A U-tube is made of capillaries of bore 2 mm and 4mm, respectively. The tube is held vertically inverted with the open ends below the surface liquid in a beaker. Calculate the difference in the levels of the meniscus in the two limbs. Take. Surface tension of liquid =48 dyne/cm density =0.8 g cm^(-3) Angle of contact between liquid and glass =0^(@) g=980 cm//s^(2).

A U-tube is made of capillaries of bore 2 mm and 4mm, respectively. Th

1.	A U-tube is made of capillaries of bore 2 mm and 4mm, respectively. The tube is held vertically inverted with the open ends below the surface liquid in a beaker. Calculate the difference in the levels of the meniscus in the two limbs. Take. Surface tension of liquid =48 dyne/cm density =0.8 g cm^(-3) Angle of contact between liquid and glass =0^(@) g=980 cm//s^(2).
Answer» Solution : LET, `P_(p), P_(Q), P_(S) and P_(T)` be the pressure at points, P,Q,R and T RESPECT. We know that the pressure on the concave side of the liquid is greater than its other side by `2sigma//R` So, here `P_(Q)=P_(P)-(2sigma)/(r_(1))` `rArr P_(P)=P_(Q)+(2sigma)/(r_(1))` `and P_(S)=P_(T)+(2sigma)/(r_(2))` But `P_(P)=P_(S)` `therefore P_(Q)+(2 sigma)/(r_(1))=P_(T)+(2sigma)/r_(2)` `rArr P_(T)-P_(Q)=2 sigma (1/r_(1)-1/r_(2))` `rArr hpg=2 sigma (1/r_(1)-1/r_(2))` `rArr h=(2sigma)/(pg) (1/r_(1)-1/r_(2))` `=(2 xx 49)/(0.8 xx 980) ((1)/(0.2) -1/(0.4))` `=(2 xx 49)/(0.8 xx 980) (1/(0.4))` =0.3125 cm =3.125 mm