Answer

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**Hint:**From ohm’s law, the definition of 1 ohm resistance can be defined. The resistance of a substance is directly proportional to its length.

**Formula used:**In this solution we will be using the following formulae;

\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.

\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.

**Complete Step-by-Step Solution:**

From the ohm law equation, the definition of 1 ohm, could be derived.

The ohm’s law equation is can be written as

\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.

Hence, for 1 ohms, we can write

\[1\Omega = \dfrac{{1{\text{V}}}}{{1{\text{A}}}}\]

Hence, 1 ohm’s could be defined as the resistance to a 1 A current which causes 1 volt potential difference or voltage drop. Or it can be defined as the resistance value which would allow only 1 A of current to flow when the component is connected to a 1 volt source.

For the second part of the question, a student possesses a resistance wire of 1 ohms which has a length 50 cm. to what length should he stretch it uniformly to obtain a resistance wire of 4 ohms.

Recall that the resistance of a wire can be given as

\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.

Hence, if the all other properties are kept constant,

\[\dfrac{R}{l} = k\] where \[k\] is a constant.

Then,

\[\dfrac{{{R_1}}}{{{l_1}}} = \dfrac{{{R_2}}}{{{l_2}}}\]

\[ \Rightarrow \dfrac{1}{{50}} = \dfrac{4}{{{l_2}}}\]

Hence, by cross multiplication,

\[{l_2} = 50 \times 4 = 250cm\]

**Hence, the length of the wire has to be stretched to 250 cm.**

**Note:**In actuality, the wire would not need to stretch that far before the resistance actually attains 4 ohms as desired. This is due to the Poisson phenomenon, which is the fact that as the length stretches, the cross sectional area of the wire will reduce, hence creating a double effect of increment of resistivity (the smaller the cross sectional area, the lower the resistance).

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