Surface Areas & Volume
Mathematics Class Tenth
NCERT Exercise 13.2
Unless stated otherwise, take `pi = 22/7`
Question: 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of `pi`.
Solution:
Given, Radius of cone, `r=1\ cm`
Height of cone `=1\ cm`
Radius of hemisphere, `r=1cm`
Volume of solid =?
Volume of cone `=1/3 pi r^2 h`
`=1/3 pi (1\ cm)^2 xx 1 cm`
`= 1/3 pi cm^3`
Volume of hemisphere `= 2/3 pi r^3`
`= 2/3 pi (1\ cm)^3`
`=2/3 pi cm^3`
Thus, volume of the solid = volume of cone + volume of hemisphere
` = 1/3 pi cm^3 + 2/3 pi cm^3`
`=pi (1/3+2/3) cm^3`
`= pi ((1+2)/3) cm^3`
`= pi 3/3 cm^3`
`= pi cm^3`
Thus, volume of the given solid `= pi cm^3` Answer
Question: 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is `3\ cm` and its length is `12\ cm`. If each cone has a height of `2\ cm`, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Given, length of the model `=12\ cm`
Height of each cone ` = 2\ cm`
Diameter of cylindrical model `=3\ cm`
Thus, radius of cylinder `=1.5\ cm`
And radius of cone` = 1.5\ cm`
Volume of air contained in the model =?
Here volume air contained in the model = volume of the model
Height of the cylinder = total height of the model – `2xx` height of one cone
`= 12\ cm - 2 xx 2\ cm`
`= (12 - 4) cm`
∴ height or length of cylinder (excluding cones) `=8\ cm`
Now, Volume of the cylinder `= pi r^2 h`
`= pi xx (1.5\ cm)^2 xx 8\ cm`
`= pi xx 2.25\ cm^2 xx 8\ cm`
`= 18\ pi \ cm^3`
Volume of cone ` = 1/3 pi r^2 h`
`= 1/3 pi xx (1.5\ cm)^2 xx 2 cm`
`= 1/3 pi xx 2.25\ cm^2 xx 2\ cm`
`= 1/3 pi 4.5 cm^3`
`= 1.5 pi cm^3`
∴ volume of two (2) cone `= 2xx 1.5 pi\ cm^3`
`= 3 pi\ cm^3`
Now, volume of the model = volume of cylinder + volume of 2 (two) cones
`= 18\ pi cm^3 + 3 pi\ cm^3`
`= pi (18+3) cm^3`
`= 22/7 xx 21 cm^3`
`= 22 xx 3\ cm^3`
`= 66 cm^3`
Thus, volume of air contained in the model `= 66\ cm^3` Answer
Question: 3. A gulab jamun , contains sugar syrup up to about `30%` of its volume. Find approximately how much syrup would be found in 45 gulab jamums, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm each.
Solution:
Given, length of gulab jamum `=5\ cm`
Diameter of gulab jamun `= 2.8\ cm`
Thus, radius of gulaba jamun `=2.8/2 = 1.4\ cm`
Radius of hemispherical ends of gulab jamun `=1.4\ cm`
Since, there are two hemispherical ends of each gulab jamun, thus length of two hemisphere, i.e. diameter of one sphere `=2.8 \ cm`
Therefore, length of cylindrical portion of gulab jamun = total length of gulab jamun – length of two hemisphere
`= 5 \ cm - 2.8\ cm`
Thus, length of cylindrical portion of gulab jamun (excluding hemispherical ends) `=2.2\ cm`
And given, radius of gulab jamun, `r = 1.4\ cm`
Now, volume of cylindrical portion of gulab jamun `= pi r^2 h`
`=pi xx (1.4\ cm)^2 xx 2.2\ cm`
`=pi xx 1.96\ cm^2 xx 2.2\ cm`
Or, volume of cylindrical portion of gulab jamun `=4.312 pi \ cm^3`
Volume of hemisphere `= 2/3 pi r^3`
`=2/3 pi (1.4\ cm)^3`
`= pi 2/3 xx 2.744 cm^3`
`= pi 1.83 cm^3 `
Or, Volume of hemisphere `= pi 1.83 cm^3 `
Thus, volume of 2 (two) hemisphere `= 2 xx pi 1.83 cm^3 = 3.659 pi cm^3`
Now, volume of one (1) gulab jamun = volume of cylindrical portion + volume of two hemisphere
`= 4.312 pi cm^3 + 3.659 cm^3`
`=pi (4.312+3.659) cm^3`
`=22/7 xx 7.971 cm^3`
Volume of one gulab jamun `=25.05 cm^3`
Thus, volume of syrup in one gulab jamun `=30% xx 25.05 cm^3`
[∵ given one gulab jamun contains about 30% of syrup]
`=30/100 xx 25.05 cm^3`
Or, Volume of syrup in one gulab jamun `=7.515 cm^3`
Thus, syrup contains in 45 gulab jamun `= 45xx 7.515 cm^3`
`= 338.175 cm^2 ~~ 338 cm^3`
Thus, volume of syrup in 45 gulab jamun `~~338\ cm^3` Answer
Question: 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are `15\ cm` by `10\ cm` by `3.5\ cm`. The radius of each of the depressions is `0.5\ cm` and depth is `1.4\ cm`. Find the volume of wood in the entire stand.
Solution:
Given, length `(l)`, width `(b)` and depth `(h)` of the cuboid `=15\ cm, 10\ cm` and `3.5\ cm` respectively
Number of depression in wooden cuboid `=4`
Radius of each depression, `r=0.5\ cm`
Depth of each depression, `h = 1.4\ cm`
Thus, volume of wooden stand =?
Volume of entire stand = volume of wooden cuboid – volume of 4 depressions in conical shape
Now, volume of cuboid `= l xx b xx h`
`= 15\ cm xx 10\ cm xx 3.5\ cm`
Or, Volume of wooden cuboid `= 525\ cm^3`
Now, Volume cone `=1/3 pi r^2h`
∴ volume of one conical depression `=1/3 pi (0.5\ cm)^2 xx 1.4\ cm`
`= 1/3 xx 22/7 xx 0.25\ cm^2 xx 1.4\ cm`
`=1/3 xx 22xx 0.25 xx 0.2\ cm^3`
Or, volume of one depression `=0.156\ cm^3`
∴ volume of four (4) depressions `=4 xx 0.156 cm^3 = 0.624 cm^3`
Now, volume of entire wooden stand = volume of wooden cuboid – volume of 4 (four) conical depression
`= 525 - 0.624 = 524.376 cm^3`
`~~ 524\ cm^3`
Thus, volume of given wooden stand `~~ 524\ cm^3` Answer
Question: 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Given, height of cone `=8\ cm`
Radius of cone `=5\ cm`
Radius of spherical lead `=0.5\ cm`
Water flows out when lead shots `=1/4` of total water in cone
Thus, volume of `1/4(th)` water =?
Volume of cone `=1/3 pi r^2 h`
∴ volume of water in cone =1/3 pi (5\ cm)^2 xx 8\ cm`
`=1/3 xx pi xx 25\ cm^2 xx 8\ cm`
`=200/3 pi\ cm^3`
Thus, volume of `1/4(th)` of water `=1/4 xx 200/3 pi\ cm^3`
`=50/3 pi cm^3`
∴ Volume of `1/4(th)` of water `=50/3 pi cm^3`
Now, volume of sphere `=4/3 pi r^3`
Thus, volume of one spherical lead `=4/3 xx pi xx (0.5\ cm)^3`
`=4/3 pi 0.125 cm^3=0.5/3 pi\ cm^3`
Or, volume of one spherical lead `=0.5/3 pi\ cm^3`
Thus, number of leads `=(text{volume of ? water})/(text{volume of one lead})`
`=50/3 pi cm^3 -: 0.5/3 pi\ cm^3`
`=(50pi)/3 xx 3/(0.5pi) =100`
Thus, total number of leads `=100` Answer
Question: 6. A solid iron pole consists of a cylinder of height `220\ cm` and base diameter `24\ cm`, which is surmounted by another cylinder of height `60\ cm` and radius `8\ cm`. Find the mass of the pole, given that `1\ cm^3` of iron has approximately `8\ g` mass. (Use `pi =3.14`)
Solution:
Given, height of lower cylindrical pole, `h=220\ cm`
Diameter of lower pole `=24\ cm`
∴ radius of lower pole, `r=24/2=12\ cm`
Height of upper cylindrical pole `h=60\ cm`
Radius of upper cylindrical pole `r=8\ cm`
`1\ cm^3` pole contains mass of iron `=8\ g`
Thus, mass of entire pole =?
Volume cylinder `= pi r^2h`
Thus volume of pole = volume of lower pole + volume of upper pole
`=pi xx (12\ cm)^2 xx 220\ cm + pi xx (8\ cm)^2 xx 60\ cm`
`=pi(144 cm^2 xx 220 cm+64\ cm^2 xx 60\ cm)`
`=pi (31680 cm^3+3840 cm^3)`
`= 35520\ pi\ cm^3`
Thus volume of entire pole `= 35520\ pi\ cm^3`
Now, since `1\ cm^3` of pole contains `8\ g` of mass
∴ `35520\ pi\ cm^3` of pole contains `= 35520\ pixx 8\ g` of mass
`=35520 xx 3.14 xx 8\ g`
[∵ given `\ pi=3.14`]
`=892262.4\ g`
`=892.2624\ kg ~~ 892.26\ kg`
Thus, approximately mass of pole `=892.26\ kg` Answer
Question: 7. A solid consisting of a right circular cone of height `120\ cm` and radius `60\ cm` standing on a hemisphere of radius `60\ cm` is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is `60\ cm` and its height is `180\ cm`.
Solution:
Given, height of cone `=120\ cm`
Radius of hemisphere `=60\ cm`
Height of cylinder `=180\ cm`
Radius of cylinder `=60\ cm`
Volume of water left in cylinder after immersing the solid=?
Volume of cylinder `=pi r^2 h`
`= pi (60\ cm)^2 xx 180\ cm`
`=pi xx 3600\ cm^2 xx 180\ cm`
Or, volume of cylinder`=pi 648000\ cm^3`
∴ volume of water in the cylinder `=pi 648000\ cm^3`
Volume of cone `=1/3 pi r^2h`
`=1/3 pi (60\ cm)^3xx120\ cm`
`=1/3 pi xx 3600\ cm^2 xx 120\ cm`
`= pi xx 1200\ cm^2 xx 120\ cm`
Or, volume of cone `= pi xx 144000\ cm^3`
Volume of hemisphere `=2/3 pi r^3`
`=2/3 pi (60\ cm)^3`
`=2/3 pi xx 216000\ cm^3`
`=2 xx 72000 pi \ cm^3`
Or volume of hemisphere `=144000 pi\ cm^3`
Thus, volume of resulting solid made of cone and hemisphere = volume of cone + volume of hemisphere
`=144000 pi \ cm^3 + 144000 pi\ cm^3`
Or, Volume of solid `=288000 pi\ cm^3`
Now, volume of water left in the cylinder = volume of water in the cylinder – volume of resulting solid
`=pi 648000\ cm^3 ? 288000 pi\ cm^3`
`=360000 xx 22/7\ cm^3`
`=1131428.571\ cm^3`
`~~1.131\ m^3`
Thus, volume of water left in the cylinder `~~1131428\ cm^3 or ~~ 1.13\ m^3` Answer
Question: 8. A spherical glass vessel has a cylindrical neck `8\ cm` long, `2\ cm` in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be `345\ cm^3`. Check whether she is correct, taking the above as the inside measurements, and `pi = 3.14`.
Solution:
Given, diameter of spherical part of vessel `= 8.5\ cm`
Thus, radius of vessel `=8.5/2 = 4.25\ cm`
Height of cylindrical neck of vessel `=8\ cm`
Diameter of neck of the vessel `=2\ cm`
∴ radius of neck `=2/2=1\ cm`
Thus, volume of water vessel contains =?
Here, volume of vessel = volume of water it contains.
Now, volume cylinder `= pi r^2 h`
∴ volume of neck `=pi (1\ cm)^2 xx 8\ cm`
`=pi xx 1\ cm^2 xx 8\ cm`
Or, volume of cylindrical neck of vessel `=8 pi\ cm^3`
Now, volume of sphere `=4/3 pi r^3`
Thus, volume of spherical part of vessel `= 4/3 pi (4.25\ cm)^3`
`=4/3 xx 76.77 pi\ cm^3`
Volume of spherical part of vessel `=102.354 pi\ cm^3`
Now volume of vessel = volume of neck + volume of spherical part
`=8 pi\ cm^3 + 102.354 pi\ cm^3`
`=pi (8+102.354)\ cm^3`
`=3.14 xx 110.354\ cm^3`
`=346.51\ cm^3`
Thus, volume of water vessel contains `=346.51\ cm^3`
Volume found by child was not correct. Correct volume is `346.51\ cm^3` Answer
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