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The volume of a cone of height 5 cm is 753.6 $c{m^3}$.Thus cone and a cylinder have equal radii and height. Find the total surface area of the cylinder. (Given$\pi = 3.14$).
$
  (a){\text{ 1281}}{\text{.12 c}}{{\text{m}}^2} \\
  (b){\text{ 1589}}{\text{.12 c}}{{\text{m}}^2} \\
  (c){\text{ 1242}}{\text{.12 c}}{{\text{m}}^2} \\
  (d){\text{ 1527}}{\text{.12 c}}{{\text{m}}^2} \\
 $

Answer
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362.1k+ views
Hint: In this problem we have been provided with a cone whose height and volume are known to us, this cone is said to have an equal radii and height with that of a cylinder, so we need to find the total surface area. Simply the use of formula for volume of cones will help us find out the radii of cones as height is given to us. Use this radii in finding out the T.S.A of the cylinder.

Volume of cone, V= 753.6 $c{m^3}$(given)
Height of cone, H = 5 cm (given)
Now using the formula for volume of cone $V = \dfrac{1}{3}\pi {r^2}h$………………. (1)
Substituting the values in equation (1) we get
$
  753.6 = \dfrac{1}{3}\pi {r^2}5 \\
   \Rightarrow 452.16 = \pi {r^2} \\
 $
Using $\pi = 3.14$ we get
$
  {r^2} = 144 \\
   \Rightarrow r = \pm 12 \\
 $
Radius of base for a cone can’t be negative thus we will take up only positive value hence r=12 cm.
Now the formula for Total surface area of cylinder ${\text{T}}{\text{.S}}{\text{.A = 2}}\pi {{\text{r}}^2} + 2\pi rH$……………….. (2)
As the radii and the height of cone are equal to that of cylinder thus substituting the values we get
${\text{T}}{\text{.S}}{\text{.A = 2}} \times {\text{3}}{\text{.14}} \times {12^2} + 2 \times 3.14 \times 12 \times 5$
On solving we get
$T.S.A = 1218.12{\text{ c}}{{\text{m}}^2}$
Thus option (a) is the right answer for this problem.

Note: Whenever we face such types of problems the key concept is simply the formula, thus it is advisable to have a good gist of all the formula especially the area, volume, T.S.A and C.S.A for various conic sections like cone, hemisphere, cylinder etc. This will help you to reach the right answer.
Last updated date: 23rd Sep 2023
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