Water flows through the tube shown. The cross | vedclass

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(D) Using the equation of continuity,$A_1 V_1 = A_2 V_2 = Q$.Given $A_1 = 5 	ext{ cm}^2$,$A_2 = 2 	ext{ cm}^2$,and $Q = 500 	ext{ cm}^3/	ext{s}$.$

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(D) Using the equation of continuity,$A_1 V_1 = A_2 V_2 = Q$.Given $A_1 = 5 	ext{ cm}^2$,$A_2 = 2 	ext{ cm}^2$,and $Q = 500 	ext{ cm}^3/	ext{s}$.$V_1 = \frac{Q}{A_1} = \frac{500}{5} = 100 	ext{ cm/s} = 1 	ext{ m/s}$.$V_2 = \frac{Q}{A_2} = \frac{500}{2} = 250 	ext{ cm/s} = 2.5 	ext{ m/s}$.Applying Bernoulli's equation: $P_1 + \frac{1}{2} ho_w V_1^2 = P_2 + \frac{1}{2} ho_w V_2^2$.$P_1 - P_2 = \frac{1}{2} ho_w (V_2^2 - V_1^2)$.The pressure difference is balanced by the mercury column: $P_1 - P_2 = h(ho_m - ho_w)g$.$h = \frac{\frac{1}{2} ho_w (V_2^2 - V_1^2)}{(ho_m - ho_w)g}$.Using $ho_w = 1000 	ext{ kg/m}^3$,$ho_m = 13600 	ext{ kg/m}^3$,$g = 10 	ext{ m/s}^2$,$V_1 = 1 	ext{ m/s}$,$V_2 = 2.5 	ext{ m/s}$.$h = \frac{0.5 	imes 1000 	imes (2.5^2 - 1^2)}{(13600 - 1000) 	imes 10} = \frac{500 	imes (6.25 - 1)}{126000} = \frac{500 	imes 5.25}{126000} = \frac{2625}{126000} 	ext{ m} \approx 0.0208 	ext{ m} = 2.08 	ext{ cm}$.Since $2.08 	ext{ cm}$ is not among the options,the correct answer is $D$.

Water flows through the tube shown. The cross-sectional areas of the wide and narrow parts are $5 \text{ cm}^2$ and $2 \text{ cm}^2$,respectively. The rate of flow is $500 \text{ cm}^3/\text{s}$. Find the difference in the mercury level of the $U$-tube in $\text{cm}$.

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