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Question: The figure shows a potentiometer in which \[R\] is a variable resistance and \[\text{AB}\] is a unif...

The figure shows a potentiometer in which RR is a variable resistance and AB\text{AB} is a uniform wire of length 100cm\text{100cm} having resistance rr. In an experiment when R=0R=0 the null point is obtained at 30cm\text{30cm} from end A\text{A}.For another value of RR, the null point is obtained at 60cm\text{60cm} from A\text{A}. Find the value RR in the second case.

Explanation

Solution

Galvanometer is a device that is used to measure presence of electric current. Null deflection in the galvanometer indicates that there is no current flowing through the galvanometer. Using this concept we can find the value of RR in the second case.

Formula used:
According to Ohm’s law we have,
V=IRV=IR
Where VV is the voltage, II is the current and RR is the resistance.

Complete step by step answer:
Given a uniform wire AB\text{AB} of length, L=100cmL=100\text{cm}.
Resistance of the whole wire is rr.
The EMF of the battery is E=10VE=10\text{V}.
Also, in the figure a galvanometer is used with a movable jockey along the wire AB\text{AB}.
In the first case, the value of RR is R=0R=0 and the null point is at l1=30cm{{l}_{1}}=30\text{cm}.
In the second case, the null point is at l2=60cm{{l}_{2}}=60\text{cm}.

Let the total current through the circuit be II.In first case, the null point is at l1=30cm{{l}_{1}}=30\text{cm} which is the length between the point A\text{A} and jockey which means there is no flow of current through the part length between point A\text{A} and jockey and the current flows through the part of wire between the jockey and point B\text{B}. The length between jockey and point B\text{B} is
l1=ll1{{l}_{1}}^{'}=l-{{l}_{1}}
Putting the values of ll and l1{{l}_{1}} we get,
l1=10030=70cm{{l}_{1}}^{'}=100-30=70\text{cm}
The wire is uniform and for 100cm\text{100cm} the resistance is rr.
For 1cm\text{1cm} the resistance will be r100\dfrac{r}{100}
For 70cm70\text{cm} the resistance will be r=r100×70=710rr'=\dfrac{r}{100}\times 70=\dfrac{7}{10}r

Now, the total resistance in the circuit is,
R1=R+r{{R}_{1}}=R+r'
Putting the values of RR and rr' we get
R1=0+7r10{{R}_{1}}=0+\dfrac{7r}{10}
R1=7r10\Rightarrow {{R}_{1}}=\dfrac{7r}{10}
Using Ohm’s law we get,
E=IR1E=I{{R}_{1}}
I=ER1\Rightarrow I=\dfrac{E}{{{R}_{1}}}
Putting the values of EE and R1{{R}_{1}} we get,
I=10(7r10)I=\dfrac{10}{\left( \dfrac{7r}{10} \right)}
Now, the denominator’s denominator is multiplied in numerator
I=10×107r\Rightarrow I=\dfrac{10\times 10}{7r}
I=1007r\Rightarrow I=\dfrac{100}{7r} (i)

Similarly, in the second case the null point is at l2=60cm{{l}_{2}}=60\text{cm} which is the length between the point A\text{A} and jockey. So, the part of the wire through which the current flows is between the jockey and point B\text{B}. The length between jockey and point B\text{B} is
l2=ll2{{l}_{2}}^{'}=l-{{l}_{2}}
Putting the values of ll and l2{{l}_{2}} we get,
l2=10060=40cm{{l}_{2}}^{'}=100-60=40\text{cm}
For 1cm\text{1cm} the resistance will be r100\dfrac{r}{100}
For 40cm40\text{cm} the resistance will be r=r100×40=410rr''=\dfrac{r}{100}\times 40=\dfrac{4}{10}r

Now, the total resistance in the circuit is,
R2=R+r{{R}_{2}}=R+r''
Putting the values of rr'' we get
R2=R+4r10{{R}_{2}}=R+\dfrac{4r}{10}
Using Ohm’s law we get,
E=IR2E=I{{R}_{2}}
I=ER2\Rightarrow I=\dfrac{E}{{{R}_{2}}}
Putting the values of EE and R2{{R}_{2}} we get,
I=10(R+4r10)I=\dfrac{10}{\left( R+\dfrac{4r}{10} \right)}

Now, the denominator’s denominator is multiplied in numerator
I=10×1010R+4r\Rightarrow I=\dfrac{10\times 10}{10R+4r}
I=10010R+4r\Rightarrow I=\dfrac{100}{10R+4r} (ii)
Equating equations (i) and (ii) we get,
1007r=10010R+4r\dfrac{100}{7r}=\dfrac{100}{10R+4r}
17r=110R+4r\Rightarrow \dfrac{1}{7r}=\dfrac{1}{10R+4r}
Cross multiplying the denominators,
10R+4r=7r\Rightarrow 10R+4r=7r
10R=7r4r\Rightarrow 10R=7r-4r
Further Simplifying we get,
10R=3r\Rightarrow 10R=3r
R=310r\therefore R=\dfrac{3}{10}r

Therefore, the value RR in the second case is 310r\dfrac{3}{10}r.

Note: Remember for a given material, the resistance of the material is directly proportional to its length, if the length increases, the resistance of the material increases and if the length decreases the resistance of the material decreases. Also, resistance is inversely proportional to the cross sectional area. In the above question the wire is uniform so the cross sectional area is constant.