Question

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of 498.96. Ifthe cost of white-washing is 2.()0 per square metre, find the:
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Answer 1

Answer:

374.22 m²

523.91 m³

Step-by-step explanation:

white-washed at the cost of 498.96

Cost of White washing = 2 Rs/m²

Area White Washed = 498.96/2 = 249.48

Surface Area of Hemisphere = 3πR² including base

Surface Area of Hemisphere without base = 2πR²

if Floor is not white washed then 2πR² = 249.48

=> 2 * (22/7 )R² =  249.48

=> R² = 39.69

=>  R = 6.3

inside surface area of the dome = 3πR² = 3 * (22/7) * 6.3² = 374.22 m²

Volume of Hemisphere = (2/3)πR³  = (2/3)(22/7)*6.3³ = 523.91 m³

R =

Answer 2

Answer:

i) Inside surface area = $S=249.48\ m^{2}$

ii) Volume inside of the dome is,

$V=523.908\ m^{3}$

Step-by-step explanation:

In the question,

i) Total cost of white washing of the inside of the dome, C = 498.96 Rs.

Cost of white washing = 2 Rs. per square metre.

Now,

We know that the Surface area of the hemisphere is given by,

$S=2\pi r^{2}$

where, r is the radius of the dome.

Now,

Total Surface Area is given by,

$Surface\ Area=\frac{Total\ Cost}{Cost\ per\ square\ metre} \\So,\\S=\frac{498.96}{2}\\ S=249.48\ m^{2}$

Therefore, the inside surface area of the dome is,

$S=249.48\ m^{2}$

ii) Surface Area is = 249.48 square metre

So,

$2\times \frac{22}{7}\times r^{2}=249.48\\ 2\times \frac{22}{7}\times r^{2}=249.48\\r^{2}=39.69\\r=6.3\ m$

So,

Radius of the dome is, r = 6.3 m

So,

Volume of the hemisphere is given by,

$V=\frac{2}{3}\pi r^{3}$

So,

Volume of air inside the dome is the volume of the hemisphere that is,

$V=\frac{2}{3} \times \frac{22}{7}\times r^{3}\\V=\frac{2}{3} \times \frac{22}{7}\times (6.3)^{3}\\V=523.908\ m^{3}$

Therefore, the Volume inside of the dome is,

$V=523.908\ m^{3}$