A cylindrical container with an internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of the wet surface of the cylinder.

Open in App

Solution

Area of the wet surface of the cylinder = $2 π r h + π r^{2}$
= $2 \times \frac{22}{7} \times 10 \times 7$ + $\frac{22}{7} \times 10 \times 10$
= 3.14 (100+140)
= 753.6

Suggest Corrections

Similar questions

Q. A cylindrical container with internal radius of its base

10

cm, contains water up to a height of

7

cm. Find the area of the wet surface of the cylinder.

Q. Find the surface area (in sq. cm.) of a cylindrical container with radius 4 cm and height 10 cm, assuming it has an open top and a closed bottom.

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water (i) displaced out of the cylinder (ii) left in the cylinder.

Q. A cylindrical tank

7 k m

in diameter, contains water to a depth of

4 m .

Find the total area of the wet surface

(π = 3.14)

Q. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder (ii) left in the cylinder. (Take π 22/7)

Join BYJU'S Learning Program

A cylindrical container with an internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of the wet surface of the cylinder.

Area of the wet surface of the cylinder = 2πrh+π r2 = 2 × 227 × 10 × 7 +227 × 10 × 10 = 3.14 (100+140) = 753.6

Area of the wet surface of the cylinder = $2 π r h + π r^{2}$
= $2 \times \frac{22}{7} \times 10 \times 7$ + $\frac{22}{7} \times 10 \times 10$
= 3.14 (100+140)
= 753.6