Step 1: Use the continuity equation.
The continuity equation for an incompressible flow is given by:
\[
A_1 v_1 = A_2 v_2
\]
Where:
- \( A_1 = \frac{\pi d_1^2}{4} \) is the cross-sectional area at the entrance,
- \( A_2 = \frac{\pi d_2^2}{4} \) is the cross-sectional area at the exit,
- \( v_1 = 3 \, \text{m/s} \) is the velocity at the entrance,
- \( d_1 = 0.25 \, \text{m} \) and \( d_2 = 0.20 \, \text{m} \) are the diameters at the entrance and exit, respectively.
Step 2: Solve for \( v_2 \).
Rearranging the continuity equation to solve for \( v_2 \):
\[
v_2 = v_1 \times \frac{A_1}{A_2} = 3 \times \frac{\left( \frac{\pi (0.25)^2}{4} \right)}{\left( \frac{\pi (0.20)^2}{4} \right)} = 3 \times \left(\frac{0.25^2}{0.20^2}\right) = 3 \times \left(\frac{0.0625}{0.04}\right) = 4.68 \, \text{m/s}
\]
Final Answer: \[ \boxed{4.68 \, \text{m/s}} \]
A cube of side 10 cm is suspended from one end of a fine string of length 27 cm, and a mass of 200 grams is connected to the other end of the string. When the cube is half immersed in water, the system remains in balance. Find the density of the cube.
Match List-I with List-II 

