Practice free Current Electricity (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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 $
An electric kettle takes 4A current at 220V How much time will it take to boil 1 kg of water from temprature $2 ^\circ C$ ? The temprature of boiling water is $10 ^ \circ C$ .
$ P \times t = mc \triangle \theta $ $ \Rightarrow t = { mc \triangle \theta \over p } = { 4200 m \triangle \theta \over p } = { 4200 \times m \times \triangle \theta \over VI } $ $ \{ \therefore water = 4200 { 1 \over kg } \times {^ \circ C} \} $ $ \Rightarrow t ={ 4200 \times 1 \times (100 -20 ) \over 220 \times 4 }= 381 sec \approx 6.3 min $
The mean free path of electrons in a metal is The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of :
Energy=2eV
Now, electric field E=
=
=
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