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Question: A student was given the experiment to measure the emf of an unknown cell using a potentiometer. He c...

A student was given the experiment to measure the emf of an unknown cell using a potentiometer. He cheated the result and wrote the observation arbitrarily and was caught. Which reading helped the teacher to arrive at the conclusion that he cheated ??

S. No.Value of rheostatNull point
1.1.R1{R_1}50cm50\,cm
2.2.R2{R_2}60cm60\,cm
3.3.R3{R_3}70cm70\,cm $$
4.4.R4{R_4}80cm80\,cm


(A)1 (B)2 (C)3 (D)4  \left( A \right)\,\,1 \\\ \left( B \right)\,\,2 \\\ \left( C \right)\,\,3 \\\ \left( D \right)\,\,4 \\\

Explanation

Solution

In the given diagram, actual value of the voltage and resistance are given. By using the equation of ohm’s law we calculate all the actual values of the resistance and also we find the cheated values in the above table.

Formulae Used:
The expression for finding the current is
I=VRI\, = \,\dfrac{V}{\operatorname{R} }
Where
II be the current of the cell, VV be the voltage of the cell and RR be the Resistance of the cell.

Complete step-by-step solution :
We know that actual voltage v=20voltsv\, = 20\,volts
Cheated voltage v=10voltsv\, = \,10\,volts
Resistance of the cells are 2Ω2\,\Omega \,and 8Ω8\,\Omega (Resistance is measured in ohms (Ω)\left( \Omega \right))
I=VR.........(1)I\, = \,\dfrac{V}{R}.........\left( 1 \right)
Substitute the known values in the equation (1)\left( 1 \right)
I=20V8+2+RI\, = \,\dfrac{{20\,V}}{{8 + 2 + R}}
Perform the arithmetic operations in the above equation we get,
I=2010+RI\, = \,\dfrac{{20}}{{10 + R}}
In the figure, we know that null point is at 8Ω8\,\Omega
Distance dd is 100cm100\,cm and length lislcml\,is\,l\,cm.
So, 8×l100i=10\dfrac{{8 \times l}}{{100}}\,i\, = \,10
Simplify the above equation we get,
li=125li\, = \,125
From above equations,
110+R=6.25\dfrac{1}{{10 + R}}\, = \,6.25
Simplify the equation we get,
R=16.2510R\, = \,\dfrac{1}{{6.25}} - 10 >0 > \,0
Hence, Iis62.5>0I\,is\,\,62.5\, > 0
So from the above option, the option (C)\left( C \right) is correct.
Therefore, the value of R1{R_1} and R2{R_2} violate the conditions.

Note:- In the question, the value of voltage is high. So we get the value of rheostat is high. But, the cheated value is half of the given value. So there must be a difference in the value of rheostat. From the table the value of rheostat is high. So, thus the value of the rheostat must be low if the voltage is low.