Let their height be h and 2h radii be x and y respectively.
Then, ?x2h = ?y2(2h)
? x2/y2 = 2/1
? x/y = ?2 / 1 = ?2 : 1
Let their radii be 2r and 3r and height 5h and 3h respectively.
? Ratio of their volumes = [?(2r)2 x 5h] / [?(3r)2 x 3h]
= 20/27 = 20 : 27
Original volume = 4/3?r3
New volume = 4/3 ? (2r)3 = 32/3 ?r3
Required increased % = [(28/3) ?r3] x [(3/4)?r3] x 100 % = 700 %
Original volume = 1/3 ?r2h;
New volume = 1/3 ?r2(2h) = 2/3?r2h
Required increased % = [?r2h/3]/[?r2h/3] x 100%
= 100%
Original area = 4? r2,
New area = 4?(2r)2 = 16 ? r2
Required increased % = [12?r2]/[4? r2] x 100 %= 300%
Original area = 6a2
New area = 6(2a)2 = 24a2
Required increased % = (18a2/6a2) x 100%
= 300%
Let their radii be x and 2x .
Ratio of their surface areas = [4?x2]/[4?(2x)2] = 1/4 = 1 : 4
Curved surface area = ?rl
= (22/7 x 6 x 28) cm2
= 528 cm2
Number of bullets = Volume of cube / Volume of 1 bullet
= (22 x 22 x 22)/(4/3 x 22/7 x 1 x 1 x 1)
= 2541.
Number of balls = Volume of big ball / Volume of 1 small ball
= [4/3 x ? x 10 x 10 x 10] / [4/3 x ? x 0.5 x 0.5 x 0.5]
= 8000
Total volume of cuboid = (10 x 5 x 2) cm3
= 100 cm3
Volume curved = 1/3 x (22/7) x 3 x 3 x 7 cm3
= 66 cm3
% of wood wasted = (100 - 66)% = 34 %
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