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Radius of sphere is increased by 50%. By how much percentage
is surface area is increased?
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Answer / aditi joshi
let original radius=100
original surface area lets say it x =4 * (22/7) * 100 *100
new radius=100 + 50% of 100=150
new surface area lets say it y = 4 * (22/7)* 150* 150
increase in surface area=((y-x)/x) * 100
i.e.
((4*(22/7)*150*150 - 4*(22/7)*100 *100) / (4*(22/7)*100*100))*100
=125
so the answer is 125%.
hope this helps you.
| Is This Answer Correct ? | 64 Yes | 17 No |
Answer / saurabh
surface area will increase by 125%
and shruti u r wrong
| Is This Answer Correct ? | 52 Yes | 17 No |
Answer / arvind yeram
r1 = initial radius
r2 = final radius
s1 = 4πr1^2
s2 = 4πr2^2
r2 = 50/100r1+r1 = 3/2r1
s2 = 4π9/4r1^2
%increase = (9/4-1)*100
= 5/4*100
= 125
| Is This Answer Correct ? | 32 Yes | 16 No |
Answer / harsh goel
let the edge of cube be=x
the edge increased by 50%
=x+50÷100
=x(1+1/2)
=3x/2
orignal surface area=6x^2
new surface area of a cube=6(3x/2)^2
=6×9x^2/4
=27x^2/2
increased in area of cube=27x^2-12x^2/2
=15x^2/2
the percentage increased=increase in area/orignal area ×100
=15×100/2×6x^2
=1500/12x^2
=125%
| Is This Answer Correct ? | 13 Yes | 5 No |
Answer / satyam
1st sphere=x and 2nd sphere=y then
x=4(22/7)×100 50=4(22/7)150
y=4(22/7)×100=4(22/7)100
area={(x-y/x))r2
={4(22/7)150-4(22/7)100/4(22/7)100}r2
={4(22/7)50/4(22/7)100}r2
={50/100}r2
=.5×r2 ;0.5×0.5=0.25 = 100 25= 125%
| Is This Answer Correct ? | 10 Yes | 3 No |
Answer / brebis
surface area increases with radius by factor 4.
i.e. 200%
| Is This Answer Correct ? | 3 Yes | 13 No |
Answer / debanjana
radius is only half of the diameter, so the diameter should
be counted which is the surface area, so it should be 100%
| Is This Answer Correct ? | 3 Yes | 14 No |
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