A wire of resistance 10 ohm is stretched 20% determinant its new resistance

							
resistance is given by, R = ρl/A
 
R = ρl²/(Al) = ρl²/V.....(1), where V is volume of wire.
 
at constant volume, V resistance of wire depends on only length of wire.
 
as it is clear in expression (1), resistance is directly proportional to square of its length.
 
now, according to question,
 
length is increased by 20%
 
so, new length of wire = l + 20% of l
 
= l + 0.2l = 1.2l
 
now, new resistance , R' = ρ(1.2l)²/V....(2)
 
from equations (1) and (2),
 
R'/R = 1.44 => R' = 1.44R
 
it is given initial resistance, R = 10ohm
 
so, R' = 1.44R = 14.4ohm
 
 
						

							
Dear student 
resistance is given by, R = ρl/A
R = ρl²/(Al) = ρl²/V.....(1), where V is volume of wire.
at constant volume, V resistance of wire depends on only length of wire.
as it is clear in expression (1), resistance is directly proportional to square of its length.
now, according to question,
length is increased by 20%
so, new length of wire = l + 20% of l
= l + 0.2l = 1.2l
now, new resistance , R' = ρ(1.2l)²/V....(2)
from equations (1) and (2),
R'/R = 1.44 => R' = 1.44R
Good Luck