A 10 ohm wire is stretched so that it's length becomes three times its original length the new resistance of the wire will we?
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Resistance ‘R = 10 ohm’ is related to length ‘L’, area of cross-section ‘A’ and resistivity ‘ρ’ as,
R = ρL/A
When L is tripled ,
As we know that the volume of the wire remains same.
old volume = new volume
AL = A'L' ....(i) ( as volume = area x length )
L' = 3L
So, from (i)
A' = A/3
So the new resistance is ,
R' = ρL'/A'
or
R' = ρ(3L)/(A/3) = 9 (ρL/A)
=> R' = 9R
so, the new resistance increases nine fold.
Now,
R' = 9 x 10ohms
thus,
R' = 90 ohms
R = ρL/A
When L is tripled ,
As we know that the volume of the wire remains same.
old volume = new volume
AL = A'L' ....(i) ( as volume = area x length )
L' = 3L
So, from (i)
A' = A/3
So the new resistance is ,
R' = ρL'/A'
or
R' = ρ(3L)/(A/3) = 9 (ρL/A)
=> R' = 9R
so, the new resistance increases nine fold.
Now,
R' = 9 x 10ohms
thus,
R' = 90 ohms
ayanizan925:
Thanks
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