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Samacheer Kalvi 10th Maths Unit Exercise 7 Solutions

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

Answers at a glance10 answers · full steps below
  1. Q14800048000 words
  2. Q2About 1604.171604.17 seconds, i.e. 2626 minutes 4444 seconds
  3. Q313πr3\dfrac{1}{3}\pi r^3 cubic units
  4. Q478247782\dfrac{4}{7} sq. cm (about 782.57782.57 cm2^2)
  5. Q5450450 coins
  6. Q6Diameter =4.8=4.8 cm
  7. Q7CSA =68=68 sq. m; cost =₹ 6800=₹\,6800
  8. Q8Thickness =2=2 cm
  9. Q9Slant height =17=17 cm
  10. Q10Volume ≈2794.18\approx 2794.18 cu. cm

Q1. The barrel of a fountain-pen cylindrical in shape, is 77 cm long …

The barrel of a fountain-pen cylindrical in shape, is 77 cm long and 55 mm in diameter. A full barrel of ink in the pen will be used for writing 330330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?

  1. Barrel: r=52r=\dfrac{5}{2} mm =0.25=0.25 cm, h=7h=7 cm.

  2. Volume of barrel =πr2h=227×0.25×0.25×7=1.375=\pi r^2h=\dfrac{22}{7}\times 0.25\times 0.25\times 7=1.375 cm3^3.

  3. Ink in bottle =15=\dfrac{1}{5} litre =10005=200=\dfrac{1000}{5}=200 cm3^3.

  4. Number of barrels =2001.375=\dfrac{200}{1.375}.

  5. Words =330×2001.375=660001.375=48000=330\times\dfrac{200}{1.375}=\dfrac{66000}{1.375}=48000.

Answer

4800048000 words

Q2. A hemi-spherical tank of radius 1.751.75 m is full of water. It …

A hemi-spherical tank of radius 1.751.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 77 litre per second. How much time will it take to empty the tank completely?

  1. r=1.75r=1.75 m =74=\dfrac{7}{4} m.

  2. Volume of water =23πr3=23×227×(74)3=53948=\dfrac{2}{3}\pi r^3=\dfrac{2}{3}\times\dfrac{22}{7}\times\left(\dfrac{7}{4}\right)^3=\dfrac{539}{48} m3^3.

  3. 11 m3=1000^3=1000 litres, so volume =53900048=11229.17=\dfrac{539000}{48}=11229.17 litres.

  4. Time =11229.177=1604.17=\dfrac{11229.17}{7}=1604.17 seconds.

  5. 1604.171604.17 s =26=26 minutes 4444 seconds (approximately).

Answer

About 1604.171604.17 seconds, i.e. 2626 minutes 4444 seconds

Q3. Find the maximum volume of a cone that can be carved out …

Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius rr units.

  1. The largest cone carved from a hemisphere of radius rr has the same base and the apex at the top of the hemisphere.

  2. So base radius of cone =r=r and height of cone =r=r.

  3. Volume =13πr2h=13πr2×r=\dfrac{1}{3}\pi r^2 h=\dfrac{1}{3}\pi r^2\times r.

  4. =13πr3=\dfrac{1}{3}\pi r^3 cubic units.

Answer

13πr3\dfrac{1}{3}\pi r^3 cubic units

Q4. An oil funnel of tin sheet consists of a cylindrical portion 1010 …

An oil funnel of tin sheet consists of a cylindrical portion 1010 cm long attached to a frustum of a cone. If the total height is 2222 cm, the diameter of the cylindrical portion be 88 cm and the diameter of the top of the funnel be 1818 cm, then find the area of the tin sheet required to make the funnel.

  1. Cylinder: r=4r=4 cm, h=10h=10 cm. Frustum: R=9R=9 cm, r=4r=4 cm, height =22−10=12=22-10=12 cm.

  2. Slant height l=h2+(R−r)2=122+52=169=13l=\sqrt{h^2+(R-r)^2}=\sqrt{12^2+5^2}=\sqrt{169}=13 cm.

  3. CSA of frustum =π(R+r)l=π(9+4)(13)=169π=\pi(R+r)l=\pi(9+4)(13)=169\pi cm2^2.

  4. CSA of cylinder =2πrh=2π(4)(10)=80π=2\pi rh=2\pi(4)(10)=80\pi cm2^2.

  5. Total area =169π+80π=249π=249×227=54787=169\pi+80\pi=249\pi=249\times\dfrac{22}{7}=\dfrac{5478}{7} cm2^2.

Answer

78247782\dfrac{4}{7} sq. cm (about 782.57782.57 cm2^2)

Q5. Find the number of coins, 1.51.5 cm in diameter and 22 mm …

Find the number of coins, 1.51.5 cm in diameter and 22 mm thick, to be melted to form a right circular cylinder of height 1010 cm and diameter 4.54.5 cm.

  1. Coin: r=1.52=0.75r=\dfrac{1.5}{2}=0.75 cm, thickness h=2h=2 mm =0.2=0.2 cm.

  2. Volume of one coin =π(0.75)2(0.2)=0.1125π=\pi(0.75)^2(0.2)=0.1125\pi cm3^3.

  3. Cylinder: R=4.52=2.25R=\dfrac{4.5}{2}=2.25 cm, H=10H=10 cm.

  4. Volume =π(2.25)2(10)=50.625π=\pi(2.25)^2(10)=50.625\pi cm3^3.

  5. Number of coins =50.625π0.1125π=450=\dfrac{50.625\pi}{0.1125\pi}=450.

Answer

450450 coins

Q6. A hollow metallic cylinder whose external radius is 4.34.3 cm and internal …

A hollow metallic cylinder whose external radius is 4.34.3 cm and internal radius is 1.11.1 cm and whole length is 44 cm is melted and recast into a solid cylinder of 1212 cm long. Find the diameter of solid cylinder.

  1. Hollow cylinder: R=4.3R=4.3 cm, r=1.1r=1.1 cm, h=4h=4 cm.

  2. Volume =π(R2−r2)h=π(4.32−1.12)(4)=π(18.49−1.21)(4)=69.12π=\pi(R^2-r^2)h=\pi(4.3^2-1.1^2)(4)=\pi(18.49-1.21)(4)=69.12\pi cm3^3.

  3. Let the radius of the solid cylinder be xx, height 1212 cm.

  4. πx2(12)=69.12π⇒x2=5.76⇒x=2.4\pi x^2(12)=69.12\pi\Rightarrow x^2=5.76\Rightarrow x=2.4 cm.

  5. Diameter =2×2.4=4.8=2\times 2.4=4.8 cm.

Answer

Diameter =4.8=4.8 cm

Q7. The slant height of a frustum of a cone is 44 m …

The slant height of a frustum of a cone is 44 m and the perimeter of circular ends are 1818 m and 1616 m. Find the cost of painting its curved surface area at ₹100₹100 per sq. m.

  1. Let the radii be RR and rr: 2πR=182\pi R=18, 2πr=162\pi r=16, l=4l=4 m.

  2. CSA =π(R+r)l=(2πR+2πr)2×l=\pi(R+r)l=\dfrac{(2\pi R+2\pi r)}{2}\times l.

  3. =18+162×4=17×4=68=\dfrac{18+16}{2}\times 4=17\times 4=68 m2^2.

  4. Cost =68×100=₹ 6800=68\times 100=₹\,6800.

Answer

CSA =68=68 sq. m; cost =₹ 6800=₹\,6800

Q8. A hemi-spherical hollow bowl has material of volume 436π3\dfrac{436\pi}{3} cubic cm. Its …

A hemi-spherical hollow bowl has material of volume 436π3\dfrac{436\pi}{3} cubic cm. Its external diameter is 1414 cm. Find its thickness.

  1. External radius R=142=7R=\dfrac{14}{2}=7 cm; let internal radius be rr.

  2. Volume of material =23π(R3−r3)=436π3=\dfrac{2}{3}\pi(R^3-r^3)=\dfrac{436\pi}{3}.

  3. 2(343−r3)=436⇒343−r3=2182(343-r^3)=436\Rightarrow 343-r^3=218.

  4. r3=125⇒r=5r^3=125\Rightarrow r=5 cm.

  5. Thickness =R−r=7−5=2=R-r=7-5=2 cm.

Answer

Thickness =2=2 cm

Q9. The volume of a cone is 1005571005\dfrac{5}{7} cu. cm. The area of …

The volume of a cone is 1005571005\dfrac{5}{7} cu. cm. The area of its base is 20117201\dfrac{1}{7} sq. cm. Find the slant height of the cone.

  1. Base area πr2=20117=14087\pi r^2=201\dfrac{1}{7}=\dfrac{1408}{7}, so r2=14087×722=64⇒r=8r^2=\dfrac{1408}{7}\times\dfrac{7}{22}=64\Rightarrow r=8 cm.

  2. Volume =13(base area)h=100557=70407=\dfrac{1}{3}(\text{base area})h=1005\dfrac{5}{7}=\dfrac{7040}{7}.

  3. 13×14087×h=70407⇒h=15\dfrac{1}{3}\times\dfrac{1408}{7}\times h=\dfrac{7040}{7}\Rightarrow h=15 cm.

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

Answer

Slant height =17=17 cm

Q10. A metallic sheet in the form of a sector of a circle …

A metallic sheet in the form of a sector of a circle of radius 2121 cm has central angle of 216°216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.

  1. Slant height of cone == radius of sector =21=21 cm.

  2. Arc length =216360×2π(21)=25.2π=\dfrac{216}{360}\times 2\pi(21)=25.2\pi cm.

  3. This equals the base circumference: 2πr=25.2π⇒r=12.62\pi r=25.2\pi\Rightarrow r=12.6 cm.

  4. h=l2−r2=441−158.76=282.24=16.8h=\sqrt{l^2-r^2}=\sqrt{441-158.76}=\sqrt{282.24}=16.8 cm.

  5. Volume =13πr2h=13×227×158.76×16.8=2794.18=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times\dfrac{22}{7}\times 158.76\times 16.8=2794.18 cm3^3.

Answer

Volume ≈2794.18\approx 2794.18 cu. cm

Quick check

  1. 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
    Show answer

    Answer: (3) 8πh29\dfrac{8\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}.

    Question 1 of 5

  2. 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
    Show answer

    Answer: (1) 43π\dfrac{4}{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.

    Question 2 of 5

  3. 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
    Show answer

    Answer: (3) 3π3\pi

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

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

    Question 3 of 5

  4. 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
    Show answer

    Answer: (1) 1212 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.

    Question 4 of 5

  5. 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
    Show answer

    Answer: (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.

    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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