When a piece of aluminium wire of finite length is drawn through series of dies to reduce it's diameter to half it's original value, it's resistance will become?
Answers
Answer:
When a piece of aluminium wire of finite length is drawn through series of dies to reduce it's diameter to half it's original value, it's resistance will become sixteen times
Explanation:
R= ρ l/A
where R is resistance of wire
ρ= resistivity of conductor
l= length of conductor
A= area of cross section
R= ρ l xA/AxA
R= p V/A²
which implies R is inversely proportional to A²
R is inversely proportional to d⁴
Rα 1/d⁴
where d is the diameter of area of cross sectio n
if d reduces to d/2
then d⁴ reduces to d⁴/16
R=16R
Hence R increases 16 times more.
Answer:
16 times
Explanation:
R0-final resistance(L0 and R0)
R- Resistance(L and R)
we take area=> A0=A/4
we take volume=> V0=V
A0(L0)=A*L
A/4(L0)=A*L
L0=4L
R=p*L/A
R0=p*L0/A0
R0=p*4L/(A/4)
R0=16(p*L/A)
R0=16R
∴ HENCE PROVED