
125 cm of potentiometer wire balances the emf of a cell and 100 cm of the wire is required for balance, If the poles of the cell are joined by $2\Omega $ resistor. Then the internal resistance of the cell is
A. $0.25\Omega $
B. $0.5\Omega $
C. $0.75\Omega $
D. $1.75\Omega $
Answer
163.8k+ views
Hint: The problem is from the current electricity section of physics. We have to apply the resistance equation to find the internal resistance of the cell. We need to know the concepts of the potentiometer to solve this problem.
Formula Used:
Internal resistance in potentiometer:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Where \[{l_1}\]= length of the potentiometer wire, \[{l_2}\]= length of the given wire, r = internal resistance of the cell and R = resistance of the resistor in the circuit.
Complete step by step solution:
The given values are \[{l_1} = 125cm\], \[{l_2} = 100cm\] and \[R = 2\Omega \].
Internal resistance in potentiometer is given as:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Substitute these values in the Internal resistance equation and we will get.
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R \\
\Rightarrow r = \dfrac{{125 - 100}}{{100}} \times 2 \\ \]
\[\therefore r = 0.25 \times 2 = 0.5\Omega \]
Hence, the correct option is option B.
Additional Information: The potentiometer is a device used to measure an unknown voltage by comparing it to a known voltage. It can be used to compare the emf of several cells and to determine the emf and internal resistance of the specified cell. The potentiometer's basic principle states that, if the wire has a uniform cross-sectional area and a uniform current flows through it, the potential drop across any part of the wire will be precisely proportional to the length of the wire.
Note: The smallest change in the potential difference that a potentiometer can detect is its sensitivity. By lowering the potential gradient, the potentiometer's sensitivity can be improved. That is done by extending the potentiometer wire's length.
Formula Used:
Internal resistance in potentiometer:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Where \[{l_1}\]= length of the potentiometer wire, \[{l_2}\]= length of the given wire, r = internal resistance of the cell and R = resistance of the resistor in the circuit.
Complete step by step solution:
The given values are \[{l_1} = 125cm\], \[{l_2} = 100cm\] and \[R = 2\Omega \].
Internal resistance in potentiometer is given as:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Substitute these values in the Internal resistance equation and we will get.
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R \\
\Rightarrow r = \dfrac{{125 - 100}}{{100}} \times 2 \\ \]
\[\therefore r = 0.25 \times 2 = 0.5\Omega \]
Hence, the correct option is option B.
Additional Information: The potentiometer is a device used to measure an unknown voltage by comparing it to a known voltage. It can be used to compare the emf of several cells and to determine the emf and internal resistance of the specified cell. The potentiometer's basic principle states that, if the wire has a uniform cross-sectional area and a uniform current flows through it, the potential drop across any part of the wire will be precisely proportional to the length of the wire.
Note: The smallest change in the potential difference that a potentiometer can detect is its sensitivity. By lowering the potential gradient, the potentiometer's sensitivity can be improved. That is done by extending the potentiometer wire's length.
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