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Question: A 250cm long wire has a diameter 1mm. It is connected at the right gap of a slide wire bridge. When ...

A 250cm long wire has a diameter 1mm. It is connected at the right gap of a slide wire bridge. When a 3Ω3\Omega resistance is connected to a left gap, the null point is obtained at 60cm. The specific resistance of the wire is
A.6.28×10−6ΩmA.\quad 6.28\times {{10}^{-6}}\Omega m
B.6.28×10−5ΩmB.\quad 6.28\times {{10}^{-5}}\Omega m
C.6.28×10−8ΩmC.\quad 6.28\times {{10}^{-8}}\Omega m
D.6.28×10−7ΩmD.\quad 6.28\times {{10}^{-7}}\Omega m

Explanation

Solution

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, R1R2=l100−l\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}. Here (l) is the balancing length. The specific resistance (ρ)(\rho ) of the wire is given by, ρ=RAl\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 (R1)({{R}_{1}}). The value of resistance connected to the left gap is given as, R1=3Ω{{R}_{1}}=3\Omega . Let’s consider the resistance connected to the right gap of the slide wire bridge be (R2)({{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=1×10−3md=1mm=1\times {{10}^{-3}}m. Hence, the radius of the wire will be r=d2=1mm2=5×10−4mr=\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, R1R2=l100−l\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{100-l}. Substituting in the values of resistances and the balancing length, we get,

R1R2=l100−l⇒3R2=60100−60⇒1R2=2040⇒R2=2Ω\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 (ρ)(\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=ρAlR=\dfrac{\rho A}{l}.

Hence, the specific resistance of the wire is given by, ρ=RAl\rho =\dfrac{RA}{l}. The cross sectional area of the wire is, πr2\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=πr2⇒A=(3.14)(5×10−4)2m2.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=R2=2Ω.R={{R}_{2}}=2\Omega .. The length of the wire is l=2.5m. Therefore, the specific resistance of the wire becomes, ρ=RAl=(2Ω)(3.14)(5×10−4)2m22.5m⇒ρ=(6.28)(25×10−8)Ωm2.5=(6.28)(10×10−8)Ωm\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 ρ=6.28×10−7Ωm\rho =6.28\times {{10}^{-7}}\Omega m

Note:
The reason behind the balancing condition of the meter bridge being R1R2=l100−l\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, R1R2=l1−l\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{l}{1-l}.