The conical container with a thin horizontal tube attached at the bottom, shown in the figure, is to be used to measure the viscosity of an oil. The flow through the tube is laminar. The discharge time needed for the oil level to drop from $h_1$ to $h_2$ is to be measured by a stopwatch. Choose the correct expression for the viscosity of oil in the container as a function of the discharge time $t$

Multiple Choice

$=\frac{A_h{d}^{2}T}{16L(\frac{R}{H}{)}^2}\frac{1}{h_1^2+h_2^2}$

$=\frac{A_h{d}^{2}T}{4L(\frac{n}{H}{)}^2}\frac{1}{h_1^2-h_2^2}$