Question

A cylindrical vessel open at the top has a base of radius 10.5cm and height 14cm find the cost of painting the inner part of the vessel at the rate of ₹10 per sq.cm take22/7 as [tex] \pi
[/tex]

Answer 1

r = 10.5 cm
h = 14 cm
inner area = inner curved surface area  + base area
               = 2πrh + πr²
               = 2×(22/7)×14×10.5 + (22/7)×10.5²
               = 1270.5 cm²
rate = ₹10 per cm²
cost = 1270.5×10 = ₹12705

Answer 2

The “cost of painting” of the “inner part” of the vessel is 12705 Rupees.

Given:

R = 10.5 cm

h = 14 cm

To find:

The “cost of painting” the “inner part” of the vessel.

Answer:

R = 10.5 cm

h = 14 cm

Inner area = Inner curve surface area + Base area

$=2 \pi r h+\pi r^{2}$

$=(2\times \frac { 22 }{ 7 } \times 10.5\times 14)+(\frac { 22 }{ 7 } \times { 10.5 }^{ 2 })$

$Inner \quad area=1270.5 \mathrm{cm}^{2}$

Given rate $= 10\quad { Rupees\quad per }\quad cm^{2}$

Cost of painting $=1270.5 \times 10 =12705 Rupees$