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Mensuration: Surface Areas and Volumes - Visualise and solve conversion problems between one solid shape and another

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The fundamental principle of conversion is that the volume remains constant when a solid of one shape is melted and recast into another shape (or multiple smaller shapes), provided there is no wastage during the process.

Diagram showing a sphere being converted to a cube with equal volumes indicated.
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When a large solid is melted to form nn identical smaller solids, the equation is Volume of larger solid=n×Volume of one smaller solidVolume \, of \, larger \, solid = n \times Volume \, of \, one \, smaller \, solid.

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In problems involving liquid flow (like water from a pipe into a tank), the 'length' of the solid formed by the liquid is equal to Speed×TimeSpeed \times Time.

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When earth is dug out from a well to form an embankment, the volume of the earth dug out equals the volume of the hollow cylindrical embankment formed around the well.

Cross-section of a well and its surrounding embankment.

📐Formulae

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Volume of a Cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

Volume of a Hemisphere: V=23πr3V = \frac{2}{3} \pi r^3

Volume of a Cuboid: V=l×b×hV = l \times b \times h

Volume of a Cube: V=a3V = a^3

Number of objects (nn): n=Volume of large solidVolume of one small solidn = \frac{Volume \, of \, large \, solid}{Volume \, of \, one \, small \, solid}

💡Examples

Problem 1:

A metallic sphere of radius 4.2 cm4.2 \, cm is melted and recast into the shape of a cylinder of radius 6 cm6 \, cm. Find the height of the cylinder.

Solution:

  1. Let rr be the radius of the sphere and R,HR, H be the radius and height of the cylinder.
  2. Given: r=4.2 cmr = 4.2 \, cm and R=6 cmR = 6 \, cm.
  3. Since the volume remains constant during recasting: Volume of sphere=Volume of cylinderVolume \, of \, sphere = Volume \, of \, cylinder.
  4. 43πr3=πR2H\frac{4}{3} \pi r^3 = \pi R^2 H.
  5. Dividing both sides by π\pi: 43×(4.2)3=62×H\frac{4}{3} \times (4.2)^3 = 6^2 \times H.
  6. 43×4.2×4.2×4.2=36×H\frac{4}{3} \times 4.2 \times 4.2 \times 4.2 = 36 \times H.
  7. 4×1.4×4.2×4.2=36H4 \times 1.4 \times 4.2 \times 4.2 = 36H.
  8. H=4×1.4×17.6436=1.4×17.649=1.4×1.96=2.744 cmH = \frac{4 \times 1.4 \times 17.64}{36} = \frac{1.4 \times 17.64}{9} = 1.4 \times 1.96 = 2.744 \, cm.

Explanation:

Because the metal is only changing shape and not quantity, we equate the volume of the original sphere to the volume of the new cylinder. Solving the linear equation for HH gives the height of the cylinder as 2.744 cm2.744 \, cm.

Problem 2:

How many silver coins, 1.75 cm1.75 \, cm in diameter and thickness 2 mm2 \, mm, must be melted to form a cuboid of dimensions 5.5 cm×10 cm×3.5 cm5.5 \, cm \times 10 \, cm \times 3.5 \, cm?

Solution:

  1. For the coin (cylindrical shape): radius r=1.752=0.875 cmr = \frac{1.75}{2} = 0.875 \, cm, thickness h=2 mm=0.2 cmh = 2 \, mm = 0.2 \, cm.
  2. Volume of one coin = πr2h=227×(0.875)2×0.2\pi r^2 h = \frac{22}{7} \times (0.875)^2 \times 0.2.
  3. Volume of cuboid = 5.5×10×3.5=192.5 cm35.5 \times 10 \times 3.5 = 192.5 \, cm^3.
  4. Let nn be the number of coins. Total volume of nn coins = Volume of cuboid.
  5. n×(227×0.875×0.875×0.2)=192.5n \times (\frac{22}{7} \times 0.875 \times 0.875 \times 0.2) = 192.5.
  6. n×0.48125=192.5n \times 0.48125 = 192.5.
  7. n=192.50.48125=400n = \frac{192.5}{0.48125} = 400.

Explanation:

A coin is modeled as a cylinder with a small height. We ensure all units are in cmcm, calculate the volume of a single coin, and then divide the target volume of the cuboid by the volume of one coin to find the total count needed.

Problem 3:

A well of diameter 3 m3 \, m is dug 14 m14 \, m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m4 \, m to form an embankment. Find the height of the embankment.

Top view of well and embankment showing radii.

Solution:

  1. Volume of earth dug from well: Radius r=32=1.5 mr = \frac{3}{2} = 1.5 \, m, Depth h=14 mh = 14 \, m. Vwell=πr2h=π×(1.5)2×14=31.5π m3V_{well} = \pi r^2 h = \pi \times (1.5)^2 \times 14 = 31.5 \pi \, m^3
  2. Dimensions of Embankment: Inner radius r=1.5 mr = 1.5 \, m, Width =4 m= 4 \, m. Outer radius R=1.5+4=5.5 mR = 1.5 + 4 = 5.5 \, m.
  3. Volume of Embankment: Let height be HH. Vemb=π(R2−r2)H=π(5.52−1.52)HV_{emb} = \pi(R^2 - r^2)H = \pi(5.5^2 - 1.5^2)H Vemb=π(30.25−2.25)H=28πHV_{emb} = \pi(30.25 - 2.25)H = 28 \pi H
  4. Equating volumes: 28πH=31.5π28 \pi H = 31.5 \pi H=31.528=1.125 mH = \frac{31.5}{28} = 1.125 \, m

Explanation:

The volume of the cylindrical well corresponds to the volume of the hollow cylinder (the embankment) formed by the displaced soil.

Problem 4:

Water in a canal, 6 m6 \, m wide and 1.5 m1.5 \, m deep, is flowing with a speed of 10 km/h10 \, km/h. How much area will it irrigate in 3030 minutes, if 8 cm8 \, cm of standing water is needed?

Perspective drawing of a canal flow.

Solution:

  1. Volume of water in 30 mins: Speed =10 km/h=10000 m/h= 10 \, km/h = 10000 \, m/h. Length of water in 30 mins30 \, mins (0.5 h0.5 \, h) =10000×0.5=5000 m= 10000 \times 0.5 = 5000 \, m. V=l×b×h=5000×6×1.5=45000 m3V = l \times b \times h = 5000 \times 6 \times 1.5 = 45000 \, m^3
  2. Area of irrigation: Let area be AA. Standing water height hstanding=8 cm=0.08 mh_{standing} = 8 \, cm = 0.08 \, m. Volume=Area×heightVolume = Area \times height 45000=A×0.0845000 = A \times 0.08 A=450000.08=562500 m2A = \frac{45000}{0.08} = 562500 \, m^2 A=56.25 hectaresA = 56.25 \, hectares

Explanation:

The volume of water flowing through the rectangular canal in a specific time is treated as a cuboid, which then forms another cuboid of a very large area and small height on the field.