In this exercise Class 10 Math Ch Surface Areas and Volumes Ex 13.3, students will learn how to calculate the volume of objects made by combining two or more basic solids such as cylinders, cones, hemispheres, and spheres. The questions in Exercise 13.2 require the use of standard volume formulas and their application in solving real-life problems. These NCERT solutions provide detailed, step-by-step explanations to help students understand the concepts and improve their accuracy in mensuration problems. A strong grasp of this topic is essential for scoring well in exams and for practical applications in daily life.
Class 10 Math Ch Surface Areas and Volumes Ex 13.3
Ex 13.3 Class 10 Maths Question 1.
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.
Solution:
Given:
- Radius of sphere = 4.2 cm
- Radius of cylinder = 6 cm
- The sphere is melted and recast into a cylinder → Volume remains the same
We are to find the height of the cylinder.
Step 1: Volume of sphere
Formula:
Volume of sphere = (4/3) × π × r³
= (4/3) × 3.1416 × (4.2)³
= (4/3) × 3.1416 × 74.088
≈ (4/3) × 3.1416 × 74.088
≈ 309.76 cm³ (approximately)
Step 2: Volume of cylinder
Formula:
Volume of cylinder = π × r² × h
= 3.1416 × 6² × h
= 3.1416 × 36 × h
= 113.0976 × h
Step 3: Equating volumes
Volume of sphere = Volume of cylinder:
309.76 = 113.0976 × h
h = 309.76 ÷ 113.0976
h ≈ 2.74 cm
Ex 13.3 Class 10 Maths Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Given:
Three metallic spheres with radii:
- r₁ = 6 cm
- r₂ = 8 cm
- r₃ = 10 cm
They are melted and recast into a single solid sphere with radius R.
Step 1: Volume of each sphere
Volume of a sphere = (4/3) × π × r³
- Volume₁ = (4/3) × π × 6³ = (4/3) × π × 216 = 288π
- Volume₂ = (4/3) × π × 8³ = (4/3) × π × 512 = 682.67π
- Volume₃ = (4/3) × π × 10³ = (4/3) × π × 1000 = 1333.33π
Step 2: Total volume
Total volume = Volume₁ + Volume₂ + Volume₃
= 288π + 682.67π + 1333.33π = 2304π
Step 3: Volume of resulting sphere
Let the radius be R. Then:
Volume = (4/3) × π × R³ = 2304π
Divide both sides by π:
(4/3) × R³ = 2304
R³ = 2304 × (3/4) = 2304 × 0.75 = 1728
Step 4: Find R
R = ∛1728 = 12 cm
Ex 13.3 Class 10 Maths Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Ex 13.3 Class 10 Maths Question 4.
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.
Solution:
Ex 13.3 Class 10 Maths Question 5.
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.
Solution:
Ex 13.3 Class 10 Maths Question 6.
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?
Solution:
Ex 13.3 Class 10 Maths Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Ex 13.3 Class 10 Maths Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Ex 13.3 Class 10 Maths Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
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