A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the fraction of the length of the rod above water

= 0.66

Hence, the fraction of the length of the rod above water = = 0.66

and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + ) g sinФ - F2 x () x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

= . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

= . 5/9 x w

Hence,

=

=

Taking common and solving for , we will get

= 0.66

Hence, the fraction of the length of the rod above water = = 0.66

and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34

i think the answer would be reserve.

amplitude

hope this