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Question Number 35267 by Tinkutara last updated on 17/May/18

Answered by ajfour last updated on 17/May/18

$$q=\left(\pi r_0^2\right)v_0=\left(\pi r^2\right)v$$$$v^2=v_0^2+2gh$$$$\Rightarrow v_0^2{\left(\frac{r_0}{r}\right)}^4=v_0^2+2gh$$$$\Rightarrow r=r_0{\left(\frac{v_0^2}{v_0^2+2gh}\right)}^{1/4}$$$$=4\times {10}^{-3}{\left(\frac{0.16}{0.16+4}\right)}^{1/4}$$$$d=2r=8\times {10}^{-3}{\left(\frac{1}{26}\right)}^{1/4}$$$$=\frac{8}{2.24}\times {10}^{-3}\approx \frac{32}{9}\times {10}^{-3}$$$$hence\left(c\right).$$

Commented by Tinkutara last updated on 17/May/18

$$Why{can}^{\prime }tweuse$$$$\rho {gh}_1+\frac{1}{2}\rho v_1^2=\rho {gh}_2+\frac{1}{2}\rho v_2^2$$$$?$$

Commented by ajfour last updated on 17/May/18

$$yeswecan.$$

Commented by Tinkutara last updated on 17/May/18

$$Butitthengives{v}_2=2.227m/s$$$$andanswer3.39\times {10}^{-3}mwhichis$$$$notexactlycorrect.$$

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