In a potentiometer experiment, the balancing length for a cell is 60 cm. When the cell is shunted by a 2Ω resistor, the balancing length becomes 40 cm. The internal resistance of the cell is: MCQ with Answer and Explanation
In a potentiometer experiment, the balancing length for a cell is 60 cm. When the cell is shunted by a 2Ω resistor, the balancing length becomes 40 cm. The internal resistance of the cell is:
A. 3Ω
B. 1Ω
C. 4Ω
D. 2Ω
Answer: Option B
Solution (By JKSSB Mock Tests)
Potentiometer measures EMF and internal resistance. Formula: r = R(L₁ - L₂)/L₂, where R is shunt resistance, L₁ initial balancing length, L₂ after shunting. Here R=2Ω, L₁=60 cm, L₂=40 cm. Thus r = 2×(60-40)/40 = 2×20/40 = 1Ω. This derives from terminal voltage V = E - Ir and potentiometer principle V ∝ balancing length. Memory tip: 'r = R(L₁/L₂ - 1)'. This advanced numerical tests potentiometer applications, common in competitive exams for experimental physics. Always verify the formula derivation to avoid misapplication.
Explanation:
Time of ascent = u/g = 4 s. Distance in last second (4th) = ½g(1)²? Actually, during last second of upward journey, velocity becomes zero. Distance = ½g(1)² = 5 m. Formula: s = u - g/2 (2n-1) with n=4? Let's compute: s_nth = u - (g/2)(2n-1). For n=4 (4th second), s = 40 - 5×7 = 40 - 35 = 5 m. Yes.
Explanation:
The relationship between phase difference (phi) and path difference (delta x) is given by the formula phi = (2pi / lambda) * delta x. Given a path difference of lambda / 2, substitute this into the formula: phi = (2pi / lambda) * (lambda / 2) = pi radians. This indicates strictly destructive interference.
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