A slender homogeneous rod of length 2L floats partly immersed in water, being supported by a string fastened to one of its ends, as shown. The specific gravity of the rod is 0.75. The length of rod that extends out of water is a)Lb)c)d)3LCorrect answer is option 'A'. Can you explain this answer?

Question

A slender homogeneous rod of length 2L floats partly immersed in water, being supported by a string fastened to one of its ends, as shown. The specific gravity of the rod is 0.75. The length of rod that extends out of water is              a)Lb)c)d)3LCorrect answer is option 'A'. Can you explain this answer?

Answer

Let's say x length of the rod is dipped into the water.
Since the buoyant force acts through the centre of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O along the line of action of T,
0=Στ_o
⇒0=wl cosθ−F_B(2l−x/2)cosθ
⇒0=ρ_rodgA(2l)(lcosθ)−ρ_watergAx(2l−x/2)cosθ
⇒0=(1/2ρ_watergAcosθ) (x²−4lx+4 (ρ_rod/ρ_water)l²) where A=cross section area
⇒x²−4lx+3l²=0
⇒x=l,3l.
x=3l is not a possible solution, hence 2l−x=2l−l=l length of the rod extends out of the water.

Answer

From chapter fluid mechanics on the basis of law of flotation it can be solve

Answer

Let's say x length of the rod is dipped into the water.
Since the buoyant force acts through the centre of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O along the line of action of T,
0=Στ_o
⇒0=wl cosθ−F_B(2l−x/2)cosθ
⇒0=ρ_rodgA(2l)(lcosθ)−ρ_watergAx(2l−x/2)cosθ
⇒0=(1/2ρ_watergAcosθ) (x²−4lx+4 (ρ_rod/ρ_water)l²) where A=cross section area
⇒x²−4lx+3l²=0
⇒x=l,3l.
x=3l is not a possible solution, hence 2l−x=2l−l=l length of the rod extends out of the water.

Answer

Let's say x length of the rod is dipped into the water.
Since the buoyant force acts through the centre of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O along the line of action of T,
0=Στ_o
⇒0=wl cosθ−F_B(2l−x/2)cosθ
⇒0=ρ_rodgA(2l)(lcosθ)−ρ_watergAx(2l−x/2)cosθ
⇒0=(1/2ρ_watergAcosθ) (x²−4lx+4 (ρ_rod/ρ_water)l²) where A=cross section area
⇒x²−4lx+3l²=0
⇒x=l,3l.
x=3l is not a possible solution, hence 2l−x=2l−l=l length of the rod extends out of the water.

Answer

Let's say x length of the rod is dipped into the water.
Since the buoyant force acts through the centre of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O along the line of action of T,
0=Στ_o
⇒0=wl cosθ−F_B(2l−x/2)cosθ
⇒0=ρ_rodgA(2l)(lcosθ)−ρ_watergAx(2l−x/2)cosθ
⇒0=(1/2ρ_watergAcosθ) (x²−4lx+4 (ρ_rod/ρ_water)l²) where A=cross section area
⇒x²−4lx+3l²=0
⇒x=l,3l.
x=3l is not a possible solution, hence 2l−x=2l−l=l length of the rod extends out of the water.

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