{"id":2881,"date":"2023-04-19T10:00:22","date_gmt":"2023-04-19T04:30:22","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2881"},"modified":"2023-04-20T10:00:19","modified_gmt":"2023-04-20T04:30:19","slug":"radius-right-circular-cylinder","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/radius-right-circular-cylinder\/","title":{"rendered":"What is the Radius of a Right Circular Cylinder"},"content":{"rendered":"

What is the Radius of a Right Circular Cylinder<\/strong><\/h2>\n

\"What<\/p>\n

If r and h denote respectively the radius of the base and height of a right circular cylinder<\/strong>, then –<\/p>\n

    \n
  1. Area of each end = \u03c0r2<\/sup><\/li>\n
  2. Curved surface area = 2\u03c0rh\u00a0= (circumference) height<\/li>\n
  3. Total surface area = 2\u03c0r (h + r) sq. units.<\/li>\n
  4. Volume = \u03c0r2<\/sup>h = Area of the base \u00d7 height
    \nwhere, \u03c0 = \\(\\frac{{22}}{{7}}\\)\u00a0or 3.14<\/li>\n<\/ol>\n

    If R and r (R > r) denote respectively the external and internal radii of a hollow right circular cylinder<\/strong>, then –
    \nLet external radius = R, Internal radius = r, height = h. Then,<\/p>\n

      \n
    1. Outer curved surface area = 2\u03c0Rh<\/li>\n
    2. Inner curved surface area = 2\u03c0rh<\/li>\n
    3. Area of each end = \u03c0 (R2<\/sup>\u00a0\u2013 r2<\/sup>)<\/li>\n
    4. Curved surface area of hollow cylinder = 2\u03c0 (R + r)\u00a0 h<\/li>\n
    5. Total surface area = 2\u03c0 (R + r) (R + h \u2013 r)<\/li>\n
    6. Volume of material = \u03c0h (R2<\/sup>\u00a0\u2013 r2<\/sup>)<\/li>\n<\/ol>\n

      Read more about Area\u00a0of a Right Circular Cone<\/a><\/p>\n

      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>\n

      Example 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>\n

      Example 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>\n

      Example 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.
      \n\"WhatSolution: \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