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.
Given length of iron rod =7 mts and and diameter of rod = 2 cms
=> r = 1/100 mts
Therefore, Volume of one rod = cu. m = cu. m = cu. m
Given volume of iron = 0.88 cu. m
Therefore, Number of rods = = 400.
capacity of a tank = volume of tank
= [ (8 x 100) x (6 x 100) x (2.5 x 100) ]/ 1000 litres
= 120000 litres
R =OA = OB = Radius of hemisphere.
= Radius of cylinder ABCD.
Height of the cylinder = BC = radius
Volume of cylinder/Volume of hemisphere =
Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] sq.cm
= (2 x 434)sq.cm = 868 sq.cm
Volume of the large cube = = 216 cu.cm.
Let the edge of the large cube be a.
So, = 216
=>a = 6 cm.
Required ratio =
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
Volume of rod = = =
Volume of one spherical ball =r/2
Volume of the spherical ball =
Number of balls=Volumeof rod/volumeof one ball
Number of blocks = =120
Let width = x
Then, height = 6x and length = 42x
42x * 6x * x = 16128
x = 4
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