Question

A gardener holds a hosepipe through which water is gushing out at a rate of $4{\text{ kg/sec}}$ with the speed of $2{\text{ m/sec}}$. The moment the speed of water is increased by $3{\text{ m/sec}}$, the gardener will experience a jerk of:

(A) 20 N in backward direction
(B) 18 N in forward direction
(C) 10 N in backward direction
(D) 10 N in forward direction

Answer 1

In the question, we need to determine the force exerted by the hosepipe on the gardener in the backward direction when the water is gushing out from the hosepipe in the forward direction. For this, we will use the relation between the rate of change of mass and change in the velocity of the flowing rate of water which is given as $F = \dfrac{{dm}}{{dt}}\left( {\vartriangle v} \right)$.

Complete step by step answer:

The backward force on an object is given as the product of the rate of change of the mass of the applied object and the change in the velocity of the applied object. Mathematically, $F = \dfrac{{dm}}{{dt}}\left( {\vartriangle v} \right)$

Here, the rate at which the water is gushing out is the rate of change in the mass of the water and change in the velocity of the water is given as $\vartriangle v = 2 + 3 = 5{\text{ m/sec}}$.

Hence, the backward force experienced by the gardener is given as:

$$F = \dfrac{{dm}}{{dt}}\left( {\vartriangle v} \right) \\\ = 4 \times 5{\text{ }}\left( {{\text{kg/sec}}} \right){\text{ \times }}\left( {{\text{m/sec}}} \right) \\\ = 20{\text{ kg - m/se}}{{\text{c}}^{\text{2}}} \\\ = 20{\text{ N}} \\\$$

The jerk experienced by the gardener when the rate of flow of water increases by 3 meters per seconds of which water is gushing out at a rate of $4{\text{ kg/sec}}$ is 20 Newton in the backward direction.

Option A is correct.

Note: Students should ensure that the units should be in the defined form only if it is not given in the question then convert the units to the defined form.