Class 10th, Surface and Volume, 13.5 (Solution)

Class 10th, Surface Areas and Volumes

 Exercise: 13.5 (Solution), Q1 to Q4

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Q1. 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:

Given that,

Diameter of cylinder = 10 cm

So, radius of the cylinder (r) 10/2 cm = 5 cm

∴ Length of wire in completely one round = 2πr = 3.14 x 5 cm = 31.4 cm

It is given that diameter of wire = 3 mm = 3/10 cm

∴ The thickness of cylinder covered in one round = 3/10 m

Hence, the number of turns(rounds) of the wire to cover 12 cm will be-

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

40 x 31.4 cm = 1256 cm

Radius of the wire = 0.3/2 = 0.15 cm

Volume of wire = Area of cross-section of wire × Length of wire

= π(0.15)2 × 1257.14

= 88.898 cm3

We know,

Mass = Volume × Density

= 88.898 × 8.88

= 789.41 gm

Q2. 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. (Choose value of π as found appropriate.)

Solution:

Draw the diagram as follows:

Let us consider the △ABA

Here,

AS = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

Now, AD/CA = AB/CB

By putting the value of CA, AB and CB we get,

AD = 2/5 cm

We also know,

DB/AB = AB/CB

So, DB = 9/5 cm

As, CD = BC – DB,

CD = 16/5 cm

Now, volume of double cone will be

Solving this we get,

V = 30.14 cm3

The surface area of the double cone wil be

= 52.75 cm2

Q3. A cistern, internally measuring 150 cm × 120 cm × 100 cm, has 129600 cm3 of 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 being 22.5 cm × 7.5 cm × 6.5 cm?

Solution:

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600

= 1850400 cm3

Now, let the number of bricks placed be “n”

So, volume of n bricks will be = n × 22.5 × 7.5 × 6.5

Now as each brick absorbs one-seventeenth of its volume, the volume will be

= n/(17) × (22.5 × 7.5 × 6.5)

For the condition given in the question,

The the volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern

Or, n × 22.5 × 7.5 × 6.5 = 1850400 + n/(17) × (22.5 × 7.5 × 6.5)

Solving this we get,

n = 1792.41

Q4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Solution:

From the question, it is clear that

Total volume of 3 rivers = 3 × [(Surface area of a river) × Depth]

Given,

Surface area of a river = [1072 × (75/1000)] km

And,

Depth = (3/1000) km

Now, volume of 3 rivers = 3 × [1072 × (75/1000)] × (3/1000)

= 0.72 km3

Now, volume of rainfall = total surface area × total height of rain

= 9.7 km3

For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.

But, 9.7 km3 ≠ 0.72 km3 

its the answers

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