MHT CET · Physics · Rotational Motion
An inclined plane makes an angle of \(30^{\circ}\) with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ( \(\mathrm{g}=\) acceleration due to gravity, \(\sin 30^{\circ}=0.5\) )
- A \(\frac{2 \mathrm{~g}}{3}\).
- B \(\frac{5 g}{14}\)
- C \(\frac{\mathrm{g}}{3}\)
- D \(\frac{5 \mathrm{~g}}{7}\)
Answer & Solution
Correct Answer
(B) \(\frac{5 g}{14}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \quad a=\frac{g \sin \theta}{\left(1+\frac{K^2}{R^2}\right)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}\right)} \\ & \therefore \quad a=\frac{5 g}{7} \times 0.5=\frac{5 g}{14}\end{aligned}\)
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