Advertisements
Advertisements
Question
A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Advertisements
Solution
Volume of rectangular block = 49 × 44 × 18 cm3
= 38808 cm3 ...(1)
Let r be the radius of sphere
∴ Volume = `4/3pir^3`
= `4/3 xx 22/7 xx r^3`
= `88/21 r^3` ...(2)
From (1) and (2)
`88/21 r^3 = 38808`
`=> r^3 = 38808 xx 21/88 = 441 xx 21`
`=>` r3 = 9261
`=>` r = 21 cm
Radius of sphere = 21 cm
APPEARS IN
RELATED QUESTIONS
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
On a map drawn to a scale of 1: 50,000, a rectangular plot of land ABCD has the following dimensions. AB = 6 cm; BC = 8 cm and all angles are right angles. Find:
1) the actual length of the diagonal distance AC of the plot in km.
2) the actual area of the plot in sq. km.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
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.

Find the radius of a sphere whose surface area is 154 cm2.
Find the surface area and volume of sphere of the following radius. (π = 3.14 )
9 cm
The total area of a solid metallic sphere is 1256 cm2. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate: the number of cones recasted [π = 3.14]
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at ₹ 0.14 per cm2
The total surface area of a hemisphere is how many times the square of its radius
A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have a maximum of 2156 cm3 of ball bearings. Find the:
- maximum number of ball bearings that each box can have.
- mass of each box of ball bearings in kg.
(Use π = `22/7`)
