Section A — 1 Mark Direct Formula Base Easy
Q1. Find slant height of cone with r=3, h=4.
Q2. Find TSA of a solid hemisphere of radius 10cm. (Use
$\pi=3.14$).
Q3. The radius of a sphere is 2r. What is its volume?
Q4. Surface area of a sphere of radius 7cm?
Q5. Area of the base of a cone is $154 cm^2$. Find radius.
Q6. Find CSA of cone whose base radius is 7cm and slant
height is 10cm.
Q7. Volume of a cone is $\frac{1}{3}\pi r^2 h$. If height
is doubled, volume becomes?
Q8. Difference between TSA and CSA of a solid hemisphere
of radius r?
Section B — Core Word Problems Medium
Q9. A conical tent is 10m high and base radius 24m. Find
slant height and cost of canvas at ₹70 per $m^2$.
Q10. Find the radius of a sphere whose surface area is
$154 cm^2$.
Q11. The radius of a spherical balloon increases from 7cm
to 14cm as air is pumped. Find ratio of surface areas in the two cases.
Q12. A hemispherical bowl made of brass has inner diameter
10.5cm. Find cost of tin-plating it on the inside at ₹16 per $100 cm^2$.
Q13. Find the volume of a right circular cone with radius
6cm, height 7cm.
Q14. The capacity of a conical vessel with height 12cm and
slant height 13cm in litres?
Q15. A capsule of medicine is in the shape of a sphere of
diameter 3.5mm. How much medicine ($mm^3$) is needed to fill it?
Section C — NCERT Deep Dive Hard
- Q16. What length of tarpaulin 3m wide will be required to
make a conical tent of height 8m and base radius 6m? Assume extra length for margins is 20cm.
(Use $\pi=3.14$).
- Q17. A joker's cap is in the form of a right circular cone
of base radius 7cm and height 24cm. Find area of sheet required to make 10 such caps.
- Q18. A hemispherical bowl is made of steel, 0.25cm thick.
The inner radius of the bowl is 5cm. Find the outer curved surface area of the bowl.
- Q19. A right circular cylinder just encloses a sphere of
radius r. Find (i) surface area of sphere, (ii) CSA of cylinder, (iii) ratio of the areas
obtained. (Standard NCERT proof).
- Q20. A dome of a building is in the form of a hemisphere.
From inside, it was white-washed at the cost of ₹4989.60. If cost is ₹20 per sqm, find (i)
inside SA, (ii) volume of air inside the dome.
- Q21. Twenty-seven solid iron spheres, each of radius r and
surface area S, are melted to form a sphere with surface area $S'$. Find (i) radius $r'$ of new
sphere, (ii) ratio of S to $S'$.
Section D — Competency & Board Level Hard
- Q22. A right triangle ABC with sides 5cm, 12cm and 13cm is
revolved about the side 12cm. Find the volume of the solid so obtained. (Forms a cone).
- Q23. If the same triangle in Q22 is revolved about the side
5cm, find the volume. Find the ratio of volumes in Q22 and Q23.
- Q24. Assuming the earth to be a sphere, and the moon also.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction
of the volume of the earth is the volume of the moon?
- Q25. How many litres of milk can a hemispherical bowl of
diameter 10.5cm hold?
- Q26. Water flows at the rate of 10m/minute through a
cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter
at the base is 40cm and depth 24cm? (Wait: Cylinder pipe concept might be outside syllabus, try
applying volume equalization).
- Q27. The radius and slant height of a cone are in ratio
4:7. If its CSA is $792cm^2$, find its radius.
- Q28. The internal and external diameters of a hollow
hemispherical vessel are 24cm and 25cm. The cost of painting $1cm^2$ surface is ₹0.05. Find
total cost. (Inner CSA + Outer CSA + Ring Area).
- Q29. Radius of a sphere is increased by 10%. Prove that its
volume will increase by 33.1%.
- Q30. A wooden bookshelf has external dimensions
85x110x25cm, thickness 5cm. Out of syllabus (Cuboid), skip or replace: A spherical balloon's
volume increases from $V_1$ to $V_2$ when radius doubles, find $V_1 : V_2$. (Ans: 1:8).
✅ Key Checkpoints: Q1: l=5 | Q2: $3 \times 3.14 \times 100 = 942$ | Q3:
$\frac{32}{3}\pi r^3$ | Q8: $\pi r^2$ (area of circular base) | Q9: l=26m, CSA=1961.14 $m^2$,
Cost=₹137280. | Q11: 1:4 | Q16: l=10, CSA=188.4, length=188.4/3 = 62.8m + 0.2 = 63m. | Q19: 1:1
ratio. | Q22: Cone with r=5, h=12. V=$100\pi$. | Q23: Cone with r=12, h=5. V=$240\pi$. Ratio = 5:12.
| Q24: 1/64.