Question

The cross-sectional area of one limb of a U-tube manometer (figure shown below) is made 500 times larger than the other, so that the pressure difference between the two limbs can be determined by measuring h on one limb of the manometer. The percentage error involved is

• A. 1.0
• B. 0.5
• C. 0.2
• D. 0.05

Answer 1

The correct option is C 0.2

Method I:
Volume drop in large limb
= Volume rise in smaller limb

$A_1\Delta h=A_{2}h$

$or\frac{\Delta h}{h}=\frac{A_2}{A_1}=\frac{1}{500}$

$\%error=\frac{\Delta h}{h}\times 100=\frac{1}{500}\times 100=0.2$

Method II:

Given that:

$A_1=500A_2$

By volume conservation

$A_1\Delta h=A_{2}h$

$\Delta h=\frac{A_2}{A_1}h=\frac{A_{2}h}{500A_2}$

$\Delta h=\frac{h}{500}$

Actual pressure difference,

$\Delta p_{actual}=\rho g(h+\Delta h)$

$=\rho gh(1+\frac{1}{500})$

$\%error=\frac{\Delta {\rho }_{actual}-\Delta {\rho }_{measured}}{\Delta {\rho }_{measured}}\times 100$

$=\frac{(1+\frac{1}{500})-1}{1}\times 100$

$=\frac{1}{500}\times 100=0.2\%$