Question

Water is flowing steadily with certain initial velocity vertically down from a tap of area of cross-section $1.8\text{ cm}^2$. At a distance 25 cm below the tap, area of cross-section of the stream of the water is $1.2\text{ cm}^2$. The mass of the water flowing from the tap in a time of 10 s is (Acceleration due to gravity $=10\text{ ms}^{-2}$)}

• A. $4.8\text{ kg}$
• B. $1.2\text{ kg}$
• C. $3.6\text{ kg}$
• D. $7.2\text{ kg}$

Hint:

For liquid streams: \[ A_1v_1=A_2v_2 \] and use Bernoulli's principle or kinematics to relate velocities.

Answer 1

The Correct Option is A

Concept: For steady flow, \[ A_1v_1=A_2v_2 \] and \[ v_2^2=v_1^2+2gh \]

Step 1: Apply continuity equation.
\[ 1.8v_1=1.2v_2 \] \[ v_2=1.5v_1 \]

Step 2: Apply equation of motion.
\[ (1.5v_1)^2=v_1^2+2(10)(0.25) \] \[ 2.25v_1^2=v_1^2+5 \] \[ 1.25v_1^2=5 \] \[ v_1=2\,m/s \]

Step 3: Calculate mass flow rate.
\[ A_1=1.8\times10^{-4}m^2 \] Volume in 10 s: \[ V=A_1v_1t \] \[ =(1.8\times10^{-4})(2)(10) \] \[ =3.6\times10^{-3}m^3 \] Mass: \[ m=\rho V \] \[ =1000(3.6\times10^{-3}) \] \[ =3.6kg \] \[ \boxed{3.6kg} \]