Class 10 Mathematics

Surface Area Volume

Exercise 13.3 (NCERT)

Question 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution: Radius of sphere = 4.2 cm, radius of cylinder = 6 cm

Volume of sphere

`=(4)/(3) πr^3`

`=(4)/(3) πxx4.2^3`

Volume of cylinder

`=πr^2h`
`=πxx6^2xxh`

Since volume of cylinder = Volume of sphere

Hence, height of cylinder

`h=(4xxπxx4.2^3)/(3xxπxx6^2)`

`=2.744 cm`


Question 2: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution: Radii of spheres = 6 cm, 8 cm, 10 cm

Volume of sphere

`=(4)/(3) πr^3`

Total volume of three spheres

`=(4)/(3) π(6^3+8^3+10^3)`

`=(4)/(3) π(616+512+1000)`

`=(4)/(3) πxx1728`

Hence radius of biggest sphere

`=3_sqrt(1728)=12 cm`


Question 3: A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution: Radius of well = 3.5 m, depth of well = 20 m

Dimensions of rectangular platform `= 22  m xx 14  m`

Volume of earth dug out

`=πr^2h`
`=πxx3.5^2xx20=770 m^3`

Area of top of platform = Area of Rectangle – Area of Circle

(because circular portion of mouth of well is open)

`=22xx14-πxx3.5^2`
`=308-38.5=269.5 m^2`

`text(Height) = text(Volume)/(Area)`

`=(770)/(269.5)=2.85 m`

Question 4: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution: Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

Radius of circular embankment `= 4 + 1.5 = 5.5` m

Volume of earth dug out

`=πr^2h`
`=πxx1.5^2xx14=31.5π m^3`

Area of top of platform = (Area of bigger circle – Area of smaller circle)

`=π(R^2-r^2)`
`=π(5.5^2-1.5^2)=28π`

`text(Height) = text(Volume)/(Area)`

`=(31.5π)/(28π)=1.125 m`


Question 5: A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution: Radius of cylinder = 6 cm, height of cylinder = 15 cm

Radius of cone = 3 cm, height of cone = 12 cm

Radius of hemispherical top on ice cream = 3 cm

Volume of cylinder

`=πr^2h`
`=πxx6^2xx15=540π cm^3`

Volume of cone

`=(1)/(3)xxπxx3^2xx12`

`=36π cm^3`

Volume of hemisphere

`=(2)/(3) πr^3`

`=(2)/(3)xxπxx3^3=18π cm^3`

Volume of ice cream

`=(36+18) π=54π cm^3`

Hence, number of ice creams = Volume of cylinder/Volume of ice cream

`=(540π)/(54π)=10`

Question 6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

Volume of coin

`=πr^2h`
`=πxx0.875^2xx0.2`
`=0.48125 cm^3`

Volume of cuboid `=5.5xx1.xx3.5=192.5 cm^3`

Number of coins

`=(192.5)/(0.48125)=400`

Question 7: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution: Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

Volume of cylinder

`=πr^2h`
`=πxx18^2xx32`

Volume of cone = Volume of cylinder

Volume of cone

`=(1)/(3) πr^2xx24`

Hence, radius of cone can be calculated as follows:

`r^2=(3xxπxx18^2xx32)/( πxx24)`

Or, `r^2=18^2xx2^2`
Or, `r=36 cm`

Now, slant height of conical heap can be calculated as follows:

`l=sqrt(h^2+r^2)`

`=sqrt(24^2+36^2)`

`=sqrt(576+1296)`

`=sqrt(1872)=36sqrt13 cm`

Question 8: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution: Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes `= 1.5 xx 6 xx 5000 = 45000` cubic m

`text(Area) = text(Volume)/(Height)`

`=(45000)/(0.08)=562500 m^2`

Question 9: A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution: Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

Volume of water in 1 hr through pipe

`=πr^2h`
`=πxx0.1^2xx3000`
`=30π m^3`

Volume of tank

`=πr^2h`
`=πxx5^2xx2=50π m^3`

Time taken to fill the tank = Volume of tank/Volume of water in 1 hr

`=(50π)/(30π)=1 hr 40 mi\n`


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Exercise 13.1

Exercise 13.2

Exercise 13.4

Exercise 13.5