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Current Electricity question

2017 · 9 Apr · Shift 1 · Q70
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Current Electricity question

2017 · 9 Apr · Shift 1 · Q70

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 Ω\OmegaΩ and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is : JEE Main 2017 (Online) 9th April Morning Slot Physics - Current Electricity Question 288 English
  1. A
    335 Ω{{33} \over 5}\,\Omega533​Ω
  2. B
    6  Ω\,\OmegaΩ
  3. C
    7  Ω\,\OmegaΩ
  4. D
    203 Ω{{20} \over 3}\,\Omega320​Ω
View written solutionFree

Correct answer: D

  1. Meter bridge balance condition

For a meter bridge at balance point,

left gap resistanceright gap resistance=l100−l\frac{\text{left gap resistance}}{\text{right gap resistance}}=\frac{l}{100-l}right gap resistanceleft gap resistance​=100−ll​

where lll is the balancing length from end AAA.

Let the resistance in the other gap be QQQ.

  1. Initial condition

Initially,

So,

PQ=6040=32\frac{P}{Q}=\frac{60}{40}=\frac{3}{2}QP​=4060​=23​

Thus,

4Q=32\frac{4}{Q}=\frac{3}{2}Q4​=23​ Q=83 ΩQ=\frac{8}{3}\,\OmegaQ=38​Ω
  1. After adding unknown resistance RRR in series with PPP

Now the left gap resistance becomes

P+R=4+RP+R=4+RP+R=4+R

and the new balance length is

l=80 cml=80\text{ cm}l=80 cm

Hence,

4+RQ=8020=4\frac{4+R}{Q}=\frac{80}{20}=4Q4+R​=2080​=4

Substitute Q=83 ΩQ=\frac{8}{3}\,\OmegaQ=38​Ω:

4+R8/3=4\frac{4+R}{8/3}=48/34+R​=4 4+R=4⋅83=3234+R=4\cdot \frac{8}{3}=\frac{32}{3}4+R=4⋅38​=332​ R=323−4=323−123=203 ΩR=\frac{32}{3}-4=\frac{32}{3}-\frac{12}{3}=\frac{20}{3}\,\OmegaR=332​−4=332​−312​=320​Ω
  1. Check with options
R=203 ΩR=\frac{20}{3}\,\OmegaR=320​Ω

So the correct option is D.

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