A glass plate of length $10 c m$ breadth $4 c m$ and thickness $0.4 c m$ weighs $20 g$ in air. It is held vertically with long side horizontal and half the plate immersed in water. What will be its apparent weight? Surface tension of water $= 70 J d y n / c m$ .

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Solution

The forces acting on the plate are weight of the plate vertically downwards
$W = m g = 20 g f$
i. Buoyant force due to liquid,
$F_{b} = V ρ_{1} g$
$= (\frac{l b t}{2}) ρ_{1} g = (\frac{10 \times 4 \times 0.4}{2}) \times 1 g f$
$= 8 g f$
ii. Force due to surface tension, vertically downward
$F = T \times perimeter of plate in contact with water$
$= \frac{70}{980} \times 2 (10 + 0.4) g f = 1.5 g f$
iii. Apparent weight of the plate $= (W + F) - F$
$= (20 + 1.5) - 8 = 13.5 g f$ .

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A glass plate of length 10 cm breadth 4 cm and thickness 0.4 cm weighs 20 g in air. It is held vertically with long side horizontal and half the plate immersed in water. What will be its apparent weight? Surface tension of water =70 J dyn/cm.

A glass plate of length $10 c m$ breadth $4 c m$ and thickness $0.4 c m$ weighs $20 g$ in air. It is held vertically with long side horizontal and half the plate immersed in water. What will be its apparent weight? Surface tension of water $= 70 J d y n / c m$ .