Question

A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2m from the person. In the following, the state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

List I	List II
P	Lift is accelerating vertically up	1	d =1.2m
Q	Lift is accelerating vertically down with an acceleration less than the gravitational acceleration	2	d >1.2m
R	Lift is moving vertically up with constant speed	3	d <1.2m
S	Lift is free falling	4	No water leaks out of the jar

A. P-2, Q-3, R-2, S-4
B. P-2, Q-3, R-1, S-4
C. P-1, Q-1, R-1, S-4
D. P-2, Q-3, R-1, S-1

Answer 1

Hint: Determine the velocity of efflux and the time to fall down the vertical distance. It will help in determining the horizontal distance it will travel. From finding the expression for horizontal distance, obtain the horizontal distance for all four given cases. Compare them and find the correct option.

Formula used:
Velocity of efflux is,
  \[v=\sqrt{2{{g}_{eff}}l}\] ……………… (1)
Where, \[{{g}_{eff}}\] is the effective gravity of the system.
                 \[l\] is the height of the jar.

Distance travelled by an object in time t is given by,
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\] ………………… (2)
Where, u is the initial velocity, a is the acceleration, t is the time of motion.

Complete step by step answer:
The velocity of efflux from a jar of height \[l\] is,
\[v=\sqrt{2{{g}_{eff}}l}\]
Now by having the velocity, we can calculate the distance it will travel in the horizontal direction.
We can consider the water as a projectile motion.
So, there is an acceleration working on the water particles in vertical direction which is given by,
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
But, The initial vertical velocity of water is 0.
So, \[u=0\] and \[a\] is given by \[a={{g}_{eff}}\] .
Hence, we can write equation (2) in the following manner,
\[\begin{align}
  & h=\dfrac{1}{2}{{g}_{eff}}{{t}^{2}} \\
 & \Rightarrow t=\sqrt{\dfrac{2h}{{{g}_{eff}}}} \\
\end{align}\]
So, time taken by the water molecule to reach ground is,
\[t=\sqrt{\dfrac{2h}{{{g}_{eff}}}}\]
Hence, the horizontal distance is given by,
\[\begin{align}
  & d=vt \\
 & =\sqrt{2{{g}_{eff}}l}\times \sqrt{\dfrac{2h}{{{g}_{eff}}}} \\
 & =2\sqrt{hl}=1.2m \\
\end{align}\]
The Quantity is independent of the \[{{g}_{eff}}\] .
Now, if we consider case P, the lift is moving up with an acceleration. i.e. \[{{g}_{eff}}=g+a\]
But, from the expression for horizontal distance,
\[d=2\sqrt{hl}=1.2m\]
For case Q, where the lift is decelerating downwards. i.e. \[{{g}_{eff}}=g-a\].
But, from the expression for horizontal distance, which is independent of effective acceleration due to gravity, distance will be,
\[d=2\sqrt{hl}=1.2m\]
For case R, where the lift is moving vertically up with constant speed. i.e. \[{{g}_{eff}}=g\].
But, from the expression for horizontal distance, which is independent of effective acceleration due to gravity, distance will be the same as other cases. i.e.
\[d=2\sqrt{hl}=1.2m\]
So, we can conclude that the range will be the same for the first three cases i.e. \[{{d}_{P}}={{d}_{Q}}={{d}_{R}}\].

When the lift is falling freely, there won’t be any efflux. This is because the acceleration due to gravity becomes zero in case of a free fall. So, no water will leak out of the water.

Hence the correct choice is option (C).

Note:
You should remember this condition that the horizontal range of the water jet does not depend on the gravitational acceleration of the system. You should not think that the range will be more because the velocity of efflux is more when effective acceleration is increased. You should work out the expression of the final quantity to determine the final option.