Two glass plates are separated by water. If surface tension of water is 75 dyne cm-1 and area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is

Question

Two glass plates are separated by water. If surface tension of water is 75 dyne cm-1 and area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is (A) 102 dyne (B) 104 dyne (C) 105 dyne (D) 106 dyne

Tardigrade Admin · Accepted Answer

Required force to separate the two glass plates, F = (2SA/t) Here, S = 75 dyn cm-1 , A = 8 cm2 t = 0.12 mm = 0.012 cm ∴ : : F = (2 × 75 × 8 /0.012 ) = 100000 F = 105 dyne

Q. Two glass plates are separated by water. If surface tension of water is $75 d y n e c m^{- 1}$ and area of each plate wetted by water is $8 c m^{2}$ and the distance between the plates is $0.12 mm$ , then the force applied to separate the two plates is

$1 0^{2}$ dyne

$1 0^{4}$ dyne

$1 0^{5}$ dyne

$1 0^{6}$ dyne

Required force to separate the two glass plates,
$F = \frac{2 S A}{t}$
Here, $S = 75 d y n c m^{- 1}, A = 8 c m^{2}$
$t = 0.12 mm = 0.012 c m$
$∴ F = \frac{2 \times 75 \times 8}{0.012} = 100000$
$F = 1 0^{5} d y n e$

Q. Two glass plates are separated by water. If surface tension of water is 75dynecm−1 and area of each plate wetted by water is 8cm2 and the distance between the plates is 0.12mm, then the force applied to separate the two plates is

102 dyne

104 dyne

105 dyne

106 dyne