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A person of height $1.65 \mathrm{~m}$ is standing upright. The additional external force required by blood vessel of length $1 \mathrm{~cm}$, diameter $1 \mathrm{~mm}$ at feet to balance the pressure compared to similar blood vessel in head is (Density of blood $=1.1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$0.57 \mathrm{~N}$
Pressure difference $\Delta \mathrm{p}=\rho \mathrm{gh}$ and force exerted
$$
\begin{aligned}
& \mathrm{F}=\Delta \mathrm{P} . \mathrm{A}=\rho \mathrm{gh} .2 \pi \mathrm{r}(\mathrm{r}+\mathrm{h}) \\
& =1.1 \times 10^3 \times 10 \times 1.65 \times 2 \times \frac{22}{7} \times 0.5 \times 10^{-3} \\
& \left(0.5 \times 10^{-3}+\mathrm{h}\right) \\
& =18150 \times 2 \times \frac{22}{7} \times 0.5 \times 10^{-3}\left(0.5 \times 10^{-3}+10 \times\right. \\
& \left.10^{-3}\right) \\
& \therefore \mathrm{F} \cong 0.57 \mathrm{~N} .
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{F}=\Delta \mathrm{P} . \mathrm{A}=\rho \mathrm{gh} .2 \pi \mathrm{r}(\mathrm{r}+\mathrm{h}) \\
& =1.1 \times 10^3 \times 10 \times 1.65 \times 2 \times \frac{22}{7} \times 0.5 \times 10^{-3} \\
& \left(0.5 \times 10^{-3}+\mathrm{h}\right) \\
& =18150 \times 2 \times \frac{22}{7} \times 0.5 \times 10^{-3}\left(0.5 \times 10^{-3}+10 \times\right. \\
& \left.10^{-3}\right) \\
& \therefore \mathrm{F} \cong 0.57 \mathrm{~N} .
\end{aligned}
$$
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