Question

A 250cm long wire has a diameter 1mm. It is connected at the right gap of a slide wire bridge. When a $3\Omega $ resistance is connected to a left gap, the null point is obtained at 60cm. The specific resistance of the wire is
\[A.\quad 6.28\times {{10}^{-6}}\Omega m\]
\[B.\quad 6.28\times {{10}^{-5}}\Omega m\]
\[C.\quad 6.28\times {{10}^{-8}}\Omega m\]
\[D.\quad 6.28\times {{10}^{-7}}\Omega m\]

Answer 1

Hint: The question is based on the wheatstone bridge concept. The meter bridge uses the same concept. The balancing condition for the meter bridge is given by, $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}$. Here (l) is the balancing length. The specific resistance $(\rho )$ of the wire is given by, $\rho =\dfrac{RA}{l}$.

Step by step solution:
Let’s understand what is given in the question. The resistance in the left gap is given by $({{R}_{1}})$. The value of resistance connected to the left gap is given as, ${{R}_{1}}=3\Omega $. Let’s consider the resistance connected to the right gap of the slide wire bridge be \[({{R}_{2}})\]. The null point is given to be at 60cm. Hence, the balancing length (l=60cm). The total length of the wire (L) is given by, L=250cm=2.5m.

Since, the diameter of the wire is given as \[d=1mm=1\times {{10}^{-3}}m\]. Hence, the radius of the wire will be \[r=\dfrac{d}{2}=\dfrac{1mm}{2}=5\times {{10}^{-4}}m\].

Now, we will use the balancing condition for the meter bridge. The formula of meter bridge is given by, $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}$. Substituting in the values of resistances and the balancing length, we get,

$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}\Rightarrow \dfrac{3}{{{R}_{2}}}=\dfrac{60}{100-60}\Rightarrow \dfrac{1}{{{R}_{2}}}=\dfrac{20}{40}\Rightarrow {{R}_{2}}=2\Omega $.

For a wire of specific resistance $(\rho )$and Cross sectional area of the wire (A) and the length of the wire (l), the resistance of the wire (R) is given by, $R=\dfrac{\rho A}{l}$.

Hence, the specific resistance of the wire is given by, $\rho =\dfrac{RA}{l}$. The cross sectional area of the wire is, $\pi {{r}^{2}}$. Substituting in the value of the radius of the wire here, we find the cross sectional area of the wire becomes, $A=\pi {{r}^{2}}\Rightarrow A=(3.14){{(5\times {{10}^{-4}})}^{2}}{{m}^{2}}.$

The resistance of this wire, as we found out earlier, is $R={{R}_{2}}=2\Omega .$. The length of the wire is l=2.5m. Therefore, the specific resistance of the wire becomes, \[\rho =\dfrac{RA}{l}=\dfrac{(2\Omega )(3.14){{(5\times {{10}^{-4}})}^{2}}{{m}^{2}}}{2.5m}\Rightarrow \rho =\dfrac{(6.28)(25\times {{10}^{-8}})\Omega m}{2.5}=(6.28)(10\times {{10}^{-8}})\Omega m\].
Therefore, the specific resistance of the wire is \[\rho =6.28\times {{10}^{-7}}\Omega m\]

Note:
The reason behind the balancing condition of the meter bridge being $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}$is, because all the lengths in the meter bridge is taken in cm. Hence the value of l is in cm.

If we consider, the lengths value being taken in terms of meter, then the balancing condition becomes, $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{1-l}$.