Current Electricity

JEE Main level, single correct, medium

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Question 304 of 380

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P=4ΩP=4\Omega and the neutral point NN is at 60cm60cm from AA . Now an unknown resistance RR is connected in series to PP and the new position of the neutral point is at 80cm80cm from AA . The value of unknown resistance RR is -

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Answer: (C) 203Ω\dfrac{20}{3}\Omega

Given initially a resistance P=4ΩP=4\Omega is connected in the left gap and let QQ is connected in the right gap of meter bridge and the null point is obtained at 60cm60cm,
From the balanced condition of the Wheatstone bridge
PQ=RS...(1)\dfrac{P}{Q}=\dfrac{R}{S}...\left(1\right)
Let ρ\rho is the resistance per unit length of meter bridge wire than
For wire ANAN, P=ρl=40ρ(∵l=40cm)P=\rho l=40\rho \left(∵l=40cm\right)
And for NB wire
Q=ρ(100−l)Q=ρ(100−60)=40ρQ=\rho \left(100-l\right) Q=\rho \left(100-60\right)=40\rho
On substituting the values from equation (1)\left(1\right)
460=Q40\dfrac{4}{60}=\dfrac{Q}{40}
⇒Q=166=83Ω\Rightarrow Q=\dfrac{16}{6}=\dfrac{8}{3}\Omega
Now an unknown resistance RR is connected in series with the 4Ω4\Omega resistance and the null point is obtained at l=80cml=80cm
Now, P=ρl=80lP=\rho l=80l
and Q=ρ(100−l)Q=ρ(100−80)=20ρQ=\rho \left(100-l\right) Q=\rho \left(100-80\right)=20\rho
Then from equation (1)\left(1\right)
4+R80=Q20⇒4+R80=8320(∵Q=83)⇒4+R80=860⇒R=203Ω\dfrac{4+R}{80}=\dfrac{Q}{20} \Rightarrow \dfrac{4+R}{80}=\dfrac{\dfrac{8}{3}}{20}\left(∵Q=\dfrac{8}{3}\right) \Rightarrow \dfrac{4+R}{80}=\dfrac{8}{60} \Rightarrow R=\dfrac{20}{3}\Omega

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