Sphere and Hemisphere
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.
Surface area of sphere
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

Volume of sphere
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.
Alternatively
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*}
\( \therefore \text {Volume of a sphere} (V) = \frac {4 \pi r^3} {3} = \frac {\pi d^3} {6} \text {cubic units.} \)
 Curved surface area of hemisphere = 2\(\pi\)r^{2}square units.
 The surface area of a sphere (SA) =d^{2}, if the diameter is given.
 Volume of a sphere(v) = \(\frac{4\ (\pi) r^3}{3}\) = \(\frac{\ (\pi)d^3}{6}\) cubic units.
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*}
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}
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}
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*}
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}
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}
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}
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}
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*}
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*}
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*}
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*}
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*}
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}
\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}

The total surface area of a sphere is 22176 sq cm.Find the diameter of the sphere.
45 cm
84 cm
65 cm
75 cm

If the total surface area of a solid sphere is 616 cm^{2}. what will be its radius?
7 cm
12 cm
10 cm
15 cm

Find the circumference of the sphere whose volume is (frac{1375}{21})cm ^{3}.
12.5 cm
10.5 cm
9.5 cm
15.71 cm

Find the circumference of the sphere whose volume is(frac{3773}{21})cm^{3}.
29 cm
22 cm
12 cm
19 cm

If the volume of a spherical object is (frac{9π}{2})cm^{3}, find its diameter.
5 cm
3 cm
10 cm
2 cm

The surface area of a sphere is equal to the area of the curved surface of a right circular cylinder.If the height and diameter of the cylinder is 16 cm each, what is the diameter of the sphere?
14 cm
16 cm
9 cm
12 cm

The surface of a lemon is equal to the surface area of a cylindrical tincontaining paint. If the height of the tin is 4 cm and its radius of the lemon ?
4 cm
1 cm
2 cm
3 cm

The surface area of a volleyball is equal to the curved surface area of a cylindrical tin. If the radius of volleyball is 7 cm and the radius of the can is 14 cm, what is the height of the can?
5 cm
10 cm
7 cm
8 cm

The metallic solid sphere of radius 3 cm is melted to form a solid cylinder of diameter 6 cm.Find the height of the cylinder.
4 cm
7 cm
6 cm
5 cm

A solid sphere of radius 18 cm is melted down and drawn out into long cylindrical wire of uniform thickness of 6 mm.Find the length of the wire.
800 m
855 m
864 m
850 m

A solid sphere of radius 6 cm is melted and a solid cylinder of radius 3 cm is formed.Find the height of the cylinder so formed.
32 cm
30cm
35 cm
25 cm

The diameter of a hemisphere is 84 cm, find its total surface area and volume.
16632 cm^{2},155232 cm^{3}
13904 cm^{2}, 155238 cm^{3}
16635 cm^{2}, 155132 cm^{3}
15532 cm^{2},155243 cm^{3}

The radius of a hemisphere is 21 cm, find its total surface area and volume.
4159 cm^{2}
4055 cm^{2}
4170 cm^{2}
4158 cm^{2}

Find the total surface area of a hemisphere having curved surface area 1232 cm^{2}.
1848 cm^{2}
1845cm^{2}
1846cm^{2}
1849 cm^{2}

You scored /14
Any Questions on Sphere and Hemisphere ?
Please Wait...
Discussions about this note
Forum  Time  Replies  Report 

sushilatotal surface area of a sphere 
Jan 20, 2017 
0 Replies View Replies 

sushilatotal surface area of a sphere 
Jan 20, 2017 
0 Replies View Replies 