Question
In experiment of the potentiometer wire $\mathrm{AB}$ of length $100 \mathrm{~cm}$ has a resistance of $10 \Omega$. It is connected in series with a resistance $\mathrm{R}$ and a cell of emf 2 volts and of negligible internal resistance. A source emf $10 \mathrm{mV}$ is balanced against a length of $40 \mathrm{~cm}$ of the potentiometer wire. What is the value of the external resistance?(A) $900 \Omega$(B) $820 \Omega$(C) $790 \Omega$(D) $670 \Omega$
Step 1
This can be calculated by dividing the total resistance of the wire by its total length. So, the resistance per unit length (r) is given by: \[ r = \frac{R_{total}}{L_{total}} = \frac{10 \Omega}{100 cm} = 0.1 \Omega/cm \] Show more…
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A potentiometer wire of length $100 \mathrm{~cm}$ has a resistance of $10 \Omega$. It is connected in series with resistance (shown in figure) and a cell of emf $2 \mathrm{~V}$ and negligible resistance. A source of emf $10 \mathrm{mV}$ is balanced against a length of $40 \mathrm{~cm}$ of potentiometer wire. What is the value of $R_{1}$ ? (a) $526.67 \Omega$ (b) $790 \Omega$ (c) $1580 \Omega$ (d) zero
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