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 cm2

radius of the sphere (r) = ?

By formula,

A = 4$$\pi$$r2

or, r2 = $$\frac A{4\pi}$$

or, r2 = $$\frac {616 \times 7}{4 \times 22}$$

or, r2 = 49

∴ r = 7cmAns

Suppose,

Then,

Surface Area of Sphere = 4$$\pi$$r2

or, $$\pi$$ = 4$$\pi$$r2

or, r2 = $$\frac {\pi}{4\pi}$$

or, r2 = $$\frac 14$$

or, r = $$\sqrt {\frac 14}$$

∴ r = $$\frac 12$$cmAns

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, $$\frac 43$$ $$\pi$$r3 = $$\frac {\pi}6cm^3$$

or, r3 = $$\frac {\pi}6$$ $$\times$$ $$\frac 3{4\pi}cm^3$$

or, r3 = $$\frac 1{2^3}cm^3$$

∴ r = $$\frac 12cm$$

Hence, the diameter of the marble = 2 $$\times$$ r = 2 $$\times$$ $$\frac 12$$ cm = 1 cmAns

Let r be the radius of the sphere.

Volume = $$\frac 43$$$$\pi$$r3

or, 38808 = $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3

or, r3 = 38808 $$\times$$ $$\frac {21}{88}$$

or, r3 = 441 $$\times$$ 21

or, r3 = (21)3

∴ r = 21Ans

Here,

Volume of sphere = $$\frac {3773}{21}cm^3$$

or, $$\frac 43$$$$\pi$$r3 = $$\frac {3773}{21}cm^3$$

or, $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3 = $$\frac {3773}{21}cm^3$$

or, $$\frac {88}{21}r^3$$ = $$\frac {3773}{21}cm^3$$

or, r3 = $$\frac {3773}{21}$$ $$\times$$ $$\frac {21}{88}cm^3$$

or, r3 = $$\frac {343}{8}cm^3$$

or, r3 = ($$\frac 73$$cm)3

∴ r = $$\frac 72$$cm

Hence, Circumference of sphere = 2$$\pi$$r = 2 $$\times$$ $$\frac {22}7$$ $$\times$$ $$\frac 72$$ cm = 22 cmAns

Here,

Volume of spherical solid = $$\frac 43$$$${\pi}cm^3$$

or, $$\frac 43$$$${\pi}r^3$$ = $$\frac 43$$$${\pi}cm^3$$

or, r3 = $$\frac {\frac {4\pi}{3}}{\frac {4\pi}{3}}cm^3$$

or, r3 = 1 cm3

∴ 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, r3 =$$\frac {1372\pi}{3}$$ $$\times$$ $$\frac 3{4\pi}$$

or, r3 = 343

or, r3 = 73

∴ 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 cm2

or, 4$${\pi}r^2$$ = 616 cm2

or, $${\pi}r^2$$ = $$\frac {616}4 cm^2$$

∴$${\pi}r^2$$ = 154 cm2

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, r2 = $$\frac {243\pi}{3\pi}$$

or, r2 = 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, r2 = $$\frac 1{(4\pi)^2}$$

∴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,

r1 = $$\frac {6cm}2$$ = 3cm

r2 = $$\frac {8cm}2$$ = 4cm

r3 = $$\frac {10cm}{2}$$ = 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, R3 = r13 + r23 + r33

or, R3 = (33 + 43 + 53)

or, R3 = 27 + 64 + 125

or, R3 = 216

or, R = $$\sqrt [3]{216}$$

∴ R = 6 cm

Hence, the diameter of new sphere (d) = 2R = 2 $$\times$$ 6 cm = 12 cmAns

\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*}

Let x be the radius of the first sphere.

Then,

Radius of second sphere is $$\frac r4$$

Volume of first sphere (V1) = $$\frac 43$$ $${\pi}r^3$$

Volume of second sphere (V2) = $$\frac 43$$ $$\pi$$ ($$\frac r4$$)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, V1 :V2 = 1 : 64Ans
The text passage contains multiple mathematical questions and their corresponding solutions, primarily dealing with the calculations of volumes and surface areas of spheres and hemispheres.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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 cm2

radius of the sphere (r) = ?

By formula,

A = 4$$\pi$$r2

or, r2 = $$\frac A{4\pi}$$

or, r2 = $$\frac {616 \times 7}{4 \times 22}$$

or, r2 = 49

∴ r = 7cmAns

Suppose,

Then,

Surface Area of Sphere = 4$$\pi$$r2

or, $$\pi$$ = 4$$\pi$$r2

or, r2 = $$\frac {\pi}{4\pi}$$

or, r2 = $$\frac 14$$

or, r = $$\sqrt {\frac 14}$$

∴ r = $$\frac 12$$cmAns

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, $$\frac 43$$ $$\pi$$r3 = $$\frac {\pi}6cm^3$$

or, r3 = $$\frac {\pi}6$$ $$\times$$ $$\frac 3{4\pi}cm^3$$

or, r3 = $$\frac 1{2^3}cm^3$$

∴ r = $$\frac 12cm$$

Hence, the diameter of the marble = 2 $$\times$$ r = 2 $$\times$$ $$\frac 12$$ cm = 1 cmAns

Let r be the radius of the sphere.

Volume = $$\frac 43$$$$\pi$$r3

or, 38808 = $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3

or, r3 = 38808 $$\times$$ $$\frac {21}{88}$$

or, r3 = 441 $$\times$$ 21

or, r3 = (21)3

∴ r = 21Ans

Here,

Volume of sphere = $$\frac {3773}{21}cm^3$$

or, $$\frac 43$$$$\pi$$r3 = $$\frac {3773}{21}cm^3$$

or, $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3 = $$\frac {3773}{21}cm^3$$

or, $$\frac {88}{21}r^3$$ = $$\frac {3773}{21}cm^3$$

or, r3 = $$\frac {3773}{21}$$ $$\times$$ $$\frac {21}{88}cm^3$$

or, r3 = $$\frac {343}{8}cm^3$$

or, r3 = ($$\frac 73$$cm)3

∴ r = $$\frac 72$$cm

Hence, Circumference of sphere = 2$$\pi$$r = 2 $$\times$$ $$\frac {22}7$$ $$\times$$ $$\frac 72$$ cm = 22 cmAns

Here,

Volume of spherical solid = $$\frac 43$$$${\pi}cm^3$$

or, $$\frac 43$$$${\pi}r^3$$ = $$\frac 43$$$${\pi}cm^3$$

or, r3 = $$\frac {\frac {4\pi}{3}}{\frac {4\pi}{3}}cm^3$$

or, r3 = 1 cm3

∴ 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, r3 =$$\frac {1372\pi}{3}$$ $$\times$$ $$\frac 3{4\pi}$$

or, r3 = 343

or, r3 = 73

∴ 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 cm2

or, 4$${\pi}r^2$$ = 616 cm2

or, $${\pi}r^2$$ = $$\frac {616}4 cm^2$$

∴$${\pi}r^2$$ = 154 cm2

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, r2 = $$\frac {243\pi}{3\pi}$$

or, r2 = 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, r2 = $$\frac 1{(4\pi)^2}$$

∴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,

r1 = $$\frac {6cm}2$$ = 3cm

r2 = $$\frac {8cm}2$$ = 4cm

r3 = $$\frac {10cm}{2}$$ = 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, R3 = r13 + r23 + r33

or, R3 = (33 + 43 + 53)

or, R3 = 27 + 64 + 125

or, R3 = 216

or, R = $$\sqrt [3]{216}$$

∴ R = 6 cm

Hence, the diameter of new sphere (d) = 2R = 2 $$\times$$ 6 cm = 12 cmAns

\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*}

Let x be the radius of the first sphere.

Then,

Radius of second sphere is $$\frac r4$$

Volume of first sphere (V1) = $$\frac 43$$ $${\pi}r^3$$

Volume of second sphere (V2) = $$\frac 43$$ $$\pi$$ ($$\frac r4$$)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, V1 :V2 = 1 : 64Ans
user
The text passage contains multiple mathematical questions and their corresponding solutions, primarily dealing with the calculations of volumes and surface areas of spheres and hemispheres.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS