Advertisements
Advertisements
Question
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm2.
`["Assume "pi=22/7]`
Solution
Inner radius (r) of hemispherical bowl = `(10.5/2) cm` = 5.25 cm
Surface area of hemispherical bowl = 2πr2
= `[2xx22/7xx(5.25)^2]cm^2`
= 173.25 cm2
Cost of tin-plating 100 cm2 area = ₹ 16
Cost of tin-plating 173.25 cm2 area = `₹ (16 xx 173.25)/100`
= ₹ 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is ₹ 27.72.
APPEARS IN
RELATED QUESTIONS
Find the surface area of a sphere of radius 14 cm.
`["Assume "pi=22/7]`
Eight metallic spheres; each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of the solid sphere formed.
The cross-section of a tunnel is a square of side 7 m surmounted by a semi-circle as shown in the adjoining figure. The tunnel is 80 m long.
Calculate:
- its volume,
- the surface area of the tunnel (excluding the floor) and
- its floor area.
If a hollow sphere of internal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.
Find the surface area and volume of sphere of the following radius. (π = 3.14)
4 cm
Find the radius of a sphere whose surface area is equal to the area of the circle of diameter 2.8 cm
Find the volume of the hollow sphere whose inner diameter is 8 cm and the thickness of the material of which it is made is 1 cm .
A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.
