CBSE Class 10 Mathematics Surface Area and Volume Notes Set 06

Conversion of Solids from One Shape to Another Shape

There are many times in daily life when we are required to convert one solid into another solid of a different shape. When one solid is converted into another solid of a different shape, the whole of the material is transformed but its volume remains the same.

If a solid is converted into more than one smaller solid of another shape, then the number of smaller objects is given by the formula \( \text{Number of smaller objects} = \frac{\text{Volume of the larger object}}{\text{Volume of the smaller object}} \).

Examples

Example 1. Ravindra cut a cylindrically shaped solid pipe, having a diameter and height as 2 cm and 18 cm respectively, into 2 equal parts. He then took one piece and rolled it into 2 similar balls. Find the diameter of the ball.
Answer:
Radius of the cylinder \( = 1 \) cm. Height of each part \( = \frac{18}{2} = 9 \) cm.
Volume of one cylindrical part \( = \) volume of 2 balls, that is, \( \pi(1)^2 \times 9 = 2 \times \frac{4}{3}\pi r^3 \).
\( r^3 = \frac{9 \times 3}{8} = \frac{27}{8} \Rightarrow r = \frac{3}{2} = 1.5 \) cm.
Hence, the diameter of each ball is 3 cm.

Example 2. A hemispherical tank of radius 3 m is full of milk. It is connected to a pipe through which the milk is emptied at the rate of \( \frac{1000}{7} \) litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank \( r = 3 \) m. Volume of milk \( = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 3^3 = \frac{22}{7} \times 2 \times 3^2 \) m³ \( = \frac{22}{7} \times 2 \times 9 \times 1000 \) litres.
Time taken \( = \frac{\frac{22}{7} \times 18 \times 1000}{\frac{1000}{7}} = 22 \times 18 = 396 \) seconds.
Hence, the tank is emptied in 396 seconds.

Example 3. The dimensions of the base of a rectangular tank are 225 m and 162 m. With what speed must water flow into it through an aperture 60 cm by 45 cm so that the level may be raised 20 cm in 5 hours?
Answer:
Volume of water that flows into the tank in 5 hours \( = 225 \times 162 \times \frac{20}{100} \) m³. So the volume in 1 hour \( = \frac{1}{5} \times 225 \times 162 \times \frac{20}{100} = 1458 \) m³.
Area of cross-section of the aperture \( = \frac{60}{100} \times \frac{45}{100} = \frac{27}{100} \) m².
Let the speed be \( x \) m/h. Then volume per hour \( = \frac{27}{100} \times x = 1458 \Rightarrow x = 5400 \) m/h.
Hence, the speed of the water is 5400 m/h = 5.4 km/h.

Example 4. In a fair, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for Rs. 15 each. Calculate the amount of money the stall keeper would receive by selling the juice completely.
Answer:
Volume of juice in the vessel \( = \pi \times 15 \times 15 \times 32 \) cm³. Volume of juice in each glass \( = \pi \times 3 \times 3 \times 8 \) cm³.
Number of glasses \( = \frac{\pi \times 15 \times 15 \times 32}{\pi \times 3 \times 3 \times 8} = 100 \).
Money received \( = 100 \times 15 = \) Rs. 1500.
Hence, the stall keeper receives Rs. 1500.

Example 5. A metallic sphere of diameter 3 cm is melted to cast into a solid wire of length 18 m. Find the thickness of the wire.
Answer:
Radius of the sphere \( = \frac{3}{2} \) cm. Volume of the sphere \( = \frac{4}{3}\pi\left(\frac{3}{2}\right)^3 = \frac{9}{2}\pi \) cm³.
Length of the wire \( = 18 \) m \( = 1800 \) cm. Let the radius of the wire be \( r \) cm. Volume of the wire \( = \pi r^2 \times 1800 \) cm³.
Volume remains the same: \( \frac{9}{2}\pi = \pi r^2 \times 1800 \Rightarrow r^2 = \frac{9}{2 \times 1800} = \frac{1}{400} \Rightarrow r = \frac{1}{20} \) cm.
Thickness of the wire \( = \) diameter \( = 2r = 0.1 \) cm.
Hence, the thickness of the wire is 0.1 cm.