MHT CET · Physics · Rotational Motion
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 \(\sin 30^{\circ}=0.5\) )
- A \(\frac{5 \mathrm{~g}}{14}\)
- B \(\frac{5 \mathrm{~g}}{4}\)
- C \(\frac{2 \mathrm{~g}}{3}\)
- D \(\frac{\mathrm{g}}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{5 \mathrm{~g}}{14}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & a=\frac{g \sin \theta}{\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}\right)} \\ \therefore \quad & a=\frac{5 \mathrm{~g}}{7} \times\left(\frac{1}{2}\right)=\frac{5 \mathrm{~g}}{14}\end{aligned}\)
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