d
The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, \(m_{1}=8 \,kg\)

Larger mass, \(m_{2}=12\, kg\)

Tension in the string \(=T\)

Mass \(m_{2},\) owing to its weight, moves downward with acceleration \(a,\) and mass \(m_{1}\) moves upward.

Applying Newton's second law of motion to the system of each mass:

For mass \(m_{\underline{1}}:\) The equation of motion can be written as:

\(T-m_{1} g =m a\)

For mass \(m_{2}\) : The equation of motion can be written as:

\(m_{2} g -T=m_{2} a\)

Adding above equations , we get:

\(\left(m_{2}-m_{1}\right) g =\left(m_{1}+m_{2}\right) a\)

\(\therefore a=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) g\)

\(=\left(\frac{12-8}{12+8}\right) \times 10=\frac{4}{20} \times 10=2 m / s ^{2}\)

Therefore, the acceleration of the masses is \(2 \;m / s ^{2}\)

Substituting the value of \(a\) in equation ( \(i i\) ), we get:

\(m_{2} g -T=m_{2}\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) g\)

\(T=\left(m_{2}-\frac{m_{2}^{2}-m_{1} m_{2}}{m_{1}+m_{2}}\right) g\)

\(=\left(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right) g\)

\(=\left(\frac{2 \times 12 \times 8}{12+8}\right) \times 10\)

\(=\frac{2 \times 12 \times 8}{20} \times 10=96\, N\)

Therefore, the tension in the string is \(96\, N\).