Question

A given copper wire is stretched to reduce its
diameter to half its previous value. What would be
its new resistance ?
​

Answer 1

Answer:

Resistance=Resistivity×length/ area

Explanation:

area=πr^2

=πD/2×D/2

According to the question diameter is half to its previous value then

Resistance=Resistivity×length/πD^2/16

Resistance= 16 Resistivity×length /πD^2

new resistance= 16 times previous resistance

Answer 2

Concept:

Resistance (R): The resistance offered to the flow of current is known as resistance.

The SI Unit of resistance is the ohm

Mathematically, resistance can be written as:

$R=\frac{pl}{A}$

Calculation:

Given that: $d_2=d_1/2$

So $d_1/d_2=2$

$A_1/A_2=((\pi d^21)/4)/(\pi d^22)/4)=2^2=4$

$A_2=A_1/4$

As we know that the resistance of the wire is,

$R=\frac{pl}{A}$

When the wire is stretched then, then its length will increase automatically. But the volume of wire will be the same.

$\therefore$ The volume of original wire=volume of new wire

$A_1/I_1=A_2/I_2$

$(\pi d^2_1I_1)4=(\pi d^2_2I_1)4$

$\Rightarrow=I_1/I_2=d^2_2/d_1^2=(1/2)^2=1/4$

So $I_2-4I_1$

The resistance of the wire in 1st case:

$R_1=R=\frac{pl_1}{A_1}$.....(1)

The resistance of the wire in 1st case:

$R_2=\frac{pl_2}{A_2}=\frac{p4l_1}{A_1/4}=\frac{16pl_1}{A_1}$.....(2)

Divide equations 1 and 2, we get

$\frac{R}{R_2}=\frac{\frac{pl_1}{A_1} }{16pl_1/A_1} =\frac{1}{16}$

$R_2=16R$

Hence, option 4 is correct.

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