(C) Since the surface is smooth,mechanical energy is conserved throughout the motion.Let the initial height be $h_1 = 100 \, m$ and the final height b

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(C) Since the surface is smooth,mechanical energy is conserved throughout the motion.Let the initial height be $h_1 = 100 \, m$ and the final height be $h_2 = 20 \, m$.The initial velocity $u = 0 \, m/s$.According to the law of conservation of energy:$PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$$mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2$$mg(h_1 - h_2) = \frac{1}{2}mv^2$$v = \sqrt{2g(h_1 - h_2)}$Substituting the values $(g = 10 \, m/s^2)$:$v = \sqrt{2 \times 10 \times (100 - 20)}$$v = \sqrt{20 \times 80}$$v = \sqrt{1600}$$v = 40 \, m/s$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Medium

$A$ spherical ball of mass $20 \, kg$ is stationary at the top of a hill of height $100 \, m$. It slides down a smooth surface to the ground,then climbs up another hill of height $30 \, m$ and finally slides down to a horizontal base at a height of $20 \, m$ above the ground. The velocity attained by the ball is ............... $m/s$.

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A
$10$
B
$10\sqrt{30}$
C
$40$
D
$20$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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