MCQ
A solid sphere is rolling down an inclined plane. Then the ratio of its translational kinetic energy to its rotational kinetic energy is
- ✓$2.5$
- B$1.5$
- C$1$
- D$0.4$
✓
Answer
Correct option: A.
$2.5$
a
$\frac{\mathrm{K}_{\mathrm{T}}}{\mathrm{K}_{\mathrm{R}}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}^{2} \mathrm{K}^{2} / \mathrm{R}^{2}}=\frac{5}{2},\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)$ for solid sphere $=\frac{2}{5}$
$\frac{\mathrm{K}_{\mathrm{T}}}{\mathrm{K}_{\mathrm{R}}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}^{2} \mathrm{K}^{2} / \mathrm{R}^{2}}=\frac{5}{2},\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)$ for solid sphere $=\frac{2}{5}$
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