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

2024 · 27 Jan · Shift 1 · Q78
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. /2024 · 27 Jan · Shift 1 · Q78

Current Electricity question

2024 · 27 Jan · Shift 1 · Q78

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A wire of length 10 cm10 \mathrm{~cm}10 cm and radius 7×10−4 m\sqrt{7} \times 10^{-4} \mathrm{~m}7​×10−4 m connected across the right gap of a meter bridge. When a resistance of 4.5Ω4.5 \Omega4.5Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm60 \mathrm{~cm}60 cm from the left end. If the resistivity of the wire is R×10−7Ωm\mathrm{R} \times 10^{-7} \Omega \mathrm{m}R×10−7Ωm, then value of R\mathrm{R}R is :
  1. A
    63
  2. B
    70
  3. C
    66
  4. D
    35
View written solutionFree

Correct answer: C

  1. Use the meter bridge balance condition

For a meter bridge at balance,

PQ=l100−l\frac{P}{Q} = \frac{l}{100-l}QP​=100−ll​

where:

  • P=4.5 ΩP = 4.5\,\OmegaP=4.5Ω (left gap)
  • Q=Q =Q= resistance of the wire in the right gap
  • l=60 cml = 60\,\text{cm}l=60cm

So,

4.5Q=6040=32\frac{4.5}{Q} = \frac{60}{40} = \frac{3}{2}Q4.5​=4060​=23​

Hence,

Q=4.5×23=3 ΩQ = 4.5 \times \frac{2}{3} = 3\,\OmegaQ=4.5×32​=3Ω

So the resistance of the given wire is

Rwire=3 ΩR_{\text{wire}} = 3\,\OmegaRwire​=3Ω
  1. Apply resistance formula

Resistance of a wire is

R=ρLAR = \rho \frac{L}{A}R=ρAL​

Thus,

ρ=RAL\rho = R\frac{A}{L}ρ=RLA​

Given:

  • Length of wire: 10 cm=0.1 m10\,\text{cm} = 0.1\,\text{m}10cm=0.1m
  • Radius: r=7×10−4 mr = \sqrt{7}\times 10^{-4}\,\text{m}r=7​×10−4m

Cross-sectional area:

A=πr2=π(7×10−4)2A = \pi r^2 = \pi \left(\sqrt{7}\times 10^{-4}\right)^2A=πr2=π(7​×10−4)2 A=π⋅7×10−8A = \pi \cdot 7 \times 10^{-8}A=π⋅7×10−8

Now,

ρ=3×7π×10−80.1\rho = 3 \times \frac{7\pi \times 10^{-8}}{0.1}ρ=3×0.17π×10−8​ ρ=3×7π×10−7\rho = 3 \times 7\pi \times 10^{-7}ρ=3×7π×10−7 ρ=21π×10−7 Ωm\rho = 21\pi \times 10^{-7}\,\Omega\text{m}ρ=21π×10−7Ωm

Using π=227\pi = \frac{22}{7}π=722​,

ρ=21×227×10−7\rho = 21 \times \frac{22}{7} \times 10^{-7}ρ=21×722​×10−7 ρ=66×10−7 Ωm\rho = 66 \times 10^{-7}\,\Omega\text{m}ρ=66×10−7Ωm

Therefore,

R=66\mathrm{R} = 66R=66
  1. Check options
  • A: 63
  • B: 70
  • C: 66
  • D: 35

Correct option is C.

PreviousNext

More from Current Electricity

  • A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :2024 · MCQ
  • Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1​,V2​ and V3​. Choose the correct option. Includes diagram2024 · MCQ
  • A current of 200μA deflects the coil of a moving coil galvanometer through 60∘. The current to cause deflection through 10π​ radian is :2024 · MCQ
  • Wheatstone bridge principle is used to measure the specific resistance (S1​) of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is ; S1​=X(Lπr2​)…2024 · MCQ
  • The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24Ω is applied. The resistance of galvanometer coil will be :2024 · MCQ
  • A galvanometer having coil resistance 10Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8 A, the value of the shunt should be:2024 · MCQ
  • The electric current through a wire varies with time as I=I0​+βt, where I0​=20 A and β=3 A/s. The amount of electric charge crossed through a section of the wire in 20 s is :2024 · MCQ
  • In the given circuit, the current in resistance R 3​ is : Includes diagram2024 · MCQ