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41. A cylindrical vessel, without lid, has to be tin-coated including both of its sides. If the radius of its base is 1/2 m and its height is 1.4 m, calculate the cost of tin-coating at the rate of र 50 per 100 cm². (Use

straight pi

= 3.14)

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42. 10 cylindrical pillars of a building have to be painted. If the diameter of each pillar is 50 cm and the height 4 m, what will be the cost of painting at the rate of र 14 per square metre?

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43. A cylinder 3 m high, is open at the top. The circumference of its base is 22 cm. Find its total surface area.

open parentheses Take space straight pi space equals space 22 over 7 close parentheses

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44. Twenty cylindrical pillars of a building are to be cleaned. If the diameter of a pillar is 0.5 m and height is 4 m, what will be the cost of cleaning them at the rate of र 3 per m². (Take

straight pi

= 3.14)

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45. Find the curved surface area of a closed cylindrical petrol storage tank that is 3.8 m in diameter and 4.9 m high.

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46. The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square metre.

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47. The cost of painting the total outside surface of a closed cylindrical tank at 60 paise per sq. d.m. is र 237.60. The height of the tank is 6 times the radius of the base of the tank. Find its volume correct to two decimal places.

Curved surface area

equals fraction numerator 237.60 over denominator 0.60 end fraction

= 396 dm²Let the radius of the base of the tank be r dm. Then, height = 6 r dm
∴ Curved surface area = 2πRH

$equals 2. straight pi. straight r. left parenthesis 6 straight r right parenthesis equals 2.22 over 7. straight r left parenthesis 6 straight r right parenthesis equals 264 over 7 straight r squared space dm squared$
According to the question,

$264 over 7 r squared equals 396 rightwards double arrow space space r squared equals fraction numerator 396 cross times 7 over denominator 264 end fraction equals 21 over 2 rightwards double arrow space r equals square root of 21 over 2 end root space d m therefore space space space space v equals πr squared straight h equals 22 over 7.21 over 2. open parentheses 6 square root of 21 over 2 end root close parentheses$

= 198 x 3.24 = 641.52 dm³

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48. Savitri had to make a model of a cylindrical kaleidoscope for her science project. She went to buy the chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take

straight pi equals 22 over 7.

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49. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4. Calculate the ratio of their curved surface areas.

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