Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = 154/(4 × 22/7)
⇒ r² = 49/4
⇒ r = 7/2 = 3.5 cm

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
                                                    = (1/64)/(1/4)
                                          = 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
                              = (5 + 0.25) cm  = 5.25 cm
Outer curved surface = 2πR²
                                  = (2 × 22/7 × 5.25 × 5.25) cm²
                                  = 173.25 cm²

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
                                               = 2π × r × 2r
                                               = 4πr²
(iii) Ratio of the areas = 4πr²:4πr² = 1:1

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
                                         = (4 × 2.5 × 1.5)  cm³ = 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15)  cm³ = 180 cm³

Dimensions of water tank = 6m x  5m x 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =lbh m³
 =(6 x 5 x 4.5)m3=135 m³
Therefore , the tank can hold = 135 x 1000 litres[Since 1m3=1000litres]
= 135000 litres of water.

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth) 
= 380/(10×8) m
= 4.75m

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
 = (8×6×3) m³
 = 144 m³
Rate of digging = rs 30 per m³
Total cost of digging the pit =  rs(144 × 30) =  rs 4320