Class 10 Mathematics

Surface Area Volume

Exercise 13.5 (NCERT Book)

Question 1: A copper wire, 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution: For copper wire: Diameter = 3 mm = 0.3 cm

For cylinder; length h = 12 cm, d = 10 cm

Density of copper = 8.88 gm cm3

Curved surface area of cylinder can be calculated as follows:

`=2πrh`
`=πxx10xx12=120π cm^2`


Length of wire can be calculated as follows:

`text(Length)=text(Area)/text(Width)`

`=(120π)/(0.3)=400π=1256 cm`

Now, volume of wire can be calculated as follows:

`V=πr^2h`
`=3.14xx0.15^2xx1256`
`=88.7364 cm^2`

Mass can be calculated as follows:

`text(Mass) = text(Density)\ xx \text(Volume)`

`= 8.88 xx 88.7364`

`= 788  g` (approx)

Question 2: A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

10 surface area volume exercise solution

Solution: In triangle ABC;

`AC^2 = AB^2 + BC^2`

Or, `AC^2 = 3^2 + 4^2`

Or, `AC^2 = 9 + 16 = 25`

Or, `AC = 5  cm`


In triangle ABC and triangle BDC;

`∠ABC=∠BDC` (Right angle)
`∠BAC=∠DBC`
Hence, `ΔABC≃ΔBDC`

So, we get following equations:

`(AB)/(AC)=(BD)/(BC)`

Or, `3/5=(BD)/(4)`

Or, `BD=(3xx4)/(5)=2.4`

In triangle BDC;

`DC^2 = BC^2 – BD^2`

Or, `DC^2 = 4^2 – 2.4^2`

Or, `DC^2 = 16 – 5.76 = 10.24`

Or, `DC = 3.2`

10 surface area volume exercise solution

From above calculations, we get following measurements for the double cone formed:

Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

Volume of cone

`=1/3\πr^2h`

`=1/3xx3.14xx2.4^2xx1.8=10.85184 cm^2`

Curved surface area of cone

`= πrl`
`=3.14xx2.4xx3=22.608 cm^2`

Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm

Volume of cone

`=1/3\ πr^2h`

`=1/3xx3.14xx2.4^2xx3.2=19.2916 cm^3`


Curved surface area of cone

`= πrl`
`=3.14xx2.4xx4=30.144 cm^2`

Total volume `= 19.29216 + 10.85184 = 30.144` cm3

Total surface area `= 30.144 + 22.608 = 52.752` cm2

Question 3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3 water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution: Volume of cistern = length x width x depth

`= 150 xx 120 xx 110= 1980000` cm3

Vacant space = Volume of cistern – Volume of water

`= 1980000 – 129600 = 1850400` cm3

Volume of brick = length x width x height

`= 22.5 xx 7.5 xx 6.5`

Since the brick absorbs one seventeenth its volume hence remaining volume will be equal to 16/17 the volume of brick

Remaining volume

`=22.5xx7.5xx6.5xx16/17`

Number of bricks = Remaining volume of cistern/remaining volume of brick

`=(1850400xx17)/(22.5xx7.5xx6.5xx16)`

`=(31456800)/(17550)=1792`

Question 4: An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

10 surface area volume exercise solution

Solution: Curved surface area of cylinder

`=2πrh`
`= πxx8xx10=80 π`

Slant height of frustum can be calculated as follows:

`l=sqrt(h^2+(r_1-r_2)^2)`

`=sqrt(12^2+(9-4)^2)`

`=sqrt(144+25)=sqrt(169)=13 cm`

Curved surface area of frustum

`= π(r_1+r_2)l`
`= π(9+4)xx13=169 π`

Total curved surface area

`=169 π+80 π`
`=249xx3.14=781.86 cm^3`


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Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4