question_answer

A cube of mass 30 g wettable by water floats on the surface of water. Each face of the cube is 4 cm long. Surface tension of water = 70 dynes/cm. The distance of the lower face of the cube from the surface of water is \[(g=980\,\,cm\,\,{{s}^{-2}}):\]

A) 1.9 cm

B) 1.93 cm

C) 1.95 cm

D) 1.98 cm

Correct Answer: C

Solution :

Mass of the cube = 30g, density of water=\[\rho =1gm/c{{m}^{3}},\] surface tension of water \[=\sigma =70\,\,dyne/cm\]

a = length of each face of the cube = 4 cm, let y be the distance of the lower face of the cube from the surface of water.

When the cube floats freely in water F (force due to surface tension)

Upward forces = downward force

\[{{F}_{B}}={{a}^{2}}y\rho g\]

\[F=4a\sigma \]

\[{{F}_{B}}=F+mg\]

\[{{a}^{2}}y\rho g=4a\sigma +mg\]

\[\therefore y=\frac{mg+4a\sigma }{{{a}^{2}}\rho g}=\frac{30\,\,\times \,\,980+4\,\,\times \,\,4\,\,\times \,\,70}{4\,\,\times \,\,4\,\,\times \,\,1\,\,\times \,\,980}\]\[=\frac{29400+1120}{15680}=\frac{30520}{15680}=1.95\,\,cm\]