Question

Two capillary tubes of length 15 cm and 5 cm and radii 0.06 cm and 0.02 cm respectively are connected is series. If the pressure difference accros the end faces is equal to pressure of 14 cm high water column then find the pressure difference across the (i) first tube and (ii) second tube.

• A. 0.5,14.5 cm of water column
• B. 1.5,12.5 cm of water column
• C. 2.5,11.5 cm of water column
• D. 3.5,10.5 cm of water column

Answer 1

The correct option is A 0.5,14.5 cm of water column

$v=\frac{P_1\times {\sin }^{4}x\pi }{8{\eta }^x{\ell }_1}=\frac{P_2\times {91}_2^4\times \pi }{8\eta \times {\ell }_2}$

$v=\frac{\pi \times (15-h)\times \mathrm{sg}\times r_1^4}{8\eta \times l_1}$

$v=\frac{\pi \times \left(h\mathrm{sg}\right)\times r_2^4}{8\eta \times l_2}$

$\frac{\pi \times (15-h)\times {\int }_g\times r_1^4}{8\eta \times l_1}=\frac{\pi \times \left(h{\rho }_g\right)\times r_2^4}{8\eta \times l_2}$

$\frac{(15-h)\times r_1^4}{l_1}=\frac{h\times r_2^4}{l_2}$

$(15-h)\times \frac{10\cdot 614}{15}=\frac{h\times 10.0214}{5}$

$h=14.46cm$

(i) The pressure difference across the first tobe $=15-14.46$

$=0.5cm$.

(ii) The pressure difference across the second tube $=14.46cm$.