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.
Solution: It can be
observed that 1 round of wire will cover 3 mm height of cylinder.
Length of wire required in 1 round = Circumference of base of
cylinder= 2pr =
2p × 5 = 10p
Number of round = Height of cylinder/Demeter of wire =
12/0.3= 40 round
Length of wire in 40 rounds = 40 × 10p = 40x10 x3.14=1256cm=12.56m
Radius of wire = 0.3/2 = 0.15cm
Volume of wire = Area of
cross-section of wire × Length of wire = p (0.15)2 ×
1256 = 3.14 x 0.0225 x 1256 = 88.736 cm3
Mass = Volume × Density = 88.736 ×
8.88 = 787.979 gm
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 p as
found appropriate.)
Solution: The double cone so formed by
revolving this right-angled triangle ABC about its hypotenuse is shown in the
figure.
AC = √ (32 + 42) = 5 cm
Area of DABC = ½ x AC x r =1/2 x BC x AB
Þ 5 x r = 3 x 4 Þ 12/5=2.4cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3 p r2H + /3 p r2h = 1/3 p r2 (H + h)
= 1/3 x 3.14 x 2.4 x 2.4
x 5 = 30.14 cm3
Surface area of double cone = Surface area of cone 1 +
Surface area of cone 2
= p r L + p r l
= p r ( L + l )
= 3.14 x 2.4 (4+3)
= 3.14 x 2.4 x7
= 52.75 cm2
Question 3: A cistern, internally measuring 150 cm × 120
cm × 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 × 7.5 cm × 6.5 cm?
Solution: Volume of cistern = 150 × 120 × 110 = 1980000 cm3
Volume to be filled in cistern = 1980000 – 129600 = 1850400
cm3
Volume of n bricks = 22.5 × 7.5 × 6.5 = 1096.875
As each brick absorbs one-seventeenth of its volume,
Þ Volume
absorbed by 1 bricks = 1/17 x 1096.875
=64.52
Actual volume of bricks without water = 1096.875 - 64.52
=1032.375
Number of bricks = 1850400/1032.375 =1792.37
Therefore, 1792 bricks were placed in the cistern.
Q 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 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: Area of the valley, A = 97280 km2
Level in the rise of water in the valley, h = 10 cm =
(10/100000) km = (1/10000) km
Thus, amount of rain fall in 14 day = Ah = 97280 km2 ×
(1/10000) km = 9.828 km3
Amount of rain fall in 1 day = 9.828 /14 = 0.702km3
Volume of water in 3 rivers = length × breadth × height
=
1072 km × 75 m × 3 m
= 1072 km × (75/1000) km × (3/1000) km = 0.2412 x3
=
0.7236 km3
This shows that the amount of rain fall is approximately
equal to the amount of water in three rivers.
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 the given figure).
Solution: Radius (R)
of upper circular end of frustum
=18/2 = 9 cm
Radius (r) of lower circular end of frustum
= Radius of
circular end of cylindrical part
= 8/2
= 4 cm
Height (H) of frustum
= 22 − 10 = 12 cm
Height (h) of cylindrical = 10 cm
Slant height (l) of frustum
= √(R-r)2 + H2 =√(9-4)2 + 122 =13 cm
Area of tin sheet required = CSA of frustum t + CSA of
cylindrical
= p (R+r)l + 2prh = p [{(9+4)x13} + {2x4x10}]=22/7{169+80}=782.57 cm2
Question 6: Derive the
formula for the volume of the frustum of a cone
Solution: Let ABC be a
cone. A frustum DECB is cut by a plane parallel to its base. Let R and r be the
radii of the ends of the frustum of the cone and h be the height of the frustum
of the cone.
In D
ABG and D
ADF, DF||BG
<A = <A and <AGB = <AFD
∴ D ABG ∼ D ADF (AA similarity)
DF/BG = AF/AG =AD/AB
Þ r/R =
H-h/H=L-l/L
Þif r/R =
H-h/HÞ H =(Rh/R-r)
Volume of frustum of cone = Volume of cone ABC − Volume
of cone ADE
= 1/3xpR2H - 1/3xpr2 (H-h)
= 1/3p
[R2H - r2(H-h)]
= 1/3p [R2(Rh/R-r) - r2{(Rh/R-r)-h}]
= 1/3p [(R3h/R-r) - r2{(Rh –Rh+rh) /(R-r)}]
= 1/3p [(R3h - r3 h) /(R-r)]
= 1/3ph [(R3 - r3 ) /(R-r)]
= 1/3ph [{(R - r )( R2 + r2 -Rr)} /(R-r)]
= 1/3ph ( R2 + r2 -R r)
Further Study links
X Maths Surface Areas and Volumes Test paper-1
X Maths Surface Areas and Volumes Test paper-2
