Question 1: The curved surface area of a right circular cylinder of height 14 cm is 88 cm^{2}. Find the diameter of the base of the cylinder.

**Answer:** CSA of cylinder = 88 sq cm, h = 14 cm

CSA of cylinder `= 2π rh`

Or, `2 π r xx 14 = 88`

Or, `2r = (88 xx 7)/(22 xx 14) = 2`

Or, diameter `= 2 cm`

Question 2: It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

**Answer:** Diameter = 140 cm so, r = 70 cm, h = 100 cm

CSA of cylinder `= 2 π rh= 2 xx (22/7) xx 70 xx 100 = 44000 sq cm`

Area of top and bottom `= 2 π r^2= 2 xx (22/7) xx 70 xx 70 = 30800 sq cm`

Total surface area of cylinder `= 44000 + 30800= 74800 sq cm = 7.48 sq m`

Question 3: A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its

(i) inner curved surface area,

**Answer:** h = 77 cm, r_{1} = 2 cm, r_{2} = 2.2 cm

Inner curved surface area `= 2 π rh= 2 xx (22/7) xx 2 xx 77 = 968 sq cm`

(ii) outer curved surface area,

**Answer:** Outer curved surface area `= 2 π rh= 2 xx (22/7) xx 2.2 xx 77 = 1064.8 sq cm`

(iii) total surface area.

**Answer:** Area of top `= π r_2^2 - π r_1^2``= π (r_2^2 - r_1^2)``= π (r_2 + r_1)(r_2 - r_1)``= π (2.2 + 2)(2.2 – 2)``= (22/7) xx 4.2 xx 0.2 = 2.64` = area of bottom

Hence, total surface area `= 968 + 1064.8 + 2.64 + 2.64 = 2038.04 sq cm`

Question 4: The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m^{2}.

**Answer:** d = 84 cm, h = 120 cm

CSA `= π dh`

`= (22/7) xx 84 xx 120= 31680 sq cm = 3.168 sq m`

Hence, area of playground `= 3.168 xx 500 = 1584 sq m`

Question 5: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m^{2}.

**Answer:** d = 50 cm = 0.5 m, h = 3.5 m

CSA of cylinder `= π dh`

`= (22/7) xx 0.5 xx 3.5 = 5.5 sq m`

`Co\st\ = Ar\ea\xxRa\te`

`= 5.5 xx 12.50 = Rs. 68.75`

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