The roller is in a form of a (Right Circular) Cylinder.
For this cylinder,
  d = 0.9 m,

∴ r = \(\displaystyle \frac {d}{2}\) = \(\displaystyle \frac {0.9}{2}\)

∴ r = 0.45 m,
Also, h (i.e. length) = 1.4 m,

Now,
 Area pressed in one revolution
= Curved Surface Area of the cylinder
= 2πrh

= 2 × \(\displaystyle \frac {22}{7}\) × 0.45 ×1.4 ... (i)

∴ Area pressed in 500 revolutions

= 2 ×\(\displaystyle \frac {22}{7}\) × 0.45 ×1.4 × 500

= 1980 m² ... (ii)

∴ The area pressed in 500 revolutions is 1980 m²

Here, outer dimensions of the tank:
 l = 60.4 cm,
 b = 40.4 cm,
 h = 40.2 cm,
and thickness of the glass sheet
= 2 mm = 0.2 cm

∴ Inner dimensions of the tank:
 l = 60.4 − 0.2 − 0.2 = 60 cm,
 b = 40.4 − 0.2 − 0.2 = 40 cm,
 h = 40.2 − 0.2 = 40 cm

Now, the tank is in the shape of a cuboid.
Volume of a cuboid:
 V = l × b × h
∴ V = 60 × 40 × 40
∴ V = 96000 cm³ ... (i)

The maximum volume of water that will be contained in the tank is 96000 cm³.

Here, r : h = 5 : 12, V = 314 m³, h = ?, l = ?

Let, r = 5x,
∴ h = 12x

Now,
 Volume of a cone:

 V = \(\displaystyle \frac {1}{3}\) πr²h

∴ 314 = \(\displaystyle \frac {1}{3}\) × 3.14 × 5x × 5x × 12x

∴ \(\displaystyle \frac {314 \times 3}{3.14 \times 5 \times 5 \times 12}\) = x³

∴ 1 = x³

∴ \(\displaystyle \sqrt [3]{1}\) = x

∴ 1 = x
i.e. x = 1 ... (i)

Now,
 r = 5x,
∴ r = 5 × 1
∴ r = 5 m ... (ii)

And,
 h = 12x,
∴ h = 12 × 1
∴ h = 12 m ... (iii)

Also,
 l² = h² + r²
∴ l² = 12² + 5²
∴ l² = 144 + 25
∴ l² = 169
∴ l = \(\displaystyle \sqrt {169}\)
∴ l = 13 m ... (iv)

∴ The perpendicular height of that cone is 12 m and its slant height is 13 m.

Here, V = 904.32 cm³, r = ?

 Volume of a sphere:

  V = \(\displaystyle \frac {4}{3}\) πr³

∴ \(\displaystyle 904.32 = \frac {4}{3}\) × 3.14 × r³

∴ \(\displaystyle \frac {904.32 \times 3}{4 \times 3.14}\) = r³

∴ 216 = r³
∴ \(\displaystyle \sqrt [3]{216}\) = r
∴ 6 = r
i.e. r = 6 cm ... (i)

∴ The radius of that sphere is 6 cm.

Here, S_t = 864 cm², V = ?

 Total surface area of a cube:
 S_t = 6l²
∴ 864 = 6l²

∴ \(\displaystyle \frac {864}{6}\) = l²

∴ 144 = l²
∴ \(\displaystyle \sqrt {144}\) = l
∴ 12 = l
i.e. l = 12 cm ... (i)

Now,
 Volume of a cube:
 V = l³
∴ V = 12³
∴ V = 1728 cm³ ... (ii)

∴ The volume of that cube is 1728 cm³.

Here, S = 154 cm², V = ?

 Surface area of a sphere:
 S = 4πr²

∴ 154 = 4 × \(\displaystyle \frac {22}{7}\) × r²

∴ \(\displaystyle \frac {154 \times 7}{4 \times 22}\) = r²

∴ \(\displaystyle \frac {49}{4}\) = r²

∴ \(\displaystyle \sqrt {\frac {49}{4}}\) = r

∴ \(\displaystyle \frac {7}{2}\) = r

i.e. r = \(\displaystyle \frac {7}{2}\) cm ... (i)

Now,
 Volume of a sphere:

 V = \(\displaystyle \frac {4}{3}\) πr³

∴ V = \(\displaystyle \frac {4}{3}\) × \(\displaystyle \frac {22}{7}\) × \(\displaystyle \frac {7}{2}\) × \(\displaystyle \frac {7}{2}\) × \(\displaystyle \frac {7}{2}\)

∴ V = \(\displaystyle \frac {539}{3}\)

∴ V = 179.666 ... ≈ 179.67 cm³ ... (ii)

∴ The volume of that sphere is 179.67 cm³.

Here, S_t = 616 cm², l = 3r, l = ?

Now, l = 3r

 ∴ \(\displaystyle \frac {l}{3}\) = r

i.e. r = \(\displaystyle \frac {l}{3}\) ... (i)

Now,
 Total Surface Area of a cone:
 S_t = πr(l + r)

∴ 616 = \(\displaystyle \frac {22}{7}\) × \(\displaystyle \frac {l}{3}\) × \(\displaystyle \left[ l + \frac {l}{3} \right]\)

∴ 616 = \(\displaystyle \frac {22}{7}\) × \(\displaystyle \frac {l}{3}\) × \(\displaystyle \frac {4l}{3}\)

∴ \(\displaystyle \frac {616 \times 7 \times 3 \times 3}{22 \times 4}\) = l²

∴ 441 = l²
∴ \(\displaystyle \sqrt {441}\) = l
∴ 21 = l
i.e. l = 21 cm ... (ii)

∴ The slant height of that cone is 21 cm

d = 4.2 m,

∴ r = \(\displaystyle \frac {d}{2}\) = \(\displaystyle \frac {4.2}{2}\)

∴ r = 2.1 m,
 h (i.e. depth) = 10 m, S_c = ?
 Rate = ₹ 52 per sq. m, Cost = ?

The well is in the form of a cylinder.
Inner Surface Area of a cylinder:
 S_c = 2πrh

∴ S_c = 2 × \(\displaystyle \frac {22}{7}\) × 2.1 × 10

∴ S_c = 2 × \(\displaystyle \frac {22}{7}\) × 21
∴ S_c = 132 cm² ... (i)

Now, the cost of plastering the well from inside
= Area × Rate
= 132 × 52
= ₹ 6864... (ii)

∴ The inner surface area of the well is 132 cm² and the cost of plastering it from inside is ₹ 6,864/-

The roller will be in the form of a (Right Circular) Cylinder.
For this cylinder:
h (i.e. length) = 2.1 m,
d = 1.4 m

∴ r = \(\displaystyle \frac {d}{2}\) = \(\displaystyle \frac {1.4}{2}\)

∴ r = 0.7 m

Now,
 Area pressed in one revolution
= Curved Surface Area of the cylinder
= 2πrh

= 2 × \(\displaystyle \frac {22}{7}\) × 0.7 × 2.1 ... (i)

∴ Area pressed in 500 revolutions

= 2 ×\(\displaystyle \frac {22}{7}\) × 0.7 × 2.1 × 500

= 2 ×\(\displaystyle \frac {22}{7}\) × 7 × 21 × 5

= 4620 m² ... (i)

Also, the cost of levelling
= Area × Rate
= 4620 × 7
= 32340... (ii)

∴ The area of the ground levelled by the road roller is 4620 m² and the cost of levelling is ₹ 32,340/-