CBSE Class 10 Maths: Surface Areas and Volumes - Exercise 13.3

Understanding Conversion of Solids and Volume Conservation

In Exercise 13.3, the central theme revolves around the idea that when a solid is melted, recast, or reshaped from one form into another, its volume remains constant. This is a fundamental principle in geometry and physics. Imagine you have a certain amount of clay shaped as a cylinder. If you break it down and then reform it into several small spheres, the total amount of clay, and thus its total volume, does not change. What does change is the surface area. The surface area of the new shape(s) will almost certainly be different from the original solid's surface area.

To solve problems in this exercise, you'll need a strong grasp of the volume formulas for basic three-dimensional shapes such as cylinders, cones, spheres, and hemispheres. The general approach is to equate the volume of the original solid (or collection of solids) to the volume of the new solid (or collection of solids). For instance, if a spherical metal ball is melted and recast into several smaller cylindrical wires, then:

Volume of original sphere = Sum of the volumes of all new cylindrical wires.

This principle allows us to find unknown dimensions (like radius, height, or number of objects) of the new solid(s) if we know the dimensions of the original solid(s). Careful attention to units is also paramount, ensuring all measurements are in the same unit (e.g., cm or m) before calculations.

Key Volume Formulas for CBSE Class 10

Volume of a Cuboid

The space occupied by a cuboid is given by the product of its length, breadth, and height. Formula: V = l × b × h cubic units.

Volume of a Cylinder

The space occupied by a cylinder is the product of the area of its circular base and its height. Formula: V = πr²h cubic units, where r is the radius and h is the height.

Volume of a Cone

The space occupied by a cone is one-third the volume of a cylinder with the same base radius and height. Formula: V = (1/3)πr²h cubic units.

Volume of a Sphere

The space occupied by a sphere. Formula: V = (4/3)πr³ cubic units, where r is the radius.

Volume of a Hemisphere

The space occupied by half a sphere. Formula: V = (2/3)πr³ cubic units, where r is the radius.

Worked Examples from NCERT Exercise 13.3

• Example 1: Sphere recast into a Cylinder A 20 cm diameter spherical metal ball is melted and recast into a cylindrical wire of diameter 0.8 cm. Find the length of the wire. Step 1: Identify given information and target. For the sphere: Diameter = 20 cm, so Radius (R) = 10 cm. For the cylindrical wire: Diameter = 0.8 cm, so Radius (r) = 0.4 cm. We need to find its length (h). Step 2: Apply the principle of volume conservation. Volume of Sphere = Volume of Cylinder (4/3)πR³ = πr²h Step 3: Substitute values and solve for h. (4/3)π(10)³ = π(0.4)²h (4/3) × 1000 = (0.16)h (4000/3) = 0.16h h = (4000/3) / 0.16 h = (4000/3) × (100/16) h = (400000 / 48) = (25000 / 3) cm h ≈ 8333.33 cm Step 4: Convert to a more suitable unit if necessary (e.g., meters). h = 8333.33 cm = 83.33 meters Final answer: The length of the cylindrical wire is approximately 83.33 meters.
• Example 2: Well dug out, earth spread to form a platform A 20 m deep well with diameter 7 m is dug, and the earth taken out of it is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. Step 1: Identify given information. For the well (which is a cylinder): Depth (h_well) = 20 m, Diameter = 7 m, so Radius (r_well) = 3.5 m. For the platform (which is a cuboid): Length (l) = 22 m, Breadth (b) = 14 m. We need to find its Height (h_platform). Step 2: Apply the principle of volume conservation. Volume of Earth dug out = Volume of Platform Volume of Cylinder = Volume of Cuboid π(r_well)²h_well = l × b × h_platform Step 3: Substitute values and solve for h_platform. (22/7) × (3.5)² × 20 = 22 × 14 × h_platform (22/7) × (7/2)² × 20 = 22 × 14 × h_platform (22/7) × (49/4) × 20 = 22 × 14 × h_platform 22 × (7/4) × 20 = 22 × 14 × h_platform 22 × 7 × 5 = 22 × 14 × h_platform 770 = 308 × h_platform h_platform = 770 / 308 h_platform = 2.5 m Final answer: The height of the platform is 2.5 meters.

Exam Tips and Avoiding Common Mistakes

To excel in problems involving conversion of solids, keep these tips in mind:

1. Units, Units, Units! Always ensure all dimensions are in the same unit before performing calculations. If some are in cm and others in m, convert them to a common unit. Mistakes here are very common.
2. Radius vs. Diameter: Be careful whether a given value is the radius or the diameter. Remember, radius = diameter/2. Many students accidentally use diameter directly in formulas that require radius.
3. Correct Formulas: Double-check that you are using the correct volume formula for each specific shape involved. A small error in the formula can lead to a completely wrong answer.
4. Simplify Carefully: During calculations, try to simplify by canceling out common terms like π before multiplying. This makes the arithmetic easier and reduces the chance of calculation errors.
5. Read the Question Thoroughly: Understand what is being asked. Are you finding a dimension, the number of smaller objects, or something else? A clear understanding prevents misdirection.
6. Show All Steps: Even if you make a calculation mistake, showing your steps clearly can earn you partial marks from the examiner. Follow the logical flow of equating volumes.

Practice Questions with Solutions

• Q: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. A: Step 1: Identify given values and goal. For sphere: R = 4.2 cm. For cylinder: r = 6 cm, h = ? Step 2: Equate volumes. Volume of sphere = Volume of cylinder (4/3)πR³ = πr²h Step 3: Substitute and solve for h. (4/3) × (4.2)³ = (6)² × h (4/3) × 74.088 = 36h 98.784 = 36h h = 98.784 / 36 h = 2.744 cm Final answer: The height of the cylinder is 2.744 cm.
• Q: A well of diameter 3 m is dug 14 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 m to form an embankment. Find the height of the embankment. A: Step 1: Identify volumes. Earth from well (cylinder) forms an embankment (hollow cylinder). For well: r_well = 3/2 = 1.5 m, h_well = 14 m. For embankment: Inner radius (r_inner) = r_well = 1.5 m. Width of ring = 4 m. So, outer radius (r_outer) = 1.5 + 4 = 5.5 m. Height of embankment (h_embankment) = ? Step 2: Equate volumes. Volume of earth from well = Volume of embankment Volume of cylinder = Volume of hollow cylinder π(r_well)²h_well = π(r_outer² - r_inner²)h_embankment Step 3: Substitute values and solve for h_embankment. π(1.5)²(14) = π((5.5)² - (1.5)²)h_embankment (2.25) × 14 = (30.25 - 2.25)h_embankment 31.5 = 28h_embankment h_embankment = 31.5 / 28 h_embankment = 1.125 m Final answer: The height of the embankment is 1.125 meters.
• Q: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? A: Step 1: Convert all units to cm. Identify volumes. For one silver coin (cylinder): Diameter = 1.75 cm, so radius (r) = 1.75/2 = 0.875 cm. Thickness (h) = 2 mm = 0.2 cm. For cuboid: l = 5.5 cm, b = 10 cm, H = 3.5 cm. Step 2: Calculate volumes. Volume of one coin = πr²h = π(0.875)²(0.2) = π × 0.765625 × 0.2 = 0.153125π cm³ Volume of cuboid = l × b × H = 5.5 × 10 × 3.5 = 192.5 cm³ Step 3: Equate total volume of coins to volume of cuboid. Let 'n' be the number of coins. n × (Volume of one coin) = Volume of cuboid n × 0.153125π = 192.5 Step 4: Solve for n (using π ≈ 22/7). n × 0.153125 × (22/7) = 192.5 n × 0.48125 = 192.5 n = 192.5 / 0.48125 n = 400 Final answer: 400 silver coins must be melted.
• Q: A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream. A: Step 1: Identify given values and components. Convert units. For cylinder: Diameter = 12 cm, so R = 6 cm. Height (H) = 15 cm. For each cone: Diameter = 6 cm, so r = 3 cm. Height of conical part (h_cone) = 12 cm. Hemisphere on top: radius (r_hemi) = r = 3 cm. Step 2: Calculate volumes. Volume of ice cream in cylinder = πR²H = π(6)²(15) = π × 36 × 15 = 540π cm³ Volume of one cone with hemispherical top = Volume of conical part + Volume of hemispherical part = (1/3)πr²h_cone + (2/3)πr³ = (1/3)π(3)²(12) + (2/3)π(3)³ = (1/3)π × 9 × 12 + (2/3)π × 27 = 36π + 18π = 54π cm³ Step 3: Find the number of cones. Number of cones = (Volume of ice cream in cylinder) / (Volume of one cone with hemispherical top) Number of cones = 540π / 54π Number of cones = 10 Final answer: 10 such cones can be filled with ice cream.

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