Water is flowing through the tube depicted in the figure. The...

Given:
Cross-sectional area of wide portion, $A_1 = 5 \text{ cm}^2$
Cross-sectional area of narrow portion, $A_2 = 2 \text{ cm}^2$
Rate of flow of water, $Q = 500 \text{ cm}^3 \text{ s}^{-1}$
Density of water, $\rho_w = 1 \text{ g cm}^{-3}$
Density of mercury, $\rho_m = 13.6 \text{ g cm}^{-3}$
Acceleration due to gravity, $g = 980 \text{ cm s}^{-2}$

Velocity of water in the wide portion:
$v_1 = \dfrac{Q}{A_1} = \dfrac{500}{5} = 100 \text{ cm s}^{-1}$

Velocity of water in the narrow portion:
$v_2 = \dfrac{Q}{A_2} = \dfrac{500}{2} = 250 \text{ cm s}^{-1}$

According to Bernoulli's principle for a horizontal tube:
$P_1 + \dfrac{1}{2}\rho_w v_1^2 = P_2 + \dfrac{1}{2}\rho_w v_2^2$
$\Rightarrow P_1 - P_2 = \dfrac{1}{2}\rho_w (v_2^2 - v_1^2)$

Substituting the values:
$P_1 - P_2 = \dfrac{1}{2} \times 1 \times (250^2 - 100^2)$
$P_1 - P_2 = \dfrac{1}{2} \times (62500 - 10000) = 26250 \text{ dyne cm}^{-2}$

The pressure difference is also given by the difference in mercury levels in the U-tube. Since water is present above the mercury in both arms, the gauge pressure difference is:
$P_1 - P_2 = h(\rho_m - \rho_w)g$

Substituting the values:
$26250 = h(13.6 - 1) \times 980$
$26250 = h \times 12.6 \times 980$
$26250 = 12348 h$
$h = \dfrac{26250}{12348} \approx 2.1258 \text{ cm}$

Rounding to two decimal places, we get $h \approx 2.13 \text{ cm}$.

Answer: $2.13 \text{ cm}$