GSEB Class 8 Maths Solutions Chapter 11 Mensuration Exercise 11.4

Question 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume?
(a) To find how much it can hold?
(b) Number of cement bags required to plaster it?
(c) To find the number of smaller tanks that can be filled with water from it?
Answer:
(a) Volume
(b) Surface area
(c) Volume
In simple words: When you want to know how much liquid a tank holds, you find its volume. If you need to cover its outside with something like plaster, you find its surface area.

Exam Tip: Remember, volume is for space inside (capacity), while surface area is for the covering on the outside. This distinction is crucial for many real-world problems.

Question 2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Answer:
In a cylinder's volume calculation, the radius is squared, which means it has a more significant impact than height. Cylinder B has a larger radius (7 cm) compared to cylinder A (3.5 cm), even though its height is smaller. Therefore, cylinder B will likely have a larger volume than cylinder A.
Let's verify this by calculating.
Volume of cylinder A:
Radius \( r = \frac{7}{2} \) cm
Height \( h = 14 \) cm
Volume of cylinder A \( = \pi r^2 h \)
\( = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 14 \text{ cm}^3 \)
\( = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14 \text{ cm}^3 \)
\( = 11 \times 7 \times 7 \text{ cm}^3 \)
\( = 539 \text{ cm}^3 \)
Volume of cylinder B:
Radius \( r = \frac{14}{2} = 7 \) cm
Height \( h = 7 \) cm
Volume of cylinder B \( = \pi r^2 h \)
\( = \frac{22}{7} \times (7)^2 \times 7 \text{ cm}^3 \)
\( = \frac{22}{7} \times 7 \times 7 \times 7 \text{ cm}^3 \)
\( = 22 \times 7 \times 7 \text{ cm}^3 \)
\( = 1078 \text{ cm}^3 \)
Thus, cylinder B indeed has a greater volume.

Now, let's check their surface areas.
Surface area of cylinder A:
Surface area \( = 2 \pi r (r + h) \)
\( = 2 \times \frac{22}{7} \times \frac{7}{2} \times \left[\frac{7}{2} + 14\right] \text{ cm}^2 \)
\( = 22 \times \left[\frac{7}{2} + \frac{28}{2}\right] \text{ cm}^2 \)
\( = 22 \times \frac{35}{2} \text{ cm}^2 \)
\( = 11 \times 35 \text{ cm}^2 \)
\( = 385 \text{ cm}^2 \)
Surface area of cylinder B:
Surface area \( = 2 \pi r (r + h) \)
\( = 2 \times \frac{22}{7} \times 7 \times [7 + 7] \text{ cm}^2 \)
\( = 2 \times 22 \times [14] \text{ cm}^2 \)
\( = 44 \times 14 \text{ cm}^2 \)
\( = 616 \text{ cm}^2 \)
Thus, the cylinder with larger capacity (more volume) also possesses a greater surface area.
In simple words: Cylinder B has a bigger radius, which makes its volume much larger. When we calculate both volumes, cylinder B is indeed bigger. For surface area, cylinder B also turns out to be larger. So, the cylinder with more space inside also has more outer surface.

Exam Tip: For volume calculations, remember that the radius is squared, meaning it affects the volume much more than the height. Always calculate both volume and surface area to confirm if there is a direct relationship between them in a given problem.

Question 3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.
Answer: Let the height of the cuboid be \( h \) cm.
The formula for the volume of a cuboid is Base area \( \times \) Height \( = \) Volume.
Given: Base area \( = 180 \text{ cm}^2 \)
Volume \( = 900 \text{ cm}^3 \)
So, \( 180 \times h = 900 \)
To find \( h \), we divide the volume by the base area:
\( h = \frac{900}{180} \)
\( h = 5 \) cm
Hence, the needed height of the cuboid is \( 5 \) cm.
In simple words: We know the base area and total volume of a cuboid. To find its height, we simply divide the total volume by the base area. This gives us 5 cm.

Exam Tip: Always remember the basic formula for the volume of a cuboid: Volume = Length x Width x Height. Since Base Area = Length x Width, it simplifies to Volume = Base Area x Height, which is helpful for these types of problems.

Question 4. A cuboid is of dimensions 60 cm \( \times \) 54 cm \( \times \) 30 cm. How many small cubes with side 6 cm can he placed in the given cuboid?
Answer: First, calculate the volume of the large cuboid.
Volume of the cuboid \( = 60 \text{ cm} \times 54 \text{ cm} \times 30 \text{ cm} \)
\( = 97200 \text{ cm}^3 \)
Next, calculate the volume of one small cube.
Volume of the small cube \( = (6 \times 6 \times 6) \text{ cm}^3 \)
\( = 216 \text{ cm}^3 \)
To find the required number of cubes, we divide the volume of the cuboid by the volume of one small cube:
Required number of cubes \( = \frac{\text{Volume of the cuboid}}{\text{Volume of the small cube}} \)
\( = \frac{97200 \text{ cm}^3}{216 \text{ cm}^3} \)
\( = 450 \)
Alternatively, we can divide each dimension of the cuboid by the side of the cube and then multiply the results:
\( = \left(\frac{60}{6}\right) \times \left(\frac{54}{6}\right) \times \left(\frac{30}{6}\right) \)
\( = 10 \times 9 \times 5 \)
\( = 450 \)
Thus, \( 450 \) small cubes can be placed into the given cuboid.
In simple words: To find how many small cubes fit inside a big cuboid, calculate the volume of the cuboid and the volume of one small cube. Then, divide the big volume by the small volume to get the total number of cubes that can fit.

Exam Tip: When fitting smaller objects into a larger one, calculate the volume of both the larger container and the smaller object. The number of smaller objects is simply the total volume divided by the volume of one small object. Ensure all units are consistent before performing calculations.

Question 5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?
Answer: Given the volume of the cylinder as \( 1.54 \text{ m}^3 \) and the diameter of the base as \( 140 \text{ cm} \).
First, convert all measurements to the same unit, preferably meters.
Diameter \( = 140 \text{ cm} \)
Convert diameter to meters: \( 140 \text{ cm} = 1.40 \text{ m} \)
Radius \( r = \frac{\text{Diameter}}{2} = \frac{1.40}{2} = 0.7 \text{ m} \)
Let the height of the cylinder be \( h \) m.
The formula for the volume of a cylinder is \( V = \pi r^2 h \).
Substitute the known values:
\( 1.54 = \frac{22}{7} \times (0.7)^2 \times h \)
\( 1.54 = \frac{22}{7} \times 0.7 \times 0.7 \times h \)
\( 1.54 = 22 \times 0.1 \times 0.7 \times h \)
\( 1.54 = 1.54 \times h \)
To find \( h \), divide \( 1.54 \) by \( 1.54 \):
\( h = \frac{1.54}{1.54} \)
\( h = 1 \text{ m} \)
So, the height of the cylinder is \( 1 \) meter.
In simple words: We are given the volume and diameter of a cylinder. First, convert all units to meters. Then, use the cylinder volume formula \( V = \pi r^2 h \). Plug in the known values for volume and radius, and then solve for the height \( h \).

Exam Tip: It is crucial to maintain consistent units throughout the calculation. Always convert all dimensions to a single unit (e.g., all to meters or all to centimeters) before applying any formulas to avoid errors.

Question 6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in lit tank?
Answer: The milk tank is cylindrical. Its length acts as the height of the cylinder.
Given: Radius \( r = 1.5 \text{ m} \)
Height \( h = 7 \text{ m} \)
First, calculate the volume of the tank.
Volume of the tank \( = \pi r^2 h \)
\( = \frac{22}{7} \times (1.5)^2 \times 7 \text{ m}^3 \)
\( = \frac{22}{7} \times 1.5 \times 1.5 \times 7 \text{ m}^3 \)
\( = 22 \times 1.5 \times 1.5 \text{ m}^3 \)
\( = 22 \times 2.25 \text{ m}^3 \)
\( = 49.5 \text{ m}^3 \)
Now, convert the volume from cubic meters to liters. We know that \( 1 \text{ m}^3 = 1000 \) liters.
Quantity of milk in the tank \( = 49.5 \times 1000 \) liters
\( = 49500 \) liters
Thus, the tank can hold \( 49500 \) liters of milk.
In simple words: To find how much milk a cylindrical tank holds, calculate its volume using the formula \( \pi r^2 h \). Once you have the volume in cubic meters, multiply it by 1000 to convert it to liters, as 1 cubic meter is equal to 1000 liters.

Exam Tip: Remember that the "length" of a cylindrical tank is its height in volume calculations. Always include the conversion factor \( 1 \text{ m}^3 = 1000 \) liters when asked for the quantity in liters.

Question 7. If each of a cube is doubled,
(i) How many times will its surface area increase?
(ii) How many times will its volume increase?
Answer: Let the original edge (side length) of the cube be \( x \).
If each edge is doubled, the new edge length will be \( 2x \).

(i) Increase in surface area:
Original surface area (S.A.) of a cube \( = 6x^2 \)
Increased surface area (S.A.) of the new cube \( = 6(2x)^2 \)
\( = 6 \times (4x^2) \)
\( = 24x^2 \)
To find how many times it increased, divide the new surface area by the original surface area:
\( \frac{24x^2}{6x^2} = 4 \)
So, the surface area increases by 4 times.

(ii) Increase in volume:
Original volume of a cube \( = x^3 \)
Increased volume of the new cube \( = (2x)^3 \)
\( = 8x^3 \)
To find how many times it increased, divide the new volume by the original volume:
\( \frac{8x^3}{x^3} = 8 \)
So, the volume increases by 8 times.
In simple words: If you double the side of a cube, its surface area becomes 4 times bigger because surface area depends on the square of the side. Its volume becomes 8 times bigger because volume depends on the cube of the side.

Exam Tip: For problems involving changes in dimensions, remember that surface area scales with the square of the dimension change (e.g., doubling side = \( 2^2=4 \) times S.A.), while volume scales with the cube of the dimension change (e.g., doubling side = \( 2^3=8 \) times volume). This is a common pattern for similar figures.

Question 8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir?
Answer: First, find the total capacity of the reservoir in liters.
Volume of the reservoir \( = 108 \text{ m}^3 \)
We know that \( 1 \text{ m}^3 = 1000 \) liters.
Capacity of the reservoir in liters \( = 108 \times 1000 \) liters
\( = 108000 \) liters
Next, determine the rate at which water is poured into the reservoir.
Rate of pouring \( = 60 \) liters per minute.
To find the total time in minutes to fill the reservoir, divide the total capacity by the pouring rate:
Time in minutes \( = \frac{\text{Total capacity}}{\text{Rate of pouring}} \)
\( = \frac{108000 \text{ liters}}{60 \text{ liters/minute}} \)
\( = 1800 \) minutes
Finally, convert the time from minutes to hours. We know that \( 1 \) hour \( = 60 \) minutes.
Time in hours \( = \frac{1800 \text{ minutes}}{60 \text{ minutes/hour}} \)
\( = 30 \) hours
Thus, it will take \( 30 \) hours to fill the reservoir.
In simple words: First, change the reservoir's volume from cubic meters to liters. Then, divide the total liters by how many liters go in per minute to find the total minutes. Finally, divide the total minutes by 60 to get the time in hours.

Exam Tip: This problem requires careful unit conversion. Always convert the volume to the same unit as the flow rate (liters) and ensure the final time unit (hours) matches what the question asks for. Break down the problem into smaller, manageable steps.