WBBSE Class 10 Maths Mensuration Chapter 3 Right Circular Cone Multiple Choice Questions

WBBSE Class 10 Maths Mensuration Chapter 3 Right Circular Cone Multiple Choice Questions

Example 1. If the slant height of a right circular cone is 15 cm. and the length of the base diameter is 16 cm, then the lateral surface are is

1. 60π cm²
2. 68π cm²
3. 120π cm²
4. 130π cm²

Solution: Lateral surface area = πrl

= π.\(\frac{16}{2}\).15 = 120 π cm²

∴ Answer is 3. 120π cm²

Example 2. If the ratio of the volumes of two right circular cones is 1: 4 and the ratio of the radii of their bases is 4: 5, then the ratio of the their height is

1. 1:5
2. 5: 4
3. 25: 16
4. 25: 64

Solution: \(\frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2}=\frac{1}{4}\)

⇒ or, \(\left(\frac{4}{5}\right)^2 \cdot \frac{h_1}{h_2}=\frac{1}{4}\)

⇒ or, \(\frac{h_1}{h_2}=\frac{25}{64}\)

∴ Answer is 4. 25: 64

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 3. Keeping the radius of a right circular cone same if the height of it is increased twice, the volume of it will be increased by

1. 100%
2. 200%
3. 300%
4. 400%

Solution: \(\frac{\frac{1}{3} \pi r^2(2 h)}{\frac{1}{3} \pi r^2 h}=2\)

⇒ Increase = (200- 100)% = 100%

∴ Answer is 1. 100%

Combination Of Solids Mcqs With Solutions Class 10

Example 4. If each of radius and height of a cone is the volume increased by twice of its length will be

1. 3 times
2. 4 times
3. 6 times
4. 8 times of previous one

Solution: \(\frac{\frac{1}{3} \pi(2 r)^2 \cdot 2 h}{\frac{1}{3} \pi r^2 h}=8\)

∴ Answer is 4. 8 times of previous one

Example 5. If the length of the radius of a cone is \(\frac{r}{2}\) units and slant height of it is 2l unit, then the total surface area is

1. 2πr (l + r) sq. u
2. πr\((l+\frac{r}{4})\) sq. u
3. πr (h + r) sq. u
4. 2πrl sq. u

Solution: Answer is 2. πr\((l+\frac{r}{4})\) sq. u

Example 6. Length of the radius and slant height of a cone are 1.5 m and 2 mt respectively. The area of curved sharface is

1. 2π sq. m
2. 3π sq. m
3. 4π sq. m
4. 5π sq. m

Solution: Required area = π x (1.5) x 2 sq. m = 3π sq. m

∴ Answer is 2. 3π sq. m

Wbbse Class 10 Maths Mensuration Notes

Example 7. Height and length of the radius of a cone are 21 cm and 12 cm. Its volume will be

1. 4710 cu. cm
2. 9504 cu. cm
3. 3168 cu cm
4. none of these

Solution: Volume = \(\frac{1}{3}\) x \(\frac{22}{7}\) x(12)² x 21 cu. cm

= 3168 cu. cm

∴ Answer is 3. 3168 cu cm

Example 8. Ratio of volumes and length of the diameters are 1: 4 and 4: 5. Ratio of height will be

1. 5: 8
2. 25: 8
3. 5: 64
4. 25: 64

Solution: \(\frac{V_1}{V_2}=\frac{1}{4}=\frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2{ }^2 h_2}=\left(\frac{r_1}{r_2}\right)^2 \times \frac{h_1}{h_2}\)

∴ \(\frac{h_1}{h_2}=\frac{25}{64}\), \(h_1: h_2=\frac{25}{64}\)

∴ Answer is 4. 25: 64

Class 10 Maths Chapter 3 Mensuration Solutions

Example 9. Length of the radius of a cone is increased by 20%. Keeping height the same. The percentage increase of volume will be

1. 44%
2. 33%
3. 22%
4. 11%

Solution: Initial volume = \(\frac{1}{3}\)πr²h cu. unit

Changed volume = \(\frac{1}{3} \pi\left(\frac{120 r}{100}\right)^2 \cdot h\) cu. unit

= \(\frac{1}{3} \pi \frac{36 r^2}{25} h\) cu. unit

Volume increased = \(\frac{\frac{1}{3} \pi \frac{36 r^2}{25} h-\frac{1}{3} \pi r^2 h}{\frac{1}{3} \pi r^2 h} \times 100 \%\)

= \(\frac{36-25}{25} \times 100 \%\) = 44%

∴ Answer is 1. 44%

Example 10. Volume of the biggest cone made from a cube of length h unit is

1. \(\frac{\pi}{4} h^3 \text { cu. u }\)
2. \(\frac{\pi}{8} h^3 \text { cu. u }\)
3. \(\frac{\pi}{12} h^3 \text { cu. u }\)
4. \(\frac{\pi}{16} h^3 \text { cu. u }\)

Solution: Volume = \(\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h \text { cu. } \mathrm{u}=\frac{\pi}{12} h^3 \text { cu. u }\)

∴ Answer is 3. 22%