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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Physics
Chapter
Capacitance

JEE Advanced 2008 · Paper 2 · Physics · Capacitance

A parallel plate capacitor C with plates of unit area and...

A parallel plate capacitor CC with plates of unit area and separation dd is filled with a liquid of dielectric constant K=2K=2. The level of liquid is d3\frac{d}{3} initially. Suppose the liquid level decreases at a constant speed vv, the time constant as a function of time tt is A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant

Options

  1. A6ε0R5d+3vt\frac{6 \varepsilon_0 R}{5 d+3 v t}
  2. B(15d+9vt)ε0R2d2−3dvt−9v2t2\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}
  3. C6ε0R5d−3vt\frac{6 \varepsilon_0 R}{5 d-3 v t}
  4. D(15d−9vt)ε0R2d2+3dvt−9v2t2\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}

Correct answer

A. 6ε0R5d+3vt\frac{6 \varepsilon_0 R}{5 d+3 v t}

Step-by-step solution

After time tt, thickness of liquid will remain \left(\frac{d}{3}-v t\right). Now, time constant as function of timetimeτc=CR=ε0(1)⋅R(d−d3+vt)+d/3−vt2 Applying C=ε0Ad−t+tk)=6ε0R5d+3vt\begin{aligned} \tau_c & =C R \\ &\left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+ v t\right)+\frac{d / 3-v t}{2} } \quad \quad \text{ Applying } C=\frac{\varepsilon_0 A}{d-t+ \frac{t}{k} }\right) \\ & = \frac{6 \varepsilon_0 R}{5 d+3 v t} \end{aligned}$$ \therefore correct option is (a).
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