Maharashtra Board Class 8 Maths Chapter 16 Surface Area and Volume Set 16.3 Solutions

Question 1. Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.
(i) r = 10.5 cm, h = 8 cm
(ii) r = 2.5 m, h = 7 m
(iii) r = 4.2 cm, h = 5 cm
(iv) r = 5.6 cm, h = 5 cm
Answer:
(i) Given: r = 10.5 cm and h = 8 cm
To find: Volume of the cylinder
Volume of the cylinder = \( \pi r^2 h \)
= \( \frac{22}{7} \times 10.5 \times 10.5 \times 8 \)
= \( 22 \times 1.5 \times 10.5 \times 8 \)
= \( 2772 \) cc
∴ The volume of the cylinder is \( 2772 \) cc.

(ii) Given: r = 2.5 m and h = 7 m
To find: Volume of the cylinder
Volume of the cylinder = \( \pi r^2 h \)
= \( \frac{22}{7} \times 2.5 \times 2.5 \times 7 \)
= \( 22 \times 2.5 \times 2.5 \)
= \( 137.5 \) cu.m
∴ The volume of the cylinder is \( 137.5 \) cu.m.

(iii) Given: r = 4.2 cm and h = 5 cm
To find: Volume of the cylinder
Volume of the cylinder = \( \pi r^2 h \)
= \( \frac{22}{7} \times 4.2 \times 4.2 \times 5 \)
= \( 22 \times 0.6 \times 4.2 \times 5 \)
= \( 277.2 \) cc
∴ The volume of the cylinder is \( 277.2 \) cc.

(iv) Given: r = 5.6 cm and h = 5 cm
To find: Volume of the cylinder
Volume of the cylinder = \( \pi r^2 h \)
= \( \frac{22}{7} \times 5.6 \times 5.6 \times 5 \)
= \( 22 \times 0.8 \times 5.6 \times 5 \)
= \( 492.8 \) cc
∴ The volume of the cylinder is \( 492.8 \) cc.
In simple words: To find the volume of a cylinder, use the formula \( V = \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height. Substitute the given values of \( r \) and \( h \) into the formula and calculate the result for each part.

🎯 Exam Tip: Ensure consistent units throughout the calculation. For example, if radius is in meters, height should also be in meters, and the resulting volume will be in cubic meters.

Question 2. How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?
Answer:
Solution:
Given: For cylindrical rod: length of rod (h) = 90 cm, and
diameter (d) = 1.4 cm
To find: Iron required to make a rod
diameter (d) = 1.4 cm
∴ radius (r) = \( \frac{d}{2} = \frac{1.4}{2} = 0.7 \) cm
Volume of rod = \( \pi r^2 h \)
= \( \frac{22}{7} \times 0.7 \times 0.7 \times 90 \)
= \( 22 \times 0.1 \times 0.7 \times 90 \)
= \( 138.60 \) cc
∴ \( 138.60 \) cc of iron is required to make the rod.
In simple words: To find the amount of iron needed, calculate the volume of the cylindrical rod using the given length (height) and diameter. First, find the radius from the diameter, then apply the cylinder volume formula.

🎯 Exam Tip: Remember that "iron needed" refers to the volume of the material. Pay attention to converting diameter to radius before applying the volume formula.

Question 3. How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?
Answer:
Solution:
Given: interior diameter of the tank (d) = 1.6 m
and depth (h) = 0.7 m
To find: Capacity of the tank
interior diameter of the tank (d) = 1.6 m
∴ Interior radius (r) = \( \frac{d}{2} = \frac{1.6}{2} \)
= \( 0.8 \) m
= \( 0.8 \times 100 \)
...[:: 1m = 100cm]
= \( 80 \)cm
h = \( 0.7 \) m = \( 0.7 \times 100 = 70 \) cm
Capacity of the tank = Volume of the tank = \( \pi r^2 h \)
= \( \frac{22}{7} \times 80 \times 80 \times 70 \)
= \( 22 \times 80 \times 80 \times 10 \)
= \( 1408000 \) cc
= \( \frac{1408000}{1000} \)
...[:1 litre = 1000 cc]
= \( 1408 \) litre
∴ The tank can hold \( 1408 \) litre of water.
In simple words: To find the water capacity, calculate the volume of the cylindrical tank. Convert diameter to radius, then meters to centimeters for consistent units, and finally apply the volume formula. Convert the final volume from cubic centimeters to liters using the given conversion factor.

🎯 Exam Tip: Unit conversions (m to cm, cc to litres) are crucial in such problems. Double-check all conversions and calculations to avoid errors.

Question 4. Find the volume of the cylinder if the circumference of the base of cylinder is 132 cm and height is 25 cm.
Answer:
Solution:
Given: Circumference of the base of cylinder = 132 cm and height (h) = 25 cm
To find: Volume of the cylinder
i. Circumference of base of cylinder = \( 2 \pi r \)
∴ \( 132 = 2 \times \frac{22}{7} \times r \)
∴ \( \frac{132 \times 7}{2 \times 22} = r \)
\( \frac{6 \times 7}{2} = r \)
∴ \( 3 \times 7 = r \)
∴ r = \( 21 \) cm
ii. Volume of the cylinder = \( \pi r^2 h \)
= \( \frac{22}{7} \times 21 \times 21 \times 25 \)
= \( 22 \times 3 \times 21 \times 25 \)
= \( 34650 \) cc
∴ The volume of the cylinder is \( 34650 \) cc.
In simple words: To find the cylinder's volume, first use the given circumference to calculate the radius of the base. Once the radius is found, use it along with the given height in the cylinder's volume formula to get the final volume.

🎯 Exam Tip: When circumference is given, it's the key to finding the radius, which is essential for calculating the volume of the cylinder.

Maharashtra Board Class 8 Maths Chapter 16 Surface Area And Volume Practice Set 16.3 Intext Questions And Activities

Question 1. Leonard Euler, discovered an interesting formula regarding the faces, vertices and edges of solid figures. Count and write the faces, vertices and edges of the following figures and complete the table. From the table verify Euler's formula, F + V = E + 2. (Textbook pg. No. 113)
Answer:

From the above table, F + V = E + 2 i.e. Euler's formula is verified.
In simple words: Euler's formula, F + V = E + 2, relates the number of faces (F), vertices (V), and edges (E) of a polyhedron. By counting these elements for various 3D shapes and substituting them into the formula, we can verify its validity.

🎯 Exam Tip: To accurately apply Euler's formula, ensure careful counting of all faces, vertices, and edges for each solid figure. A systematic approach helps avoid missing any element.