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Physics Class 12th System of Particles and Rotational Motion physics ncert exemplar solutions class 12th chapter two

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is-

Option 1 -

$\frac{2}{5}$

Option 2 -

$\frac{2}{7}$

Option 3 -

$\frac{1}{5}$

Option 4 -

$\frac{7}{10}$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

K.ER (Rotational) = $\frac{1}{2} I_{C M} ω^{2} = \frac{1}{2} \times \frac{2}{5} M R^{2} ω^{2} = \frac{M R^{2} ω^{2}}{5}$ - (1)

$K . E_{T o t a l} = K . E_{R} + K . E_{T}$

$= \frac{M R^{2} ω^{2}}{5} + \frac{1}{2} M V_{c m^{2}}$

$= \frac{7 M R^{2} ω^{2}}{10}$ - (2)

$\frac{k . E_{R}}{K . E_{T o t a l}} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$

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