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Samacheer Kalvi 10th Maths Exercise 7.3 Solutions

10th Maths Exercise 7.3 solutions (Chapter 7 Mensuration) with step-by-step answers for all 9 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below
  1. Q1Capacity of the vessel =1642.67=1642.67 cm3^3
  2. Q2Volume of the model =66=66 cm3^3
  3. Q3Volume of the remaining solid =2.464≈2.46=2.464\approx 2.46 cm3^3 (i.e. about 22 cm3^3)
  4. Q4Volume of water displaced =905.14=905.14 cm3^3
  5. Q5The capsule can hold 77.7977.79 mm3^3 of medicine
  6. Q6Surface area of the solid =332.5=332.5 cm2^2
  7. Q7 (i)4πr24\pi r^2 sq. units
  8. Q7 (ii)4πr24\pi r^2 sq. units
  9. Q7 (iii)1:11:1

Q1. A vessel is in the form of a hemispherical bowl mounted by …

A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 1414 cm and the height of the vessel is 1313 cm. Find the capacity of the vessel.

Show steps (7)

  1. Diameter =14=14 cm, so radius r=7r=7 cm.

  2. Height of hemisphere == radius =7=7 cm.

  3. Height of cylinder h=13−7=6h=13-7=6 cm.

  4. Capacity == Volume of cylinder ++ Volume of hemisphere

  5. =πr2h+23πr3=πr2(h+2r3)=\pi r^2h+\dfrac{2}{3}\pi r^3=\pi r^2\left(h+\dfrac{2r}{3}\right)

  6. =227×49×(6+143)=\dfrac{22}{7}\times 49\times\left(6+\dfrac{14}{3}\right)

  7. =154×323=49283=1642.67=154\times\dfrac{32}{3}=\dfrac{4928}{3}=1642.67

Answer

Capacity of the vessel =1642.67=1642.67 cm3^3

Q2. Nathan, an engineering student was asked to make a model shaped like …

Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 33 cm and its length is 1212 cm. If each cone has a height of 22 cm, find the volume of the model that Nathan made.

Show steps (7)

  1. Diameter =3=3 cm, so radius r=32r=\dfrac{3}{2} cm.

  2. Height of each cone h2=2h_2=2 cm.

  3. Height of cylinder h1=12−(2+2)=8h_1=12-(2+2)=8 cm.

  4. Volume of model == Volume of cylinder +2×+2\times Volume of cone

  5. =πr2h1+2×13πr2h2=πr2(h1+2h23)=\pi r^2h_1+2\times\dfrac{1}{3}\pi r^2h_2=\pi r^2\left(h_1+\dfrac{2h_2}{3}\right)

  6. =227×94×(8+43)=\dfrac{22}{7}\times\dfrac{9}{4}\times\left(8+\dfrac{4}{3}\right)

  7. =227×94×283=66=\dfrac{22}{7}\times\dfrac{9}{4}\times\dfrac{28}{3}=66

Answer

Volume of the model =66=66 cm3^3

Q3. From a solid cylinder whose height is 2.42.4 cm and the diameter …

From a solid cylinder whose height is 2.42.4 cm and the diameter 1.41.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm3cm^3.

Show steps (6)

  1. Radius r=1.42=0.7r=\dfrac{1.4}{2}=0.7 cm, height h=2.4h=2.4 cm (cone and cylinder have the same rr and hh).

  2. Volume of remaining solid == Volume of cylinder −- Volume of cone

  3. =πr2h−13πr2h=23πr2h=\pi r^2h-\dfrac{1}{3}\pi r^2h=\dfrac{2}{3}\pi r^2h

  4. =23×227×0.7×0.7×2.4=\dfrac{2}{3}\times\dfrac{22}{7}\times 0.7\times 0.7\times 2.4

  5. =23×3.696=2.464=\dfrac{2}{3}\times 3.696=2.464 cm3^3

  6. Therefore the volume of the remaining solid is 2.462.46 cm3^3 (approximately 22 cm3^3 when rounded to the nearest whole cm3cm^3).

Answer

Volume of the remaining solid =2.464≈2.46=2.464\approx 2.46 cm3^3 (i.e. about 22 cm3^3)

Q4. A solid consisting of a right circular cone of height 1212 cm …

A solid consisting of a right circular cone of height 1212 cm and radius 66 cm standing on a hemisphere of radius 66 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 66 cm and height is 1818 cm.

A cylinder containing a pink cone of height 12 cm standing on a blue hemisphere of radius 6 cm, with the cylinder's base radius marked 6 cm.

  1. Water displaced == Volume of the solid (cone ++ hemisphere).

  2. Cone: r=6r=6 cm, h=12h=12 cm; Hemisphere: r=6r=6 cm.

  3. Volume =13πr2h+23πr3=13πr2(h+2r)=\dfrac{1}{3}\pi r^2h+\dfrac{2}{3}\pi r^3=\dfrac{1}{3}\pi r^2(h+2r)

  4. =13×227×36×(12+12)=\dfrac{1}{3}\times\dfrac{22}{7}\times 36\times(12+12)

  5. =227×12×24=63367=905.14=\dfrac{22}{7}\times 12\times 24=\dfrac{6336}{7}=905.14

Answer

Volume of water displaced =905.14=905.14 cm3^3

Q5. A capsule is in the shape of a cylinder with two hemisphere …

A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is 1212 mm and the diameter of the capsule is 33 mm, how much medicine it can hold?

Show steps (6)

  1. Diameter =3=3 mm, so radius r=32r=\dfrac{3}{2} mm.

  2. The two hemispheres take up 2r=32r=3 mm of the length.

  3. Height of cylinder h=12−3=9h=12-3=9 mm.

  4. Volume =πr2h+2×23πr3=πr2(h+4r3)=\pi r^2h+2\times\dfrac{2}{3}\pi r^3=\pi r^2\left(h+\dfrac{4r}{3}\right)

  5. =227×94×(9+2)=\dfrac{22}{7}\times\dfrac{9}{4}\times(9+2)

  6. =227×994=108914=77.79=\dfrac{22}{7}\times\dfrac{99}{4}=\dfrac{1089}{14}=77.79

Answer

The capsule can hold 77.7977.79 mm3^3 of medicine

Q6. As shown in figure a cubical block of side 77 cm is …

As shown in figure a cubical block of side 77 cm is surmounted by a hemisphere. Find the surface area of the solid.

A blue cube of side 7 cm surmounted by a pink hemisphere resting on its top face.
Show steps (6)

  1. Side of cube a=7a=7 cm; diameter of hemisphere =7=7 cm, so r=72r=\dfrac{7}{2} cm.

  2. The circular base of the hemisphere covers part of the top face, so

  3. S.A. == T.S.A. of cube ++ C.S.A. of hemisphere −- base area of hemisphere

  4. =6a2+2πr2−πr2=6a2+πr2=6a^2+2\pi r^2-\pi r^2=6a^2+\pi r^2

  5. =6×49+227×494=6\times 49+\dfrac{22}{7}\times\dfrac{49}{4}

  6. =294+38.5=332.5=294+38.5=332.5

Answer

Surface area of the solid =332.5=332.5 cm2^2

Q7. A right circular cylinder just enclose a sphere of radius rr units. …

A right circular cylinder just enclose a sphere of radius rr units. Calculate

(i)

the surface area of the sphere.

  1. Radius of the sphere =r=r units.

  2. Surface area of sphere =4πr2=4\pi r^2 sq. units.

Answer

4πr24\pi r^2 sq. units

(ii)

the curved surface area of the cylinder.

  1. The cylinder just encloses the sphere, so its radius =r=r and its height =2r=2r.

  2. C.S.A. of cylinder =2πrh=2πr(2r)=4πr2=2\pi rh=2\pi r(2r)=4\pi r^2 sq. units.

Answer

4πr24\pi r^2 sq. units

(iii)

the ratio of the areas obtained in (i) and (ii).

  1. Ratio =S.A. of sphereC.S.A. of cylinder=4πr24πr2=11=\dfrac{\text{S.A. of sphere}}{\text{C.S.A. of cylinder}}=\dfrac{4\pi r^2}{4\pi r^2}=\dfrac{1}{1}

Answer

1:11:1

Quick check

  1. A spherical ball of radius r1r_1 units is melted to make 88 new identical balls each of radius r2r_2 units. Then r1:r2r_1 : r_2 is

    1. (1) 2:12:1
    2. (2) 1:21:2
    3. (3) 4:14:1
    4. (4) 1:41:4
    Show answer

    Answer: (1) 2:12:1

    1. Volume is unchanged: 43πr13=8×43πr23\dfrac{4}{3}\pi r_1^3=8\times\dfrac{4}{3}\pi r_2^3

    2. r13=8r23⇒r1=2r2r_1^3=8r_2^3\Rightarrow r_1=2r_2

    3. r1:r2=2:1r_1:r_2=2:1.

    Question 1 of 5

  2. If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

    1. (1) 1:21:2
    2. (2) 1:41:4
    3. (3) 1:61:6
    4. (4) 1:81:8
    Show answer

    Answer: (2) 1:41:4

    1. New volume =π(r2)2h=πr2h4=\pi\left(\dfrac{r}{2}\right)^2h=\dfrac{\pi r^2h}{4}.

    2. Ratio =πr2h/4πr2h=14=\dfrac{\pi r^2h/4}{\pi r^2h}=\dfrac{1}{4}.

    Question 2 of 5

  3. A solid sphere of radius xx cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is

    1. (1) 3x3x cm
    2. (2) xx cm
    3. (3) 4x4x cm
    4. (4) 2x2x cm
    Show answer

    Answer: (3) 4x4x cm

    1. Volume of sphere == volume of cone

    2. 43πx3=13πx2h\dfrac{4}{3}\pi x^3=\dfrac{1}{3}\pi x^2h

    3. h=4xh=4x cm.

    Question 3 of 5

  4. A frustum of a right circular cone is of height 1616cm with radii of its ends as 88cm and 2020cm. Then, the volume of the frustum is

    1. (1) 3328π3328\pi cm3^3
    2. (2) 3228π3228\pi cm3^3
    3. (3) 3240π3240\pi cm3^3
    4. (4) 3340π3340\pi cm3^3
    Show answer

    Answer: (1) 3328π3328\pi cm3^3

    1. V=13πh(R2+Rr+r2)V=\dfrac{1}{3}\pi h(R^2+Rr+r^2)

    2. =13π(16)(400+160+64)=\dfrac{1}{3}\pi(16)(400+160+64)

    3. =163π(624)=3328π=\dfrac{16}{3}\pi(624)=3328\pi cm3^3.

    Question 4 of 5

  5. If two solid hemispheres of same base radius rr units are joined together along their bases, then curved surface area of this new solid is

    1. (1) 4πr24\pi r^2 sq. units
    2. (2) 6πr26\pi r^2 sq. units
    3. (3) 3πr23\pi r^2 sq. units
    4. (4) 8πr28\pi r^2 sq. units
    Show answer

    Answer: (1) 4πr24\pi r^2 sq. units

    1. Two hemispheres joined along their bases form a sphere of radius rr.

    2. CSA of sphere =4πr2=4\pi r^2 sq. units.

    Question 5 of 5
Chapter 7 formulas
  • Cuboid — Lateral surface area

    2h(l+b)2h(l+b)

  • Cuboid — Total surface area

    2(lb+bh+lh)2(lb+bh+lh)

  • Cuboid — Volume

    l×b×hl \times b \times h

  • Cube — Lateral surface area

    4a24a^{2}

  • Cube — Total surface area

    6a26a^{2}

  • Cube — Volume

    a3a^{3}

  • Right circular cylinder — Curved surface area

    2πrh2\pi r h

  • Right circular cylinder — Total surface area

    2πr(h+r)2\pi r(h+r)

  • Right circular cylinder — Volume

    πr2h\pi r^{2} h

  • Right circular cone — Curved surface area

    πrl\pi r l

  • Right circular cone — Slant height

    l=r2+h2l=\sqrt{r^{2}+h^{2}}

  • Right circular cone — Total surface area

    πrl+πr2=πr(l+r)\pi r l+\pi r^{2}=\pi r(l+r)

  • Right circular cone — Volume

    13πr2h\dfrac{1}{3}\pi r^{2} h

  • Sphere — Curved surface area

    4πr24\pi r^{2}

  • Sphere — Total surface area

    4πr24\pi r^{2}

  • Sphere — Volume

    43πr3\dfrac{4}{3}\pi r^{3}

  • Hemisphere — Curved surface area

    2πr22\pi r^{2}

  • Hemisphere — Total surface area

    3πr23\pi r^{2}

  • Hemisphere — Volume

    23πr3\dfrac{2}{3}\pi r^{3}

  • Hollow cylinder — Curved surface area

    2π(R+r)h2\pi(R+r)h

  • Hollow cylinder — Total surface area

    2π(R+r)(R−r+h)2\pi(R+r)(R-r+h)

  • Hollow cylinder — Volume

    π(R2−r2)h\pi\left(R^{2}-r^{2}\right)h

  • Hollow sphere — Outer surface area

    4πR24\pi R^{2}

  • Hollow sphere — Total surface area

    4π(R2+r2)4\pi\left(R^{2}+r^{2}\right)

  • Hollow sphere — Volume

    43π(R3−r3)\dfrac{4}{3}\pi\left(R^{3}-r^{3}\right)

  • Hollow hemisphere — Curved surface area

    2π(R2+r2)2\pi\left(R^{2}+r^{2}\right)

  • Hollow hemisphere — Total surface area

    π(3R2+r2)\pi\left(3R^{2}+r^{2}\right)

  • Hollow hemisphere — Volume

    23π(R3−r3)\dfrac{2}{3}\pi\left(R^{3}-r^{3}\right)

  • Frustum of a right circular cone — Curved surface area

    π(R+r)l\pi(R+r)l

  • Frustum of a right circular cone — Slant height

    l=h2+(R−r)2l=\sqrt{h^{2}+(R-r)^{2}}

  • Frustum of a right circular cone — Total surface area

    π(R+r)l+πR2+πr2\pi(R+r)l+\pi R^{2}+\pi r^{2}

  • Frustum of a right circular cone — Volume

    13πh[R2+r2+Rr]\dfrac{1}{3}\pi h\left[R^{2}+r^{2}+Rr\right]

Answers checked against the Samacheer Kalvi textbook · Updated

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