• # question_answer  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 $\varepsilon ?$

(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.