if the length of the wire of resistance R is increased to n times,it's mass remaining constant then its new resistance will be?
Answers
Answer:
Let l be the original length of A.
Original area of cross-section, then
Original resistance R=
A
ρl
Now, length of the wire =l
′
=nl
If A
′
be the new cross-sectional area, then l
′
A
′
=lA
(∵ volume of metal is a constant).
A
′
=
l
′
lA
=
nl
lA
=
n
A
New resistance of the wire is
R
′
=
Al
ρl
′
=
n
A
ρ(nl)
=
l
ρnl
×
A
n
=n
2
(
A
ρl
)=n
2
R
Answer:
n²R
Explanation:
Let l be the original length of the wire.
Original cross-sectional area --> A
∴ R(original) --> ρl/A (ρ is resistivity)
New length = l' = n*l => nl
Let A' be new cross-sectional area.
Thus, Al = A'l' (Volume will remain the same)
∴ A' = Al/l' => Al/nl => A/n
∴ R' = ρl'/Al => ρ(nl) ÷ (A/n)
=> (ρnl/l)*(n/A) [Dividing by a number is the same as multiplying by its reciprocal.]
n²(ρl/A) => n²R