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Current Electricity MCQs for NEET — Physics Questions with Answers

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If the voltage across a bulb decreases by 1%, then the percentage change in its power output is :

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$A wire having resistance 16 Ω is stretched to double of its original length . The new resistance of the wire will be :$

$4 Ω$

$8 Ω$

$32 Ω$

$64 Ω$

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Explanation

$R = ρ \frac{l}{A} A s v o l u m e o f t h e w i r e i s c o n s \tan t, t h e n, A l = A' l' . W h e n, l' = n l, t h e n A' = \frac{A}{n} R' = ρ \frac{l'}{A'} = ρ \frac{n l}{\frac{A}{n}} = ρ \frac{n^{2} l}{A} = n^{2} R R' = 2^{2} \times 16 = 64 Ω$

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If a resistance coil is made by joining in parallel two resistances each of 20 $Ω$ . An emf of 2V is applied across this coil for 100 seconds. The heat produced in the coil is

20 J

10 J

40 J

80 J

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$Two battries of emf E_{1}, E_{2} and internal resistance r_{1}, r_{2} are connected i n p a r a l l e l . The effective emf of the circuit across A and B is$

$\frac{E_{1} r_{1} + E_{2} r_{2}}{r_{1} + r_{2}}$

$\frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}$

$\frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} \times r_{2}}$

$\frac{E_{1} r_{1} + E_{2} r_{2}}{r_{1} - r_{2}}$

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Explanation

$I = I_{1} + I_{2} R = E_{e f f} - I R P o t e n t i a l a c r o s s u p p e r b r a n c h = E_{1} - I r_{1} P o t e n t i a l a c r o s s l o w e r b r a n c h = E_{2} - I r_{2} H e n c e, E_{e f f} = (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) R_{e q} A n d R_{e q} = \frac{r_{1} r_{2}}{(r_{1} + r_{2})} P u t t i n g t h e v a l u e o f R_{e q} E_{e f f} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}$

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A wire has resistance 24 $Ω$ . It is bent in the form of a circle. The effective resistance between the two point on the diameter of the circle is

6 $Ω$

4 $Ω$

3 $Ω$

12 $Ω$

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Explanation

$T h e r e a r e t w o h a l f s e m i - c i r c l e s b e t w e e n t h e p o i n t s o n d i a m e t e r o d t h e c i r c l e . R e s i s \tan c e o f e a c h s e m i - c i r c l e = 12 Ω B o t h s e m i - c i r c l e s a r e i n p a r a l l e l b e t w e e n t h e d i m a t e r i c a l l y o p p o s i t e p o i n t s . N e t R = \frac{12 \times 12}{12 + 12} = 6 Ω$

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The drift velocity of the electrons in a current-carrying metallic conductor is of the order of

$10^{4} m / s$

$10^{8} m / s$

$10^{0} m / s$

$10^{- 4} m / s$

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Explanation

4. $10^{- 4} m / s$

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If I be the current limit of a fuse wire of length l and radius r, then select the appropriate relation

$I α l$

$I α r^{2}$

$I α l^{0}$

$I α r^{0}$

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Explanation

$(3) T h e t o t a l h e a t p e r u n i t t i m e p r o d u c e d b y a w i r e o f r e s i s \tan c e R i s g i v e n b y : P = I^{2} R I f t h e w i r e i s c o m p l e t e l y u n i f o r m t h e n : R = ρ \frac{L}{A} H e n c e, P = ρ \frac{L}{A} I^{2} H e a t r e l e a s e d p e r u n i t l e n g t h i s : P_{L} = \frac{ρ}{A} I^{2} \Rightarrow I = \sqrt{\frac{P_{L} A}{ρ}} S o t h e c u r r e n t l i m i t o f f u s e w i r e d o e s n o t d e p e n d o n i t s l e n g t h .$

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$The resistance of a platinum wire at 0 ° C is 22.05 Ω and at 100 ° C it becomes 22.70 Ω . When the wire is heated to a temperature o f t ° C the resistance of wire becomes 24.91 Ω . The value of t is$

220

440

310

550

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Explanation

$(2) t = [\frac{R_{t} - R_{0}}{R_{100} - R_{0}}] x 100 = [\frac{24.91 - 22.05}{22.70 - 22.05}] x 100 = 440$

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$Two wire A and B of copper have equal masses and the lengths are 10 cm and 20 cm respectively . If resistance o f A be R_{A} and B be R_{B}, then the value of \frac{R_{A}}{R_{B}} is$

$\frac{1}{2}$

$2$

$\frac{1}{4}$

$4$

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Explanation

$R = ρ . \frac{l}{A} = ρ . \frac{l^{2}}{Al} = ρ . \frac{l^{2}}{V} R = ρ . \frac{{dl}^{2}}{v . d} (d = density) R = ρ . \frac{l^{2} d}{m} R \propto \frac{l^{2}}{m} \frac{R_{A}}{R_{B}} = {(\frac{10}{20})}^{2} = \frac{1}{4}$

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The sensitivity of a potentiometer can be increased by

increasing the potential gradient

decreasing the length of potentiometer wire

decreasing the current in the potentiometer wire

all of these

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Explanation

$(3) Sensitivity \propto \frac{1}{Potential gradient}$

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