A body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is 0.5 , then its retardation is ( $\mathrm{g}$ - acceleration due to gravity)

• Correct answer
  A

  $\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}$
• B

  $\left(\frac{2-\sqrt{3}}{4}\right) \mathrm{g}$
• C

  $\left(\frac{2-\sqrt{3}}{2}\right) \mathrm{g}$
• D

  $\left(\frac{2+\sqrt{3}}{2}\right) \mathrm{g}$

Solution

Retardation, $a=g(\sin \theta+\mu \cos \theta)$ $=\mathrm{g}\left[\sin 30^{\circ}+0.5\left(\cos 30^{\circ}\right)\right]$
$=g\left(\frac{1}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)=\left(\frac{1}{2}+\frac{\sqrt{3}}{4}\right) \mathrm{g}=\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}$

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