As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle $30^{\circ}$, with horizontal for $\mu_{\mathrm{s}}=0.25$, the block will just start to move for the value of F : [Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$]

• Correct answer
  A

  25.2 N
• B

  20 N
• C

  33.3 N
• D

  33.7 N

Solution

Hint: For block to move
$$ \begin{aligned} &\begin{aligned} & \mathrm{mg}-\mathrm{F} \sin 30^{\circ}=\mathrm{N} \\ & \mathrm{f}=\mu(\mathrm{N}) \quad=\mu\left(\mathrm{mg}-\mathrm{F} \sin 30^{\circ}\right) \\ & \mu=0.25 \text { given } \end{aligned}\\ &\text { To start motion , F } \cos 30^{\circ} \geq \mu \mathrm{N}\\ &\begin{aligned} & \mathrm{F} \cos 30^{\circ} \geq \mu\left(\mathrm{mg}-\mathrm{F} \sin 30^{\circ}\right) \\ & \mathrm{F} \frac{\sqrt{3}}{2} \geq \frac{1}{4}\left(100-\mathrm{F} \frac{1}{2}\right) \quad \mathrm{F} \frac{\sqrt{3}}{2}+\mathrm{F} \frac{1}{8} \geq \frac{100}{4} \\ & \mathrm{~F}\left(\frac{4 \sqrt{3}+1}{8}\right) \geq 25, \mathrm{~F}=\frac{25 \times 8}{4 \sqrt{3}+1}=25.2 \mathrm{~N} \end{aligned} \end{aligned} $$

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