A prism shaped styrofoam of density `rho_(styrofoam) (ltrho_(water))` is held completely submerged in water. It lies with its base horizontal. The base of foam is at depth `h_(0)` below water surface and atmospheric pressure is `P_(0)` Surface of water is open to atmosphere. Styrofoam prism is held in equilibrium by the string attached symmetrically as shown (Take `rho_(styrofoam)=rho_(f), rho_(water)=rho_(w))`.
Now string is cut and styrofoam is allowed to come to surface. A point mass is to be placed symmetrically on the upper surface of styrofoam such that it is in equilibrium with its base in horizontal plane. In equilibrium position styrofoam has half of its slant length submerged. Surface tension of water is T. contact angle is 0^(@). Determine mass m to achieve equilibrium.

Options

A. `(rho_(f)gLl^(2))/2+3/4 rho_(w)gLl^(2)-2T[L+l]`
B. `(-rho_(f)gLl^(2))/2+3/8rho_(w)gLl^(2)-sqrt(2)T[L+l]`
C. `(-rho_(f)gLl^(2))/2+3/8rho_(w)gLl^(2)-(T[L+l)]/sqrt(2)`
D. None of these

Answer

B, D

Explanation

Resultant downward force due to surface tension `=sqrt(2) T [L+l]`
Buoyent force `=3/8rho_(w)gLl^(2)`
Weight `=(rho_(f)gLl^(2))/2`