Question

9. a. The Resistance of a wire is 5ohm
. If this wire is folded to make the length half and
thickness double, what will be the new value of resistance​

Answer 1

Given :

• The Resistance of a wire is 5 Ω . This wire is folded to make the length half and  thickness double

⇒ R = ρl/A = 5Ω

R' = ?

ρ' = ρ

l' = l/2

A' = 2A

To Find :

• New value of resistance​

Solution :

$\bf R=\dfrac{\rho l}{A}\\\\\implies \bf 5=\dfrac{\rho l}{A}\;[\;Given\;]$

Now we have to find the new value of resistance .

$\implies \bf R' =\dfrac{\rho\; l'}{A'}\\\\\implies \bf R'=\dfrac{\rho (\dfrac{l}{2}l)}{(2A)}\;[\;From\ Given\ Values\;]\\\\\implies \bf R'=\dfrac{1}{4}(\dfrac{\rho l}{A})\\\\\implies \underline{\bf R'=\dfrac{1}{4}(R)}\\\\\implies \bf R'=\dfrac{1}{4}(5)\;[\;Given\;]\\\\\implies \bf R'=\dfrac{5}{4}$

So , the new resistance is 1/4 times the old resistance .