Case Study Question 04

Chapter 13: Surface Areas and Volumes

Class 10

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.

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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