(i) (b): We know that, volume of cylinder \(=\pi r^{2} h\)
\(\Rightarrow 1540=\frac{22}{7} \times r^{2} \times 10 \)
\(\Rightarrow \frac{154 \times 7}{22}=r^{2} \Rightarrow r^{2}=49 \Rightarrow r=7 \mathrm{~cm}\)
\(\therefore\) Diameter of the base of cylinder \(=2 r=2 \times 7=14 \mathrm{~cm}\)
(ii) (b): We know that, volume of sphere \(=\frac{4}{3} \pi r^{3}\)
\(\Rightarrow 38808=\frac{4}{3} \times \frac{22}{7} \times r^{3} \)
\(\Rightarrow r^{3}=\frac{38808 \times 3 \times 7}{4 \times 22}=441 \times 21=(21)^{3} \Rightarrow r=21 \mathrm{~cm}\)
\(\therefore\) Diameter of sphere = 42 cm.
(iii) (d): Total volume of shape formed = Volume of cylindrical shapes + Volume of sphere
= 11 x 1540 + 38808 = 16940 + 38808 = 55748 cm³
(iv) (c): Curved surface area of one cylindrical shape \(=2 \pi r h \)
\(=2 \times \frac{22}{7} \times 7 \times 10=440 \mathrm{~cm}^{2}\)
(v) (a): Area covered by cylindrical shapes on the surface of sphere
\(=11 \times \pi r^{2}=11 \times \frac{22}{7} \times 7 \times 7=1694 \mathrm{~cm}^{2}\)

(i) (b): Lateral surface area of Hermika which is cubical in shape = 4a² = 4 x (8)² = 256 m²
(ii) (a): Diameter of cylindrical base = 42 m
\(\therefore\) Radius of cylindrical base (r) = 21 m
Height of cylindrical base (h) = 12 m
\(\therefore\) Number of bricks used \(=\frac{\frac{22}{7} \times 21 \times 21 \times 12}{0.01}\)
= 1663200
(iii) (c) : Given, diameter of Anda which is
hemispherical in shape = 42 m
\(\Rightarrow\) Radius of Anda (r) = 21 m
\(\therefore \quad \text { Volume of } A n d a =\frac{2}{3} \pi r^{3}=\frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 \)
\(=44 \times 21 \times 21=19404 \mathrm{~m}^{3}\)
(iv) (d): Given, radius of Pradakshina Path (r) = 25 m
\(\therefore\) Perimeter of path = \(2\pi r\)
\(=\left(2 \times \frac{22}{7} \times 25\right) \mathrm{m}\)
· . Distance covered by priest \(=14 \times 2 \times \frac{22}{7} \times 25\)
= 2200 m
(v) (b):\(\because\) Radius of Anda (r) = 21 m
\(\therefore\) Curved surface area of Anda = \(2\pi r^{2}\)
\(=2 \times \frac{22}{7} \times 21 \times 21=2772 \mathrm{~m}^{2}\)

(i) (a)
(ii) (d): Slant height of conical cavity \(l=\sqrt{h^{2}+r^{2}}\)
\(=\sqrt{(24)^{2}+(7)^{2}}=\sqrt{576+49}=\sqrt{625}=25 \mathrm{~cm}\)
(iii) (b): Curved surface area of conical cavity = \(\pi r l\)
\(=\frac{22}{7} \times 7 \times 25=550 \mathrm{~cm}^{2}\)
(iv) (c) : External curved surface area of cylinder 
\(=2 \pi r h=2 \times \frac{22}{7} \times 7 \times 24=1056 \mathrm{~cm}^{2}\)
(v) (a): Volume of conical cavity \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=1232 \mathrm{~cm}^{3}\)

(i) (b): Volume of cuboidal part = 1 x b x h = (20 x 15 x 5) cm³ = 1500 cm³
(ii) (c): Radius of conical depression, r = 0.6 cm
Height of conical depression, h = 2.1 cm
\(\therefore\) Total volume of conical depressions \(=3 \times \frac{1}{3} \pi r^{2} h\)
\(=\frac{22}{7} \times 0.6 \times 0.6 \times 2.1=\frac{2.376}{1000}-2.376 \mathrm{~cm}^{3}\)
(iii) (d): Volume of wood used in the entire stand = Volume of cuboidal part - Total volume of conical depressions
= 1500 - 2.376 = 1497.624 cm³
(iv) (a)
(v) (d): Cost of wood per cm³ = Rs 5
\(\therefore\) Total cost of making the pen stand
= Rs (5 x 1497.624) = Rs 7488.12

We have, radius of each coin = 3.5 cm
\(=\frac{35}{10} \mathrm{~cm}=\frac{7}{2} \mathrm{~cm}\)
Thickness of each coin \(=0.5 \mathrm{~cm}=\frac{1}{2} \mathrm{~cm}\)
So,height of cylinder made by Meera (h1) \(=12 \times \frac{1}{2}=6 \mathrm{~cm}\)
and height of cylinder made by Dhara (h2)
\(=8 \times \frac{1}{2}=4 \mathrm{~cm}\)
(i) (b): Curved surface area of cylinder made by Meera \(=2 \times \frac{22}{7} \times \frac{7}{2} \times 6=132 \mathrm{~cm}^{2}\)
(ii) (b): Required ratio
\(=\frac{\text { Curved surface area of cylinder made by Meera }}{\text { Curved surface area of cylinder made by Dhara }}\)
\(=\frac{2 \pi r h_{1}}{2 \pi r h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(iii) (a): Volume of cylinder made by Dhara \(=\pi r^{2} h_{2}\)
\(=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4=154 \mathrm{~cm}^{3}\)
(iv) (c): Required ratio
\(=\frac{\text { Volume of cylinder made by Meera }}{\text { Volume of cylinder made by Dhara }} \)
\(=\frac{\pi r^{2} h_{1}}{\pi r^{2} h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(v) (a): When two coins are shifted from Meera's cylinder to Dhara's cylinder, then length of both cylinders become equal. So, volume of both cylinders become equal