Question

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

Answer 1

given for cone:

• radius: a

• height: 2a

volume of cone:

`\sf \rightarrow (1)/(3) \pi r^2h`

`\sf \rightarrow (1)/(3) \pi (a)^2(2a)`

`\sf \rightarrow (2\pi a^3)/(3)`

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

volume of hemisphere:

`\hookrightarrow \sf (2)/(3) \pi r^3`

This is equal to the volume of cone.

`\hookrightarrow \sf (2)/(3) \pi r^3 = (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`

Answer 2

Part A

Cone

`\mathsf{volume \ of \ a \ cone=\frac13\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=\frac13\pi \cdot a^2\cdot2a=\frac23\pi a^3}`

Hemisphere

`\mathsf{volume \ of \ a \ sphere=\frac43\pi r^3}`

`\implies \mathsf{volume \ of \ a \ hemisphere=\frac12 \cdot\frac43\pi r^3=\frac23\pi r^3}`

Given

• r = a

`\implies \mathsf{volume \ of \ the \ hemisphere=\frac23\pi a^3}`

Therefore

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

Part B

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

Take log of base 10:

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

Using log law
`\log(a * 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)+\frac13\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 * 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)+\frac13\log_(10)(0.005)`

`\implies 2\cdot0.8338 + \frac13 \cdot -2.301`

`\implies 1.6676 - 0.767`

`\implies 0.9006`

Therefore,

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

Using
`\log_(a)b=c \implies a^c=b`

`\implies 6.82^2*\sqrt[3]{0.005}=10^(0.9006)`

`\implies 6.82^2*\sqrt[3]{0.005}=7.954`