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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, \(\sin 30^{\circ}=0.5\) )

  1. A \(\frac{2 \mathrm{~g}}{3}\).
  2. B \(\frac{5 g}{14}\)
  3. C \(\frac{\mathrm{g}}{3}\)
  4. D \(\frac{5 \mathrm{~g}}{7}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{5 g}{14}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \quad a=\frac{g \sin \theta}{\left(1+\frac{K^2}{R^2}\right)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}\right)} \\ & \therefore \quad a=\frac{5 g}{7} \times 0.5=\frac{5 g}{14}\end{aligned}\)