Question 2: Properties of Matter

Question

Determine the pressure difference in tube of
non-uniform cross sectional area as shown
in figure.
$\Delta P=?, d_1=5 \mathrm{~cm}, v_1=4 \mathrm{~m} / \mathrm{s}, d_2=2 \mathrm{~cm}, v_2=?$
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m0dyspnl/09b5f4e0-2eab-4b37-a5c6-6fb52c17a672/ace7f510-654b-11ef-a62c-813be17446ad/file-jaoe38c1m0dyspnm.png?format=png)

Accepted Answer

Given :
Diameter at the first end, $ d_1 = 5 \, 	ext{cm} $
Velocity at the first end, $ v_1 = 4 \, 	ext{m/s} $
Diameter at the second end, $ d_2 = 2 \, 	ext{cm} $
$ \Delta P = ? $
Step 1 : Calculate the radii
$ r_1 = \frac{d_1}{2} = 2.5 \, 	ext{cm} = 2.5 	imes 10^{-2} \, 	ext{m} $
$ r_2 = \frac{d_2}{2} = 1 \, 	ext{cm} = 10^{-2} \, 	ext{m} $
Step 2 : Apply the principle of continuity
$ A_1 v_1 = A_2 v_2 $
$ \pi r_1^2 v_1 = \pi r_2^2 v_2 $
$ \left(2.5 	imes 10^{-2}ight)^2 	imes 4 = \left(10^{-2}ight)^2 \cdot v_2 $
$ v_2 = 25 \, 	ext{m/s} $
Step 3 : Use Bernoulli's theorem
$ p_1 + \frac{1}{2} ho v_1^2 = p_2 + \frac{1}{2} ho v_2^2 $
$ p_1 - p_2 = \frac{1}{2} ho \left(v_2^2 - v_1^2ight) $
Given $ ho = 10^3 \, 	ext{kg/m}^3 $:
$ p_1 - p_2 = \frac{1}{2} 	imes 10^3 \left(25^2 - 4^2ight) $
$ = 304500 \, 	ext{Pa} $
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m0dyspnl/09b5f4e0-2eab-4b37-a5c6-6fb52c17a672/ace7f510-654b-11ef-a62c-813be17446ad/file-jaoe38c1m0dyspnm.png?format=png)

Answer

304200 Pa

Answer

304500 Pa

Answer

302500 Pa

Answer

303500 Pa

Properties of Matter

Determine the pressure difference in tube of non-uniform cross sectional area as shown in figure. \Delta P=?, d_1=5 \mathrm{~cm}, v_1=4 \mathrm{~m} / \mathrm{s}, d_2=2 \mathrm{~cm}, v_2=?

Choose the correct answer: