Find the ratio of resistances of two copper rods X and Y of lengths 30 cm and 10 cm respectively and having radii 2 cm and 1 cm respectively.

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Solution

Step 1: Given data
Length of wire $X = 30 c m$
Length of wire $Y = 10 c m$
Radius of wire $X = 2 c m$
Radius of wire $Y = 1 c m$
Step 2: Formula for resistance
$R = \frac{ρ l}{A}$ [ where $ρ =$ Specific Resistance, $A =$ Cross-sectional area, $l =$ length of wire ]
Step 3: Calculating the cross-sectional area of both the wired
The radius of the rod $X$ , $r_{1} = 2 c m$
The cross-sectional area is calculated as
$A = π r^{2}$ [ where $A =$ Cross-sectional area, $r =$ radius ]
The cross-sectional area of the rod $X$
$A_{1} = π {r_{1}}^{2} A_{1} = π {(2 c m)}^{2} A_{1} = 4 π c m^{2}$
Similarly radius of the rod $Y, r_{2} = 1 c m$
The cross-sectional area of the rod $Y$
$A_{2} = π {(1 c m)}^{2} A_{2} = π c m^{2}$
Step 4: Calculating the ratio
$\frac{R_{1}}{R_{2}} = (\frac{\frac{l_{1}}{A_{1}}}{\frac{l_{2}}{A_{2}}}) \frac{R_{1}}{R_{2}} = (\frac{\frac{l_{1}}{l_{2}}}{\frac{A_{1}}{A_{2}}})$
Putting value of all these terms
$\frac{R_{1}}{R_{2}} = (\frac{\frac{30 c m}{10 c m}}{\frac{4 π c m^{2}}{π c m^{2}}}) \frac{R_{1}}{R_{2}} = (\frac{3}{4}) r a t i o o f r e s i s \tan c e = 3 : 4$
Hence the ratio of resistance of two copper rods $X$ and $Y$ is $3 : 4$

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