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A raindrop falling from a height $h$ above ground, attains a near terminal velocity when it has fallen through a height $(3 / 4) h$. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?
Options:
- A
- B
- C
- D
Solution:
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The correct answer is:
P.E. is maximum when drop start falling at $t=0$ as it fall is P.E. decrease gradually to zero. So, it rejects the graph (a), (c) and (d).
K.E. at $t=0$ is zero as drop falls with zero velocity, its velocity increases (gradually), hence, first KE also increases. After sometime speed (velocity) is constant this is called terminal velocity, so, KE also become constant. It happens when it falls $\left(\frac{3}{4}\right)$ height or remains at $\left(\frac{4}{4}\right)$ from ground, then PE decreases continuously as the drop is falling continuously. The variation in PE and KE is best represented by (b).
K.E. at $t=0$ is zero as drop falls with zero velocity, its velocity increases (gradually), hence, first KE also increases. After sometime speed (velocity) is constant this is called terminal velocity, so, KE also become constant. It happens when it falls $\left(\frac{3}{4}\right)$ height or remains at $\left(\frac{4}{4}\right)$ from ground, then PE decreases continuously as the drop is falling continuously. The variation in PE and KE is best represented by (b).
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