# Case Study Question 05

## Chapter 13: Surface Areas and Volumes

### Class 10

Classroom Activity

To make the learning process more interesting, creative and innovative, Amayra’s class teacher brings clay in the classroom, to teach the topic – Surface Areas and Volumes. With clay, she forms a cylinder of radius 6 cm and height 8 cm. Then she moulds the cylinder into a sphere and asks some questions to students. Question.1.
The radius of the sphere so formed is

(a) 4 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

(b) : We know that, volume of cylinder = \pi r^{2}h
⇒ 1540 = \frac{22}{7}\times r^{2} \times 10
⇒ \frac{154\times 7}{22}= r^{2}
⇒ r^{2} = 49
⇒ r = 7
Therefore, Diameter of the base of cylinder = 2r = 2 \times 7 = 14 cm

Question.2.
The volume of the sphere so formed is

(a) 905.14 cm^{3}
(b) 903.27 cm^{3}
(c) 1296.5 cm^{3}
(d) 1156.63 cm^{3}

(a) : Volume of sphere = \frac{4}{3} \pi R^{3}
= \frac{4}{3}\times \frac{22}{7}\times 6 \times 6\times 6
= 905.14 cm^{3}

Question.3.
Find the ratio of the volume of sphere to the volume of cylinder.

(a) 2 : 1
(b) 1 : 2
(c) 1 : 1
(d) 3 : 1

(c) : Since, Volume of sphere = Volume of cylinder
Therefore, Required ratio = 1 : 1

Question.4.
Total surface area of the cylinder is

(a) 528 cm^{2}
(b) 756 cm^{2}
(c) 625 cm^{2}
(d) 636 cm^{2}

(a) : Total surface area of the cylinder = 2 \pi r(r+h)
= 2\times \frac{22}{7} \times 6(6+8)
= 2\times \frac{22}{7} \times 6 \times 14
= 528 cm^{2}

Question.5.
During the conversion of a solid from one shape to another the volume of new shape will

(a) be increase
(b) be decrease
(c) remain unaltered
(d) be double

(c) : Volume of new shape will remain same.

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