Volume and Surface Area Questions Answers

• 22. A hollow spherical metallic ball has an external diameter 6 cm and is 1/2 cm thick. The volume of metal used in the metal is:

  1. \begin{aligned} 47\frac{1}{5} cm^3 \end{aligned}
  2. \begin{aligned} 47\frac{3}{5} cm^3 \end{aligned}
  3. \begin{aligned} 47\frac{7}{5} cm^3 \end{aligned}
  4. \begin{aligned} 47\frac{9}{5} cm^3 \end{aligned}

  Answer And Explanation

  Answer: Option B

  Explanation:

  Please note we are talking about "Hollow" ball. Do not ignore this word in this type of question in a hurry to solve this question.
  If we are given with external radius and thickness, we can get the internal radius by subtracting them. Then the volume of metal can be obtained by its formula as,
  
  External radius = 3 cm,
  Internal radius = (3-0.5) cm = 2.5 cm
  
  \begin{aligned}
  \text{Volume of sphere =}\frac{4}{3}\pi r^3 \\
  = \frac{4}{3}*\frac{22}{7}*[3^2 - 2.5^2]cm^3 \\
  = \frac{4}{3}*\frac{22}{7}*\frac{91}{8}cm^3 \\
  = \frac{143}{3} cm^3 \\
  = 47\frac{2}{3}cm^3
  \end{aligned}

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• 23. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

  1. 4 cm
  2. 6 cm
  3. 8 cm
  4. 10 cm

  Answer And Explanation

  Answer: Option B

  Explanation:

  \begin{aligned}
  \text{Curved surface area of sphere =}\\
  \frac{4}\pi r^2 \\
  \text{Surface area of cylinder =} \\
  2\pi rh \\
  => \frac{4}\pi r^2 = 2\pi rh \\
  => r^2 = \frac{6*12}{2} \\
  => r^2 = 36 \\
  => r = 6
  \end{aligned}
  
  Note: Diameter of cylinder is 12 so radius is taken as 6.

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• 24. 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. Find the diameter of each sphere.

  1. 4 cm
  2. 6 cm
  3. 8 cm
  4. 10 cm

  Answer And Explanation

  Answer: Option A

  Explanation:

  In this type of question, just equate the two volumes to get the answer as,
  \begin{aligned}
  \text{Volume of cylinder =}\pi r^2h\\
  \text{Volume of sphere =} \frac{4}{3}\pi r^3\\
  => 12*\frac{4}{3}\pi r^3 = \pi r^2h \\
  => 12*\frac{4}{3}\pi r^3 = \pi *8*8*2 \\
  => r^3 = \frac{8*8*2*3}{12*4} \\
  => r^3 = 8 \\
  => r = 2 cm \\
  => \text{Diameter =}2*2 = 4 cm
  \end{aligned}

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• 25. A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. The percentage of the wood wasted is :

  1. 45%
  2. 56%
  3. 67%
  4. 75%

  Answer And Explanation

  Answer: Option D

  Explanation:

  We will first subtract the cone volume from wood volume to get the wood wasted.
  Then we can calculate its percentage.
  
  \begin{aligned}
  \text{Sphere Volume =}\frac{4}{3}\pi r^3 \\
  \text{Cone Volume =}\frac{1}{3}\pi r^2h\\
  
  \text{Volume of wood wasted =}\\
  \left(\frac{4}{3}\pi *9*9*9\right)-\left(\frac{1}{3}\pi *9*9*9\right) \\
  = \pi *9*9*9 cm^3 \\
  \text{Required Percentage =} \\
  \frac{\pi *9*9*9}{\frac{4}{3}\pi *9*9*9}*100 \% \\
  = \frac{3}{4}*100 \% \\
  = 75\%
  \end{aligned}

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• 26. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :

  1. \begin{aligned} 2:1 \end{aligned}
  2. \begin{aligned} 1:\sqrt{2} \end{aligned}
  3. \begin{aligned} \sqrt{2}:1 \end{aligned}
  4. \begin{aligned} \sqrt{3}:1 \end{aligned}

  Answer And Explanation

  Answer: Option C

  Explanation:

  Let the radius of hemisphere and cone be R,
  Height of hemisphere H = R.
  So the height of the cone = height of the hemisphere = R
  Slant height of the cone
  \begin{aligned}
  = \sqrt{R^2+R^2} \\
  = \sqrt{2}R \\
  \frac{\text{Hemisphere Curved surface area}}{\text{Cone Curved surface area}} = \\
  \frac{2\pi R^2}{\pi *R*\sqrt{2}R} \\
  = \sqrt{2}:1
  \end{aligned}

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• 27. There are bricks with 24 cm x 12 cm x 8 cm dimensions. Find the total number of bricks required to construct a wall 24 m long, 8 m high and 60 m thick with 10% of wall filled with mortar.

  1. 35000
  2. 40000
  3. 45000
  4. 50000

  Answer And Explanation

  Answer: Option C

  Explanation:

  So as per question,
  \begin{aligned}
  \text{Volume of wall} = (2400 * 800 * 60 ) cm^3 \\
  \text{Volume of bricks} = \text { 90% of volume of wall } \\
  = [ \frac{90}{100} * 2400 * 800 * 60 ] cm^3 \\
  \text{ Volume of 1 brick = } (24 * 12 * 8) cm^3 \\
  \text{Number of Bricks required} \\
  = \frac{ (\frac{90}{100}) * (2400 * 800 * 60)}{24 * 12 * 8} \\
  = 45000
  \end{aligned}

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