Surface Areas & Volume

NCERT Exercise 13.3

Take `pi = 22/7`, unless stated otherwise.

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:

Given, radius of sphere `=4.2\ cm`

And radius of cylinder which is re casted `=6\ cm`

Thus, height of the recasted cylinder =?

Volume of sphere `=4/3 pi r^3`

Thus, volume of the given sphere `=4/3 pi (4.2\ cm)^3`

`=4/3 pi xx 74.088 cm^3`

`=98.784 pi cm^3`

Now, volume of the cylinder `= pi r^2 h`

Thus, volume of recasted cylinder `=pi (6\ cm)^2 h`

`=pi xx36 h cm^2`

`=36 pi h cm^2`

Now, volume of the cylinder = volume of the sphere

Thus, `36pi h cm^2 = 98.784 pi cm^3`

`h = (98.784 pi cm^3)/(36 pi cm^2)`

`=>h = 2.744\ cm`

Thus, height of the resulting cylinder `=2.744\ cm` Answer

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

Solution:

Given, radius of the sphere1 `=6\ cm`

And radius of the sphere2 `=8\ cm`

And radius of the sphere3 `=10\ cm`

Thus, radius of the resulting sphere =?

Volume of sphere `=4/3 pi r^3`

Thus, volume of sphere1 `=4/3 pi (6\ cm)^3`

`=4/3 xx 216\ pi\ cm^3`

And volume of sphere2 `=4/3 pi (8\ cm)^3`

`=4/3 xx 512\ pi\ cm^3`

And Volume of sphere3 `=4/3 pi (10\ cm)^3`

`=4/3 xx 1000\ pi\ cm^3`

Thus, volume of recasted sphere = Volume of shpere1 + volume of sphere2 + volume of sphere3

`=4/3 xx 216\ pi\ cm^3 + 4/3 xx 512\ pi\ cm^3 + 4/3 xx 1000\ pi\ cm^3`

`=4/3 pi (216+512+1000) cm^3`

`=4/3\ 1728\ pi cm^3`

Now, volume of the recasted sphere `= 4/3 pi r^3`

`=>4/3 pi r^3=4/3\ 1728\ pi cm^3`

`=>r^3 = (4//3\ 1728\ pi\ cm^3)/(4//3 pi)`

`=>r^3 =1728\ cm^3`

`=>r =root3\1728 cm^3`

`=>r = 12 cm`

Thus, volume of resulting sphere `=12\ cm` Answer

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 22m by 14 m. Find the height of the platform.

Solution:

Given, height of cylindrical well `=20\ m`

Diameter of well `=7\ m`

Thus, radius of well `=7/2=3.5\ m`

Length of resulting platform in cuboid shape `=22\ m`

Breadth of resulting platform `=14\ m`

Thus, height of resulting platform in the shape of cuboid =?

Now, volume of cylinder `=pi r^2 h`

`=22/7 xx (3.5\ m)^2 h`

`=22/7 xx 3.5 xx3.5xx20\ m^3`

`=22xx0.5xx3.5xx 20\ m^3`

`=770\ m^3`

Or, Volume of well `=770\ m^3`

Now, volume of cuboid `=lxxbxxh`

Thus, volume of resulting platform `=22\ m xx 14\ mxx h`

`=>770\ m^3 = 308\ h\ m^2`

[∵ volume of well = volume of platform]

`=>h = (770\ m^3)/(308\ m^2)`

`=>2.5\ m`

Thus, height of the resulting platform `=2.5\ m` Answer

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.

Given, Depth, i.e. height of the well `=14\ m`

Diameter of the well `=3\ m`

Thus, radius of well `=1.5\ m`

Since, earth taken out from well is spread around the well in the form of circular ring, thus radius of the inner circular ring = radius of well `=1.5\ m`

Width of the circular ring of embankment`=4\ m`

Thus, Radius of outer ring of embankment`=4\ m+1.5\ m=5.5\ m`

∴ height of the circular embankment =?

Now, Volume of a cylinder `=pi r^2h`

Thus, volume of earth taken out from well = volume of well `=pi (1.5\ m)^2xx14\ m`

`=pi\ 2.25xx14\ m^3`

Or, volume of earth taken out from well `=pi\ 31.5 m^3`

Now, volume of circular embankment = Volume of outer ring of embankment – Volume of inner circle of embankment

`=pi (5.5\ m)^2\ h - pi(1.5\ m)^2\ h`

[∵ volume of cylinder `=pir^2h`]

`=pi\ h(30.25-2.25)m^2`

`=pi\ hxx 28\ m^2`

`=>pi\ 31.5 m^3 = pi\ hxx 28\ m^2`

[∵ volume of circular embankment = volume of earth dug out from well]

`=>h = (pi\ 31.5\ m^3)/(pi\ 28\ m^2)`

`=> h = 1.125\ m`

Thus, height of the resulting embankment `=1.125\ m` Answer

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:

Given, height of the circular cylinder `=15\ cm`

And diameter of the cylinder `=12\ cm`

Thus, radius of the cylinder `=12/2=6\ cm`

Again given, height of cone `=12\ cm`

And diameter of the cone `=6\ cm`

Thus, radius of the cone `=6/2=3\ cm`

Thus, number of cones filled with the icecream in the given circular cylinder =?

Here volume of the ice cream = volume of the circular cylinder

And volume of the cone = volume of the icecream filled in cone

Now, volume of cylinder `=pi r^2h`

Thus, volume of the given circular cylinder `= pi (6\ cm)^2 xx 15\ cm`

`=pi xx 36 xx15 cm^3`

`=pi\ 540\ cm^3`

Thus, volume of ice cream in cylinder `= pi\ 540\ cm^3`

Now, since diameter of the cone `=6\ cm`

Thus, radius of cone `=6/2 = 3\ cm`

Thus, radius of hemisphere above the cone `=3\ cm`

Now, volume of cone `=1/3 pi r^2h`

Thus, volume of given cone `=1/3 pi (3\ cm)^2xx12\ cm`

`=1/3xx pi xx 3xx3xx12\ cm^3`

Or, volume of cone `=pi\ 36\ cm^3`

Thus, volume of the ice cream in the cone excluding hemisphere above the cone `=`=pi\ 36\ cm^3`

Now, volume of hemisphere `=2/3 pi r^3`

Thus, volume of ice cream in the hemisphere above the cone `=2/3 pi (3\ cm)^3`

`=2/3 xx pi xx 3 xx3xx3\ cm^3`

Or, volume of ice cream of the hemisphere above the cone`=18\ pi\ cm^3`

Thus, Total volume of ice cream in a cone = volume of cone + volume of hemisphere

`=pi\ 36\ cm^3 + pi\ 18\ cm^3`

`=pi(36+18)\ cm^3`

Thus, volume of ice cream in one cone `=54\ pi\ cm^3`

Thus, number of cones filled with ice cream `=(text{volume of ice cream in cylinder})/(text{volume of ice cream in one cone})`

`=( pi\ 540\ cm^3)/( 54\ pi\ cm^3)=10`

Thus, number of cones filled with ice cream `=10` Answer

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

Solution:

Given, length of cuboid, `l=10\ cm`

Width of cuboid, `b=5.5\ cm`

And, height of cuboid, `h=3.5\ cm`

Thickness of coin `h=2\ mm =0.2\ cm`

Diameter of coin `=1.75\ cm`

Thus, radius of coin, `r=1.75/2=0.875\ cm`

Thus, Number of required to form cuboid of given dimensions =?

Now, Volume of cuboid `=lxxbxxh`

Thus, volume of given cuboid `=10\ cmxx5.5\ cmxx3.5\ cm`

`=192.5\ cm^3`

Now, volume of cylinder `=pi r^2h`

Since, a coin has a cylindrical shape

Thus, volume of one coin `=pi xx (0.875\ cm)^2xx0.2\ cm`

`=22/7xx0.766 xx 0.2\ cm^3`

`=0.48125\ cm^3`

Thus, number of coins required to form the given cuboid `=(text{volume of cuboid})/(text{volume of one coin})`

`=(192.5\ cm^3)/(0.48125\ cm^3)`

`=400`

Thus, number of coins required to form cuboid of given dimenstion `=400` Answer

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

Solution:

Given, height of cylindrical bucket `=32\ cm`

And radius of cylindrical bucket `=18\ cm`

Height of conical heap `=24\ cm`

Thus, radius `(r)` and slant height `(l)` of conical heap =?

Volume of cylinder `=pi r^2h`

Thus, volume of given cylindrical bucket `=pi (18\ cm)^2\ 32\ cm`

`=pi xx 324 xx 32\ cm^3`

Or, Volume of bucket `=pi\ 18xx18xx32\ cm^3`

Now, volume of cone `=1/3 pi r^2h`

Thus, volume of given conical heap `=1/3pi r^2xx24 cm`

Or, volume of conical heap `=8r^2\ pi\ cm`

Now, since conical heap is formed by sand emptied out from the bucket.

Thus, volume of the conical heap = volume of sand of bucked = volume of bucket

∴ `8r^2\ pi\ cm = pi xx 324 xx 32\ cm^3`

`=> r^2 = (pi xx 324 xx 32\ cm^3)/(8\ pi\ cm)`

`=>r^2 = 324 xx 4\ cm^2`

`=>r = sqrt(18xx18xx2xx2\ cm^2)`

`=>r =18 xx2\ cm = 36\ cm`

Thus, radius of conical heap `=36\ cm`

Now slant height of cone, `l=sqrt(h^2+r^2)`

Thus, slant height of conical heap, `l=sqrt((24\ cm)^2+(36\ cm)^2)`

`=>l=sqrt((576+1296)cm^2)`

`=>l = sqrt(1872\ cm^2)`

`=>l = sqrt(12xx12xx13\ cm^2)`

`=>12sqrt(13)\ cm`

Thus, radius and slant height of conical heap are equal to `36\ cm` and `12sqrt(13)\ cm` respectively. Answer

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:

Given, width of canal `=6\ m`

And depth of canal `=1.5\ m`

Speed of water `=10\ km//h`

Height of standing water `=8\ cm`

Thus, area of field irrigated in 30 minute = ?

Now, area of cross section of canal = length `xx` depth

`=6\ m xx 1.5\ m = 9\ m^2`

Given, in 1 hour or 60 minute length covered by water = 10 km or `10 xx 1000\ m`

∴ In `1` minute length covered by water `=(10xx1000\ m)/60`

∴ in `30` minute water `=(10xx1000\ m)/60 xx 30`

`=(10xx1000\ m)/2`

`=5xx1000 = 5000\ m`

Thus, length of water in 30 minute =5000 m

Thus, volume of water in 30 minute = area of cross section of canal `xx` length of water

`=9\ m^2xx5000\ m`

`=45000\ m^3`

Or, volume of water in 30 minute `=45000\ m^3`

Now, according to question, height of water `=8\ cm`

`=8/100\ m =0.08\ m`

Now, Area `=(text{Volume})/(text{Height})`

Thus, area of field `=(45000\ m^3)/(0.08\ m)`

`=562500\ m^2`

`= 562500/10000` hectare

[∵ 1 hectare = 10000 square meter]

`=56.25` hectare

Thus, in given time `562500\ m^2` or `56.25` hectare of field is irrigated. Answer

Question: 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10m 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:

Given diameter of pipe `=20\ cm =20/100 m=0.2\ m`

Thus, radius of pipe `=0.2/2=0.1\ m`

Depth, i.e. height of the cylindrical tank `=2\ m`

Diameter of the cylindrical tank `=10\ m`

Thus, radius of cylindrical tank `=10/2=5\ m`

Speed of water `=3km//h =3xx1000=3000\ m//h`

∵ In 60 minute length covered by water `=3000\ m`

∴ In 1 minute length covered by water `=(3000\ m)/60=50\ m`

Now, volume of water discharge in 1 minute through pipe

= area of cross section of pipe `xx` length of water

`=pi (0.1\ m)^2 xx 50\ m`

`=pi xx 0.01xx50\ m^3`

Or, volume of water discharge through pipe in 1 minute `=pi\ 0.5\ m^3`

Now, Volume of cylinder `=pi r^2h`

Thus, volume of cylindrical tank `=pi (5\m)^2 xx 2\ m`

`=pi xx 25\ m^2xx2\ m`

Or, volume of cylindrical tank `=pi\ 50\ m^3`

∵ Time taken to fill `pi\ 0.5\ m^3` water `=1` minute

∴ Time taken to fill `1\ m^3` of water `=1/(pi\ 0.5\ m^3)` minute

∴ Time taken to fill ` pi\ 50\ m^3` of water `=1/(pi\ 0.5\ m^3) xx pi\ 50\ m^3` minute

`=(50\ m^3)/(0.5\ m^3) =100` minute

Thus, cylindrical tank will be filled in 100 minute. Answer

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