$A$ spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid,then the diameter of the body is

  • A
    $2 \sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
  • B
    $2 \sqrt{\frac{6 \sigma}{g(2 \rho-d)}}$
  • C
    $2 \sqrt{\frac{4 \sigma}{g(2 \rho-d)}}$
  • D
    $2 \sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$

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$A$ pin or a needle floats on the surface of water,the reason for this is:

What is the ratio of the surface tension force acting on the curved part to that on the flat part?

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Surface tension is exhibited by liquids due to the force of attraction between the molecules of the liquid. The surface tension decreases with an increase in temperature and vanishes at the boiling point. Given that the latent heat of vaporization for water $L_v = 540 \text{ kcal/kg}$,the mechanical equivalent of heat $J = 4.2 \text{ J/cal}$,density of water $\rho_w = 10^3 \text{ kg/m}^3$,Avogadro's number $N_A = 6.0 \times 10^{26} \text{ molecules/kmol}$,and the molecular weight of water $M_A = 18 \text{ kg/kmol}$.
$(a)$ Estimate the energy required for one molecule of water to evaporate.
$(b)$ Show that the intermolecular distance for water is $d = \left( \frac{M_A}{N_A \rho_w} \right)^{1/3}$ and find its value.
$(c)$ $1 \text{ g}$ of water in the vapour state at $1 \text{ atm}$ occupies $1601 \text{ cm}^3$. Estimate the intermolecular distance at the boiling point in the vapour state.
$(d)$ During vaporisation,a molecule overcomes a force $F$,assumed constant,to go from an intermolecular distance $d$ to $d'$. Estimate the value of $F$.
$(e)$ Calculate $\frac{F}{d}$,which is a measure of the surface tension.

The length of a needle floating on the surface of water is $1.5\,cm$. The force in addition to its weight required to lift the needle from the water surface will be...... $N$ (surface tension of water $= 7.5\,N/cm$).

It is easy to wash clothes in hot water because its

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