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JEE Main 2018 (15 Apr, Shift 1) · Wheatstone Bridge & Metre Bridge

Wheatstone Bridge & Metre Bridge question from JEE Main 2018 (15 Apr, Shift 1)

Q74JEE Main 2018 · 15 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

Figure

In a meter bridge, as shown in the figure, it is given that resistance Y=12.5 ΩY = 12.5\ \Omega and that the balance is obtained at a distance 39.5 cm39.5\text{ cm} from end AA (by Jockey J). After interchanging the resistances XX and YY, a new balance point is found at a distance l2l_2 from end A.A. What are the value of XX and l2l_2?

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Answer: (a)

  1. XY=39.560.5⇒X=12.5×39.560.5=8.16 Ω\dfrac{X}{Y} = \dfrac{39.5}{60.5} \Rightarrow X = 12.5 \times \dfrac{39.5}{60.5} = 8.16\ \Omega.
  2. After interchanging, the balance length from A becomes the old right length: l2=60.5 cml_2 = 60.5\text{ cm}.
  3. So the answer is (a) 8.16 Ω8.16\ \Omega and 60.5 cm60.5\text{ cm}.
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