Let width = x
Then, height = 6x and length = 42x
42x * 6x * x = 16128
x = 4
Number of blocks = =120
Volume of rod = = =
Volume of one spherical ball =r/2
Volume of the spherical ball =
Number of balls=Volumeof rod/volumeof one ball
Volume required in the tank = (200 x 150 x 2) cu.m = 60000 cu.m
Length of water column flown in1 min =(20 x 1000)/60 m =1000/3 m
Volume flown per minute = 1.5 x 1.25 x (1000/3) cu.m= 625 cu.m.
Required time = (60000/625)min = 96min
From the given data,
let the length, breadth and height of the cuboid are m, n, r
m x n = 12
n x r = 20
r x m = 15
Hence, m x n x n x r x r x m = 12 x 20 x 15
mnr = sqrt of (12x20x15) = 60 cub.cm.
v1=
v2=
% Increase in volume = (V2-V1) / (V2*100)
Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.
As speed is same, area of cross sections should be same.
Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12
i.e
Here R = R and R1 = 12/2 = 6
R = 14.696
=> D = X = 14.696 * 2 = 29.3938 cm.
Let l, b the sides of the base, so that the area of the base = lb
remains constant.
Let ?h? be the height of the solid.
Volume of the solid, V = l× b× h.
Since the volume is to be increased by 50%
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