In the diagram shown, the normal reaction force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is (Given $g=10 \mathrm{~ms}^{-2}$) (Consider the surface, to be smooth):

• Correct answer
  A

  $25 \mathrm{~N}$
• B

  $39 \mathrm{~N}$
• C

  $6 \mathrm{~N}$
• D

  $10 \mathrm{~N}$

Solution

Total force along inclined plane, $\begin{aligned} f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\ f & =-60+6 g \sin 30^{\circ}+18 \\ f & =-60+6 \times 10 \times \frac{1}{2}+18 \\ f & =12 \mathrm{~N} \text { (downwards) } \\ \text {acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$ As per question, the normal force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is, $\begin{aligned} \mathrm{N}-18-10 \sin 30^{\circ} & =m a \\ \mathrm{~N}-18-10 \times \frac{1}{2} & =1 \times 2 \\ \mathrm{~N} & =2+23 \\ \mathrm{~N} & =25 \mathrm{~N} \end{aligned}$

Hence, the correct answer is option: 'A'.