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Samacheer Kalvi 10th Maths Exercise 7.4 Solutions

10th Maths Exercise 7.4 solutions (Chapter 7 Mensuration) with step-by-step answers for all 8 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below
  1. Q1Height of the cylinder =36=36 cm
  2. Q222 hours
  3. Q3Height of water in the cylindrical flask =h3x2=\dfrac{h}{3x^{2}} units
  4. Q4Internal diameter =6=6 cm
  5. Q5Volume of water left in the sump =1812000=1812000 cm3^3 =1.812=1.812 m3^3
  6. Q6Height of the cylinder =43=1.33=\dfrac{4}{3}=1.33 cm
  7. Q7Thickness of the cylinder =1=1 cm
  8. Q8100%100\% of the juice can be transferred

Q1. An aluminium sphere of radius 1212 cm is melted to make a …

An aluminium sphere of radius 1212 cm is melted to make a cylinder of radius 88 cm. Find the height of the cylinder.

  1. Volume of sphere == Volume of cylinder.

  2. 43πR3=πr2h\dfrac{4}{3}\pi R^{3}=\pi r^{2}h

  3. 43π(12)3=π(8)2h\dfrac{4}{3}\pi (12)^{3}=\pi (8)^{2}h

  4. 43×1728=64 h\dfrac{4}{3}\times 1728 = 64\,h

  5. 2304=64 h⇒h=362304 = 64\,h \Rightarrow h = 36

Answer

Height of the cylinder =36=36 cm

Q2. Water is flowing at the rate of 1515 km per hour through …

Water is flowing at the rate of 1515 km per hour through a pipe of diameter 1414 cm into a rectangular tank which is 5050 m long and 4444 m wide. Find the time in which the level of water in the tanks will rise by 2121 cm.

Show steps (6)

  1. Radius of pipe =142=7=\dfrac{14}{2}=7 cm =7100=\dfrac{7}{100} m.

  2. Speed =15=15 km/hr =15000=15000 m/hr.

  3. Volume of water through pipe in 1 hour =πr2×15000=\pi r^{2}\times 15000

  4. =227×7100×7100×15000=231=\dfrac{22}{7}\times\dfrac{7}{100}\times\dfrac{7}{100}\times 15000 = 231 m3^3.

  5. Volume needed in tank =50×44×21100=462=50\times 44\times \dfrac{21}{100}=462 m3^3.

  6. Time =462231=2=\dfrac{462}{231}=2 hours.

Answer

22 hours

Q3. A conical flask is full of water. The flask has base radius …

A conical flask is full of water. The flask has base radius rr units and height hh units, the water is poured into a cylindrical flask of base radius xrxr units. Find the height of water in the cylindrical flask.

  1. Volume of water in cone == Volume of water in cylinder.

  2. 13πr2h=π(xr)2H\dfrac{1}{3}\pi r^{2}h = \pi (xr)^{2}H

  3. 13r2h=x2r2H\dfrac{1}{3}r^{2}h = x^{2}r^{2}H

  4. H=h3x2H = \dfrac{h}{3x^{2}}

Answer

Height of water in the cylindrical flask =h3x2=\dfrac{h}{3x^{2}} units

Q4. A solid right circular cone of diameter 1414 cm and height 88 …

A solid right circular cone of diameter 1414 cm and height 88 cm is melted to form a hollow sphere. If the external diameter of the sphere is 1010 cm, find the internal diameter.

Show steps (8)

  1. Cone: r=142=7r=\dfrac{14}{2}=7 cm, h=8h=8 cm.

  2. Volume of cone =13π(7)2(8)=392π3=\dfrac{1}{3}\pi (7)^{2}(8)=\dfrac{392\pi}{3} cu.cm.

  3. Sphere: external radius R=102=5R=\dfrac{10}{2}=5 cm, internal radius =r=r.

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

  5. 43π(125−r3)=392π3\dfrac{4}{3}\pi (125 - r^{3}) = \dfrac{392\pi}{3}

  6. 4(125−r3)=392⇒125−r3=984(125-r^{3}) = 392 \Rightarrow 125-r^{3}=98

  7. r3=27⇒r=3r^{3}=27 \Rightarrow r = 3 cm.

  8. Internal diameter =2×3=6=2\times 3 = 6 cm.

Answer

Internal diameter =6=6 cm

Q5. Seenu's house has an overhead tank in the shape of a cylinder. …

Seenu's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 22 m ×1.5\times 1.5 m ×\times 11 m. The overhead tank has its radius of 6060 cm and height 105105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.

  1. Volume of sump =2×1.5×1=3=2\times 1.5\times 1 = 3 m3^3 =3000000=3000000 cm3^3.

  2. Overhead tank: r=60r=60 cm, h=105h=105 cm.

  3. Volume =πr2h=227×60×60×105=1188000=\pi r^{2}h = \dfrac{22}{7}\times 60\times 60\times 105 = 1188000 cm3^3.

  4. Water left =3000000−1188000=1812000=3000000 - 1188000 = 1812000 cm3^3.

  5. =1.812=1.812 m3^3.

Answer

Volume of water left in the sump =1812000=1812000 cm3^3 =1.812=1.812 m3^3

Q6. The internal and external diameter of a hollow hemispherical shell are 66 …

The internal and external diameter of a hollow hemispherical shell are 66 cm and 1010 cm respectively. If it is melted and recast into a solid cylinder of diameter 1414 cm, then find the height of the cylinder.

  1. Hemispherical shell: r=62=3r=\dfrac{6}{2}=3 cm, R=102=5R=\dfrac{10}{2}=5 cm.

  2. Volume =23π(R3−r3)=23π(125−27)=196π3=\dfrac{2}{3}\pi (R^{3}-r^{3}) = \dfrac{2}{3}\pi (125-27)=\dfrac{196\pi}{3} cu.cm.

  3. Cylinder radius =142=7=\dfrac{14}{2}=7 cm.

  4. π(7)2h=196π3\pi (7)^{2}h = \dfrac{196\pi}{3}

  5. 49h=1963⇒h=4349h = \dfrac{196}{3} \Rightarrow h = \dfrac{4}{3}

Answer

Height of the cylinder =43=1.33=\dfrac{4}{3}=1.33 cm

Q7. A solid sphere of radius 66 cm is melted into a hollow …

A solid sphere of radius 66 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 55 cm and its height is 3232 cm, then find the thickness of the cylinder.

Show steps (6)

  1. Volume of sphere =43π(6)3=288π=\dfrac{4}{3}\pi (6)^{3}=288\pi cu.cm.

  2. Hollow cylinder: R=5R=5 cm, height =32=32 cm, internal radius =r=r.

  3. π(R2−r2)h=288π\pi (R^{2}-r^{2})h = 288\pi

  4. (25−r2)×32=288(25 - r^{2})\times 32 = 288

  5. 25−r2=9⇒r2=16⇒r=425 - r^{2} = 9 \Rightarrow r^{2}=16 \Rightarrow r = 4 cm.

  6. Thickness =R−r=5−4=1=R-r = 5-4 = 1 cm.

Answer

Thickness of the cylinder =1=1 cm

Q8. A hemispherical bowl is filled to the brim with juice. The juice …

A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50%50\% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.

Show steps (7)

  1. Let the radius of the hemispherical bowl be RR.

  2. Volume of juice =23πR3=\dfrac{2}{3}\pi R^{3} cu.units.

  3. Diameter is same, so the cylinder also has radius RR.

  4. Radius is 50%50\% more than height: R=h+50100h=3h2R = h + \dfrac{50}{100}h = \dfrac{3h}{2}, so h=2R3h = \dfrac{2R}{3}.

  5. Volume of cylinder =πR2h=πR2×2R3=23πR3=\pi R^{2}h = \pi R^{2}\times \dfrac{2R}{3}=\dfrac{2}{3}\pi R^{3}.

  6. Both volumes are equal, so all the juice is transferred.

  7. Percentage =23πR323πR3×100=100%=\dfrac{\frac{2}{3}\pi R^{3}}{\frac{2}{3}\pi R^{3}}\times 100 = 100\%

Answer

100%100\% of the juice can be transferred

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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