Case Study Question 04

Mathematics

Chapter 13: Surface Areas and Volumes

Class 10

Read the following passage carefully and answer the following questions:

Science Project

Arun a `10^{th}` standard student makes a project on corona virus in science for an exhibition in his school. In this project, he picks a sphere which has volume 38808 `cm^{3}` and 11 cylindrical shapes, each of volume 1540 `cm^{3}` with length 10 cm.Case Study 04

Question.1.
Diameter of the base of the cylinder is

(a) 7 cm
(b) 14 cm
(c) 12 cm
(d) 16 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.
Diameter of the sphere is

(a) 40 cm
(b) 42 cm
(c) 21 cm
(d) 20 cm

(b) : We know that, volume of sphere `= \frac{4}{3}\pi r^{3}`
⇒ `38808 = \frac{4}{3} \times \frac{22}{7} \times r^{3}`
⇒ `r^{3}=\frac{38808\times3\times7}{4\times22}`
⇒ `r^{3}=441 \times 21`
⇒ `r^{3}= (21)^{3}`
⇒ `r = 21` cm
Therefore, Diameter of sphere = 42 cm

Question.3.
Total volume of the shape formed is

(a) 85541 `cm^{3}`
(b) 45738 `cm^{3}`
(c) 24625 `cm^{3}`
(d) 55748 `cm^{3}`

(d) : Total volume of shape formed = Volume of cylindrical shapes + Volume of sphere
= `11 \times 1540 + 38808`
= `16940 + 38808`
= `55748 cm^{3}`

Question.4.
Curved surface area of the one cylindrical shape is

(a) 850 `cm^{2}`
(b) 221 `cm^{2}`
(c) 440 `cm^{2}`
(d) 540 `cm^{2}`

(c) : Curved surface area of one cylindrical shape = `2\pi rh`
= `2\times \frac{22}{7} \times 7 \times 10`
= `440 cm^{2}`

Question.5.
Total area covered by cylindrical shapes on the surface of sphere is

(a) 1694 `cm^{2}`
(b) 1580 `cm^{2}`
(c) 1896 `cm^{2}`
(d) 1470 `cm^{2}`

(a) : Area covered by cylindrical shapes on the surface of sphere = `11 \times \pi r^{2}`
= `11 \times \frac{22}{7} \times 7 \times 7`
= `1694 cm^{2}` 

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