Q114 / 150 · MHT CET 2024 (11 May Shift 1) · Physics

An inclined plane makes an angle $30^{\circ}$ with horizontal. A solid sphere rolls down from the top of the inclined plane from rest without slipping has a linear acceleration along the plane equal to (where g is acceleration due to gravity) (given $\sin 30^{\circ}=0.5$ )

Correct answer
A
$\frac{5 \mathrm{~g}}{14}$
B
$\frac{5 \mathrm{~g}}{4}$
C
$\frac{2 \mathrm{~g}}{3}$
D
$\frac{\mathrm{g}}{3}$

Solution

$\begin{aligned} & a=\frac{g \sin \theta}{\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}\right)} \\ \therefore \quad & a=\frac{5 \mathrm{~g}}{7} \times\left(\frac{1}{2}\right)=\frac{5 \mathrm{~g}}{14}\end{aligned}$

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

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