Question

Two wires of the same length are made of brass and silver but wires are different in diameter. Diameter of the brass wire is 4times the diameter of the silver wire and the resistivity of the brass is 5 times greater than the resistivity of the sliver. If ${{R}_{B}}$ denotes the resistance of the brass wire and ${{R}_{S}}$ denotes the resistance of the silver wire, which of the following is true?
$\begin{align}
  & A.\text{ }{{R}_{B}}=\dfrac{5}{16}{{R}_{S}} \\
 & B.\text{ }{{R}_{B}}=\dfrac{4}{5}{{R}_{S}} \\
 & C.\text{ }{{R}_{B}}=\dfrac{5}{4}{{R}_{S}} \\
 & D.\text{ }{{R}_{B}}=\dfrac{5}{2}{{R}_{S}} \\
 & E.\text{ }{{R}_{B}}=\dfrac{16}{5}{{R}_{S}} \\
\end{align}$

Answer 1

Hint: In order to solve this question we will use a formula which states relation between resistance, resistivity and area. And by comparing both the resistance value using the formula which will give a relation between both the wires resistance.
Formula used:
$\begin{align}
  & R=\dfrac{\rho l}{A} \\
 & A=\dfrac{\pi }{4}{{d}^{2}} \\
\end{align}$

Complete answer:
To know the difference between both materials (brass and silver) we will use the formula for the resistance.
$R=\dfrac{\rho l}{A}.....\left( 1 \right)$
Where, R = resistance of a wire.
ρ = resistivity of a material
l = Length of the wire.
A = area of the resistance.
Now we know that cross sectional area of wire is given as,
$A=\dfrac{\pi }{4}{{d}^{2}}....\left( 4 \right)$
Now substitute value of equation (2) in the equation (1)
$\begin{align}
  & R=\dfrac{\rho .l}{\dfrac{\pi }{4}{{d}^{2}}} \\
 & R=\dfrac{4.l.\rho }{\pi {{d}^{2}}}.....\left( 3 \right) \\
\end{align}$
It is given in the question that ${{R}_{B}}$denotes the resistance of the brass.
${{R}_{B}}=\dfrac{4l}{\pi }\times \dfrac{{{\rho }_{B}}}{{{d}_{B}}^{2}}....\left( 3 \right)$
And ${{R}_{S}}$denotes the resistance of the sliver
${{R}_{S}}=\dfrac{4l}{\pi }\times \dfrac{{{\rho }_{S}}}{{{d}_{S}}^{2}}....\left( 4 \right)$
Now it is given that the diameter of brass wire is 4 times the diameter of the silver wire, therefore,
${{d}_{B}}=4ds...\left( 5 \right)$
And it is given that the resistivity of brass is 5times greater than the resistivity of the silver.
${{\rho }_{B}}=5{{\rho }_{S}}...\left( 6 \right)$
Now substitute value of the equation (5) and (6) in equation (3)
\[\begin{align}
  & {{R}_{B}}=\dfrac{4l}{\pi }\times \dfrac{5{{\rho }_{S}}}{{{\left( 4ds \right)}^{2}}} \\
 & {{R}_{B}}=\dfrac{5}{16}\left( \dfrac{4l}{\pi }\times \dfrac{{{\rho }_{S}}}{d{{s}^{2}}} \right)....\left( 7 \right) \\
\end{align}\]
Now substitute the values of the equation (4) in the equation (7)
$\therefore {{R}_{B}}=\dfrac{5}{16}{{R}_{S}}$

Therefore the correct option is (A) .

Additional information:
Ohm’s law:
Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, while all the physical conditions and temperature remain constant.

Note:
As an approach to solve this question first we write the formula for the resistance stating relation between resistivity diameter and the resistance after that we used relation or difference between the diameter and the resistivity of both the materials (brass and the sliver) which eventually gives us difference between both materials resistance.