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}