A cylindrical wire is compressed in length by 10%. The
percentage decrease in the resistance will be?????
Answers were Sorted based on User's Feedback
Answer / ali asghar
As we know
R= p*L/A
if resistance decreased by 10 % it mean new resistance is
L'=0.9L
And Area is increased by 10 % because volume will remain constant so new area will be
A' = 1.1A
so
R'=p* .9L/1.1A =.9R/1.1=0.81R
so change in resistance is = 1-0.81=0.91=0.91*100 %= 19 %
so resistance decreased by 19%
Is This Answer Correct ? | 55 Yes | 9 No |
Answer / aurangzaib
percentage decrease in the resistance will be 19%
Is This Answer Correct ? | 32 Yes | 0 No |
Answer / ashlesha
R=p*l/A
Length is decreased by 19% so
L'= 90%of l
=0.9l
Volume remains the same and area increased by 10%
So, new area=1.1A
R=0.9l/1.1A=081R
Change in resistance=1-0.81
=0.19
0.19=19%
Is This Answer Correct ? | 6 Yes | 0 No |
Answer / aamir ali
Wire is compressed it means volume will be constant (V= Area
x Length) if length is decreased by 10% then area will
increase by 10% (Because volume is constant) suppose
l=length, a=area R=Resistance p=rho, Then R=p*(l/a)
Now if length is compressed by 10% then length=0.9*l and
area=1.1*a
Resistance=p*(.9l/1.1a)=0.8R
This means that Resistance will be decreased by 20%
Hope I answered your Question
Is This Answer Correct ? | 15 Yes | 24 No |
Answer / akash
I think ,the basic equaction
Is that. R directly prposnal to l
So r=10%
Is This Answer Correct ? | 1 Yes | 15 No |
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