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JEE Main 2019 (11 Jan, Shift 2) · Wheatstone Bridge & Metre Bridge

Wheatstone Bridge & Metre Bridge question from JEE Main 2019 (11 Jan, Shift 2)

Q70JEE Main 2019 · 11 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 Ω\Omega resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) Ω\Omega such that the null point shifts back to its initial position is:

Figure

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Answer: (d)

  1. R1R2=4060⇒R1=23R2\dfrac{R_1}{R_2} = \dfrac{40}{60} \Rightarrow R_1 = \dfrac{2}{3}R_2.
  2. Adding 10 Ω10\ \Omega to R1R_1 moves the null point to 50 cm: R1+10=R2R_1 + 10 = R_2.
  3. 23R2+10=R2⇒R2=30 Ω, R1=20 Ω\dfrac{2}{3}R_2 + 10 = R_2 \Rightarrow R_2 = 30\ \Omega,\ R_1 = 20\ \Omega.
  4. Need 30∥R=2030 \parallel R = 20: 30R30+R=20⇒R=60 Ω\dfrac{30R}{30+R} = 20 \Rightarrow R = 60\ \Omega.
  5. So the answer is (d) 60 Ω60\ \Omega.
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