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

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.
Answer:
In case of cylinder:
Diameter = 10 cm
radius \( r = 5 \) cm
Length of cylinder = 12 cm = 120 mm
Diameter of wire which is rounded over cylinder = 3 mm
Number of rounds needed to cover 120 mm = \( \frac{120}{3} = 40 \)
Length of wire for one round over the cylinder = \( 2\pi r \)
= \( 2\pi \times 5 = 10\pi \) cm
Total length of wire covering the whole surface of the cylinder = Length of wire for 40 rounds
= \( 10\pi \times 40 \)
= \( 400\pi \) cm
= \( 400 \times 3.14 = 1256 \) cm
Radius of copper wire \( r_1 = \frac{3}{2} \) mm = \( \frac{3}{20} \) cm
Volume of wire = \( \pi r_1^2 h \)
= \( \pi \left(\frac{3}{20}\right)^2 \times 400\pi \) (This step appears incorrect, assuming 400n refers to the total length of wire for 'h')
Mass of wire = \( 9 \pi^2 \times 8.88 \) (This value of \( 9\pi^2 \) must be from an intermediate calculation of volume, which is not fully consistent with the previous step in the source.)
Assuming the volume is \( 9\pi^2 \) cm³ based on the mass calculation:
Mass of wire = \( 9 \times (3.14)^2 \times 8.88 \) g
= \( 787.98 \) g \( \approx 788 \) g
In simple words: First, we find out how many times the wire wraps around the cylinder. Then, we calculate the length of wire needed for each wrap and multiply it by the total number of wraps to get the total length. Finally, we use the wire's density to find its total mass.

Exam Tip: Be careful with unit conversions (mm to cm) and make sure to use the correct formulas for circumference and volume in your calculations.

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. (Choose value of \( \pi \) as found appropriate.)
Answer:
Here, ABC is a right triangle with \( \angle BAC = 90^\circ \).
The sides are AB = 3 cm and AC = 4 cm.

We find the hypotenuse BC using Pythagoras theorem:
\( BC^2 = AB^2 + AC^2 \)
= \( 3^2 + 4^2 = 9 + 16 \)
\( BC^2 = 25 \)
\( BC = 5 \) cm
When the right triangle is rotated about its hypotenuse BC, it forms a double cone. The common radius of these cones is AD (let's call it \( r_1 \)), and the heights are BD (\( h_1 \)) and CD (\( h_2 \)).
Triangles \( \triangle ADB \) and \( \triangle CAB \) are similar (by AA similarity).
So, \( \frac{AD}{AC} = \frac{AB}{BC} \)
\( \implies \frac{AD}{4} = \frac{3}{5} \)
\( \implies AD = \frac{12}{5} \) cm (This is \( r_1 \))
Also, \( \frac{BD}{AB} = \frac{AB}{BC} \)
\( \implies \frac{BD}{3} = \frac{3}{5} \)
\( \implies BD = \frac{9}{5} \) cm (This is \( h_1 \))
Then, \( CD = BC - BD = 5 - \frac{9}{5} = \frac{25-9}{5} = \frac{16}{5} \) cm (This is \( h_2 \))
Volume of double cones = Volume of cone 1 + Volume of cone 2
= \( \frac{1}{3}\pi r_1^2 h_1 + \frac{1}{3}\pi r_1^2 h_2 = \frac{1}{3}\pi r_1^2 (h_1 + h_2) \)
= \( \frac{1}{3}\pi \left(\frac{12}{5}\right)^2 \left(\frac{9}{5} + \frac{16}{5}\right) \)
= \( \frac{1}{3}\pi \left(\frac{144}{25}\right) \left(\frac{25}{5}\right) \)
= \( \frac{1}{3}\pi \left(\frac{144}{25}\right) (5) \)
= \( \frac{1}{3}\pi \frac{144}{5} \)
= \( \frac{144\pi}{15} = \frac{48\pi}{5} \) cm³
Using \( \pi = 3.14 \):
Volume = \( \frac{48 \times 3.14}{5} = \frac{150.72}{5} = 30.144 \) cm³
Surface area of the double cones = Curved Surface Area of cone 1 + Curved Surface Area of cone 2
= \( \pi r_1 l_1 + \pi r_1 l_2 \) (where \( l_1 = AB \) and \( l_2 = AC \))
= \( \pi r_1 (l_1 + l_2) \)
= \( \pi \left(\frac{12}{5}\right) (3 + 4) \)
= \( \pi \left(\frac{12}{5}\right) (7) \)
= \( \frac{84\pi}{5} \) cm²
Using \( \pi = 3.14 \):
Surface Area = \( \frac{84 \times 3.14}{5} = \frac{263.76}{5} = 52.752 \) cm²
In simple words: When a right triangle spins around its longest side, it creates two cones joined at their bases. We first find the length of that longest side. Then, we calculate the common radius and the individual heights of the two cones. Finally, we use these values to determine the combined volume and total surface area of this new shape.

Exam Tip: Always remember that when a right triangle revolves around its hypotenuse, the perpendicular from the right angle to the hypotenuse becomes the common radius of the two cones formed.

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 brick being 22.5 cm x 7.5 cm x 6.5 cm?
Answer:
Volume of cistern = \( l \times b \times h \)
= \( 150 \times 120 \times 110 \) cm³
= \( 1980000 \) cm³
Volume of water already in the cistern = \( 129600 \) cm³
Volume of cistern to be filled = Total volume of cistern - Volume of water
= \( 1980000 - 129600 \)
= \( 1850400 \) cm³
Volume of one brick = \( l \times b \times h \)
= \( 22.5 \times 7.5 \times 6.5 \)
= \( 1096.875 \) cm³
Water absorbed by one brick = \( \frac{1}{17} \) of its own volume = \( \frac{1096.875}{17} \) cm³
Let \( n \) bricks be needed to fill the cistern to the brim.
Total volume of \( n \) bricks = \( n \times 1096.875 \) cm³
Total water absorbed by \( n \) bricks = \( n \times \frac{1096.875}{17} \) cm³
According to the question, the volume of water to be filled plus the volume of water absorbed by bricks should equal the volume displaced by the solid part of the bricks:
Volume to be filled + Water absorbed by \( n \) bricks = Volume of \( n \) bricks (solid part, i.e., not absorbed)
This logic is slightly off in the source; the correct logic is that the volume of the space to be filled must be equal to the volume of the bricks *minus* the water they absorb. So, the effective volume of one brick is its total volume minus the absorbed water.
Effective volume of one brick = \( 1096.875 - \frac{1096.875}{17} = 1096.875 \left(1 - \frac{1}{17}\right) = 1096.875 \times \frac{16}{17} \) cm³
Total effective volume of \( n \) bricks = \( n \times 1096.875 \times \frac{16}{17} \)
This total effective volume must fill the remaining space in the cistern.
So, \( n \times 1096.875 \times \frac{16}{17} = 1850400 \)
\( \implies n = \frac{1850400 \times 17}{1096.875 \times 16} \)
\( \implies n = \frac{31456800}{17550} = 1792.41 \)
Since we can only put a whole number of bricks, the number of bricks that can be put in the cistern without overflowing the water is 1792.
In simple words: First, we find out how much space is left in the cistern. Then, we figure out the actual space each brick takes up, considering it soaks up some water. Finally, we divide the remaining space by the space each brick effectively occupies to find how many bricks fit.

Exam Tip: In problems involving absorption or displacement, distinguish between the total volume of an object and its effective volume (the volume it displaces after considering any absorption).

Question 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 7280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers and each 1072 km long, 75 m wide and 3 m deep.
Answer:
Area of valley = \( 7280 \) km²
Rainfall = \( 10 \) cm = \( \frac{10}{100} \) metre = \( \frac{10}{100 \times 1000} \) km = \( \frac{1}{1000} \) km
Volume of rainfall = Area of river valley \( \times \) Rainfall
= \( 7280 \times \frac{1}{1000} \)
= \( 7.28 \) km³
The source calculation appears to have an error with the rainfall conversion to km, using \( \frac{10}{100 \times 1000} \). Let's re-evaluate based on units.
\( 10 \) cm = \( 0.1 \) m = \( 0.0001 \) km
Volume of rainfall = \( 7280 \text{ km}^2 \times 0.0001 \text{ km} = 0.728 \text{ km}^3 \)
Volume of three rivers = \( 3 \times l \times b \times h \)
Length of one river = \( 1072 \) km
Width of one river = \( 75 \) m = \( \frac{75}{1000} \) km
Depth of one river = \( 3 \) m = \( \frac{3}{1000} \) km
Volume of one river = \( 1072 \times \frac{75}{1000} \times \frac{3}{1000} \)
= \( 1072 \times 0.075 \times 0.003 \)
= \( 1072 \times 0.000225 \)
= \( 0.2412 \) km³
Volume of three rivers = \( 3 \times 0.2412 \) km³ = \( 0.7236 \) km³
Hence, the total rainfall (0.728 km³) and the normal water of three rivers (0.7236 km³) are approximately equivalent.
In simple words: We first calculate the total volume of rainwater that fell on the valley. Then, we find the volume of water in three rivers. By comparing these two volumes, we show that they are nearly the same.

Exam Tip: Pay close attention to unit conversions (cm, m, km) in such problems to avoid errors in calculations. Always convert all measurements to a consistent unit (e.g., km) before performing multiplication.

Question 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 figure).
Answer:

Here are the measurements for the frustum part:
Radius of the top (large) circle \( r_1 = \frac{18}{2} = 9 \) cm
Radius of the bottom (small) circle \( r_2 = \frac{8}{2} = 4 \) cm
Total height of funnel = 22 cm
Height of cylindrical portion = 10 cm
Height of frustum portion \( h = 22 - 10 = 12 \) cm
Slant height of frustum \( l = \sqrt{(r_1 - r_2)^2 + h^2} \)
= \( \sqrt{(9 - 4)^2 + 12^2} \)
= \( \sqrt{5^2 + 12^2} \)
= \( \sqrt{25 + 144} = \sqrt{169} = 13 \) cm
The area of tin sheet required = Curved surface area of the cylindrical portion + Curved surface area of the frustum portion
Curved surface area of cylindrical portion = \( 2\pi r_2 h_c \) (where \( h_c \) is height of cylinder)
= \( 2 \times \pi \times 4 \times 10 = 80\pi \) cm²
Curved surface area of frustum = \( \pi (r_1 + r_2)l \)
= \( \pi (9 + 4) \times 13 \)
= \( \pi \times 13 \times 13 = 169\pi \) cm²
Total area = \( 80\pi + 169\pi = 249\pi \) cm²
Using \( \pi = \frac{22}{7} \):
Total area = \( 249 \times \frac{22}{7} = \frac{5478}{7} = 782 \frac{4}{7} \) cm²
In simple words: To find the total tin sheet needed, we add the curved surface area of the cylindrical part and the curved surface area of the cone's frustum. We calculate these two areas separately using the given dimensions and then combine them for the final answer.

Exam Tip: Remember to calculate the slant height of the frustum first before attempting to find its curved surface area. Also, ensure you only use the relevant curved surface areas, as the top and base of the funnel are not covered by the tin sheet, except for the base of the cylinder which connects to the frustum.

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 are explained.
Answer:
Let \( r_1 \) and \( r_2 \) be the radii of the circular bases, where \( r_1 > r_2 \). Let \( h \) be the height of the frustum of a cone.

Let \( h_1 \) be the height of the large cone LPQ and \( (h_1 - h) \) be the height of the small cone LRS. Let \( l_1 \) be the slant height of cone LPQ and \( (l_1 - l) \) be the slant height of cone LRS. From \( \triangle LNM \) and \( \triangle LTM \) (The image refers to \( \triangle LMQ \) and \( \triangle LNS \)), which are similar (by AA similarity), where M and N are the feet of perpendiculars from L to PQ and RS respectively.
\( \frac{LM}{LN} = \frac{MQ}{NS} = \frac{LQ}{LS} \)
\( \implies \frac{h_1}{h_1 - h} = \frac{r_1}{r_2} = \frac{l_1}{l_1 - l} \)
From \( \frac{h_1}{h_1 - h} = \frac{r_1}{r_2} \):
\( \frac{h_1}{h_1 - h} = \frac{r_1}{r_2} \)
\( \implies r_2 h_1 = r_1 (h_1 - h) \)
\( \implies r_2 h_1 = r_1 h_1 - r_1 h \)
\( \implies r_1 h = r_1 h_1 - r_2 h_1 = h_1 (r_1 - r_2) \)
\( \implies h_1 = \frac{r_1 h}{r_1 - r_2} \) ... (1)
From \( \frac{l_1}{l_1 - l} = \frac{r_1}{r_2} \):
\( \frac{l_1}{l_1 - l} = \frac{r_1}{r_2} \)
\( \implies r_2 l_1 = r_1 (l_1 - l) \)
\( \implies r_2 l_1 = r_1 l_1 - r_1 l \)
\( \implies r_1 l = r_1 l_1 - r_2 l_1 = l_1 (r_1 - r_2) \)
\( \implies l_1 = \frac{r_1 l}{r_1 - r_2} \) ... (2)

(i) Lateral surface area or curved surface area (CSA) of a frustum of cone = CSA of large cone - CSA of small cone
= \( \pi r_1 l_1 - \pi r_2 (l_1 - l) \)
Substitute \( l_1 = \frac{r_1 l}{r_1 - r_2} \) and \( l_1 - l = \frac{r_2 l}{r_1 - r_2} \) (from \( l_1/(l_1-l) = r_1/r_2 \implies l_1-l = l_1 r_2/r_1 \implies (l_1-l) = (\frac{r_1 l}{r_1-r_2}) \frac{r_2}{r_1} = \frac{r_2 l}{r_1-r_2} \)):
= \( \pi r_1 \left(\frac{r_1 l}{r_1 - r_2}\right) - \pi r_2 \left(\frac{r_2 l}{r_1 - r_2}\right) \)
= \( \frac{\pi l}{r_1 - r_2} (r_1^2 - r_2^2) \)
= \( \frac{\pi l}{r_1 - r_2} (r_1 - r_2)(r_1 + r_2) \)
= \( \pi l (r_1 + r_2) \) sq. units

(ii) Slant height of frustum \( l = \sqrt{(r_1 - r_2)^2 + h^2} \)
(iii) Total surface area of frustum of a cone = CSA of frustum of cone + Area of top circle + Area of bottom circle
= \( \pi (r_1 + r_2)l + \pi r_1^2 + \pi r_2^2 \) sq. units
In simple words: To find the formulas for a cone frustum's surface area, we imagine it as a large cone with a smaller cone cut off the top. We use similar triangles to find relationships between heights, radii, and slant heights. Then, we subtract the smaller cone's curved area from the larger cone's curved area to get the frustum's curved area. For total surface area, we add the areas of the two circular bases.

Exam Tip: Clearly draw and label your diagram, ensuring all dimensions for the larger cone, smaller cone, and frustum are correctly identified before starting the derivations. Similarity of triangles is key here.

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.
Answer:
Let \( r_1 \) and \( r_2 \) be the radii of the two circular bases, where \( r_1 > r_2 \). Let \( h \) be the height of the frustum of a cone.

Let \( h_1 \) be the height of the large cone LPD and \( (h_1 - h) \) be the height of the small cone LRS. From \( \triangle LNM \) and \( \triangle LTM \) (The image refers to \( \triangle LMN \) and \( \triangle LNS \)), which are similar (by AA similarity), where M and N are the feet of perpendiculars from L to PQ and RS respectively. \( \frac{LM}{LN} = \frac{MQ}{NS} \)
\( \implies \frac{h_1}{h_1 - h} = \frac{r_1}{r_2} \)
\( \implies r_2 h_1 = r_1 (h_1 - h) \)
\( \implies r_2 h_1 = r_1 h_1 - r_1 h \)
\( \implies r_1 h = h_1 (r_1 - r_2) \)
\( \implies h_1 = \frac{r_1 h}{r_1 - r_2} \) ... (1)
Volume of a frustum of a cone = Volume of large cone LPD - Volume of small cone LRS
= \( \frac{1}{3} \pi r_1^2 h_1 - \frac{1}{3} \pi r_2^2 (h_1 - h) \)
Substitute \( h_1 = \frac{r_1 h}{r_1 - r_2} \) and \( h_1 - h = \frac{r_2 h}{r_1 - r_2} \) (from \( h_1/(h_1-h) = r_1/r_2 \implies h_1-h = h_1 r_2/r_1 \implies (h_1-h) = (\frac{r_1 h}{r_1-r_2}) \frac{r_2}{r_1} = \frac{r_2 h}{r_1-r_2} \)):
= \( \frac{1}{3} \pi r_1^2 \left(\frac{r_1 h}{r_1 - r_2}\right) - \frac{1}{3} \pi r_2^2 \left(\frac{r_2 h}{r_1 - r_2}\right) \)
= \( \frac{1}{3} \pi h \left( \frac{r_1^3}{r_1 - r_2} - \frac{r_2^3}{r_1 - r_2} \right) \)
= \( \frac{1}{3} \pi h \left( \frac{r_1^3 - r_2^3}{r_1 - r_2} \right) \)
Using the identity \( (a^3 - b^3) = (a - b)(a^2 + ab + b^2) \):
= \( \frac{1}{3} \pi h \left( \frac{(r_1 - r_2)(r_1^2 + r_1 r_2 + r_2^2)}{r_1 - r_2} \right) \)
= \( \frac{1}{3} \pi h (r_1^2 + r_1 r_2 + r_2^2) \) cu. units.
In simple words: To get the volume of a frustum, we imagine it as a large cone from which a smaller cone has been removed from the top. We use similar triangles to find the height of the imaginary small cone. Then, we subtract the volume of the small cone from the volume of the large cone. After simplifying the expression, we get the standard formula for the frustum's volume.

Exam Tip: Remember the algebraic identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) as it is crucial for simplifying the frustum volume derivation. Ensure all variables are correctly defined and used consistently throughout the derivation.