NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4

Question 18

Step 1: Write down given data.

• Radius of pipe = 1 cm = 0.01 m
• Speed of water in pipe = 80 cm/s = 0.8 m/s
• Time = half an hour = 30 min = 1800 s
• Radius of tank = 40 cm = 0.40 m

Step 2: Find volume of water flowing into the tank in 1800 s.

Volume flow per second through pipe = Cross-sectional area × velocity

\( A_{pipe} = \pi r^2 = \pi (0.01)^2 = \pi (0.0001) = 0.0001\pi \, m^2 \)

Flow per second = \( A_{pipe} \times v = 0.0001\pi \times 0.8 = 0.00008\pi \, m^3/s \)

Total volume in 1800 s = \( 0.00008\pi \times 1800 = 0.144\pi \, m^3 \)

Step 3: Relate this volume to the tank.

Tank volume = Base area × height rise

Base area of tank = \( \pi R^2 = \pi (0.40)^2 = 0.16\pi \, m^2 \)

Step 4: Find rise in water level.

Rise = \( \dfrac{\text{Volume of inflow}}{\text{Base area}} = \dfrac{0.144\pi}{0.16\pi} = 0.90 \, m \)

Step 5: Convert to cm.

\(0.90 \, m = 90 \, cm\)

Final Answer: Rise in water level = 90 cm.