A cylindrical wire of length L and radius r has a resistance R. The resistance of another wire of same material but having four times its length and

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RRB ALP - 2018

A cylindrical wire of length $L$ and radius $r$ has a resistance $R$. The resistance of another wire of same material but having four times its length and half its radius will be:

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RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)

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$8R$

$R$

$12R$

$16R$

The correct answer is

$16R$

The resistance ($R$) of a wire depends on its material (resistivity $\rho$), length ($l$), and cross-sectional area ($A$). The formula is:

Resistance Formula Basics

The relationship is given by:

$R = \rho \frac{l}{A}$

For a cylindrical wire with radius $r$, the cross-sectional area is $A = \pi r^2$. Substituting this, the resistance becomes:

$R = \rho \frac{l}{\pi r^2}$

Initial Wire Parameters

We are given:

Length = $L$
Radius = $r$
Resistance = $R$

Using the formula, the initial resistance is:

$R = \rho \frac{L}{\pi r^2}$

Calculating New Resistance

For the second wire:

Length ($l_2$) = $4L$ (four times the original length)
Radius ($r_2$) = $r/2$ (half the original radius)

First, calculate the new cross-sectional area ($A_2$):

$A_2 = \pi r_2^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4}$

Now, calculate the new resistance ($R_2$) using the resistance formula:

$R_2 = \rho \frac{l_2}{A_2}$

Substitute the values for $l_2$ and $A_2$:

$R_2 = \rho \frac{4L}{\frac{\pi r^2}{4}}$

Simplify the expression:

$R_2 = \rho \frac{4L \times 4}{\pi r^2} = \rho \frac{16L}{\pi r^2}$

Recognize that $\rho \frac{L}{\pi r^2}$ is the original resistance $R$. Therefore:

$R_2 = 16 \left( \rho \frac{L}{\pi r^2} \right) = 16R$

The new resistance is $16R$.

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4. Resistivity of the wire is directly proportional to the length of the
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