JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 6)
A uniform sphere of mass 500 g rolls without
slipping on a plane horizontal surface with its
centre moving at a speed of 5.00 cm/s. Its
kinetic energy is :
8.75 × 10–3 J
1.13 × 10–3 J
8.75 × 10–4 J
6.25 × 10–4 J
Explanation
K.E = $${1 \over 2}m{V^2} + {1 \over 2}{I_{cm}}{\omega ^2}$$
= $${1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}$$
= $${1 \over 2}m{V^2} + {1 \over 5}m{V^2}$$
= $${7 \over {10}}m{V^2}$$
= $${7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}$$
= $${{35} \over 4} \times {10^{ - 4}}$$
= 8.75 × 10–4 J
= $${1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}$$
= $${1 \over 2}m{V^2} + {1 \over 5}m{V^2}$$
= $${7 \over {10}}m{V^2}$$
= $${7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}$$
= $${{35} \over 4} \times {10^{ - 4}}$$
= 8.75 × 10–4 J
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