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JEE Main202224 Jun 2022Evening ShiftPhysicsCurrent ElectricityActual

A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 25 V battery and an external resistance 30 Ω . A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is x 10 . The value of x is _____ .

Correct answer

0

Step-by-step solution

Since the given resistance is in series, so, total resistance R = 20 + 30 = 50   Ω Given, Potential difference = 25   V Current   I = V R = 25 50 = 0 . 5   A Also, Potential gradient = V pd L = I R L Where, V p d = potential drop at wire L = Length of wire Potential gradient = 20 × 0 .5 10 = 1   V   m - 1 So, Emf of the cell is E = Potential gradient × Balancing Length E =   1 ×   250 100 E =   2 . 5   V = 25 10   V Thus, x = 25 .

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