Question

Cross-section of an object (having same section normal to the paper) submerged into a fluid consists of a square of sides 2m and triangle as shown in the figure. The object is hinged at point P that is one metre below the fluid free surface. If the object is to be kept in the position as shown in the figure, the value of 'x' should be

• A. $2\sqrt{3}$ m
• B. $4\sqrt{3}$ m
• C. 4m
• D. 8m

Answer 1

The correct option is A $2\sqrt{3}$ m

The vertical force on the surface bounded by square and triangle would be respectively,

$F_1$ = 4(S-1)${\gamma }_{\omega }$ - for square edge and acting at 1m from P

$F_2$ = x(S-1)${\gamma }_{\omega }$ - for inclined edge of triangle and acting at x/3 from P

Taking moments of both the forces about P, we get

$F_1\times 1=F_2\times \frac{x}{3}$

$\Rightarrow 4(S-1){\gamma }_{\omega }=x(S-1){\gamma }_{\omega }\frac{x}{3}$

$\Rightarrow$ $x^2$ = 12

$\Rightarrow$ $x=2\sqrt{3}$ m