Question

A tank containing a liquid has turns of wire wrapped around it, causing it to act as an inductor. The liquid content of the tank can be measured by using its inductance to determine the height of the liquid level in the tank. The inductance of the tank changes from a value of $L_0$ corresponding to a relative permeability of $1$ when the tank is empty to value $L_f$ corresponding to a relative permeability $X_m$ (relative permeability of liquid) when the tank is full. The appropriate electronic circuit can determine the inductance correct upto $5$ significant figures and thus the effective relative permeability of the combined air and liquid within the rectangular has height $D$. The height of the liquid level in the tank is $d$. Ignore the fringing effects. Assume tank is fitted with $Hg{X}_{Hg}=2.9\times {10}^5$.

Express $d$ as a function of $L$, inductance corresponding to a certain liquid height $L_0,L_f$ and $D$.

• A. $d=\frac{(L-L_0)D}{L_f-L_0}$
• B. $d=\frac{LD}{L_f-L_0}$
• C. $d=\frac{(L-L_0)D}{L_f}$
• D. None of these

Answer 1

Answer: (A)

$d=\frac{(L-L_0)D}{L_f-L_0}$

Inductance of empty tank=$L_0$

Relative permeability=1

Inductance of full tank=$L_f$

Relative permeability=$X_m$

Now,$L_f-L_0\alpha D$

and $L_f-L_0\alpha \frac{1}{d}$

therefore, $(L_f-L_0)\alpha \frac{D}{d}$

or$(L_f-L_0)=k\frac{D}{d}$ [k is the probability constant]

Now when the tank is filled partly with liquid and partly empty,Inductance=$L$

$K=L-L_0$

$L_f-L_0=(L-L_0)\frac{D}{d}$

$d=\frac{(L-L_0)D}{(L_f-L_0)}$