Search

Samacheer Kalvi 10th Maths Exercise 7.5 Solutions

10th Maths Exercise 7.5 solutions (Chapter 7 Mensuration) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1
    1. 136π136\pi cm2^2
  2. Q2
    1. 4πr24\pi r^2 sq. units
  3. Q3
    1. 1212 cm
  4. Q4
    1. 1:41:4
  5. Q5
    1. 8πh29\dfrac{8\pi h^2}{9} sq.units
  6. Q6
    1. 1120π1120\pi cm3^3
  7. Q7
    1. made 1818 times
  8. Q8
    1. 3π3\pi
  9. Q9
    1. 4x4x cm
  10. Q10
    1. 3328π3328\pi cm3^3
  11. Q11
    1. frustum of a cone and a hemisphere
  12. Q12
    1. 2:12:1
  13. Q13
    1. 43π\dfrac{4}{3}\pi
  14. Q14
    1. 1:21 : 2
  15. Q15
    1. 3:1:23:1:2

Q1. The curved surface area of a right circular cone of height 1515 …

The curved surface area of a right circular cone of height 1515 cm and base diameter 1616 cm is

  1. (1) 60π60\pi cm2^2
  2. (2) 68π68\pi cm2^2
  3. (3) 120π120\pi cm2^2
  4. (4) 136π136\pi cm2^2

  1. r=162=8r=\dfrac{16}{2}=8 cm, h=15h=15 cm.

  2. l=r2+h2=64+225=289=17l=\sqrt{r^2+h^2}=\sqrt{64+225}=\sqrt{289}=17 cm.

  3. CSA =πrl=π×8×17=136π=\pi r l=\pi\times 8\times 17=136\pi cm2^2.

Answer
  1. 136π136\pi cm2^2

Q2. If two solid hemispheres of same base radius rr units are joined …

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

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

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

Answer
  1. 4πr24\pi r^2 sq. units

Q3. The height of a right circular cone whose radius is 55 cm …

The height of a right circular cone whose radius is 55 cm and slant height is 1313 cm will be

  1. (1) 1212 cm
  2. (2) 1010 cm
  3. (3) 1313 cm
  4. (4) 55 cm

  1. l2=r2+h2l^2=r^2+h^2

  2. 132=52+h2⇒h2=169−25=14413^2=5^2+h^2\Rightarrow h^2=169-25=144

  3. h=12h=12 cm.

Answer
  1. 1212 cm

Q4. If the radius of the base of a right circular cylinder is …

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

  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}.

Answer
  1. 1:41:4

Q5. The total surface area of a cylinder whose radius is 13\dfrac{1}{3} of …

The total surface area of a cylinder whose radius is 13\dfrac{1}{3} of its height is

  1. (1) 9πh28\dfrac{9\pi h^2}{8} sq.units
  2. (2) 24πh224\pi h^2 sq.units
  3. (3) 8πh29\dfrac{8\pi h^2}{9} sq.units
  4. (4) 56πh29\dfrac{56\pi h^2}{9} sq.units

  1. Given r=h3r=\dfrac{h}{3}.

  2. TSA =2πr(r+h)=2πh3(h3+h)=2\pi r(r+h)=2\pi\dfrac{h}{3}\left(\dfrac{h}{3}+h\right)

  3. =2πh3×4h3=8πh29=2\pi\dfrac{h}{3}\times\dfrac{4h}{3}=\dfrac{8\pi h^2}{9}.

Answer
  1. 8πh29\dfrac{8\pi h^2}{9} sq.units

Q6. In a hollow cylinder, the sum of the external and internal radii …

In a hollow cylinder, the sum of the external and internal radii is 1414 cm and the width is 44 cm. If its height is 2020 cm, the volume of the material in it is

  1. (1) 5600π5600\pi cm3^3
  2. (2) 1120π1120\pi cm3^3
  3. (3) 56π56\pi cm3^3
  4. (4) 3600π3600\pi cm3^3

  1. R+r=14R+r=14, width R−r=4R-r=4, h=20h=20.

  2. Volume of material =π(R2−r2)h=π(R+r)(R−r)h=\pi(R^2-r^2)h=\pi(R+r)(R-r)h

  3. =π×14×4×20=1120π=\pi\times 14\times 4\times 20=1120\pi cm3^3.

Answer
  1. 1120π1120\pi cm3^3

Q7. If the radius of the base of a cone is tripled and …

If the radius of the base of a cone is tripled and the height is doubled then the volume is

  1. (1) made 66 times
  2. (2) made 1818 times
  3. (3) made 1212 times
  4. (4) unchanged

  1. New volume =13π(3r)2(2h)=13π⋅9r2⋅2h=\dfrac{1}{3}\pi(3r)^2(2h)=\dfrac{1}{3}\pi\cdot 9r^2\cdot 2h

  2. =18×13πr2h=18\times\dfrac{1}{3}\pi r^2h.

  3. So the volume is made 1818 times.

Answer
  1. made 1818 times

Q8. The total surface area of a hemi-sphere is how much times the …

The total surface area of a hemi-sphere is how much times the square of its radius.

  1. (1) π\pi
  2. (2) 4π4\pi
  3. (3) 3π3\pi
  4. (4) 2π2\pi

  1. TSA of hemisphere =3πr2=3\pi r^2.

  2. So it is 3π3\pi times the square of the radius.

Answer
  1. 3π3\pi

Q9. A solid sphere of radius xx cm is melted and cast into …

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

  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.

Answer
  1. 4x4x cm

Q10. A frustum of a right circular cone is of height 1616 cm …

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

  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.

Answer
  1. 3328π3328\pi cm3^3

Q11. A shuttle cock used for playing badminton has the shape of the …

A shuttle cock used for playing badminton has the shape of the combination of

  1. (1) a cylinder and a sphere
  2. (2) a hemisphere and a cone
  3. (3) a sphere and a cone
  4. (4) frustum of a cone and a hemisphere

  1. The cork base is a hemisphere and the feather part widens like a frustum of a cone.

  2. So it is frustum of a cone and a hemisphere.

Answer
  1. frustum of a cone and a hemisphere

Q12. A spherical ball of radius r1r_1 units is melted to make 88 …

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

  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.

Answer
  1. 2:12:1

Q13. The volume (in cm3^3) of the greatest sphere that can be cut …

The volume (in cm3^3) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 11 cm and height 55 cm is

  1. (1) 43π\dfrac{4}{3}\pi
  2. (2) 103π\dfrac{10}{3}\pi
  3. (3) 5π5\pi
  4. (4) 203π\dfrac{20}{3}\pi

  1. The greatest sphere has radius equal to the base radius =1=1 cm.

  2. V=43π(1)3=43πV=\dfrac{4}{3}\pi(1)^3=\dfrac{4}{3}\pi cm3^3.

Answer
  1. 43π\dfrac{4}{3}\pi

Q14. The height and radius of the cone of which the frustum is …

The height and radius of the cone of which the frustum is a part are h1h_1 units and r1r_1 units respectively. Height of the frustum is h2h_2 units and radius of the smaller base is r2r_2 units. If h2:h1=1:2h_2 : h_1 = 1 : 2 then r2:r1r_2 : r_1 is

  1. (1) 1:31 : 3
  2. (2) 1:21 : 2
  3. (3) 2:12 : 1
  4. (4) 3:13 : 1

  1. The small cone cut off has height h1−h2h_1-h_2 and radius r2r_2.

  2. By similar triangles r2r1=h1−h2h1\dfrac{r_2}{r_1}=\dfrac{h_1-h_2}{h_1}.

  3. Given h2=h12h_2=\dfrac{h_1}{2}, so r2r1=h1−h12h1=12\dfrac{r_2}{r_1}=\dfrac{h_1-\frac{h_1}{2}}{h_1}=\dfrac{1}{2}.

Answer
  1. 1:21 : 2

Q15. The ratio of the volumes of a cylinder, a cone and a …

The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is

  1. (1) 1:2:31:2:3
  2. (2) 2:1:32:1:3
  3. (3) 1:3:21:3:2
  4. (4) 3:1:23:1:2

  1. Same diameter and same height ⇒h=2r\Rightarrow h=2r.

  2. Cylinder =πr2(2r)=2πr3=\pi r^2(2r)=2\pi r^3

  3. Cone =13πr2(2r)=23πr3=\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3, Sphere =43πr3=\dfrac{4}{3}\pi r^3

  4. Ratio =2:23:43=6:2:4=3:1:2=2:\dfrac{2}{3}:\dfrac{4}{3}=6:2:4=3:1:2.

Answer
  1. 3:1:23:1:2
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

Spotted a mistake? Email us at [email protected] and we will fix it.