Original area = 6a2
New area = 6(2a)2 = 24a2
Required increased % = (18a2/6a2) x 100%
= 300%
? 2?rh = 264
? 2 x (22/7) x r x 14 = 264
? r = 3
So, Volume = ?r2h
= (22/7) x 3 x 3 x 14 cm3
= 396 cm3
Radius of sphere = 9 cm
Volume of sphere = [ 4/3 x ? x (9)3] cm3 = 972? cm3
Radius of wire = 0.2 mm = 2/(10 x 10) cm = 1/50 cm
Let the length of wire be = L cm
Then, 972? = ? x (1/50)2 x L
? L = (972 x 50 x 50 ) cm
?972? = ? x (1/50)2 x L
? L = (972 x 50 x 50) cm
? Length of wire = (972 x 50 x 50)/100 m = 24, 300 m
Let the number of spheres be N
Then, N x 4/3 ? x (3)3 = ? x (2)2 x 45
? 36N = 180
? N = 180/36 = 5
? Circumference of the girth = 440 cm
? 2?r = 440
? r = 440 / (2 x 7/22) = 70 cm
Thus, Outer radius = 70 cm
Inner radius = (70 - 4) cm = 66 cm
Volume of iron = ?[(70)2 - (66)2] x 63 cm3
= (22/7) x 136 x 4 x 63 cm3
= 107712 cm 3
? 4/3 ?r3 = ?r2 h
? h = 4/3 r
? Height = 4/3 times its radius.
Original area = 4? r2,
New area = 4?(2r)2 = 16 ? r2
Required increased % = [12?r2]/[4? r2] x 100 %= 300%
Original volume = 1/3 ?r2h;
New volume = 1/3 ?r2(2h) = 2/3?r2h
Required increased % = [?r2h/3]/[?r2h/3] x 100%
= 100%
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 %
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
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
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