• 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

 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$
You will be redirected in 3 sec 