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(D) Given,height of the inclined plane,$h = 7 \ m$.Acceleration due to gravity,$g = 10 \ m/s^2$.By the law of conservation of energy,the potential ene

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$10 \ m/s$

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(D) Given,height of the inclined plane,$h = 7 \ m$.Acceleration due to gravity,$g = 10 \ m/s^2$.By the law of conservation of energy,the potential energy lost by the solid sphere equals the total kinetic energy gained (translational + rotational).$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and the condition for rolling without slipping is $\omega = \frac{v}{R}$.Substituting these values:$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{2}{5}mR^2ight) \left(\frac{v}{R}ight)^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$$mgh = \frac{7}{10}mv^2$$v^2 = \frac{10}{7}gh$$v = \sqrt{\frac{10}{7} 	imes 10 	imes 7} = \sqrt{100} = 10 \ m/s$.

$A$ solid sphere rolls down from the top of an inclined plane,$7 \ m$ high,without slipping. Its linear speed at the foot of the plane is $\left(g = 10 \ m/s^2\right)$

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