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The external resistance of a circuit is $ \eta $ times higher than the internal resistance of the source. The ratio ofthe potential difference across the terminals of the source to its emf is...........
$ R = \eta .r , V = IR ,\epsilon = I (R + r ) $
A wire has resistance $R_O$, What will its resistance if it is stretched to double, What is its length ?
$ R \alpha (n) l $
The temperature coefficient of resistance of a wire is $ 1.25 \times 10^{-3} / ^\circ c $ . At 300 K, its resistance is What is the temperature at which its resistance becomes $ 2 \Omega $ ?
$ R = ( 1 + \alpha \triangle T ) and T_f = T_0 + \triangle T $
If a copper wire is stretched to make it $ \eta = 0.1 \% $ longer, then what is the percentage change in its resistance ?
$ { dR \over R } = { 2dl \over l } $ (As $ \eta $ = 0.1% is small, dl = 0.1 %)
The reciprocal of resistivity is called
The reciprocal of resistivity (specific resistance) is conducting $ \sigma $ = knowledge base
A 10 m wire potentiometer is connected to an accumulator of steady voltage. A 7.8 m length of it balances the emf of a cell on 'open-circuit'. When cell delivers current through a conductor of resistance $10 \Omega $ it is balanced against 7.0 m of the same poleutomelic. What is the internal resistance of the cell ?
Using the formula desired and used in the experiment $ r = R \left( l_1 - l_2 \over l_2 \right) $
A potentiometer is preferred over a voltmeter to measure the emf of a cell because.....
Potentiometer measures the emf at null deflection of the galvanometer. $ \epsilon -Ir = V $ when $ I = 0 \Rightarrow V = \varepsilon $
If the deflection of the galvanometer is m the same direction at both the ends of the potentio meter wire, then point out the statement which is difinetly not the possible reason for the.
If connecting wire in the galvanometer circuit is broken, the galvanometer will show null deflection, as no current passes through it.
A potentiometer wire, which is 4 m long is connected to the terminals of a battery of steady voltage. A leclanche cell gives a null point at 1m if the length of the potentiometer wire be increased by 1 m, What is the new position of the null point ?
Emf of cell = Potential gradient x null point on potentiometer wire, Hence , $ { V_0 \over 4 } \times 1 = { V_0 \over 5} \times l $
A battery of 2V and internal resistance 1 is used to send a current through a potentiometer wire of length 200 cm and resistance $4 \Omega $. What is the potential gradieut of the wire ?
$ i = { 2 v \over 1 \Omega + 4 \Omega } = { 2 \over 5 } A = 0.4 A $ Potential difference across wire = $ (0.4A)(4\Omega )= 1.6V$ Potential gradient = Poteutial diff . across wire / lenght of potentiometer wire $ = {1.6 v \over 200 cm } = 8 \times 10^{-3} V /cm $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.