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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Current Electricity

BITSAT 2022 · Physics · Current Electricity

In the arrangement shown in figure, when the switch S_2 is open,...

In the arrangement shown in figure, when the switch S2S_2 is open, the galvanometer, shows no deflection for l=50cml=50 \mathrm{ cm} when the switch S2S_2 is closed, the galvanometer shows no deflection for l=0.416ml=0.416 \mathrm{ m}. The internal resistance (r)(r) of 6V6 \mathrm{ V} cell is In the arrangement shown in figure, when the switch S_2 is open, the galvanometer, shows no deflection for l=50 cm when

Options

  1. A2Ω2 \Omega
  2. B3Ω3 \Omega
  3. C5Ω5 \Omega
  4. D9Ω9 \Omega

Correct answer

A. 2Ω2 \Omega

Step-by-step solution

When S2S_2 open. Assume resistance of XY=RX Y=R. Resistance of wire per unit length, x=RL=RΩ\mathrmm−1∵I=E0/Rx=\frac{R}{L}=R \Omega \mathrmm^{-1}\because \quad I=E_0 / R Now, the potential drop across 50cm50 \mathrm{ cm} length is 6V6 \mathrm{ V}, So E0R×R×50100=6⇒ and E0=12V\begin{array}{cc} \frac{E_0}{R} \times R \times \frac{50}{100}=6 \\ \Rightarrow \text{ and } E_0=12 \mathrm{ V} \end{array} When S2S_2 closed, potential drop across 0.416cm0.416 \mathrm{ cm} length, V_1=E0R×\frac{E_0}{R} \times R ×0.416=12×0.416\times0.416=12 \times0.416\approx5V5 \mathrm{ V} Hence, E−Ir=5V\quad E-I r=5 \mathrm{ V}⇒6−Ir=5∵I=510∴ and  and 6−5 and =510r⇒r=2Ω\begin{aligned} \Rightarrow& & 6-I r & =5 \\ & \because & I & =\frac{5}{10}\\ \therefore \text{ and } \text{ and } 6-5 \text{ and } =\frac{5}{10} r \\ \Rightarrow & & r & =2 \Omega \end{aligned}$
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