Physics teacher Mr. Sharma conducts viva-voce for board practical and asks the following two questions from every student. |
(i) Why a potentiometer be preferred over a voltmeter for the measurement of emf of a cell? |
(ii) Why should a six wire potentiometer be preferred over a three wire potentiometer? The student A who could not answer any question was teacher's ward. However another student B who answered both questions correctly was not teacher's ward. Mr. Sharma awarded full marks to student B. Answer the following questions on the basis of given informations. |
(a) Which values are displayed by Mr. Sharma? |
(b) Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor \[R=10\Omega \] is found to be 60 cm, while that with the unknown resistance X is 70 cm. Determine the value of X. What might you do, if you failed to find a balance point with the given cell \[\varepsilon ?\] |
Answer:
(i) Honesty, courage, concern for teaching profession are the values displayed by Mr. Sharma. (ii) Here, \[R=10\Omega ,\]\[{{l}_{1}}=60cm,\] X = ?, \[{{l}_{2}}=70cm.\]Let, \[{{\varepsilon }_{1}}\] and \[{{\varepsilon }_{2}}\] be the potential drops across R and X, respectively and I be the current in potentiometer wire. Then, \[{{\varepsilon }_{2}}/{{\varepsilon }_{1}}=IX/IR=X/R\] But, \[{{\varepsilon }_{2}}/{{\varepsilon }_{1}}={{l}_{2}}/{{l}_{1}}\] \[\therefore \] \[X/R={{l}_{2}}/{{l}_{1}}\] or \[X=({{l}_{2}}/{{l}_{1}}).R\] \[\Rightarrow \] \[X=\frac{70}{60}\times 10=11.66\Omega \] We will rub on the potentiometer wire AB to find the balance part.
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