Let original radius = r and height = h
Original volume = (1/3)?r2h
New radius = 2r and new height = 2h
? New volume = 1/3 x ?(2r)2 x 2h = 8/3?r2h
Required ratio = 8/3?r2h : 1/3?r2h = 8 : 1
Volume of Earth dug out = 5 x 4.5 x 2.1 = 47.25 m2
Area over which Each is spread = 13.5 x 25 - 5 x 4.5
= 33.75 - 220 = 11.75 m
Rise in level = 47.25/11.75 = 4.02 m
Let r be the radius of the sphere and a be the side of the cube
? 4? r2 = 6a2
r = ?6 a / 2??
? r3 = 6?6a3 / 8???
? Ratio of volume = (4/3 ?r3)/a3
? Required ratio = ?6 / ??
The volume that is needed to built by the bricks
= Volume of outer dimensions of sump - Volume of sump
= 6.2 x 5.2 x 4.2 - 6 x 5 x 4
= 135.408 - 120 = 15.405 m3
Volume of one brick = 20/100 x 10/100 x 5/100 = 0.01 m3
? Total number of bricks required to build the sump
= 15.408 / 0.001 = 15408
Total area of four walls = 2h(l + b)
Perimeter of the celling = 2l + 2b = 2(l + b)
Perimeter of the ceiling in percentage = 2(l +b ) x 100/2h(l + b) %
= 100/h %
Volume of the tank = 2.6 cu m
Base area of the tank = 6500 cm = 0.65 sq m
? Depth of the tank = 26/0.65 = 4m
The decrease in volume is given [ x + y + xy/100 ] %
Here , x = y = -24
= [(-24) + (-24) + (-24) x (-24)/100] %
= [-48 + 576/100] % = -48 + 5.76 = - 42.24 %
(-ve sign denotes decrease in volume)
The net decrease in the surface area of the cube is given by
[x + y + xy/100] %
Here, x = y = 19%
= [-19 + (-19) + (-19) x (-19)/100] %
= [-38 + 361/100] %
= [-38 + 3.61] %= -34.39% (-ve sign implies for decrease)
From the formula the net change in volume is given as [x + y + z + (xy + yz + zx)/100 + xyz/(100)2 ] %
Here, x = 200, y = z = -50 (-ve sign for decrease)
= [ 200 + (-50) + (-50) + (200) x (-50) + (-50) x (-50) + (-50) x (200)/100 + (200) x (-50) x (-50)/(100)2 ] %
= 100 + (-10000 + 2500 - 10000)/100 + 500000/100 x 100 ] %
= [ 100 + (-17500)/100 + 50]%
= (150 - 175)% = -25% (-ve sign implies for a decrease )
So, the volume is decreased by 25%.
Volume of the spherical ball = Volume of the water displaced
? 4/3 ?r3 = ?(12)2 x 6.75
? r3 = 144 x 6.75 x 3/4 = 729
? r = 9 cm
Volume of the sphere = Volume of the water displaced
Let the required height to which the water rises be h.
Then, ?r12h = 4/3 ?r23
? 16 h = 4/3 x 27 ? h = 36/16 = 9/4 cm
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