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plot grapb t/v V/s V\begin{array}{l}t / \mathrm{v}=1 \times 0^{-4} \mathrm{v}+0.01 \\P=100 \mathrm{k} \mathrm{Pa}\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_{2} 100 \mathrm{ml}t=4 S.b) filteration rate =\frac{d v}{d t}+\frac{1}{B+2 k x} (from the table) at end\begin{aligned}d y /\left.d 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}d y /\left.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}iii) New \frac{f}{v} V_{s} V equation.\begin{array}{l}\Delta P=25 \mathrm{kPa} \\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 \\\text { time }=\frac{0.1 \times 1000}{d x /\left.d b\right|_{F}} \\=\frac{0.1 \times 1000}{5} \\=205 \\\end{array} ...