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Surface Areas and Volumes — CBSE Class 10 Maths Important Questions

13 hand-picked CBSE Class 10 Maths important questions for Surface Areas and Volumes, each with a full model answer — the formats and topics most likely to appear in your board exam.

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Surface Areas and Volumes — CBSE Class 10 Maths Important Questions

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Quick answer

For a combined solid, volumes always add, but for surface area you add only the exposed curved/plane surfaces (never the hidden joining faces). For a cone always find the slant height first: l=r2+h2l=\sqrt{r^2+h^2}l=√r^2+h^2. When a solid is melted and recast, volume stays constant.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Surface Areas and Volumes

This chapter combines the standard solids (cylinder, cone, sphere, hemisphere) into real objects such as toys, tents, capsules, ice-cream cones and containers, and also covers melting and recasting one solid into another. Boards almost always set one case-study (4 marks) and one 5-mark question here. The single most common mistake is forgetting the slant height l=r2+h2l=\sqrt{r^2+h^2}l=√r^2+h^2 for cones and wrongly including hidden faces in surface area.

Surface area of a combination of solidsVolume of a combination of solidsSlant height of a cone $l=\sqrt{r^2+h^2}$Conversion of solids (melting and recasting)Flow and capacity problems

Key concepts & formulas

Standard formulas

Cylinder: CSA =2πrh=2\pi rh=2π rh, Volume =πr2h=\pi r^2h=π r^2h. Cone: CSA =πrl=\pi rl=π rl, Volume =13πr2h=\dfrac{1}{3}\pi r^2h=1/3π r^2h. Sphere: SA =4πr2=4\pi r^2=4π r^2, Volume =43πr3=\dfrac{4}{3}\pi r^3=4/3π r^3. Hemisphere: CSA =2πr2=2\pi r^2=2π r^2, TSA =3πr2=3\pi r^2=3π r^2, Volume =23πr3=\dfrac{2}{3}\pi r^3=2/3π r^3.

Combining solids

Volume of a combined solid === sum of the volumes of its parts. Surface area === sum of the exposed surfaces only; the shared or joined faces are not counted.

Slant height

For any cone (or the conical part of a solid), l=r2+h2l=\sqrt{r^2+h^2}l=√r^2+h^2. It is needed for CSA of a cone =πrl=\pi rl=π rl. Compute it before finding surface area.

Melting and recasting

When one solid is melted and moulded into another, the material (volume) is unchanged: Volume of old solid === Volume of new solid. This gives the equation to find the unknown dimension.

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Important questions with answers

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Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The total surface area of a solid hemisphere of radius 7 cm7\ \text{cm}7 cm is:

  1. (a)

    462 cm2462\ \text{cm}^2462 cm^2

  2. (b)

    308 cm2308\ \text{cm}^2308 cm^2

  3. (c)

    154 cm2154\ \text{cm}^2154 cm^2

  4. (d)

    231 cm2231\ \text{cm}^2231 cm^2

Show model answer

Answer: (a) 462 cm2462\ \text{cm}^2462 cm^2

TSA of a solid hemisphere =3πr2=3×227×49=462 cm2.=3\pi r^2=3\times\dfrac{22}{7}\times 49=462\ \text{cm}^2.=3π r^2=3×22/7× 49=462 cm^2.

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Q2MCQEasy1 mark

A solid sphere of radius 3 cm3\ \text{cm}3 cm is melted and recast into small spheres each of radius 1 cm1\ \text{cm}1 cm. The number of small spheres formed is:

  1. (a)

    272727

  2. (b)

    999

  3. (c)

    333

  4. (d)

    181818

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Answer: (a) 272727

Number =43π(3)343π(1)3=271=27.=\dfrac{\frac{4}{3}\pi(3)^3}{\frac{4}{3}\pi(1)^3}=\dfrac{27}{1}=27.=4/3π(3)^34/3π(1)^3=27/1=27.

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Q3MCQModerate1 mark

The curved surface area of a cone of radius 7 cm7\ \text{cm}7 cm and height 24 cm24\ \text{cm}24 cm is:

  1. (a)

    550 cm2550\ \text{cm}^2550 cm^2

  2. (b)

    704 cm2704\ \text{cm}^2704 cm^2

  3. (c)

    154 cm2154\ \text{cm}^2154 cm^2

  4. (d)

    275 cm2275\ \text{cm}^2275 cm^2

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Answer: (a) 550 cm2550\ \text{cm}^2550 cm^2

l=72+242=625=25 cm.l=\sqrt{7^2+24^2}=\sqrt{625}=25\ \text{cm}.l=√7^2+24^2=√625=25 cm. CSA =πrl=227×7×25=550 cm2.=\pi rl=\dfrac{22}{7}\times 7\times 25=550\ \text{cm}^2.=π rl=22/7× 7× 25=550 cm^2.

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Q4MCQModerate1 mark

A metallic sphere of radius 6 cm6\ \text{cm}6 cm is melted and drawn into a solid cylinder of base radius 4 cm4\ \text{cm}4 cm. The height of the cylinder is:

  1. (a)

    18 cm18\ \text{cm}18 cm

  2. (b)

    12 cm12\ \text{cm}12 cm

  3. (c)

    24 cm24\ \text{cm}24 cm

  4. (d)

    27 cm27\ \text{cm}27 cm

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Answer: (a) 18 cm18\ \text{cm}18 cm

43π(6)3=π(4)2h⇒288π=16πh⇒h=18 cm.\dfrac{4}{3}\pi(6)^3=\pi(4)^2h\Rightarrow 288\pi=16\pi h\Rightarrow h=18\ \text{cm}.4/3π(6)^3=π(4)^2h 288π=16π h h=18 cm.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonEasy1 mark

Assertion (A): The total surface area of a solid hemisphere of radius rrr is 2πr22\pi r^22π r^2.

Reason (R): The curved surface area of a hemisphere is 2πr22\pi r^22π r^2.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (d) A is false but R is true.

The curved surface area of a hemisphere is 2πr22\pi r^22π r^2 (R is true), but its total surface area also includes the flat circular base, giving 2πr2+πr2=3πr22\pi r^2+\pi r^2=3\pi r^22π r^2+π r^2=3π r^2; hence A is false.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

A cone and a cylinder have equal bases and equal heights. Write the ratio of their volumes. Also, if the volume of the cylinder is 360 cm3360\ \text{cm}^3360 cm^3, find the volume of the cone.

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Volume of coneVolume of cylinder=13πr2hπr2h=13\dfrac{\text{Volume of cone}}{\text{Volume of cylinder}}=\dfrac{\frac{1}{3}\pi r^2h}{\pi r^2h}=\dfrac{1}{3}Volume of cone/Volume of cylinder=1/3π r^2hπ r^2h=1/3, so the ratio is 1:3.1:3.1:3.

Volume of cone =13×360=120 cm3.=\dfrac{1}{3}\times 360=120\ \text{cm}^3.=1/3× 360=120 cm^3.

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Q7Very ShortModerate2 marks

Two cubes each of volume 64 cm364\ \text{cm}^364 cm^3 are joined end to end. Find the surface area of the resulting cuboid.

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Each cube has side =643=4 cm.=\sqrt[3]{64}=4\ \text{cm}.=[3]64=4 cm. The cuboid formed has dimensions 8×4×4 cm.8\times 4\times 4\ \text{cm}.8× 4× 4 cm.

Surface area =2(lb+bh+hl)=2(8×4+4×4+4×8)=2(32+16+32)=2×80=160 cm2.=2(lb+bh+hl)=2(8\times 4+4\times 4+4\times 8)=2(32+16+32)=2\times 80=160\ \text{cm}^2.=2(lb+bh+hl)=2(8× 4+4× 4+4× 8)=2(32+16+32)=2× 80=160 cm^2.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

A toy is in the form of a cone mounted on a hemisphere of the same radius 3.5 cm3.5\ \text{cm}3.5 cm. The total height of the toy is 15.5 cm15.5\ \text{cm}15.5 cm. Find the total surface area of the toy.

CBSE Class 10 Maths — Surface Areas and Volumes: A toy is in the form of a cone mounted on a hemisphere of the same radius 3.5\ \text{cm}. The total height of the toy is 15.5\ \tex
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Radius r=3.5 cm.r=3.5\ \text{cm}.r=3.5 cm. Height of cone =15.5−3.5=12 cm.=15.5-3.5=12\ \text{cm}.=15.5-3.5=12 cm.

Slant height l=r2+h2=3.52+122=12.25+144=156.25=12.5 cm.l=\sqrt{r^2+h^2}=\sqrt{3.5^2+12^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5\ \text{cm}.l=√r^2+h^2=√3.5^2+12^2=√12.25+144=√156.25=12.5 cm.

TSA =CSA of cone+CSA of hemisphere=πrl+2πr2=πr(l+2r)=\text{CSA of cone}+\text{CSA of hemisphere}=\pi rl+2\pi r^2=\pi r(l+2r)=CSA of cone+CSA of hemisphere=π rl+2π r^2=π r(l+2r)
=227×3.5×(12.5+7)=11×19.5=214.5 cm2.=\dfrac{22}{7}\times 3.5\times(12.5+7)=11\times 19.5=214.5\ \text{cm}^2.=22/7× 3.5×(12.5+7)=11× 19.5=214.5 cm^2.

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Q9Short AnswerModerate3 marks

A cylindrical container of radius 6 cm6\ \text{cm}6 cm and height 15 cm15\ \text{cm}15 cm is full of ice cream, which is to be distributed among children in cones of radius 3 cm3\ \text{cm}3 cm and height 12 cm12\ \text{cm}12 cm, each having a hemispherical top of the same radius. Find the number of such cones that can be filled.

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Volume of cylinder =πr2h=π×62×15=540π cm3.=\pi r^2h=\pi\times 6^2\times 15=540\pi\ \text{cm}^3.=π r^2h=π× 6^2× 15=540π cm^3.

Volume of one cone +++ hemisphere =13π(3)2(12)+23π(3)3=36π+18π=54π cm3.=\dfrac{1}{3}\pi(3)^2(12)+\dfrac{2}{3}\pi(3)^3=36\pi+18\pi=54\pi\ \text{cm}^3.=1/3π(3)^2(12)+2/3π(3)^3=36π+18π=54π cm^3.

Number of cones =540π54π=10.=\dfrac{540\pi}{54\pi}=10.=540π/54π=10.

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Q10Short AnswerHOTS3 marks

A solid iron cuboidal block of dimensions 4.4 m×2.6 m×1 m4.4\ \text{m}\times 2.6\ \text{m}\times 1\ \text{m}4.4 m× 2.6 m× 1 m is recast into a hollow cylindrical pipe of internal radius 30 cm30\ \text{cm}30 cm and thickness 5 cm5\ \text{cm}5 cm. Find the length of the pipe.

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Volume of block =4.4×2.6×1=11.44 m3.=4.4\times 2.6\times 1=11.44\ \text{m}^3.=4.4× 2.6× 1=11.44 m^3.

Internal radius =0.30 m=0.30\ \text{m}=0.30 m, external radius =0.30+0.05=0.35 m.=0.30+0.05=0.35\ \text{m}.=0.30+0.05=0.35 m.

Volume of pipe =π(R2−r2)H=227(0.352−0.302)H=227(0.0325)H.=\pi(R^2-r^2)H=\dfrac{22}{7}\big(0.35^2-0.30^2\big)H=\dfrac{22}{7}(0.0325)H.=π(R^2-r^2)H=22/7(0.35^2-0.30^2)H=22/7(0.0325)H.

Set equal to 11.4411.4411.44: 227×0.0325×H=11.44⇒0.10214 H=11.44⇒H=112 m.\dfrac{22}{7}\times 0.0325\times H=11.44\Rightarrow 0.10214\,H=11.44\Rightarrow H=112\ \text{m}.22/7× 0.0325× H=11.44 0.10214\,H=11.44 H=112 m.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

From a solid cylinder of height 2.4 cm2.4\ \text{cm}2.4 cm and diameter 1.4 cm1.4\ \text{cm}1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find (i) the total surface area and (ii) the volume of the remaining solid.

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Radius r=0.7 cmr=0.7\ \text{cm}r=0.7 cm, height h=2.4 cmh=2.4\ \text{cm}h=2.4 cm, slant height l=0.72+2.42=0.49+5.76=6.25=2.5 cm.l=\sqrt{0.7^2+2.4^2}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5\ \text{cm}.l=√0.7^2+2.4^2=√0.49+5.76=√6.25=2.5 cm.

(i) TSA =CSA of cylinder+area of base+CSA of cone=2πrh+πr2+πrl=πr(2h+r+l)=\text{CSA of cylinder}+\text{area of base}+\text{CSA of cone}=2\pi rh+\pi r^2+\pi rl=\pi r(2h+r+l)=CSA of cylinder+area of base+CSA of cone=2π rh+π r^2+π rl=π r(2h+r+l)
=227×0.7×(4.8+0.7+2.5)=2.2×8.0=17.6 cm2.=\dfrac{22}{7}\times 0.7\times(4.8+0.7+2.5)=2.2\times 8.0=17.6\ \text{cm}^2.=22/7× 0.7×(4.8+0.7+2.5)=2.2× 8.0=17.6 cm^2.

(ii) Volume =πr2h−13πr2h=23πr2h=23×227×0.49×2.4=2.464 cm3.=\pi r^2h-\dfrac{1}{3}\pi r^2h=\dfrac{2}{3}\pi r^2h=\dfrac{2}{3}\times\dfrac{22}{7}\times 0.49\times 2.4=2.464\ \text{cm}^3.=π r^2h-1/3π r^2h=2/3π r^2h=2/3×22/7× 0.49× 2.4=2.464 cm^3.

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Q12Long AnswerHOTS5 marks

A farmer connects a pipe of internal diameter 20 cm20\ \text{cm}20 cm from a canal into a cylindrical tank of diameter 10 m10\ \text{m}10 m and depth 2 m2\ \text{m}2 m. If water flows through the pipe at the rate of 3 km/h3\ \text{km/h}3 km/h, in how much time will the tank be filled completely?

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Radius of tank =5 m=5\ \text{m}=5 m, depth =2 m.=2\ \text{m}.=2 m. Volume of tank =π(5)2(2)=50π m3.=\pi(5)^2(2)=50\pi\ \text{m}^3.=π(5)^2(2)=50π m^3.

Pipe radius =10 cm=0.1 m.=10\ \text{cm}=0.1\ \text{m}.=10 cm=0.1 m. In 111 hour water flows a length of 3 km=3000 m.3\ \text{km}=3000\ \text{m}.3 km=3000 m.

Volume of water per hour =π(0.1)2×3000=30π m3.=\pi(0.1)^2\times 3000=30\pi\ \text{m}^3.=π(0.1)^2× 3000=30π m^3.

Time =50π30π=53 hours=53×60=100 minutes.=\dfrac{50\pi}{30\pi}=\dfrac{5}{3}\ \text{hours}=\dfrac{5}{3}\times 60=100\ \text{minutes}.=50π/30π=5/3 hours=5/3× 60=100 minutes.

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Case-based questions (4 marks)

Q13Case-basedHOTS4 marks

A metal grain-silo is in the shape of a cylinder surmounted by a cone, as shown. The common radius is 1.5 m1.5\ \text{m}1.5 m, the cylindrical part is 7 m7\ \text{m}7 m high and the conical top is 2 m2\ \text{m}2 m high.

CBSE Class 10 Maths — Surface Areas and Volumes: A metal grain-silo is in the shape of a cylinder surmounted by a cone, as shown. The common radius is 1.5\ \text{m}, the cylindrica

(i) Find the slant height of the conical top.

(ii) Find the total volume of the silo.

(iii) Find the area of the metal sheet required to make the curved surfaces (cylinder and cone), ignoring the base.

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(i) l=r2+h2=1.52+22=2.25+4=6.25=2.5 m.l=\sqrt{r^2+h^2}=\sqrt{1.5^2+2^2}=\sqrt{2.25+4}=\sqrt{6.25}=2.5\ \text{m}.l=√r^2+h^2=√1.5^2+2^2=√2.25+4=√6.25=2.5 m.

(ii) Volume =πr2hcyl+13πr2hcone=227(1.5)2(7)+13⋅227(1.5)2(2)=\pi r^2h_{\text{cyl}}+\dfrac{1}{3}\pi r^2h_{\text{cone}}=\dfrac{22}{7}(1.5)^2(7)+\dfrac{1}{3}\cdot\dfrac{22}{7}(1.5)^2(2)=π r^2h_cyl+1/3π r^2h_cone=22/7(1.5)^2(7)+1/3·22/7(1.5)^2(2)
=49.5+4.71=54.21 m3.=49.5+4.71=54.21\ \text{m}^3.=49.5+4.71=54.21 m^3.

(iii) Curved area =2πrhcyl+πrl=2×227×1.5×7+227×1.5×2.5=66+11.79=77.79 m2.=2\pi rh_{\text{cyl}}+\pi rl=2\times\dfrac{22}{7}\times 1.5\times 7+\dfrac{22}{7}\times 1.5\times 2.5=66+11.79=77.79\ \text{m}^2.=2π rh_cyl+π rl=2×22/7× 1.5× 7+22/7× 1.5× 2.5=66+11.79=77.79 m^2.

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

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Frequently asked questions

  • Are these Surface Areas and Volumes important questions free?
    Yes. All 13 CBSE Class 10 Maths important questions for Surface Areas and Volumes are free, with full model answers and no login required.
  • Do these Surface Areas and Volumes questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.
  • How should I practise the Surface Areas and Volumes important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Surface Areas and Volumes?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

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