Subject: Compulsory Mathematics
We are so much familiar with spherical objects. The sphere is also a solid object whose each point in the outer surface is equidistance from the fixed point inside it. Such a fixed point is called the centre of the sphere. The constant distance is called the radius of the sphere. A solid object such as a globe, volleyball, toy ball, table tennis ball, marble etc is the example of a sphere.
The figure as shown alongside is a sphere. The fixed point 'O' inside it is the centre which is equidistance from each point P on the surface of the sphere. So, OP = r is the radius of the sphere.
If we cut a sphere through its diameter, there are two half spheres called the hemisphere and the cross section is called the great circle. The radius of the sphere is same as the radius of the great circle. The centre of the sphere and its great circle. The centre of the sphere and its great circle is same.
The surface area of a sphere is the area of its outer part, which is a smooth curved surface.
The surface area of a sphere is given by SA = 4πr^{2} where r is the radius of the sphere. The total surface area of hemisphere = 2πr^{2}+ πr^{2 } = 3πr^{2} square unit.
Note

The volume of a sphere means the space that it occupies. We can measure the volume of sphere experimentally. Fill up the measuring cylinder with the water level in the cylinder. The difference of two levels is the volume of the sphere.
Choose a sphere of given diameter (d) =10 cm (say).
Immerse of the whole sphere into the water in the measuring cylinder, the water level is raised by 523.33 ml. Therefore its volume is 523.33 cm^{3}. From this experiment,
The diameter (d) = 10cm
The volume of sphere (V) = 523.33 cm^{3}
Let us make the ratio
\begin{align*} 6V:d^3 &= \frac {6V} {d^3} \\ &= \frac {6 \times 523.33} {10^3} \\ &= \frac {3140} {1000} \\ &= 3.14 \end{align*}
$$ (\because \pi = \frac {22} {7} = 3.14) $$
\begin{align*} \therefore \frac {6V} {d^3} &= \pi \\ or, V &= \frac {\pi (2r)^3} {6} \\ &= \frac {4 \pi r^3} {3} \\ \end{align*}
\( \therefore \text {Volume of a sphere} (V) = \frac {4 \pi r^3} {3} = \frac {\pi d^3} {6} \text {cubic units.} \)
Find the total surface area of the given solid hemisphere. In which the radius of the base of hemisphere is 15 cm.
Here,
r = 15 cm
\begin{align*} \text{Total Surface Area of hemisphere} &= 3{\pi}r^2\\ &= 3 \times \frac {22}7 \times 15 \times 15\\ &= \frac {14850}{7}\\ &= 2121.43 cm^2_{Ans}\\ \end{align*}
If the total surface area of a solid sphere is 616 cm^{2}, what will be its radius?
Surface Area of the sphere (A) = 616 cm^{2}
radius of the sphere (r) = ?
By formula,
A = 4\(\pi\)r^{2}
or, r^{2} = \(\frac A{4\pi}\)
or, r^{2} = \(\frac {616 \times 7}{4 \times 22}\)
or, r^{2} = 49
∴ r = 7cm_{Ans}
The surface area of a sphere is \(\pi\) sq. cm., find its radius.
Suppose,
radius of sphere = r
Then,
Surface Area of Sphere = 4\(\pi\)r^{2}
or, \(\pi\) = 4\(\pi\)r^{2}
or, r^{2} = \(\frac {\pi}{4\pi}\)
or, r^{2} = \(\frac 14\)
or, r = \(\sqrt {\frac 14}\)
∴ r = \(\frac 12\)cm_{Ans}
Find the volume of the adjoining solid hemisphere.
Here,
r = \(\frac {42 cm}{2}\) = 21 cm
\begin{align*} \text{Volume of the hemisphere (V)} &=\frac 23{\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times (21)^3\\ &= 44 \times (21)^2\\ &= 19404 cm^3_{Ans}\\ \end{align*}
The volume of a marble is \(\frac {\pi}6\) cm^{3}, find its diameter.
Here,
Volume of marble (V) = \(\frac {\pi}6cm^3\)
or, \(\frac 43\) \(\pi\)r^{3} = \(\frac {\pi}6cm^3\)
or, r^{3} = \(\frac {\pi}6\) \(\times\) \(\frac 3{4\pi}cm^3\)
or, r^{3} = \(\frac 1{2^3}cm^3\)
∴ r = \(\frac 12cm\)
Hence, the diameter of the marble = 2 \(\times\) r = 2 \(\times\) \(\frac 12\) cm = 1 cm_{Ans}
The volume of a sphere is 38808 cubic centimeter, find the radius of the sphere. (\(\pi\) = \(\frac {22}7\))
Let r be the radius of the sphere.
Volume = \(\frac 43\)\(\pi\)r^{3}
or, 38808 = \(\frac 43\) \(\times\) \(\frac {22}7\) \(\times\) r^{3}
or, r^{3} = 38808 \(\times\) \(\frac {21}{88}\)
or, r^{3} = 441 \(\times\) 21
or, r^{3} = (21)^{3}
∴ r = 21_{Ans}
Find the circumference of the sphere whose volume is \(\frac {3773}{21}cm^3.\)
Here,
Volume of sphere = \(\frac {3773}{21}cm^3\)
or, \(\frac 43\)\(\pi\)r^{3} = \(\frac {3773}{21}cm^3\)
or, \(\frac 43\) \(\times\) \(\frac {22}7\) \(\times\) r^{3} = \(\frac {3773}{21}cm^3\)
or, \(\frac {88}{21}r^3\) = \(\frac {3773}{21}cm^3\)
or, r^{3} = \(\frac {3773}{21}\) \(\times\) \(\frac {21}{88}cm^3\)
or, r^{3} = \(\frac {343}{8}cm^3\)
or, r^{3} = (\(\frac 73\)cm)^{3}
∴ r = \(\frac 72\)cm
Hence, Circumference of sphere = 2\(\pi\)r = 2 \(\times\) \(\frac {22}7\) \(\times\) \(\frac 72\) cm = 22 cm_{Ans}
Find the radius of a spherical solid, whose volume is \(\frac 43\)\({\pi}cm^3\).
Here,
Volume of spherical solid = \(\frac 43\)\({\pi}cm^3\)
or, \(\frac 43\)\({\pi}r^3\) = \(\frac 43\)\({\pi}cm^3\)
or, r^{3} = \(\frac {\frac {4\pi}{3}}{\frac {4\pi}{3}}cm^3\)
or, r^{3} = 1 cm^{3}
∴ r = 1 cm
Hence, the radius of spherical solid is 1 cm._{Ans}
The volume of a sphere is 36\({\pi}cm^3\), find its surface area.
Here,
Volume of sphere (V) = 36\({\pi}cm^3\)
Diameter of a sphere (d) = ?
We know that,
V = \(\frac 16\)\({\pi}d^3\)
or, d = \(\sqrt [3]\frac{6V}{d}\)
or, d = \(\sqrt [3]\frac {6 \times 36\pi}{\pi}\)
or, d = \(\sqrt [3]{6 \times 6 \times 6}\)
∴ d = 6cm
\begin{align*} \text{The Surface Area (A)} &= {\pi}d^2\\ &= \frac {22}7 \times (6cm)^2\\ &= \frac {22}7 \times 36 cm^2\\ &= 113.14cm^2_{Ans}\\ \end{align*}
Find the total surface area of a sphere whose volume is \(\frac {1372}{3}\pi {cm^3}\).
Here,
Volume (V) = \(\frac {1372\pi}{3}cm^3\)
Total Surface Area (S) = ?
By Formula,
V = \(\frac 43\)\({\pi}r^3\)
or, \(\frac {1372\pi}{3}\) =\(\frac 43\)\({\pi}r^3\)
or, r^{3} =\(\frac {1372\pi}{3}\) \(\times\) \(\frac 3{4\pi}\)
or, r^{3} = 343
or, r^{3} = 7^{3}
∴ r = 7
We know that,
\begin{align*} S &= 4{\pi}r^2\\ &= 4 \times \frac {22}7 \times 7^2 cm^2\\ &= 616 cm^2_{Ans}\\ \end{align*}
The total surface area of a solid sphere made of earth is 616 square cm. Two hemispheres are formed when it is cut into two equal parts. Find the total surface area of each hemisphere.
Here,
Surface Area of sphere = 616 cm^{2}
or, 4\({\pi}r^2\) = 616 cm^{2}
or, \({\pi}r^2\) = \(\frac {616}4 cm^2\)
∴\({\pi}r^2\) = 154 cm^{2}
We know that,
\begin{align*} \text{Total Surface Area of a hemisphere} &= 3{\pi}r^2\\ &= 3 \times 154 cm^2\\ &= 462 cm^2_{Ans}\\ \end{align*}
The total surface area of a hemisphere is 243 \({\pi} cm^2\). Find its volume.
Here,
Total Surface Area of hemisphere (S) = 243 \({\pi}cm^2\)
Volume (V) = ?
We know that,
S = 3\({\pi}r^2\)
or, 243\(\pi\) = 3\({\pi}r^2\)
or, r^{2} = \(\frac {243\pi}{3\pi}\)
or, r^{2} = 81
∴ r = 9 cm
Now,
\begin{align*} V &= \frac 23 {\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times 9^3\\ &= 1527.43 cm^3_{Ans}\\ \end{align*}
If the surface area of a marble is \(\frac 1{4\pi}\)sq. cm, find its volume.
Here,
4\({\pi}r^2\) = \(\frac 1{4\pi}\)
or, r^{2} = \(\frac 1{(4\pi)^2}\)
∴r = \(\frac 1{4\pi}\)
Now,
\begin{align*} \text{Volume (V)} &= \frac 43 {\pi}r^3\\ &= \frac {4\pi}3 (\frac 1{4\pi})^3\\ &= \frac 13 \times \frac 1{16\pi^2}\\ &= \frac 1{44\pi^2}cm^3_{Ans}\\ \end{align*}
Three spheres of diameters 6 cm, 8 cm, and 10 cm. are melted and formed a single sphere. Find the diameter of the sphere.
Here,
r_{1} = \(\frac {6cm}2\) = 3cm
r_{2} = \(\frac {8cm}2\) = 4cm
r_{3} = \(\frac {10cm}{2}\) = 5cm
Suppose,
radius of new sphere = R
Now,
Volume of new sphere = sum of the volume of first, second and third spheres
or, \(\frac 43\)\({\pi}R^3\) = \(\frac 43\)\({\pi}r_1^3\) +\(\frac 43\)\({\pi}r_2^3\) +\(\frac 43\)\({\pi}r_3^3\)
or, R^{3} = r_{1}^{3} + r_{2}^{3} + r_{3}^{3}
or, R^{3} = (3^{3} + 4^{3} + 5^{3})
or, R^{3} = 27 + 64 + 125
or, R^{3} = 216
or, R = \(\sqrt [3]{216}\)
∴ R = 6 cm
Hence, the diameter of new sphere (d) = 2R = 2 \(\times\) 6 cm = 12 cm_{Ans}
How many balls of each radius 1 cm can be made by melting a big ball whose diameter is 8 cm?
\begin{align*} \text{Volume of small ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}1^3\\ &= \frac 43 {\pi}cm^3\\ \end{align*}
\begin{align*} \text{Radius of big ball} &= \frac {8cm}2\\ &= 4 cm\\ \end{align*}
\begin{align*} \text{Volume of big ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}4^3\\ &= \frac 43 {\pi}64cm^3\\ \end{align*}
Now,
\begin{align*} \text{Number of small balls that can be made from the big ball} &= \frac {\frac 43 {\pi} \times 64}{\frac 43 {\pi}}\\ &= 64_{Ans}\\ \end{align*}
If the radius of one sphere is \(\frac 14\) that of a second sphere, find the ratio of their volume.
Let x be the radius of the first sphere.
Then,
Radius of second sphere is \(\frac r4\)
Volume of first sphere (V_{1}) = \(\frac 43\) \({\pi}r^3\)
Volume of second sphere (V_{2}) = \(\frac 43\) \(\pi\) (\(\frac r4\))^{3} = \(\frac {4\pi}{3}\)\(\times\) \(\frac {r^3}{64}\)
Now,
\(\frac {V_2}{V_1}\) = \(\frac {\frac {4\pi}{3} \times \frac {r^3}{64}}{\frac 43 {\pi}r^3}\) = \(\frac 1{64}\)
Hence, V_{1} :_{}V_{2} = 1 : 64_{Ans}
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