(i) Cost of whitewashing the dome from inside = Rs 498.96
Cost of white washing 1m2 area = Rs 2
Therefore, CSA of the inner side of dome =\(\frac{4989.60}{20} \)= 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
\(2\pi r^2\) = 249.48 m2
\(⇒\) r2 = \(\frac{249.48}{2\pi}\) m2
\(⇒\) r2 = 249.48 \(÷\) (2 ×\(\frac{ 22}{7}\))
\(⇒\) r2 = 39.69
\(⇒\) r = \(\sqrt{39.69}\)
\(⇒\) r = 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
\(=\frac{ 2}{3}\pi\)r3
\(= \frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 m × 6.3 m × 6.3 m
= 523.9 m3 (approximately)
Therefore, the volume of air inside the dome is 523.9 m3 .
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?