Question

The force/acting after time t on the bottom of a beaker of area A when water of density ρfalls from a tap at a height h at uniform rate v m³ s^-1 is

• A. $\mathrm { v } \left[ g t + 2 g \left( \mathrm {~h} - \frac { \mathrm { vt } } { \mathrm { A } } \right) \right]$
• B. $v \rho [ g t + \sqrt { 2 g h } ]$
• C. vρgh
• D. $\mathrm { vg } \left[ g t + \sqrt { 2 g \left( h + \frac { v t } { A } \right) } \right]$

Answer 1

Answer: (A)

$\mathrm { v } \left[ g t + 2 g \left( \mathrm {~h} - \frac { \mathrm { vt } } { \mathrm { A } } \right) \right]$

Answer 2

Height through which water falls in time dt after t second is (h – h'), where h' is the height of water already collected.

Mass of water collected = vρdt

and velocity of water = $\sqrt { 2 g \left( \mathrm {~h} - \mathrm { h } ^ { \prime } \right) }$

Total force = weight of water already collected + impact force

= vtρg + vρ2g (h – h')

= vρ $\left[ g t + 2 g \left( h - \frac { v t } { \mathrm {~A} } \right) \right]$