Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Current Electricity question

2021 · 26 Feb · Shift 2 · Q55
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Current Electricity
  5. /2021 · 26 Feb · Shift 2 · Q55

Current Electricity question

2021 · 26 Feb · Shift 2 · Q55

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A wire of 1 Ω\OmegaΩ has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :
  1. A
    76%
  2. B
    12.5%
  3. C
    25%
  4. D
    56%
View written solutionFree

Correct answer: D

  1. Given:

    • Initial resistance: R1=1 ΩR_1 = 1\,\OmegaR1​=1Ω
    • Initial length: L1=1 mL_1 = 1\,\text{m}L1​=1m
    • Final length increases by 25%25\%25%

    So, L2=1.25L1L_2 = 1.25L_1L2​=1.25L1​

  2. Use the relation for resistance: R=ρLAR = \rho \frac{L}{A}R=ρAL​

    When a wire is stretched, its volume remains constant: A1L1=A2L2A_1L_1 = A_2L_2A1​L1​=A2​L2​

    Therefore, A2=A1L1L2=A111.25=0.8A1A_2 = A_1\frac{L_1}{L_2} = A_1\frac{1}{1.25} = 0.8A_1A2​=A1​L2​L1​​=A1​1.251​=0.8A1​

  3. Find the new resistance: R2=ρL2A2R_2 = \rho \frac{L_2}{A_2}R2​=ρA2​L2​​

    Divide by R1=ρL1A1R_1 = \rho \frac{L_1}{A_1}R1​=ρA1​L1​​: R2R1=L2/A2L1/A1=L2L1⋅A1A2\frac{R_2}{R_1} = \frac{L_2/A_2}{L_1/A_1} = \frac{L_2}{L_1}\cdot \frac{A_1}{A_2}R1​R2​​=L1​/A1​L2​/A2​​=L1​L2​​⋅A2​A1​​

    Substitute values: R2R1=1.25×10.8=1.25×1.25=1.5625\frac{R_2}{R_1} = 1.25 \times \frac{1}{0.8} = 1.25 \times 1.25 = 1.5625R1​R2​​=1.25×0.81​=1.25×1.25=1.5625

    Hence, R2=1.5625R1R_2 = 1.5625R_1R2​=1.5625R1​

  4. Percentage change in resistance: % change=R2−R1R1×100\%\text{ change} = \frac{R_2 - R_1}{R_1} \times 100% change=R1​R2​−R1​​×100

    =(1.5625−1)×100=0.5625×100=56.25%= (1.5625 - 1)\times 100 = 0.5625 \times 100 = 56.25\%=(1.5625−1)×100=0.5625×100=56.25%

    To the nearest integer, 56%56\%56%

  5. Option check:

    • A: 76%76\%76% ❌
    • B: 12.5%12.5\%12.5% ❌
    • C: 25%25\%25% ❌
    • D: 56%56\%56% ✅

Final Answer: 56%\boxed{56\%}56%​ (Option D)

PreviousNext

More from Current Electricity

  • Five identical cells each of internal resistance 1 Ω and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?2021 · MCQ
  • First, a set of n equal resistors of 10 Ω each are connected in series to a battery of emf 20V and internal resistance 10 Ω. A current I is observed to flow. Then, the n resistors are connected in parallel to the same…2021 · Numerical
  • For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :2021 · MCQ
  • The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x : 8. The value of x is ​. Includes diagram2021 · Numerical
  • In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sections having radii r1 and r2 (r2 < r1). Choose the correct option as one moves from P to Q : Includes diagram2021 · MCQ
  • The resistance of a conductor at 15 ∘ C is 16 Ω and at 100 ∘ C is 20 Ω. What will be the temperature coefficient of resistance of the conductor?2021 · MCQ
  • For the circuit shown, the value of current at time t = 3.2 s will be ​ A. [Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)] Includes diagram2021 · Numerical
  • Consider a galvanometer shunted with 5 Ω resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?2021 · MCQ