Question

A fat hose pipe is held horizontally by a fireman. It delivers water through a constricting nozzle $1\,lit\,{\sec ^{ - 1}}$ . If by increasing the pressure, the water is delivered at $2\,lit\,{\sec ^{ - 1}}$ , the fireman has to:
A. Push forward
B. Push forward $4$ times hard
C. Push forward $8$ times hard
D. Push backward $4$ times hard

Answer 1

Hint: In order to solve this question, firstly we will find the force due to the initial velocity of water and its mass equivalent to product of density of water and its volume and then we will find force after an increase in pressure and will compare both equations.

Complete step by step answer:
Let us first calculate the initial mass of flowing water which is given as $m = \rho \times V$ and initially we have a volume $V = 1\,litre$
So we get mass $m = \rho \to (i)$ (in its respective SI unit)
Now, force $F = m \times \dfrac{v}{t}$
And initial velocity of flowing water is given as $v = 1lit\,{\sec ^{ - 1}}$ and in one second so,
$F = m$
Or we can write,
$F = \rho \to (ii)$
Now, in final case after increasing the pressure, the new velocity let’s say $v' = 2lit\,{\sec ^{ - 1}}$
Final force say $F' = m'\dfrac{{v'}}{t}$
Where $m'$ is the final mass which is given as of volume two litres
$m' = \rho \times v'$
$\Rightarrow m' = 2\rho $ Using this we get final force as,
$\Rightarrow F' = 4\rho \to (iii)$
Now, from equations $(ii)$ and $(iii)$ we can see that,
$\therefore F' = 4F$
Which means the fireman has to apply force four times forward.

Hence,the correct answer is option B.

Note:Remember, density is the mass per unit volume of a substance. Density of water is a constant numerical value which is equal to $1\,g\,c{m^{ - 3}}$ . And here Force will need to apply in forward direction because we get the magnitude of force positive if it were negative it implies firemen have to push the same amount of force in backward direction.