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

Exercise 13.5

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 \[\text{c}{{\text{m}}^{3}}\].

Ans: Diameter of the copper wire = \[3\text{ mm}\approx 0.3\text{ cm}\]

So, the radius of the copper wire = \[\frac{0.3}{2}\text{ cm = 0}\text{.15 cm}\]

Diameter of the cylinder = 10cm.

So, the radius of the cylinder (r)= 5cm.

Length of the cylinder = 12cm.

One round of wire around the cylinder covers 3mm height of the cylinder.

Total number of rounds = \[\frac{\text{Total height of the cylinder}}{\text{Diameter of wire}}\]

\[\Rightarrow \frac{12}{\text{0}\text{.3}}\text{ = 40 rounds}\]

Length of wire in one complete round = \[2\pi r\]

\[=2\pi \times 5\]

\[=10\pi \text{ }cm\]

Hence, length of wire in 40 rounds (l) =  \[40\times 10\pi \text{ }cm\]

\[=1256\text{ cm}\]

Volume of wire = area of cross section of wire * length

\[=\text{ }\pi {{r}^{2}}\times l\]

\[=\text{ 3}\text{.14}\times {{(0.15)}^{2}}\times 1256\]

\[=\text{ }88.898\text{ c}{{\text{m}}^{3}}\]

Mass of the wire = volume of wire * density

\[=\text{ }88.898\times 8.88\]

\[=\text{ }789.41\text{gm}\]

The length of the wire \[=1256\text{ cm}\] and mass of the wire \[=\text{ }789.41\text{gm}\].

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. (Choose value of \[\text{ }\!\!\pi\!\!\text{ }\] as found appropriate.)

Ans: Two sides of the right triangle , base (b) = 3cm and length (l) = 4cm.

Hypotenuse of the right triangle using Pythagoras theorem: \[\sqrt{{{b}^{2}}+{{l}^{2}}}\]

=\[\sqrt{{{3}^{2}}+{{4}^{2}}}\]

=\[5\text{ cm}\]

Two cone is formed because of revolution of right triangle about its hypotenuse.

Area of \[\Delta ABC\] = \[\frac{1}{2}\times l\times b\]

\[\Rightarrow \frac{1}{2}\times AC\times OB=\frac{1}{2}\times 4\times 3\]

\[\Rightarrow \frac{1}{2}\times 5\times OB=6\text{ c}{{\text{m}}^{2}}\]

\[\Rightarrow OB=\frac{12}{5}cm\]

Volume of double cone  = volume of upper cone + volume of lower cone

\[=\frac{\pi }{3}\times B{{O}^{2}}\times A\text{O }+\text{ }\frac{\pi }{3}\times B{{O}^{2}}\times C\text{O}\]

\[=\frac{\pi }{3}\times B{{O}^{2}}\times (A\text{O }+C\text{O)}\]

\[=\frac{\pi }{3}\times B{{O}^{2}}\times AC\] (AO+CO=AC)

\[=\frac{1}{3}\times \frac{22}{7}\times {{\left( \frac{12}{5} \right)}^{2}}\times 5\]

\[=30.14\text{ c}{{\text{m}}^{3}}\]

Surface area of the double cone = Surface area of upper cone + Surface area of lower cone

\[=\pi rl\text{ + }\pi rb\]

\[=\pi \frac{12}{5}\times \text{4 + }\pi \frac{12}{5}\times 3\]

\[=\pi \left( \frac{36}{5}+\frac{48}{5} \right)\]

\[=\frac{22}{7}\left( \frac{84}{5} \right)\]

\[=52.75\text{ c}{{\text{m}}^{2}}\]

Volume of double cone \[=30.14\text{ c}{{\text{m}}^{3}}\]

Surface area of the double cone \[=52.75\text{ c}{{\text{m}}^{2}}\]

3. A cistern, internally measuring $\text{150 cm  }\!\!\times\!\!\text{  120 cm  }\!\!\times\!\!\text{  110 cm}$, has \[\text{129600c}{{\text{m}}^{\text{3}}}\] 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 brick being $\text{22}\text{.5 cm  }\!\!\times\!\!\text{  7}\text{.5 cm  }\!\!\times\!\!\text{  6}\text{.5 cm}$? 

Ans: Dimension of cistern = \[150\text{cm }\times \text{ 120cm }\times \text{ 110cm}\]

So, the volume of cistern = \[1980000\text{c}{{\text{m}}^{3}}\].

Volume of water = \[129600\text{c}{{\text{m}}^{3}}\].

Volume of cistern filled = \[1980000\text{c}{{\text{m}}^{3}}\text{ - }129600\text{c}{{\text{m}}^{3}}\]

\[=1850400\text{c}{{\text{m}}^{3}}\]

Dimension of bricks = \[22.5\text{cm }\times \text{ 7}\text{.5cm }\times \text{ 6}\text{.5cm}\]

Volume of bricks = \[1096.875\text{c}{{\text{m}}^{3}}\]

Let the number of porous bricks placed in cistern be n.

Volume of n bricks = n * Volume of brick

\[=n\times 1096.875\text{c}{{\text{m}}^{3}}\]

It is given that each brick absorbs one-seventeenth of its own volume. 

Then, the volume absorbed by n bricks =  \[\frac{n\times 1096.875}{17}\text{c}{{\text{m}}^{3}}\].

According to the question,

The volume of n bricks = volume absorbed by n bricks + Volume to be filled in cistern

\[\Rightarrow n\times 1096.875\text{ =  }\frac{n\times 1096.875}{17}\text{  + 1850400 }\]

\[\Rightarrow \frac{16n\times 1096.875}{17}\text{  = 1850400 }\]

\[\Rightarrow n\text{  = 1792}\text{.41 }\]

Therefore, the number of required bricks = 1792.

4. 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 \[\text{7280K}{{\text{m}}^{\text{2}}}\], 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.

Ans: The area of valley = \[\text{7280K}{{\text{m}}^{\text{2}}}\].

Rainfall in valley = \[10cm\text{ }\approx \text{ }\frac{10}{100\times 1000}km\]

Volume of rain = \[\frac{\text{7280}\times \text{10}}{100\times 1000}k{{m}^{3}}\]

\[=0.728k{{m}^{3}}\]

Length of the river = 1072km

Breadth of the river = \[75m\text{ }\approx \text{ 0}\text{.075}km\]

Depth of the river = \[3m\text{ }\approx \text{ 0}\text{.003}km\]

Volume of a river = \[\text{(1072}\times \text{0}\text{.075}\times \text{0}\text{.003)}k{{m}^{3}}\]

\[=0.2412\text{ }k{{m}^{3}}\]

Volume of three river \[=3\times 0.2412\text{ }k{{m}^{3}}\]

\[=0.7236\text{ }k{{m}^{3}}\]

Hence,the total rainfall was approximately equivalent to the addition to the normal water of three rivers.

5. 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 the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure).

Ans:

Height of the cylinder = 10cm.

Total height = 22cm.

Height of the frustum (h) = Total height  - Height of the cylinder 

\[=(22-10)cm\]

\[=12cm\]

Diameter of lower base = 8cm.

Therefore, radius (r) = 4cm.

Diameter of top funnel = 18cm.

Therefore, the radius (R) = 9cm.

Slant height of the frustum (l) =  $\sqrt{{{(R-r)}^{2}}+{{h}^{2}}}$

$=\sqrt{{{(9-4)}^{2}}+{{12}^{2}}}$

$=\sqrt{{{5}^{2}}+{{12}^{2}}}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$=13cm$

Area of thin sheet which is required = curved surface area of frustum + curved surface area of cylinder.

\[=\pi (r+R)l\text{ + 2}\pi \text{rh}\]

\[=\frac{22}{7}\times (4+9)\times 13\text{ + 2}\times \frac{22}{7}\times 4\times 10\]

\[=\frac{22}{7}\times 249\]

\[=782.57\text{ c}{{\text{m}}^{2}}\]

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.

Ans: 

Let ABC be a cone of height \[{{h}_{1}}\] and slant height of \[{{l}_{1}}\].The radius of base is \[{{r}_{1}}\]. 

From the cone ABC a frustum DECB is cut by a plane parallel to its base. The height of frustum is  h and slant height is l. The radius of upper end is \[{{r}_{2}}\].

Considering the \[\Delta ABG\] and \[\Delta ADF\],

\[BG\parallel DF\] and \[\angle A\] is common in both . Hence, \[\Delta ABG\sim \Delta ADF\].

Therefore, \[\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}\]

\[\Rightarrow \frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{h}_{1}}-h}{{{h}_{1}}}=\frac{{{l}_{1}}-l}{{{l}_{1}}}\]

\[\Rightarrow \frac{\text{l}}{{{l}_{1}}}=1-\frac{{{r}_{2}}}{{{r}_{1}}}\]

After rearranging,

\[\Rightarrow {{l}_{1}}=\frac{{{r}_{1}}l}{{{r}_{1}}-{{r}_{2}}}\]

Curve surface area of frustum DECB = Curve surface area of cone ABC - Curve surface area of ADE.

\[=\pi {{r}_{1}}{{l}_{1}}\text{ - }\pi {{r}_{2}}({{l}_{1}}-l)\]

\[=\pi {{r}_{1}}\left( \frac{{{r}_{1}}l}{{{r}_{1}}-{{r}_{2}}} \right)\text{ - }\pi {{r}_{2}}\left[ \left( \frac{{{r}_{1}}l}{{{r}_{1}}-{{r}_{2}}} \right)-l \right]\]

\[=\frac{\pi {{r}_{1}}^{2}l}{{{r}_{1}}-{{r}_{2}}}\text{ - }\pi {{r}_{2}}\left( \frac{{{r}_{1}}l-{{r}_{1}}l+{{r}_{2}}l}{{{r}_{1}}-{{r}_{2}}} \right)\]

\[=\frac{\pi {{r}_{1}}^{2}l}{{{r}_{1}}-{{r}_{2}}}\text{ - }\frac{\pi {{r}_{2}}^{2}l}{{{r}_{1}}-{{r}_{2}}}\]

\[=\pi l\left( \frac{{{r}_{1}}^{2}-{{r}_{2}}^{2}}{{{r}_{1}}-{{r}_{2}}}\text{ } \right)\]

\[=\pi l\left( \frac{\left( {{r}_{1}}+{{r}_{2}} \right)\left( {{r}_{1}}-{{r}_{2}} \right)}{{{r}_{1}}-{{r}_{2}}}\text{ } \right)\]

Curve surface area of frustum \[=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l\]

The total surface area of frustum = Curve surface area of frustum + Area of upper circular end + Area of the lower circular end

\[=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l\text{ }+\text{ }\pi {{r}_{2}}^{2}\text{ + }\pi {{r}_{1}}^{2}\]

The total surface area of frustum \[=\pi \left[ \left( {{r}_{1}}+{{r}_{2}} \right)l\text{ }+\text{ }{{r}_{2}}^{2}\text{ + }{{r}_{1}}^{2} \right]\].

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.

Ans: 

Let ABC be a cone of height \[{{h}_{1}}\] and slant height of \[{{l}_{1}}\].The radius of base is \[{{r}_{1}}\]. 

From the cone ABC a frustum DECB is cut by a plane parallel to its base. The height of frustum is  h and slant height is l. The radius of upper end is \[{{r}_{2}}\].

Considering the \[\Delta ABG\] and \[\Delta ADF\],

\[BG\parallel DF\] and \[\angle A\] is common in both triangle. Hence, \[\Delta ABG\sim \Delta ADF\].

Therefore, \[\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}\]

\[\Rightarrow \frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{h}_{1}}-h}{{{h}_{1}}}=\frac{{{l}_{1}}-l}{{{l}_{1}}}\]

\[\Rightarrow \frac{h}{{{h}_{1}}}=1-\frac{{{r}_{2}}}{{{r}_{1}}}\]

 After rearranging,

\[\Rightarrow {{h}_{1}}=\frac{{{r}_{1}}h}{{{r}_{1}}-{{r}_{2}}}\]

The total volume of frustum of the cone  = Volume of cone ABC - Volume of cone ADE

\[=\frac{1}{3}\pi {{({{r}_{1}})}^{2}}{{h}_{1}}\text{ - }\frac{1}{3}\pi ({{r}_{2}})^2({{h}_{1}}-h)\]

\[=\left( \frac{\pi }{3} \right)\left[ {{({{r}_{1}})}^{2}}{{h}_{1}}\text{- }{{r}_{2}}^{2}({{h}_{1}}-h) \right]\]

\[=\left( \frac{\pi }{3} \right)\left[ {{r}_{1}}^{2}\left( \frac{h{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}} \right)\text{- }{{r}_{2}}^{2}\left( \frac{h{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}-h \right) \right]\]

\[=\left( \frac{\pi }{3} \right)\left[ \left( \frac{h{{r}_{1}}^{3}}{{{r}_{1}}-{{r}_{2}}} \right)\text{- }{{r}_{2}}^{2}\left( \frac{h{{r}_{1}}-h{{r}_{1}}+h{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}} \right) \right]\]

\[=\left( \frac{\pi }{3} \right)\left( \frac{h{{r}_{1}}^{3}}{{{r}_{1}}-{{r}_{2}}}-\frac{h{{r}_{2}}^{3}}{{{r}_{1}}-{{r}_{2}}} \right)\]

\[=\left( \frac{\pi }{3}h \right)\left( \frac{{{r}_{1}}^{3}-{{r}_{2}}^{3}}{{{r}_{1}}-{{r}_{2}}} \right)\]

As we all know ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)$

\[=\left( \frac{\pi }{3}h \right)\left( \frac{({{r}_{1}}-{{r}_{2}})({{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}})}{{{r}_{1}}-{{r}_{2}}} \right)\]

\[=\left( \frac{\pi }{3}h \right)({{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}})\]

So, Volume of frustum of the cone = \[\frac{1}{3}\pi h({{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}})\]

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

Opting for the NCERT solutions for Ex 13.5 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.5 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Surface Areas and Volumes textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 13 Exercise 13.5 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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