Question

Answer the Both Question in the image using the knowledge of Volume of Solid. 50 points​

Answer 1

Answer:

Part A

Cone

$\mathsf{volume \ of \ a \ cone=\dfrac13\pi r^2h}$

(where r is the radius and h is the height)

Given:

• r = a
• h = 2r = 2a

$\implies \mathsf{volume \ of \ the \ cone=\dfrac13\pi \cdot a^2\cdot2a=\dfrac23\pi a^3}$

Hemisphere

$\mathsf{volume \ of \ a \ sphere=\dfrac43\pi r^3}$

$\implies \mathsf{volume \ of \ a \ hemisphere=\dfrac12 \cdot\dfrac43\pi r^3=\dfrac23\pi r^3}$

Given

• r = a

$\implies \mathsf{volume \ of \ the \ hemisphere=\dfrac23\pi a^3}$

Therefore

volume of cone with radius a = volume of hemisphere with radius a

Part B

$6.82^2\times\sqrt[3]{0.005}$

Take log of base 10:

$\implies \log_{10}(6.82^2\times\sqrt[3]{0.005})$

Using log law $\log(a \times b)=\log a+\log b$:

$\implies \log_{10}(6.82^2)+\log_{10}(\sqrt[3]{0.005})$

Using low law $\log(a^b)=b \log a$

$\implies 2\log_{10}(6.82)+\dfrac13\log_{10}(0.005)$

Log tables

The characteristic of the logarithm of a number is the exponent of 10 in its scientific notation.

The mantissa is found using the log tables and is always prefixed by a decimal point.

The row is the first two non-zero digits of the number, and the column is the 3rd digit of the number

Use the log tables to find $\log_{10}(6.82)$:

6.82 = 6.82 × 10⁰

⇒ characteristic = 0

log table:  row 68, column 2 ⇒ mantissa 8338 ⇒ 0.8338

characteristic + mantissa = 0 + 0.8338 = 0.8338

Therefore, $\log_{10}(6.82)=0.8338$

Use the log tables to find $\log_{10}(0.005)$:

$0.005 = 5.0 \times 10^{-3}$

⇒ characteristic = -3

log table: row 50, column 0 ⇒ mantissa 6990 ⇒ 0.6990

characteristic  + mantissa = -3 + 0.6990 = -2.301

Therefore, $\log_{10}(0.005)=-2.301$

Therefore,

$2\log_{10}(6.82)+\dfrac13\log_{10}(0.005)$

$\implies 2\cdot0.8338 + \dfrac13 \cdot -2.301$

$\implies 1.6676 - 0.767$

$\implies 0.9006$

Therefore,

$\log_{10}(6.82^2\times\sqrt[3]{0.005})=0.9006$

Using $\log_{a}b=c \implies a^c=b$

$\implies 6.82^2\times\sqrt[3]{0.005}=10^{0.9006}$

$\implies 6.82^2\times\sqrt[3]{0.005}=7.954$

Answer 2

given for cone:

• radius: a

• height: 2a

volume of cone:

$\sf \rightarrow \dfrac{1}{3} \pi r^2h$

$\sf \rightarrow \dfrac{1}{3} \pi (a)^2(2a)$

$\sf \rightarrow \dfrac{2\pi a^3}{3}$

no metal wasted means no volume wasted while melting so volume: same.

volume of hemisphere:

$\hookrightarrow \sf \dfrac{2}{3} \pi r^3$

This is equal to the volume of cone.

$\hookrightarrow \sf \dfrac{2}{3} \pi r^3 = \dfrac{2\pi a^3}{3}$

$\hookrightarrow \sf a^3 = r^3$

$\hookrightarrow \sf a = r$

Therefore shown that radius of cone is similar to radius of hemisphere.

(b)

$6.82^2 *\sqrt[3]{0.005}$

$7.9535$

$7.95$