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AIIMS 2010
Exam
AIIMS
Year
2010
Subject
Physics
Chapter
Current Electricity

AIIMS 2010 · Physics · Current Electricity

A galvanometer has a current sensitivity of 1 ~mA per division. A...

A galvanometer has a current sensitivity of 1 mA1 \mathrm{~mA} per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of 500Ω500 \Omega and cell of internal resistance 1Ω1 \Omega. It gives a deflection of 5 division for shunt of 5ohm5 \mathrm{ohm} and 20 division for shunt of 25ohm25 \mathrm{ohm}. The emf of cell isA galvanometer has a current sensitivity of 1 ~mA per division. A variable shunt is connected across the galvanometer

Options

  1. A47.1 V47.1 \mathrm{~V}
  2. B57.1 V57.1 \mathrm{~V}
  3. C67.1 V67.1 \mathrm{~V}
  4. D77.1 V77.1 \mathrm{~V}

Correct answer

A. 47.1 V47.1 \mathrm{~V}

Step-by-step solution

A galvanometer has a current sensitivity of 1 ~mA per division. A variable shunt is connected across the galvanometer
Here, I=ER+r+GSG+SI=\frac{E}{R+r+\frac{G S}{G+S}} and Ig=ISG+SI_g=\frac{I S}{G+S}Ig=E(R+r)+GS(G+S)×S(G+S)∴Ig=ES(R+r)(G+S)+GS\begin{aligned}& I_g=\frac{E}{(R+r)+\frac{G S}{(G+S)}} \times \frac{S}{(G+S)}\\ & \therefore \quad I_g=\frac{E S}{(R+r)(G+S)+G S} \end{aligned}For $S=5 \mathrm{ohm}, I_g=5 \times 10^{-3} \mathrm{~A}$ and for $S=25 \mathrm{ohm}I_g=20 \times 10^{-3} \mathrm{~A}$ Hence, $5 \times 10^{-3}=\frac{E \times 5}{501(G+5)+5 G}$ and $20 \times 10^{-3}=\frac{E \times 25}{501(G+25)+25 G}$ Dividing and solving, G=88.2 \Omega From (i), we get\begin{aligned}E & =10^{-3}[501(88.2+5)+5 \times88.2] \\ & =47.1 \text{ volt } \end{aligned}$$
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