These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1.

A copper wire, 3 mm in diameter, is wound about in cylinder whose length is 12 cm, and diameter 10 cm, so as the cover the curve 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:

One round of wire covers 3 m = \(\frac { 3 }{ 10 }\) cm in thickness of the surface of the cylinder.

Length of the cylinder = 12 cm

Number of the rounds to cover 12 cm

Diameter of the cylinder = 10 cm.

Radius = 5 cm

Length of the wire in completing one round

= 2 x π x cm = 10π cm

Length of the wire in covering the whole surface = Length of the wire in completely 40 rounds

(10π x 40) cm = 400π cm

= (400 x 3.14) cm.

= 1256 cm

= 125.6 m

Radius of copper wire = \(\frac { 3 }{ 20 }\) cm

Volume of wire = π x r²h

= π x \(\frac { 3 }{ 20 }\) x \(\frac { 3 }{ 20 }\) x 400π

= 9π² cm³.

So, mass of the wire = 9π² x 8,88 per cm³.

= 787.98 gm

Length of wire = 12.56 m and mass of the wire = 787.98 gm.

Question 2.

A right triangle, whose sides 3 cm and 4 cm (other than hypotenuse) is made to resolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate),

Solution:

Let ΔABC be the right angle triangled at B whose sides AB and BC are 3 cm, 4 cm respectively.

Length of hypotenuse = \(\sqrt{(3)^{2}+(4)^{2}}\) = \(\sqrt{25}\) = 5 cm.

BO and DO are the common base of double cone formed by revolving the right triangle about AC,

Slant height = 3 cm (for cone ABD)

Slant height = 4 cm (for cone BCD)

(∵ AOD = 90 = ADC and DAC is common)

\(\frac { AO }{ 3 }\) = \(\frac { 3 }{ 4 }\) = \(\frac { OD }{ 4 }\)

⇒ AO = \(\frac { 9 }{ 5 }\) = 1.8 cm

h = 1.8 cm

CO = AC – OA

⇒ CO = 5 – 18 = 3.2 cm (H = 3.2 cm)

Similarly OD = \(\frac { 12 }{ 5 }\) 2.4 cm = r

Now, volume of double cone

Curved surface area of double cone

= πr(l_{1}) + πr(l_{2})

= πr(3) + πr(4)

– 3.14 x 2.4 (7)

= 52.75 cm².

Question 3.

A cistern, internally measuring 150 cm x 120 cm x 110 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 x 7.5 cm x 6.5 cm

Solution:

Area of cistern = 150 cm x 120 cm x 110 cm – 1980000 cm³.

Empty volume of cistern = (1980000 – 129600)cm³.

Let number of bricks dropped in the water be n

So the volume of bricks = (22.5 7.5 x 6.5) n = 1096.875 cm³.

The volume of water absorbed by each brick = \(\frac { 1 }{ 17 }\) x (1096.S95) cm³.

Volume of total bricks = (1096.875) \(\frac { n }{ 17 }\) cm³.

So, 1850400 cm³ = 1096.875

n = (1096.875) \(\frac { n }{ 17 }\)

∴ Number of bricks = 1792.

Question 4.

In one fortnight of a given month, there was rainfall of 10 cm in a river valley. If the area of one 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:

Volume of a river = 1072 km x 75 m x 3 m

= 1072000 m x 75 m 3 m

= 241200000 m³

Volume of 3 rivers

= 3 x 241200000

= 723600000 m³.

Now value of rainfall in the valley

= height of rainfall x Area

= 10 cm x 97280 km².

= \(\frac { 10 }{ 100 }\) m x 97280 x (100)² m².

= \(\frac { 10 }{ 100 }\) x 97280000000

= 9728000000 m³.

Now 9728000000 m³ > 73360000 m³.

Hence, the additional water i,e. of the 3 river cannot be equivalent to the rainfall.

Question 5.

An oil funnel made of tin sheet consist 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 (See the adjoining figure).

Solution:

Let l be the slant height of the frustum part of the funnel.

Then, l = \(\sqrt{(9-4)^{2}+12^{2}}\) = \(\sqrt{25+144}\) cm

= \(\sqrt{169}\) cm = 13 cm

Now, Tin required = Curved surface area of cylindrical portion + Curved surface area of frustum potion

= 2πr_{1}h + π(r_{1} + r_{2})h

= [2π x 4 x 10 + π (4 + 9) x 13] cm²

= (80π + 169π) cm² = 249π cm²

= \(\left(249 \times \frac{22}{7}\right)\) cm²

= \(\frac { 5478 }{ 7 }\)

= 782 \(\frac { 4 }{ 7 }\) cm².

Question 6.

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in section 13.5, using the symbols as explained.

Solution:

APO’B and APQR are similar

Now curved surface area of the fur stum is given by

= curved surface area of cone PQR (bigger)- curved surface and of cone PAB (smaller).

= πr_{1}l_{1} – πr_{2}l_{2} … (3)

∴ From (2) and (3)

Curved surface area of the frustum

= π(r_{1} + r_{2})l [Using (2)]

∴ Total surface area of the frustum curved surface area of the frustum + area of both circular bases.

= π(r_{1} + r_{2}) l + πr²_{1} + πr²_{2}

= π(r_{1} + r_{2}) l + r²_{1} + r²_{2.}

Question 7.

Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.

Solution:

Since ΔPO’B and ΔPQR are similar

Now volume of the frustum = Volume of cone PQR (bigger) – Voolume of cone PAQ (smaller)