KVPY 2020-21 SA Exam Previous Papers

Question : 67 of 80

Marks: +1, -0

Three balls, A, B and C are released and all reach the point X (shown in the figure). Balls A and B are released from two identical structures, one kept on the ground and the other at height h, from the ground as shown in the figure. They take time

t_A

and

t_B

respectively to reach X (time starts after they leave the end of the horizontal portion of the structure). The ball C is released from a point at height, h, vertically above X and reaches X in time

t_C

. Choose the correct statement.

$t_C < t_A < t_C$
$t_C = t_A = t_C$
$t_C = t_A < t_C$
$t_B < t_A = t_C$

Validate

Solution:

In option (B), it should be

t_A = t_B = t_C

For ball A
As A leaves the structure, its velocity is horizontal and vertical component of velocity is zero.

t_{A}=\√{{2 h}/{g}}

Also,

x=v t_{A}

\⇒t_{A}={x}/{V}

Note: As per energy conservation:

m g h_{1}={1}/{2} m v^{2}

\⇒v=\√{2 g h_{1}}

\⇒t_{A}={x}/{\√{2 g h_{1}}}

For ball B

As per energy conservation

v_{B}=\√{2 g h_{1}}

t_{B}={x}/{v_{B}}={x}/{\√{2 g h_{1}}}

For ball C

t_{c}=\√{{2 h}/{g}}

fall freely from rest)

\∴t_{A}=t_{B}=t_{C}

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