A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ_{w}$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
(Take $m = 1 k g, g = 10 m / s^{2}, a T = \frac{10}{4} u n i t$ and $ρ_{w} a^{2} g = 10 u n i t)$ .

Open in App

Solution

$M g + F_{S T} = B$
$m g + 4 A T = a^{2} h ρ_{w} g$
$10 + 4 \times \frac{10}{4} = 10 h$
$h = 2 m$ .

Suggest Corrections

Similar questions

Q. A cube with a mass '

m

' completely wettable by water floats on the surface of water. Each side of the cube is '

a

'. The distance

h

between the lower face of cube and the surface of the water if surface tension is

S

is given as

\frac{m g + x S a}{ρ_{w} - a^{2} g}

. Find

x

. Take density of water as

ρ_{w}

. Take angle of contact to be zero.

A cube with a mass m= 20g floats on the surface of water. Each face of the cube is 3cm long. what is the distance between lower surface of the cube and surface of the water? (Surface tension of water = 70 $\times$ $10^{3}$ N/m , and angle of contact is acute)

Q. A cube with a mass

= 20 g

wettable water floats on the surface of water. Each face of the cube is

α = 3 c m

long. Surface tension of water is

70 d y n / c m

. The distance of the lower face of the cube from the surface of water is

(g = 980 c m / s^{2})

Q. A cube of mass m =800 g floats on the surface of water. Water wets it completely. The cube is 10 cm on each edge. By what additional distance is it buoyed up or down by surface tension? Surface tension of water = 0.07 N

m^{- 1}

Join BYJU'S Learning Program

Mg+FST=Bmg+4AT=a2hρwg10+4×104=10hh=2 m.

$M g + F_{S T} = B$
$m g + 4 A T = a^{2} h ρ_{w} g$
$10 + 4 \times \frac{10}{4} = 10 h$
$h = 2 m$ .