Question

a wire of length L and cross section A is doubled on itself find the ratio of its initial resistance to its new resistance...
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Answer 1

Length of wire = L

Area of cross section = A

Resistance = ρ L/A

$\underline{ \bf{when \: it \: is \: doubled \: on \: itself}}$


Length will reduce to half .

So, new length = L/2

Area of cross section will be doubled

So, New area of cross section=2A

New Resistance = ρ L/2/2A

Ratio of initial resistance to new resistance


$= \frac{ \rho \frac{l}{a} }{ \frac{ \rho \frac{l}{2} }{2a} }$


$= \frac{ \frac{l}{a} }{ \frac{l}{4a} }$


$= \frac{4}{1}$


$= 4 : 1$



So, initial resistance was 4 times the new resistance.

Answer 2

$\huge{ANSWER}$

▶LENGTH =L

▶CROSS SECTION =A

LET RESISTANCE BE R.

SO,

➡RESISTANCE =$\rho \frac{Length }{Area of cross section}$

➡RESISTANCE= $\rho \frac{L}{A}$

GIVEN,

That the wire is being doubled on itself,

Since,

⚫On increasing the cross section the resistance increases.

R$\propto\frac{1}{A}$

▶NEW AREA OF CROSS SECTION =2A.

so, resistance might have been decreased.

⚫On decreasing length the resistance also decreases.

Since, the wire is doubled on itself the length have been decreased .

▶NEW LENGTH =L/2.

▶Let the new resistance be R1.

RESISTANCE =$\rho \frac{Length}{Area of cross section }$

R=$\rho \frac{\frac{L}{2}}{2A}$.

On dividing the original resistance by the new resistance...

$\frac{R}{R1} =\rho\frac{\frac{L}{2}}{2A}÷\rho \frac{L}{A}$

On solving ,

$\frac{R}{R1}=\rho \frac{L}{A} ÷\rho \frac{\frac{L}{2}}{2A}$

$\frac{R}{R1} =\frac{L}{A} \times \frac{4A}{L}$

Further on solving,

$R=4R1$

$\frac{R}{4}=R1$

The new resistance will be one fourth of the original resistance.