WBBSE Class 10 Maths Mensuration Chapter 2 Right Circular Cylinder Multiple Choice Questions
Example 1. If the length of radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, the ratio of their lateral surface is
- 2: 5
- 8: 7
- 10: 9
- 16: 9
Solution: Let radii of r1,r2 units and that of height be h1,h2 units respectively.
∴ Required ratio = \(\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}=\frac{2}{3} \times \frac{5}{3}=10: 9\)
∴ Answer is 3. 10: 9

Example 2. If the length of radii of two solid right circular cylinders are in the ratio 2 : 3 and their height in the ratio 5 : 3, then the ratio of their volume is
- 27: 20
- 20: 27
- 4: 9
- 9: 4
Solution: Let radii be r1,r2 units and that of height be h1,h2 units respectively.
Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions
Required ratio = \(\frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}=\left(\frac{2}{3}\right)^2 \times \frac{5}{3}=\frac{4}{9} \times \frac{5}{3}=20: 27\)
∴ Answer is 2. 20: 27
Class 10 Maths Mensuration Chapter 2 MCQs
Example 3. If volumes of two solid right circular cylinder are same and their height are in the ratio 1: 2, then the ratio of length of radii is
- 1:√2
- √2: 1
- 1: 2
- 2:1
Solution: Let radii be r1,r2 units and that of height h1,h2 units.
∴ \(\pi r_1{ }^2 h_1=\pi r_2{ }^2 h_2\)
or, \(\left(\frac{r_1}{r_2}\right)^2=\frac{h_2}{h_1}=\frac{2}{1}\)
∴ r1: r2 = √2: 1
∴ Answer is 2. √2: 1
Example 4. In a right circular cylinder, if the length of radius is halved and height is doubled, volume of cylinder will be
- Equal
- Double
- Half
- 4 times
Solution: volume = πr2h cubic unit
changed volume = \(\pi\left(\frac{r}{2}\right)^2\) .2h cubic unit = \(\frac{\pi r^2 h}{2}\) cubic units
∴ Answer is 3. Half
Example 5. If the length of radius of a right circular cylinder is doubled and height is halved, the lateral surface area will be
- Equal
- Double
- Half
- 4 times
Solution: Volume πr2h c.u
changed volume = π(2r)2.\(\frac{h}{2}\) c.u.= 2πr2h cubic units.
∴ Answer is 2. Double
Example 6. Diameter and height of two solid right circular cylinders are 8 units and 3(h – 1) units respectively and the volumes are equal, then h =?
- 9
- 2
- 17
- 19
Solution: \(\pi\left(\frac{8}{2}\right)^2 \cdot 3(h-1)=\pi\left(\frac{12}{2}\right)^2 \times(h+5)\)
or, 48 (h – 1) = 36 (h + 1) or, 12h = 228
∴ h = 19
∴ Answer is 4. 19
Wbbse Class 10 Maths Mensuration Notes
Example 7. Volume of a cylindrical tank is 6160 cubic metro and radius is 14 metro. Depth is
- 8 metre
- 10 metre
- 14 metre
- 16 metre
Solution: π (14)2.h = 6160 or, h = \(\frac{6160 \times 7}{22 \times 14 \times 14}\) = 10
∴ Answer is 2. 10 metre
Example 8. No. of coins with diameter 3 cm and 0.25 cm thickness from a cuboid of dimension 11 cm x 9 cm x 6 cm will be
- 320
- 330
- 336
- 340
Solution: No. of coins = \(\frac{11 \times 9 \times 6}{\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 0.25}\) = 336
∴ Answer is 3. 336
Class 10 Mensuration Chapter 2 Mcqs With Answers
Example 9. Area of the bane and lateral surface area of a cylinder of radius 8 cm are equal. Height will be
- 3 cm
- 4 cm
- 5 cm
- 6 cm
Solution: πr2 = 2πrh
⇒ h = \(\frac{r}{2}\) = \(\frac{8}{2}\) = 4
∴ Answer is 2. 4 cm
Example 10. Volume of the hollow cylinder of outer radius R, inner radius r and height h will be
- π (R2 + r2)h
- π (R2 – r2)h
- \(\frac{\pi}{2}\)(R2 – r2)h
- \(\frac{\pi}{2}\)(R2 + r2)h
Solution: Answer is 2. π (R2 – r2)h