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QJEEM19Ph651e: JEE Main 2019 Physics - Class 12 JEEM19Ph651e Solution: The resistance of the meter bridge AB in given figure is 4 .…

The resistance of the meter bridge AB in given figure is 4 Ω\Omega . With a cell of emf ε\varepsilon = 0.5 V and rheostat resistance Rh = 2 Ω\Omega the null point is obtained at some point J. When the cell is replaced by another one of emf ε\varepsilon = ε\varepsilon 2 the same null point J is found for Rh = 6 Ω\Omega . The emf ε\varepsilon 2 is, :

JEE Main 2019 (Online) 11th January Morning Slot Physics - Current Electricity Question 260 English

Multiple Choice Options

  • A 0.3 V
  • B 0.6 V
  • C 0.5 V
  • D 0.4 V

Step-by-Step Verified Solution

Potential gradient with Rh = 2 Ω\Omega

is (62+4)×4L=dVdL;\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};   L == 100 cm

Let null point be at ℓ\ell cm

thus ε\varepsilon 1 == 0.5V == (62+4)×4L×ℓ\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell       . . .(1)

Now with Rh == 6 Ω\Omega new potential gradient is

(64+6)×4L\left( {{6 \over {4 + 6}}} \right) \times {4 \over L} and at null point

(64+6)(4L)×ℓ=ε2\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}                           . . .(2)

dividing equation (1) by (2) we get

0.5ε2=106{{0.5} \over {{\varepsilon _2}}} = {{10} \over 6} thus ε2=0.3{\varepsilon _2} = 0.3

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