Right Circular Cylinder Example Problems with Solutions
\n<\/strong><\/h2>\n <\/p>\n
Example 1: \u00a0 \u00a0\u00a0<\/strong>The inner diameter of a circular well in 2 m and its depth is 10.5 m. Find :
\n(i)\u00a0\u00a0 the inner curved surface area of the well.
\n(ii)\u00a0 the cost of plastering this curved surface area at the rate of Rs. 35 per m2<\/sup>.
\nSolution: \u00a0 \u00a0\u00a0<\/strong>Given : Radius = \\(\\frac{1}{2} \\times 2\\,m\\)\u00a0= 1m i.e., r = 1 m and depth = 10.5 m\u00a0\u00a0\u00a0 i.e., h = 10.5 m
\n(i) \u00a0 The inner curved surface area of the well = 2\u03c0rh = 2 \u00d7 \\(\\frac{{22}}{7} \\times 1 \\times 10.5\\) m2<\/sup>\u00a0= 66 m2<\/sup>
\n(ii) The cost of plastering
\n= Area to be plastered \u00d7 Rate
\n= 66 \u00d7 Rs. 35\u00a0 = Rs. 2,310<\/p>\nExample 2: \u00a0 \u00a0\u00a0<\/strong>In a hot water heating system there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>
\nSolution: \u00a0 \u00a0\u00a0<\/strong>Given : Length of the cylindrical pipe = 28 m i.e., h = 2800 cm and,\u00a0its radius = \\(\\frac{1}{2}\\) \u00d7 diameter = \\(\\frac{1}{2}\\) \u00d7 5 cm = 2.5 cm.
\n\u2234\u00a0\u00a0\u00a0 The total radiating surface in the system
\n= curved surface area of the pipe
\n= 2\u03c0rh = 2 \u00d7 \\(\\frac{22}{7}\\)\u00a0\u00d7 2.5 \u00d7 2800 cm2<\/sup>
\n= 44000 cm2<\/sup>
\nIt is not clear from the question that how the cylindrical pipe is used. We can take the radiating surface of the system
\n= Total surface area of the pipe
\n= 2\u03c0r (r + h) = 2 \u00d7 \\(\\frac{{22}}{7} \\times 2.5\\)\u00a0(2.5 + 2800) cm2<\/sup>
\n= 44039.29 cm2<\/sup><\/p>\nExample 3: \u00a0 \u00a0\u00a0<\/strong>Find :(i)\u00a0\u00a0 the lateral or curved surface area of a cylindrical patrol storage tank that is 4.2 m in diameter and 4.5 m high.
\n(ii)\u00a0 how much steel was actually used if \\(\\frac{1}{12}\\)\u00a0of the steel actually used was wasted in making the closed tank.
\nSolution: \u00a0 \u00a0\u00a0<\/strong>(i)\u00a0 Given : r = \\(\\frac{{4.2}}{2}m\\)\u00a0= 2.1 m and h = 4.5 m
\n\u2234 \u00a0\u00a0 Curved surface area of the tank
\n= 2\u03c0rh = 2 \u00d7 \\(\\frac{{22}}{7} \\times 2.1\\)\u00a0\u00d7 4.5 m2<\/sup>\u00a0= 59.4 m2<\/sup>
\n(ii) Let the steel actually used be x m2<\/sup>
\n\u2234 \u00a0\u00a0 x \u2013\\(\\frac{1}{{12}}x\\) = 59.4 \u00a0 \u21d2 \u00a0\\(\\frac{11}{{12}}x\\) = 59.4
\n\u21d2\u00a0\u00a0 x = 59.4 \u00d7 \\(\\frac{12}{{11}}\\)\u00a0 = 64.8 m2<\/sup>
\n\u2234\u00a0\u00a0\u00a0 Actual amount of steel used = 64.8 m2<\/sup><\/p>\nExample 4: \u00a0 \u00a0\u00a0<\/strong>The figure, given alongside, shows the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame . Find how much cloth is required for covering the lampshade.
\nSolution: \u00a0 \u00a0\u00a0<\/strong>Given : The height of the lampshade = 30 cm.
\n\u2235\u00a0\u00a0\u00a0 A margin of 2.5 cm is required for folding its over and bottom of the frame ; the resulting height of the cloth required in the shape of the cylinder = (30 + 2.5 + 2.5) cm\u00a0\u00a0\u00a0\u00a0 = 35 cm.
\n\u2234 For the cloth, which must be in the
\nshape of a cylinder with height 35 cm and
\nradius \u00a0\\(\\frac{{20}}{2}cm\\) = 10 cm.
\ni.e., h = 35 cm and r = 10 cm.
\n\u2234 Area of the cloth required for covering the lampshade.
\n= 2\u03c0rh\u00a0 = 2 \u00d7 \\(\\frac{22}{7}\\)\u00a0\u00d7 10 \u00d7 35 cm2<\/sup>
\n= 2200 cm2\u00a0<\/sup>.<\/p>\n