Water flows through the tube shown. Area of c | vedclass

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Question

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}$

$\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2}

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

$\mathrm{P}_{1}+\frac{1}{2} ho \mathrm{V}_{2}^{2}=\mathrm{P}_{2}+\frac{1}{2} ho \mathrm{V}_{2}^{2}$

$\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2} ho\left(\mathrm{V}_{2}^{2}-\mathrm{v}_{1}^{2}ight)$

$\mathrm{h} 	imes 13600 	imes \mathrm{g}=\frac{1}{2} 	imes 1000\left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}ight)$

$A_{1} V_{1}=A_{2} V_{2}=\frac{d(	ext { volume })}{d t}$

$\mathrm{V}_{1}=\frac{500}{5}=100 \mathrm{cm} / \mathrm{s}=1 \mathrm{m} / \mathrm{s}$

$\mathrm{V}_{2}=\frac{500}{2}=250 \mathrm{cm} / \mathrm{s}=2.5 \mathrm{m} / \mathrm{s}$

$\mathrm{h}=\frac{\frac{1}{2} 	imes 1000(0.25-1)}{13600 	imes 10} \mathrm{m}$

$\mathrm{h}=\frac{1}{2} 	imes \frac{5.25 	imes 100}{136} \mathrm{cm}$

Water flows through the tube shown. Area of cross-section of wide and narrow part are $5$ $cm^2$ $\&$ $2$ $cm^2$. The rate of flow is $500$ $cm^3/sec$. Find difference in mercury level of $U-$ tube .......... $cm$

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