Let the fixed height of a right circular cone is h and initial radius is r
Then, initial volume of cone, V1 = (1/3)?r2h
After increasing 15% radius of a cone = (r + 3r/20) = 23r/20
New volume become, V2 = (1/3)?(23/20)2r2h
? Increasing percentage = [(V2 - V1) / V1] x 100
= {[(1/3)?r2h] / [(1/3)?r2h]} {(23/20)2 - 1} x 100
= (23/20 + 1)(23/20 - 1) x 100
= 43/20 x 3/20 x 100 = 32.25%
Given, l1/l2 = 5/7
Now, curved surface area of the first cone = ? rl1
and curved surface area of second cone = ?rl2
? Ratio = ?rl1/?rl2 = l1/l2 = 5 : 7
Let diameter, radius and height of first cone are d1, r1 and h1, respectively and that of second cone are d2, r2 and h2, respectively.
r1/r2 = d1/d2 = 3/5,
h1/h2 = ?
Given,
[(1/3)?r21h1] / [(1/3)?r22h2] = 1/3
? (3/5)2 x h1/h2 = 1/3
? h1/h2 = (1/3) x (25/9) = 25/27
Here, x = -50%, y = 200%
According to the formula Net effect
= [2x + y + x2 + 2xy/100 + x2y/1002] %
= [-100 + 200 + 2500 - 20000/100 + 500000/10000]%
= [100 - 175 + 50] % = -25%
Let radius = 5k, height = 12k
According to the question
= 1/3 x 22/7 x (5k)2 x 12k = 2200/7
? k = 1
? r = 5, h = 12
? Slant height (l) = ?r2 + h2
= ?25 + 144
= ?169
=13 cm
[(1/3)?r12h1]/[(1/3)?r22h2] = 2/3
? (r1/r2)2 x (h1/h2) = 2/3
? (1/2)2 x (h1/h2) = 2/3 [ &bacaus; r1/r2 = 1/2]
? h1/h2 = 8/3
? h1 : h2 = 8 : 3
Surface area of sphere = 6616 cm2
4?r2 = 6166
?r2 = (6166 x 7)/(4 x 22)
? r2 = 7 x 7
? r = 7 cm
? Diameter of largest circle lying on sphere = 2 x r = 2 x 7 = 14 cm
Given that, the diameter of moon is approximately one-fourth of the diameter of Earth.
Let radius on moon = r
Then, radius of Earth = 4r
? Volume of Moon/Volume of Earth
= [(4/3)?r3] / [(4/3)?(4r)3]
= [r3] / [64r3] = 1/64 = 1 : 64
According to the question,
Surface area of sphere = Surface area of hemisphere
4?r12 =3?r22
? r1/r2 = ?3/2
? Ratio in volume = [(4/3)?r13] / [(4/3)?r23]
= 3?3/8 : 1
Radius of the sphere = 16/2 = 8 cm
Volume of the sphere = (4/3) x ? x 8 x 8 x 8 cm3
Radius of each lead ball = 2/2 = 1 cm
Volume of each lead ball = Volume of sphere / Volume of lead ball
= (4/3) ? x 1 x 1 x 1 = 4?/3 cm3
? Number of lead balls = [(4/3) x ? x 8 x 8 x 8 x 3] / [4 ?]
= 8 x 8 x 8 = 512
Curved surface area of the hemisphere = 2?r2
= 2 x (22/7) x (7/2) x (7/2) = 77 sq
As bowl is to painted inside and outside.
? Total surface to be painted = 77 x 2 = 154 sq cm
? Cost of painting 154 sq cm = (5/10) x 154 = 1/2 x 154 = ? 77
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