Class 10: Cone and Sphere – Exercise 22(b)

$\displaystyle \textbf{Important Formulae:}$
$\displaystyle \text{Parameter of a sphere: Radius }(r).$
$\displaystyle \text{Volume of a sphere}=\frac{4}{3}\pi r^3$
$\displaystyle \text{Surface area of a sphere}=4\pi r^2$
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$\displaystyle \textbf{Question 1: }\text{The surface area of a sphere is }2464\text{ cm}^2.\text{ Find its volume.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Surface area of a sphere}=2464\text{ cm}^2$
$\displaystyle 4\pi r^2=2464$
$\displaystyle r^2=\frac{2464\times7}{4\times22}=196$
$\displaystyle r=14\text{ cm}$
$\displaystyle \text{Volume of the sphere}=\frac{4}{3}\pi r^3$
$\displaystyle =\frac{4}{3}\times\frac{22}{7}\times14^3$
$\displaystyle =11498.67\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the sphere}=11498.67\text{ cm}^3$
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$\displaystyle \textbf{Question 2: }\text{The volume of a sphere is }38808\text{ cm}^3.\text{ Find its diameter and surface area.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the sphere}=38808\text{ cm}^3$
$\displaystyle \frac{4}{3}\pi r^3=38808$
$\displaystyle r^3=\frac{3\times38808\times7}{4\times22}=9261$
$\displaystyle r=21\text{ cm}$
$\displaystyle \therefore \text{Diameter}=2r=42\text{ cm}$
$\displaystyle \text{Surface area}=4\pi r^2$
$\displaystyle =4\times\frac{22}{7}\times21^2$
$\displaystyle =5544\text{ cm}^2$
$\displaystyle \therefore \text{Surface area}=5544\text{ cm}^2$
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$\displaystyle \textbf{Question 3: }\text{A spherical ball of lead is melted and made into identical smaller balls, eac}$
$\displaystyle \text{ having radius equal to half the radius of the original ball. How many such balls can be made?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the original ball be }r_1$
$\displaystyle \text{Let the radius of each smaller ball be }r_2$
$\displaystyle \text{Given, }r_2=\frac{r_1}{2}\Rightarrow \frac{r_1}{r_2}=2$
$\displaystyle \text{Number of smaller balls}=\frac{\text{Volume of the original ball}}{\text{Volume of one smaller ball}}$
$\displaystyle =\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}$
$\displaystyle =\left(\frac{r_1}{r_2}\right)^3=2^3=8$
$\displaystyle \therefore \text{The number of smaller balls formed is }8.$
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$\displaystyle \textbf{Question 4: }\text{How many balls, each of radius }1\text{ cm, can be made by melting a larger} \\ \text{ball whose diameter is }8\text{ cm?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter of the larger ball}=8\text{ cm}$
$\displaystyle \therefore \text{Radius of the larger ball}=4\text{ cm}$
$\displaystyle \text{Radius of each smaller ball}=1\text{ cm}$
$\displaystyle \text{Number of smaller balls}=\frac{\text{Volume of the larger ball}}{\text{Volume of one smaller ball}}$
$\displaystyle =\frac{\frac{4}{3}\pi(4)^3}{\frac{4}{3}\pi(1)^3}$
$\displaystyle =\frac{4^3}{1^3}=64$
$\displaystyle \therefore \text{The number of smaller balls formed is }64.$
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$\displaystyle \textbf{Question 5: }\text{Eight metallic spheres, each of radius }2\text{ mm, are melted and cast into a }$
$\displaystyle \text{single sphere. Calculate the radius of the new sphere.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of spheres}=8$
$\displaystyle \text{Radius of each sphere}=2\text{ mm}$
$\displaystyle \text{Let the radius of the new sphere be }r\text{ mm}$
$\displaystyle 8\times\frac{4}{3}\pi(2)^3=\frac{4}{3}\pi r^3$
$\displaystyle 8\times8=r^3$
$\displaystyle r^3=64$
$\displaystyle r=4\text{ mm}$
$\displaystyle \therefore \text{The radius of the new sphere is }4\text{ mm}.$
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$\displaystyle \textbf{Question 6: }\text{The volume of one sphere is }27\text{ times that of another sphere. }$
$\displaystyle \text{Calculate the ratio of their: (i) radii \qquad (ii) surface areas}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the volumes be }V_1\text{ and }V_2\text{ and the corresponding radii be }r_1\text{ and }r_2.$
$\displaystyle \text{Given, }V_1=27V_2$
$\displaystyle \text{(i) }\frac{4}{3}\pi r_1^3=27\times\frac{4}{3}\pi r_2^3$
$\displaystyle \Rightarrow r_1^3=27r_2^3$
$\displaystyle \Rightarrow r_1=3r_2$
$\displaystyle \therefore r_1:r_2=3:1$
$\displaystyle \text{(ii) Let the surface areas be }S_1\text{ and }S_2\text{ respectively.}$
$\displaystyle \frac{S_1}{S_2}=\frac{4\pi r_1^2}{4\pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2$
$\displaystyle =3^2=9$
$\displaystyle \therefore S_1:S_2=9:1$
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$\displaystyle \textbf{Question 7: }\text{If the number of square centimetres on the surface of a sphere is equal to the }$
$\displaystyle \text{number of cubic centimetres in its volume, find the diameter of the sphere.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the sphere be }r\text{ cm}$
$\displaystyle \text{Given, the numerical value of surface area}=\text{the numerical value of volume}$
$\displaystyle 4\pi r^2=\frac{4}{3}\pi r^3$
$\displaystyle 1=\frac{r}{3}$
$\displaystyle r=3\text{ cm}$
$\displaystyle \therefore \text{Diameter}=2r=2\times3=6\text{ cm}$
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$\displaystyle \textbf{Question 8: }\text{A solid metal sphere is cut through its centre into }2\text{ equal parts. If its }$
$\displaystyle \text{diameter is }3.5\text{ cm,} \ \text{find the total surface area of each part, correct to two decimal places.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter of the sphere}=3.5\text{ cm}$
$\displaystyle r=\frac{3.5}{2}=1.75\text{ cm}$
$\displaystyle \text{Each part is a hemisphere.}$
$\displaystyle \text{Total surface area of a hemisphere}=2\pi r^2+\pi r^2=3\pi r^2$
$\displaystyle =3\times\frac{22}{7}\times(1.75)^2$
$\displaystyle =28.875\text{ cm}^2$
$\displaystyle \therefore \text{Total surface area of each part}=28.88\text{ cm}^2\text{, correct to two decimal places.}$
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$\displaystyle \textbf{Question 9: }\text{The internal and external diameters of a hollow hemispherical vessel are }$
$\displaystyle 21\text{ cm and }28\text{ cm respectively.} \ \text{Find: (i) internal curved surface area (ii) external curved surface area }$
$\displaystyle \text{(iii) total surface area (iv) volume of material of the vessel.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Internal radius }r=\frac{21}{2}=10.5\text{ cm}$
$\displaystyle \text{External radius }R=\frac{28}{2}=14\text{ cm}$
$\displaystyle \text{(i) Internal curved surface area}=2\pi r^2$
$\displaystyle =2\times\frac{22}{7}\times(10.5)^2=693\text{ cm}^2$
$\displaystyle \text{(ii) External curved surface area}=2\pi R^2$
$\displaystyle =2\times\frac{22}{7}\times14^2=1232\text{ cm}^2$
$\displaystyle \text{(iii) Area of the annular rim}=\pi(R^2-r^2)$
$\displaystyle =\frac{22}{7}\left(14^2-(10.5)^2\right)=269.5\text{ cm}^2$
$\displaystyle \text{Total surface area}=1232+693+269.5$
$\displaystyle =2194.5\text{ cm}^2$
$\displaystyle \text{(iv) Volume of material}=\frac{2}{3}\pi(R^3-r^3)$
$\displaystyle =\frac{2}{3}\times\frac{22}{7}\left(14^3-(10.5)^3\right)$
$\displaystyle =3323.83\text{ cm}^3$
$\displaystyle \therefore \text{The required areas are }693\text{ cm}^2,\ 1232\text{ cm}^2,\ 2194.5\text{ cm}^2$
$\displaystyle \text{and the volume of material is }3323.83\text{ cm}^3.$
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$\displaystyle \textbf{Question 10: }\text{A solid sphere and a solid hemisphere have the same total surface area.} \\ \text{Find the ratio of their volumes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radii of the sphere and hemisphere be }r_1\text{ and }r_2\text{ respectively.}$
$\displaystyle \text{Total surface area of the sphere}=4\pi r_1^2$
$\displaystyle \text{Total surface area of the hemisphere}=3\pi r_2^2$
$\displaystyle \text{Given, }4\pi r_1^2=3\pi r_2^2$
$\displaystyle 4r_1^2=3r_2^2$
$\displaystyle \frac{r_1}{r_2}=\frac{\sqrt{3}}{2}$
$\displaystyle \frac{\text{Volume of sphere}}{\text{Volume of hemisphere}}$
$\displaystyle =\frac{\frac{4}{3}\pi r_1^3}{\frac{2}{3}\pi r_2^3}$
$\displaystyle =2\left(\frac{r_1}{r_2}\right)^3$
$\displaystyle =2\left(\frac{\sqrt{3}}{2}\right)^3$
$\displaystyle =\frac{3\sqrt{3}}{4}$
$\displaystyle \therefore \text{Volume of sphere : Volume of hemisphere}=3\sqrt{3}:4$
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$\displaystyle \textbf{Question 11: }\text{Metallic spheres of radii }6\text{ cm},\ 8\text{ cm and }10\text{ cm respectively}$
$\displaystyle \text{are melted and recast into a single solid sphere. Taking }\pi=\frac{22}{7},\text{ find the surface area} \\ \text{of the solid sphere formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the new sphere be }r\text{ cm}$
$\displaystyle \text{Volume of the new sphere}=\text{sum of the volumes of the three spheres}$
$\displaystyle \frac{4}{3}\pi r^3=\frac{4}{3}\pi(6^3+8^3+10^3)$
$\displaystyle r^3=6^3+8^3+10^3$
$\displaystyle =216+512+1000=1728$
$\displaystyle r=\sqrt[3]{1728}=12\text{ cm}$
$\displaystyle \text{Surface area of the new sphere}=4\pi r^2$
$\displaystyle =4\times\frac{22}{7}\times12^2$
$\displaystyle =\frac{12672}{7}\text{ cm}^2$
$\displaystyle \approx1810.29\text{ cm}^2$
$\displaystyle \therefore \text{The surface area of the solid sphere formed is }1810.29\text{ cm}^2.$
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$\displaystyle \textbf{Question 12: }\text{The surface area of a solid sphere is increased by }21\%\text{ without changing}$
$\displaystyle \text{ its shape. Find the percentage increase in its: (i) radius (ii) volume.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the initial surface area, radius and volume be }S_1,\ r_1\text{ and }V_1\text{ respectively.}$
$\displaystyle \text{Let the final surface area, radius and volume be }S_2,\ r_2\text{ and }V_2\text{ respectively.}$
$\displaystyle \text{Given, }S_2=1.21S_1$
$\displaystyle 4\pi r_2^2=1.21\times4\pi r_1^2$
$\displaystyle \left(\frac{r_2}{r_1}\right)^2=1.21$
$\displaystyle \frac{r_2}{r_1}=\sqrt{1.21}=1.1$
$\displaystyle \therefore \text{(i) Increase in radius}=10\%$
$\displaystyle \frac{V_2}{V_1}=\left(\frac{r_2}{r_1}\right)^3=(1.1)^3=1.331$
$\displaystyle \therefore \text{(ii) Increase in volume}=(1.331-1)\times100\%=33.1\%$
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