Question
A spherical shell rolls down a plane inclined at $30^{\circ}$ to the horizontal. What is its acceleration ?

Answer

The acceleration of the spherical shell, $\mathrm{a}=\frac{3}{5} \mathrm{~g} \sin \theta=0.6 \mathrm{~g} \sin 30^{\circ}=0.6 \mathrm{~g} \times \frac{1}{2}=0.3 \mathrm{~g} \mathrm{~m}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free