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This is one of our many quizzes on Surface area of solids. Click on this to access more such competitive quizzes on Mensuration.

Surface area of solids Quiz 2

mathematics

 Calculate the total surface area of a capsule of radius $R$ and length $6R$ .<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/aebb5bd5-ce4a-45ae-b077-a12778f2df5c3497566878933697887.png"/>

 Consider the given figure to find the hidden cross-sections to simplify the calculations and hence, calculate the total surface area of the capsule.

$   \text{The total surface area of the capsule is given by;} \\   {{A}_{t}}=2\left( 2\pi {{R}^{2}} \right)+2\pi Rh \\   \text{From the diagram of the capsule the length of the cylindrical part is given by;} \\   h=6R-R-R=4R \\   \Rightarrow {{A}_{t}}=4\pi {{R}^{2}}+8\pi {{R}^{2}} \\   \therefore {{A}_{t}}=12\pi {{R}^{2}} \\ $

 Calculate the minimum area of the metal sheet required to make a cylinder that stores $4m$ gas by mass. Assume the density of the gas is $\rho $ and height of the cylinder is $h$ .

 The area of the sheet required will be equal to the total surface area of the cylinder that holds the gas and so, we have to determine the density which in turn will lead us to the total surface area.

$   \text{The density is given by;} \\   \rho =\dfrac{m}{V} \\   \Rightarrow \rho =\dfrac{4m}{V} \\   \Rightarrow V=\dfrac{4m}{\rho } \\   \Rightarrow \pi {{r}^{2}}h=\dfrac{4m}{\rho } \\   \Rightarrow r=\sqrt{\dfrac{4m}{\rho \pi h}} \\   \text{Total surface area of the cylinder is given by;} \\   A=2\pi r\left( r+h \right) \\   \Rightarrow A=2\pi {{r}^{2}}+2\pi rh \\   \Rightarrow A=2\pi \left( \dfrac{4m}{\rho \pi h} \right)+2\pi h\sqrt{\dfrac{4m}{\rho \pi h}} \\   \therefore A=\dfrac{8m}{\rho h}+4\sqrt{\dfrac{\pi mh}{\rho }} \\ $

 If the surface area of the bigger base of the frustum is four times the surface area of the smaller base, determine the amount of water that is stored in liters if the smaller base area is $A\text{ sq}\text{. m}$ and the water is filled to a height of $h\text{ m}$ . 

 The amount of water stored in the bucket can be calculated by the help of the volume of the frustum and volume of frustum is directly proportional to its surface area.

$   \text{The bigger base has a surface area of;} \\   {{A}_{1}}=4A \\   \Rightarrow \pi {{R}^{2}}=4A \\   \text{Surface area of the smaller base is given by;} \\   {{A}_{2}}=A \\   \Rightarrow \pi {{r}^{2}}=A \\   \text{Volume of the frustum is given by;} \\   V=\dfrac{1}{3}\pi h\left( {{r}^{2}}+rh+{{R}^{2}} \right) \\   V=\dfrac{1}{3}h\left( \pi {{r}^{2}}+\pi rh+\pi {{R}^{2}} \right) \\   \Rightarrow V=\dfrac{1}{3}h\left( A+\pi h\sqrt{\dfrac{A}{\pi }}+4A \right) \\   \Rightarrow V=\dfrac{1}{3}h\left( 5A+h\sqrt{A\pi } \right)\text{ }{{\text{m}}^{3}} \\   1000\text{ L}=1\text{ }{{\text{m}}^{3}} \\   \text{Amount of water is given by;} \\   V=\dfrac{h\left( 5A+h\sqrt{A\pi } \right)}{3000}\text{ L} \\ $

 Calculate the amount of molten glass in $\text{kg}$ required to make a spherical bulb if the outer surface area is $A$ and inner surface area is $a$ . Assume the outer radius to be $R$ and the thickness of the bulb surface is $t$ and density of the glass is $\rho $

 The amount of glass used to make the bulb is actually the effective volume of the bulb and the effective volume of the bulb is the difference between the volume trapped by the outer surface and volume trapped by the inner surface.

$   \text{The effective volume of the bulb will be;} \\   {{V}_{eff}}={{V}_{out}}-{{V}_{in}} \\   \text{The outer volume is given by;} \\   {{V}_{out}}={{A}_{out}}R=AR \\   \text{The inner volume is given by;} \\   {{V}_{in}}={{A}_{in}}r \\  r=R-t \\   \Rightarrow {{V}_{in}}=a\left( R-t \right) \\   \text{Substituting the values;} \\   {{V}_{eff}}=AR-a\left( R-t \right) \\   \Rightarrow {{V}_{eff}}=AR-aR+at \\   \Rightarrow {{V}_{eff}}=R\left( A-a \right)+at \\   \text{Density of the glass is given as;} \\   \rho =\dfrac{m}{{{V}_{eff}}} \\   \Rightarrow m=\rho {{V}_{eff}} \\ \therefore m=\rho \left[R \left( A-a \right)+at \right] \\ $

 If the total surface area of the bigger cone is $A$ , what is the ratio of the total surface area of the inverted bucket to the bigger cone, if $H=4R$ ? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dd102729-4371-4786-98df-335d957a5f398325926560631057164.png"/>

 The surface area of the frustum is the difference between the surface area of the bigger cone and the surface area of the smaller cone.

$   \text{The total surface area of the bigger cone is given by;} \\   A=\pi r\left( L+R \right) \\   \text{Given that }H=4R \\   L=\sqrt{{{H}^{2}}+{{R}^{2}}}=\sqrt{16{{R}^{2}}+{{R}^{2}}}=R\sqrt{17} \\   \Rightarrow L=4.12R \\   A=\pi R\left( 4.12R+R \right) \\   \Rightarrow A=5.12\pi {{R}^{2}} \\   \text{From similarity of triangles;} \\  \dfrac{h}{r}=\dfrac{H}{R} \\   \Rightarrow h=0.3H \\   \text{The total surface area of the bucket (frustum) is given by;} \\   {A}'=\pi \left( r+R \right)l+\pi \left( {{r}^{2}}+{{R}^{2}} \right) \\   l=\sqrt{{{H}^{2}}+{{R}^{2}}}-\sqrt{{{h}^{2}}+{{r}^{2}}}=\sqrt{{{H}^{2}}+{{R}^{2}}}-\sqrt{{{\left( 0.3H \right)}^{2}}+{{\left( 0.3R \right)}^{2}}} \\   \Rightarrow l=0.7\sqrt{{{H}^{2}}+{{R}^{2}}}=0.7\sqrt{16{{R}^{2}}+R} \\   \Rightarrow l=0.7\times 4.12R=2.88R \\   \Rightarrow {A}'=\pi \left( 1.3R \right)\left( 2.88R \right)+\pi \left( 1.09{{R}^{2}} \right) \\   \Rightarrow {A}'=\pi R\left[ 3.75R+1.09R \right] \\   \Rightarrow {A}'=4.84\pi {{R}^{2}} \\   \text{The ratio is given by;} \\   \dfrac{{{A}'}}{A}=\dfrac{4.84\pi {{R}^{2}}}{5.12\pi {{R}^{2}}} \\   \therefore \dfrac{{{A}'}}{A}=\dfrac{4.84}{5.12}=0.94 \\ $