Ex. 4. Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?
Sol. Volume of the wall = (2400 x 800 x 60) cu. cm.
Volume of bricks = 90% of the volume of the wall
=((90/100)*2400 *800 * 60)cu.cm.
Volume of 1 brick = (24 x 12 x 8) cu. cm.
\Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.
Ex. 5. Water flows into a tank 200 m x 160 m througb a rectangular pipe of
1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2
metres?
Sol. Volume required in the tank = (200 x 150 x 2) m3 = 60000 m3.
. .
Length of water column flown in1 min =(20*1000)/60 m =1000/3 m
Ex. 6. Tbe dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is
2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.
Sol. Volume of the metal used in the box = External Volume - Internal Volume
= [(50 * 40 * 23) - (44 * 34 * 20)]cm3
= 16080 cm3
\ Weight of the metal =((16080*0.5)/1000)kg = 8.04 kg.
Ex. 7. The diagonal of a cube is 6Ö3cm. Find its volume and surface area.
Sol. Let the edge of the cube be a.
\Ö3a = 6../3 _ a = 6.
So,Volume = a3 = (6 x 6 x 6) cm3 = 216 cm3.
Surface area = 6a2 = (6 x 6 x 6) cm2 == 216 cm2.
Ex. 8. The surface area of a cube is 1734 sq. cm.Find its volume.
Sol. Let the edge of the cube bea. Then,
6a2 = 1734 Þ a2 = 289 => a = 17 cm.
\ Volume = a3 = (17)3 cm3 = 4913 cm3.
Ex. 9. A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.
Sol. Volume of the block = (6 x 12 x 15) cm3 = 1080 cm3.
Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.
Volume of this cube = (3 x 3 x 3) cm3 = 27 cm3.
Number of cubes = 1080/27 = 40.
Ex.l0. A cube of edge 15 cm is immersed completely in a rectangular vessel containingwater . If the dimensions of the base of vessel are 20 cm x 15 cm, find the rise in water level.
Sol. Increase in volume = Volume of the cube = (15 x 15 x 15) cm3.
\Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.
Ex. 11. Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new
cube. Find the surface area of the cube so formed.
Sol. Volume of new cube = (13 + 63 + 83) cm+ = 729 cm3.
Edge of new cube = 3Ö729 cm = 9 cm.
\Surface area of the new cube = (6 x 9 x 9) cm2 = 486 cm2.
Ex. 12. If each edge of a cube is increased by 50%, find the percentage increase inIts surface area."
Sol. Let original length of each edge = a.
Then, original surface area = 6a2.
New edge = (150% of a) = (150a/100) = 3a/2
New surface area = 6x (3a/2)2 = 27a2/2
Increase percent in surface area = ((15a2) x ( 1 ) x 100)% = 125%
26a2
Ex. 13. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their
Ex. 21. Find the length of canvas 1.25 m wide required to build a conical tent ofbase radius 7 metres and height 24 metres.
Sol. Here, r = 7m and h = 24 m.
So,l = Ö(r2 + h2) = Ö(72 + 242) = Ö (625) = 25 m.
Area of canvas = Õrl=((22/7)*7*25)m2 =550 m2 .
Length of canvas = (Area/Width) = (550/1.25) m = 440 m.
Ex. 22. The heights of two right circular cones are in the ratio 1 : 2 and the perimeters of their bases are in the ratio 3 : 4. Find the ratio of their volumes.
Sol. Let the radii of their bases be r and R and their heights be h and 2h respectively.
Then,(2Õr/2ÕR)=(3/4) Þ R=(4/3)r.
\ Ratio of volumes = (((1/3)Õr2h)/((1/3)Õ(4/3r)2(2h)))=9: 32.
Ex. 23. The radii of the bases of a cylinder and a cone are in the ratio of 3 : 4 and It heights are in the ratio 2 : 3. Find the ratio of their volumes.
Sol. Let the radii of the cylinder and the cone be 3r and 4r and their heights be 2h and
3h respectively.
:. Volume of cylinder = Õ x (3r)2 * 2h = 9/8 = 9 : 8.
Volume of cone(1/3)Õr2 *3h
Ex. 24. A conical vessel, whose internal radius is 12 cm and height 50 cm, is fullof liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the liquid rises in the cylindrical vessel.
Sol. Volume of the liquid in the cylindrical vessel
Volume of 1 small lead ball = ((4/3)Õ*(1/2)*(1/2)*(1/2)) cm3 = Õ/6 cm3.
\ Number of lead balls = (288Õ*(6/Õ)) = 1728.
Ex.28.How many spherical bullets can be made out of a lead cylinder 28cm high and with radius 6 cm, each bullet being 1.5 cm in diameter ?
Sol. Volume of cylinder = (∏ x 6 x 6x 28 ) cm3= ( 9∏/16) cm3.
Number of bullet =Volume of cylinder= [(36 x 28)∏ x 16] /9∏ = 1792.
Volume of each bullet
Ex.29.A copper sphere of diameter 18cm is drawn into a wire of diameter 4 mm Find the length of the wire.
Sol. Volume of sphere = ((4∏/3) x9 x 9 x 9 ) cm3 = 972∏ cm3
Volume of sphere = (∏ x0.2 x 0.2 x h ) cm3
\ 972∏ = ∏ x (2/10) x (2/10) x h Þ h = (972 x 5 x 5 )cm = [( 972 x 5 x5 )/100 ]m
= 243m
Ex.30.Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and the radii of their bases 2.1 cm each, have been melted together and recast into a sphere. Find the diameter of the sphere.
Sol. Volume of sphere = Volume of 2 cones
= (1 ∏ x (2.102)x 4.1 + 1 ∏ x (2.1)2 x 4.3)
33
Let the radius of sphere be R
\(4/3)∏R3 = (1/3)∏(2.1)3orR = 2.1cm
Hence , diameter of the sphere = 4.2.cm
Ex.31.A Cone and a sphere have equal radii and equal volumes. Find the ratio of the sphere of the diameter of the sphere to the height of the cone.
Sol. Let radius of each be R and height of the cone be H.
Ex.32.Find the volume , curved surface area and the total surface area of a hemisphere of radius 10.5 cm.
Sol. Volume = (2∏r3/3) = ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))cm3
= 2425.5 cm3
Curved surface area = 2∏r3 = (2 x (22/7) x (21/2) x (21/2))cm2
=693 cm2
Total surface area = 3∏r3 = (3 x (22/7) x (21/2) x (21/2))cm2
=1039.5 cm2.
Ex.33.Hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl ?
Sol. Volume of bowl = ((2∏/3) x9 x 9 x 9 ) cm3 = 486∏ cm3.
Volume of 1bottle = (∏ x(3/2) x (3/2) x 4 ) cm3 = 9∏ cm3
Number of bottles = (486∏/9∏) = 54.
Ex34.A Cone,a hemisphere and a cylinder stand on equal bases and have the same height.Find ratio of their volumes.
Sol. Let R be the radius of each
Height of the hemisphere = Its radius = R.
\Height of each = R.
Ratio of volumes = (1/3)∏ R2 x R : (2/3)∏R3 : ∏ R2 x R = 1:2:3