If $m_{1}=10 \, kg, \, m_{2}=6 \, kg$ and $m_{3}=4 \, kg,$ then tension $T_{2}$ will be

• A. $10N$
• B. $20N$
• C. $32N$
• D. $40N$

Answer 1

Answer: (B)

Since, the table is frictionless, i.e., it is smooth, therefore, force on the block is given by
$F=\left(m_{1} + m_{2} + m_{3}\right)a$
$\Rightarrow \, \, a=\frac{F}{m_{1} + m_{2} + m_{3}}$
$\Rightarrow \, \, a=\frac{40}{10 + 6 + 4}=\frac{40}{20}=2 \, m \, \text{s}^{- 2}$
Now, the tension between 10 kg and 6 kg masses is given by
$T_{2}=\left(m_{2} + m_{3}\right)a$
$=\left(4 + 6\right)2=10\times 2$
$T_{2}=20 \, \text{N}$