Surface Areas & Volume
Mathematics Class Tenth
NCERT EXERCISE 13.5 (Optional)
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 `cm^3`.
Solution:
Given, Length of cylinder = 12 cm
Diameter of cylinder = 10 cm
∴ radius of cylinder `=10/2= 5\ cm`
Diameter of wire = 3 mm `=3/10=0.3\ cm`
∴ Radius of wire `=(0.3)/2=0.15\ cm`
Density of wire `=8.88\ g//cm^3`
Since, diameter of wire `= 0.3\ cm`
Thus, length of cylinder cover by wire in one (1) round = 0.3 cm
Thus, Number of rounds to cover the cylinder `=(text{height of cylinder})/(text{Diameter of wire})`
`=12/0.3 = 40` rounds
Now, Circumference of cylinder `= 2pir`
Now, length of wire in one (1) round = Circumference of cylinder
`=2pi r = 2 xx22/7xx5\ cm`
Or, Length of wire in one (1) round `= 220/7 cm`
Thus, length of wire in 40 rounds `=220/7 xx 40\ cm`
`=8800/7 = 1257.14 cm`
Area of cross section of wire `= pir^2`
`=pi(0.15\ cm)^2`
`=pi xx 0.0225\ cm^2`
`=0.0225\ pi\ cm^2`
Now, Volume of wire = Area of cross section of wire `xx` length of wire
`=0.0225xx22/7\ cm^2 xx 1257.14\ cm`
`=(622.284)/7 = 88.90\ cm^3`
∵ `1\ cm^3` of wire contains `8.88\ g` mass of wire
∴ `88.90\ cm^3` of wire contains mass of wire `=8.88 xx88.90\ g`
`=789.43\ g` of wire
Thus, mass of wire `=789.43\ g`. Answer
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)
Solution:
Given, Sides of triangle = 3 cm and 4 cm
Let assume, ABC is a right triangle, in which
`/_ ABC = 90^0`
Thus, side AC is hypotenuse
And, AB = 3 cm and BC = 4 cm
Since, ABC is a right triangle
Thus, `(AC)^2 = (BC)^2+(AB)^2`
`=>AC^2 = (4\ cm)^2 + (3\ cm)^2`
`=> AC^2 = (16+9) cm^2`
`=>AC^2 = 25\ cm^2`
`:. AC = 5\ cm`
Now, Area of `triangle` ABC `= 1/2 xx AB xx BC` -----------(i)
And, Area of `triangle` ABC `=1/2 xx AC xx OB` ------------(ii)
Thus, from equation (i) and (ii)
`1/2 xx AB xx BC = 1/2xx AC xx OB`
After substituting the values of AB, BC and AC
`=>1/2 xx 3\ cm xx 4\ cm = 1/2 xx 5\ cmxx OB`
`=>1/2 xx 12\ cm^2 = 1/2 xx 5\ cm\ OB`
`=>5\ cm\ OB = 12\ cm^2`
[∵ `1/2` is in both side, and hence cancelled]
`=>OB = (12\ cm^2)/(5\ cm)`
`=>OB = 2.4\ cm`
Let, This right triangle ABC is made to revolve about its hypotenuse AC, which forms two cone DAB and BCD
Thus, volume and surface are of this double cone =?
Here, we have, slant height of cone1, `l_1 = 4\ cm`
And slant height of cone2, `l_2 = 3\ cm`
And radius of both of the cones = OB = 2.4 cm
Now, Volume of cone `= 1/3\ pi r^2h`
Thus, volume of cone1 `=1/3\ pi(2.4)^2xx h_1`
Or volume of cone1 `=1/3\ pi xx5.76xxh_1\ cm^3`
And volume of cone2 `=1/3\ pi\ (2.4)^2xxh_2\ cm^3`
Or, volume of cone2 `=1/3\ pi\ 5.76xxh_2\ cm^3`
Thus, volume of both the cones = volume of cone1 + volume of cone2
`= 1/3\ pi xx5.76xxh_1\ cm^3 + 1/3\ pi\ 5.76xxh_2\ cm^3`
`=1/3\ pixx5.76(h_1+h_2)\ cm^3`
`=1/3\ pixx5.76(5)\ cm^3`
[∵ `h_1+h_2=AO+OC=AC=5\ cm`]
`=1/3xx3.14 xx 28.8\ cm^3`
`=(90.432)/3\ cm^3`
`=30.14\ cm^3`
Thus, volume of both of the cone `=30.14\ cm^3`
Now, curved surface area of cone `=pi \ r \ l` sq. unit
Thus, curved surface area (CSA) of both of the resulting cone = curved surface area (CSA) of cone1 + curved surface area (CSA) of cone2
`=pi\ 2.4xx4\ cm^2 + pi\ 2.4xx 3\ cm^2`
`= pi\ 2.4(4+3) cm^2`
`= 22/7xx2.4xx7\ cm^2`
`= 22xx2.4\ cm^2`
`= 52.8\ cm^2`
Thus, volume and curved surface area of resulting double cone are `30.14\ cm^3` and `52.8\ cm^2` respectively. Answer
Question: 3. A cistern, internally measuring 150 cm ? 120 cm ? 110 cm, has 129600 `cm^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 22.5 cm ? 7.5 cm ? 6.5 cm?
Solution:
Given, dimension of cistern `=150\ cmxx120\ cmxx110\ cm`
Dimension of brick `=22.5\ cmxx7.5\ cmxx6.5\ cm`
Water absorbs by one brick `=1/17xx` volume of brick
Volume of water in cister `=129600\ cm^3`
Number of bricks which full the cistern up to brim=?
Volume of cistern `=lxxbxxh`
`=150\ cmxx120\ cmxx110\ cm`
`=1980000\ cm^3`
Now volume of one brick `=lxxbxxh`
`=22.5\ cmxx7.5\ cmxx6.5\ cm`
`=1096.875\ cm^3`
As given, water absorbs by one brick `=1/17xx` volume of brick
Thus, volume remains in the brick `=1-1/17=16/17` of volume of brick
`=16/17xx1096.875\ cm^3`
`=1032.35\ cm^3`
Volume of cistern vacant after water = volume of cistern – volume of water
`=1980000\ cm^3 - 129600\ cm^3`
`=1850400\ cm^3`
∵ `1032.35\ cm^3` volume of cistern is covered by one brick
∴ `1\ cm^3` of volume of cistern is covered by `1/(1032.35\ cm^3)` bricks
∴ `1850400\ cm^3` of cistern is covered by `1/(1032.35\ cm^3)xx1850400\ cm^3` bricks
`=1792.41` bricks
Thus, number of bricks put into the cistern without overflowing of water =1792. Answer
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 each 1072 km long, 75 m wide and 3 m deep.
Solution:
Given, rainfall in the river valley = 10 cm `=10/100=01\ m`
Given, Area of valley `=7280\ km^2`
`=7280xx1000xx1000\ m^2`
Thus, volume of water due to rainfall in river valley
`=7280xx1000xx1000xx0.1\ m^3`
`=728xx1000xx1000\ m^3`
Now, given, length of the river `=1072\ km=1072xx1000\ m`
Width of the river `=75\ m`
And, depth of the river `=3\ m`
Thus, volume of water in one river `=1072xx1000xx75xx3\ m^3`
`=1072xx1000xx225\ m^3`
Or, volume of water in one river`=241200xx1000\ m^3`
∴ volume of water in three (3) rivers `=3xx` volume of water in one river
`=3xx241200xx1000\ m^3`
`=723600xx1000\ m^3`
Thus, volume of water due to rain fall in river valley `~~` volume of water in three river
`728xx1000xx1000\ m^3\ ~~\ 723600xx1000\ m^3`
`=>7xx100xx1000xx1000\ m^3 ~~ 7xx100xx1000xx1000\ m^3`
`7xx10^8\ m^3 ~~ 7xx10^8m^3` Proved
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).
Solution:
Given, height of funnel = 22 cm
And height of cylindrical portion of funnel = 10 cm
Thus, height of cone = height of funnel – height of cylindrical portion of funnel
`=22-10=12\ cm`
Again, given, diameter of cylindrical portion of funnel = 8 cm
∴ radius of cylindrical portion of funnel `= 8/2=4\ cm`
Again given, diameter of top of the funnel = 18 cm
Thus, radius of top of the funnel, i.e. bigger radius of frustum of cone `= 18/2=9\ cm`
Now, smaller radius of frustum of cone = radius of cylindrical portion of funnel = 4 cm
Area of Tin sheet required to make the funnel =?
Slant height of frustum of cone, `l =sqrt(h^2+(r_1-r^2)^2)`
`=>l=sqrt(12^2+(9-4)^2)\ cm `
`=>l=sqrt(144+(5)^2)\ cm `
`=>l = sqrt(144+25)\ cm`
`=>l = sqrt(169)\ cm`
`=> l = 13\ cm`
Now, curved surface area of frustum of cone `=pi (r_1+r_2)l`
`=pi (9+4)13\ cm^2`
`=pi xx13xx13\ cm^2`
`=169\ pi\ cm^2`
Now, curved surface area of cylinder `=2\ pi\ r\ h`
Thus, curved surface area of cylindrical part of funnel
`=2xx pi\ 4xx10\ cm^2`
`=80\ pi\ cm^2`
Thus, total curved surface area of funnel = curved surface area of frustum of cone(upper part of funnel) + curved surface area of cylindrical part of funnel
`=169\ pi\ cm^2+80\ pi\ cm^2`
`=pi(169+80)\ cm^2`
`=22/7xx249\ cm^2`
`=5478/7\ cm^2`
`=782\ 4/7\ cm^2` or `782.57\ cm^2`
Thus, area of tin sheet required to make the funnel `=782\ 4/7\ cm^2` or `782.57\ cm^2` Answer
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:
Let ABC is a cone.
From this cone, a frustum DECB is cut by a plane which is parallel to its base.
Let `r_1` is the radius of lower end of frustum.
And `r_2` is the radius of upper end of frustum
Let, `h_1` is the height of the cone.
And, `h` is the height of the frustum of the cone.
Thus, height of the cone after cutting of frustum `=h_1-h`
Now, in `triangle` ABG and `triangle` ADF,
`DF||BG`
`:. triangle ABG` ~ `triangle ADF`
∴ `(DF)/(BG)=(AF)/(AG)=(AD)/(AB)`
`=>r_2/r_1=(h_1-h)/h_1=(l_1-l)/l_1`
`=>r_2/r_1=1-l/l_1`
`=>l/l_1 = 1-r_2/r_1`
`=>l/l_1 = (r_1-r_2)/r_1`
`=>l_1/l=r_1/(r_1-r_2)`
`=>l_1 = (r_1\ l)/(r_1-r_2)` -----(i)
Now, curved surface area of frustum DECB = Curved surface area of cone ABC – Curved surface area of cone ADE
`=pi\ r_1\ l_1 - pi\ r_2(l_1-l)`
By substituting the value of `l_1` from equation (i)
`=pi\ r_1((lr_1)/(r_1-r_1)) - pi\ r_2((r_1l)/(r_1-r_2)-l)`
`=(pi\ r_1^2\ l)/(r_1-r_2)-pi\ r_2((r_1l-r_1l+r_2l)/(r_1-r_2))`
`=(pi\ r_1^2\ l)/(r_1-r_2) - (pi\ r_2^2\ l)/((r_1-r_2)`
`=pi\ l((r_1^2-r_2^2)/(r_1-r_2))`
`=pi\ l[((r_1+r_2)(r_1-r_2))/(r_1-r_2)]`
`=pi\ l (r_1+r_2)`
Thus, curved surface area (CSA) of frustum of a cone `=pi\ l (r_1+r_2)`
Now, Total surface area (TSA) of frustum = curved surface area of frustum + Area of upper circular end + Area of lower circular end
`=pi(r_1+r_2)+pir_2^2+pir_1^2`
`=pi[(r_1+r_2)l+r_1^2+r_2^2]`
∴ Total surface area (TSA) of frustum `=pi[(r_1+r_2)l+r_1^2+r_2^2]`
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:
Let ABC is a cone.
From this cone, a frustum DECB is cut by a plane which is parallel to its base.
Let `r_1` is the radius of lower end of frustum.
And `r_2` is the radius of upper end of frustum
Let, `h_1` is the height of the cone.
And, `h` is the height of the frustum of the cone.
Thus, height of the cone after cutting of frustum `=h_1-h`
Now, in `triangle` ABG and `triangle` ADF,
`DF||BG`
`:. triangle ABG` ~ `triangle ADF`
∴ `(DF)/(BG)=(AF)/(AG)`
`=>r_2/r_1=(h_1-h)/h_1`
`=>r_2/r_1 = 1-h/h_1`
`=>h/h_1 = 1-r_2/r_1`
`=>h/h_1 = (r_1-r_2)/r_1`
`=>h_1/h = r_1/(r_1-r_2)`
`=> h_1 = (r_1\ h)/(r_1-r_2)` --------(i)
Now, volume of frustum of cone = Volume of cone ABC – Volume of cone ADE
`=1/3pir_1^2h_1-1/3pir_2^2-(h_1-h)`
`=1/3\ pi[r_1^2h_1-r_2^2(h_1-h)]`
After substituting the value of `h_1` from equation (i)
`=pi/3[r_1^2((hr_1)/(r_1-r_2))-r_2^2((hr_1)/(r_1-r_2)-h)]`
`=pi/3[((hr_1^3)/(r_1-r_2))-r_2^2((hr_1-hr_1+hr_2)/(r_1-r_2))]`
`=pi/3[((hr_1^3)/(r_1-r_2))-r_2^2((hr_2)/(r_1-r_2))]`
`=pi/3[(hr_1^3)/(r_1-r_2)-(hr_2^3)/(r_1-r_2)]`
`=pi/3h[(r_1^3)/(r_1-r_2)-(r_2^3)/(r_1-r_2)]`
`=pi/3h[(r_1^3-r_2^3)/(r_1-r_2)]`
`=pi/3h[((r_1-r_2)(r_1^2+r_2^2+r_1*r_2))/(r_1-r_2)]`
`=1/3pi\ h [r_1^2+r_2^2+r_1*r_2]`
Thus, Volume of frustum of cone `=1/3pi\ h [r_1^2+r_2^2+r_1*r_2]`
Reference: