Search

Samacheer Kalvi 10th Maths Exercise 7.2 Solutions

10th Maths Exercise 7.2 solutions (Chapter 7 Mensuration) with step-by-step answers for all 10 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below
  1. Q1Height of the embankment =143=4.67=\dfrac{14}{3}=4.67 m
  2. Q2The water rises by 11 cm
  3. Q3Volume of the conical wooden piece =652190=652190 cm3^3
  4. Q4The container will be emptied in about 6363 minutes
  5. Q5Difference in volumes =32π=100.57=32\pi=100.57 cm3^3
  6. Q6Ratio of heights =5:7=5:7
  7. Q7Ratio of their volumes =64:343=64:343
  8. Q9Volume of the material =1332π=4186.29=1332\pi=4186.29 cm3^3
  9. Q10Volume =10459.43=10459.43 cm3^3 (≈10.46\approx 10.46 litres); cost of milk =₹418.38=₹418.38

Q1. A 1414 m deep well with inner diameter 1010 m is dug …

A 1414 m deep well with inner diameter 1010 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 55 m. Find the height of the embankment.

Show steps (7)

  1. Radius of the well r=102=5r=\dfrac{10}{2}=5 m, depth h=14h=14 m.

  2. Volume of earth dug out =πr2h=π×52×14=350π=\pi r^2h=\pi\times 5^2\times 14=350\pi cu.m.

  3. The embankment is a hollow cylinder of inner radius 55 m and outer radius 5+5=105+5=10 m.

  4. Let HH be its height.

  5. Volume of embankment =π(R2−r2)H=π(102−52)H=75πH=\pi(R^2-r^2)H=\pi(10^2-5^2)H=75\pi H.

  6. Earth dug out = earth spread, so 75πH=350π75\pi H=350\pi.

  7. H=35075=143=4.67H=\dfrac{350}{75}=\dfrac{14}{3}=4.67 m.

Answer

Height of the embankment =143=4.67=\dfrac{14}{3}=4.67 m

Q2. A cylindrical glass with diameter 2020 cm has water to a height …

A cylindrical glass with diameter 2020 cm has water to a height of 99 cm. A small cylindrical metal of radius 55 cm and height 44 cm is immersed completely. Calculate the raise of the water in the glass?

  1. Radius of the glass R=202=10R=\dfrac{20}{2}=10 cm.

  2. Volume of the metal cylinder =πr2h=π×52×4=100π=\pi r^2h=\pi\times 5^2\times 4=100\pi cu.cm.

  3. This volume raises the water level by h1h_1 cm.

  4. Volume of raised water =πR2h1=π×102×h1=100πh1=\pi R^2h_1=\pi\times 10^2\times h_1=100\pi h_1.

  5. 100πh1=100π⇒h1=1100\pi h_1=100\pi \Rightarrow h_1=1 cm.

Answer

The water rises by 11 cm

Q3. If the circumference of a conical wooden piece is 484484 cm then …

If the circumference of a conical wooden piece is 484484 cm then find its volume when its height is 105105 cm.

  1. Circumference of the base =2πr=484=2\pi r=484.

  2. r=484×72×22=77r=\dfrac{484\times 7}{2\times 22}=77 cm.

  3. Height h=105h=105 cm.

  4. Volume =13πr2h=13×227×77×77×105=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times\dfrac{22}{7}\times 77\times 77\times 105

  5. =13×22×11×77×105=652190=\dfrac{1}{3}\times 22\times 11\times 77\times 105=652190 cu.cm.

Answer

Volume of the conical wooden piece =652190=652190 cm3^3

Q4. A conical container is fully filled with petrol. The radius is 1010 …

A conical container is fully filled with petrol. The radius is 1010m and the height is 1515 m. If the container can release the petrol through its bottom at the rate of 2525 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.

  1. Radius r=10r=10 m, height h=15h=15 m.

  2. Volume of petrol =13πr2h=13×227×102×15=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times\dfrac{22}{7}\times 10^2\times 15

  3. =22×5007=110007=1571.43=\dfrac{22\times 500}{7}=\dfrac{11000}{7}=1571.43 cu.m.

  4. Time =Volumerate=1571.4325=62.857=\dfrac{\text{Volume}}{\text{rate}}=\dfrac{1571.43}{25}=62.857 minutes.

  5. Rounding off, time ≈63\approx 63 minutes.

Answer

The container will be emptied in about 6363 minutes

Q5. A right angled triangle whose sides are 66 cm, 88 cm and …

A right angled triangle whose sides are 66 cm, 88 cm and 1010 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

Show steps (6)

  1. The sides containing the right angle are 66 cm and 88 cm.

  2. Revolving about the 66 cm side gives a cone with r=8r=8 cm, h=6h=6 cm.

  3. V1=13π×82×6=128πV_1=\dfrac{1}{3}\pi\times 8^2\times 6=128\pi cu.cm.

  4. Revolving about the 88 cm side gives a cone with r=6r=6 cm, h=8h=8 cm.

  5. V2=13π×62×8=96πV_2=\dfrac{1}{3}\pi\times 6^2\times 8=96\pi cu.cm.

  6. Difference =128π−96π=32π=32×227=7047=100.57=128\pi-96\pi=32\pi=32\times\dfrac{22}{7}=\dfrac{704}{7}=100.57 cu.cm.

Answer

Difference in volumes =32π=100.57=32\pi=100.57 cm3^3

Q6. The volumes of two cones of same base radius are 36003600 cm3^3 …

The volumes of two cones of same base radius are 36003600 cm3^3 and 50405040 cm3^3. Find the ratio of heights.

  1. Let the common base radius be rr and heights be h1,h2h_1,h_2.

  2. V1V2=13πr2h113πr2h2=h1h2\dfrac{V_1}{V_2}=\dfrac{\frac{1}{3}\pi r^2h_1}{\frac{1}{3}\pi r^2h_2}=\dfrac{h_1}{h_2}.

  3. h1h2=36005040=57\dfrac{h_1}{h_2}=\dfrac{3600}{5040}=\dfrac{5}{7}.

Answer

Ratio of heights =5:7=5:7

Q7. If the ratio of radii of two spheres is 4:74:7 , find …

If the ratio of radii of two spheres is 4:74:7, find the ratio of their volumes.

  1. Let the radii be r1r_1 and r2r_2 with r1r2=47\dfrac{r_1}{r_2}=\dfrac{4}{7}.

  2. V1V2=43πr1343πr23=(r1r2)3\dfrac{V_1}{V_2}=\dfrac{\frac{4}{3}\pi r_1^{3}}{\frac{4}{3}\pi r_2^{3}}=\left(\dfrac{r_1}{r_2}\right)^{3}.

  3. =(47)3=64343=\left(\dfrac{4}{7}\right)^{3}=\dfrac{64}{343}.

Answer

Ratio of their volumes =64:343=64:343

Q8. A solid sphere and a solid hemisphere have equal total surface area. …

A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volume is 33:43\sqrt{3} : 4.

Show steps (7)

  1. Let r1r_1 be the radius of the sphere and r2r_2 the radius of the hemisphere.

  2. TSA of sphere =4πr12=4\pi r_1^{2}, TSA of solid hemisphere =3πr22=3\pi r_2^{2}.

  3. Given they are equal: 4πr12=3πr224\pi r_1^{2}=3\pi r_2^{2}.

  4. r12r22=34⇒r1r2=32\dfrac{r_1^{2}}{r_2^{2}}=\dfrac{3}{4}\Rightarrow \dfrac{r_1}{r_2}=\dfrac{\sqrt3}{2}.

  5. V1V2=43πr1323πr23=2(r1r2)3\dfrac{V_1}{V_2}=\dfrac{\frac{4}{3}\pi r_1^{3}}{\frac{2}{3}\pi r_2^{3}}=2\left(\dfrac{r_1}{r_2}\right)^{3}.

  6. =2×(32)3=2×338=334=2\times\left(\dfrac{\sqrt3}{2}\right)^{3}=2\times\dfrac{3\sqrt3}{8}=\dfrac{3\sqrt3}{4}.

  7. Hence V1:V2=33:4V_1:V_2=3\sqrt3:4.

Answer

Ratio of their volumes =33:4=3\sqrt{3}:4 (proved)

Q9. The outer and the inner surface areas of a spherical copper shell …

The outer and the inner surface areas of a spherical copper shell are 576π576\pi cm2^2 and 324π324\pi cm2^2 respectively. Find the volume of the material required to make the shell.

Show steps (6)

  1. Let RR and rr be the outer and inner radii.

  2. 4πR2=576π⇒R2=144⇒R=124\pi R^{2}=576\pi\Rightarrow R^{2}=144\Rightarrow R=12 cm.

  3. 4πr2=324π⇒r2=81⇒r=94\pi r^{2}=324\pi\Rightarrow r^{2}=81\Rightarrow r=9 cm.

  4. Volume of material =43π(R3−r3)=\dfrac{4}{3}\pi(R^{3}-r^{3})

  5. =43π(1728−729)=43π×999=1332π=\dfrac{4}{3}\pi(1728-729)=\dfrac{4}{3}\pi\times 999=1332\pi

  6. =1332×227=293047=4186.29=1332\times\dfrac{22}{7}=\dfrac{29304}{7}=4186.29 cu.cm.

Answer

Volume of the material =1332π=4186.29=1332\pi=4186.29 cm3^3

Q10. A container open at the top is in the form of a …

A container open at the top is in the form of a frustum of a cone of height 1616 cm with radii of its lower and upper ends are 88 cm and 2020 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹40₹40 per litre.

Show steps (6)

  1. For the frustum: h=16h=16 cm, R=20R=20 cm, r=8r=8 cm.

  2. Volume =πh3(R2+Rr+r2)=\dfrac{\pi h}{3}(R^{2}+Rr+r^{2})

  3. =13×227×16×(400+160+64)=\dfrac{1}{3}\times\dfrac{22}{7}\times 16\times(400+160+64)

  4. =13×227×16×624=732167=10459.43=\dfrac{1}{3}\times\dfrac{22}{7}\times 16\times 624=\dfrac{73216}{7}=10459.43 cu.cm.

  5. 10001000 cm3^3 = 11 litre, so the milk =10.4594=10.4594 litres.

  6. Cost =10.4594×40=₹418.38=10.4594\times 40=₹418.38 (approximately).

Answer

Volume =10459.43=10459.43 cm3^3 (≈10.46\approx 10.46 litres); cost of milk =₹418.38=₹418.38

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

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