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plot grapb t / v e V\begin{array}{l}t /=1 \times b^{-4} \gamma+0.01 \\P=100 \mathrm{kPa} .\end{array}i)\begin{array}{l}t / v=k v+B \\k \propto \frac{1}{\Delta P} \text { and } B \propto \frac{1}{\Delta p} \\\text { So when } \Delta p=50 \mathrm{kPa} \\t / v=\frac{1 \times 10^{-4}}{0.5} v+\frac{0.01}{0.5}\end{array}put V=100 \mathrm{ml}t=45b) filteration rate =\frac{d v}{d t} \cdot 1 / B+2 k x (from the table) at end\begin{aligned}d v /\left.a t\right|_{F} & =\frac{1}{\frac{0.01}{0.5}+\frac{2 \times 1 \times 10^{-4} \times}{0.5}} 200 \\& =20\end{aligned} at start (from the table)\begin{aligned}\left.d \frac{a t}{}\right|_{F} & =\frac{1}{\frac{0.01}{0.5}+\frac{2 \times 1 \times 10^{-4} \times 1}{0.5}} \\& =49.02\end{aligned}(ii) New \frac{1}{V} V_{S} V equation.\begin{array}{l}\Delta p=25 k p a \\f / v=\frac{1 \times 10^{-4}}{0.25} \times v+\frac{0.01}{.25} \\\frac{d v}{d t}=\frac{1}{\frac{0.01}{0.25}+\frac{2 \times 1 \times 10^{-4}}{0.25} \times 200}=5\end{array}time: \left.\frac{0.1 \times 1000}{d y / a b}\right|_{F}\begin{array}{l}=\frac{0.1 \times 1000}{5} \\=205\end{array} -------------------------------------------------------------------------------------------- Really iam trying my best to answer your questions. So, Please dont dislike because my cf score is less. If you really didnt get the answer for your question, comment below so that i will check and rectify it. Once again iam requesting you please dont dislike my answers, as i am trying my best for you. ...