Question

The dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of 498.96. If the cost of white washing is Rs. 2 per square meter, find the
a) inside surface area of the dome
b)volume of the air inside the dome
(π=22/7)
plz solve this....

Answer 1

Cost of white washing = 498.96
Cost of white washing per square meter = Rs.2
Inner CSA of the dome = 498.96/2
= 249.48 square meters

voume of the air inside it =Volume of the done

Volume of the dome =2/3 × pi× r^3_____(1)
CSA = 2 × pi × r^2
249.48square meters = 2× pi× r^2
r^2 = 249.48/ 2×7/22
r^2=39.69
r= 6.3 square metres________(2)

substituting eq (2)in eq (1)
volume of the air = 2/3 × 22/7 ×(6.3)^3
= 44/21× 250.047
= 2.09 × 250.047
= 522.59
=522.6(approximately)



hope it helps...

Answer 2

Step-by-step explanation:

(i) We know that

Area whitewashed × Cost of white wash = Total cost

Area whitewashed ×2=498.96

Area whitewashed =

2

498.96

Area whitewashed =249.48 m

2

Now,

Area of whitewashed is curved surface area of hemisphere (as only walls are whitewashed, not floor)

Therefore,

Inner surface area of dome =249.48 m

2

(ii) Volume of air inside dome = Volume of hemisphere

=

3

2

πr

3

Let the radius of dome be 'r' m

Surface area of dome =249.48 m

2

2πr

2

=249.48

2×

7

22

×r

2

=249.48

r

2

=39.69

r=

39.69

r=6.3 m

Volume of air inside the dome = Volume of hemisphere

=

3

2

πr

3

=

3

2

×

7

22

×6.3

3

=523.9 m

3