Question

a rod is hinged at point O. what part of length should be submerged inside water so that it remains in equilibrium, if specific gravity is 0.50? options are in pic☝️

please help​

Answer 1

The length that should be submerged inside water so that it remains in equilibrium is 1/2

Therefore, option (A) is correct.

Step-by-step explanation:

Let the mass per unit length of the rod is m, its total length L and the submerged length is l

Let the density of the material of the rod be $\rho$

If the cross sectional area of the rod is A

Then

The submerged mass will displace water equal to its volume

If the density of water is $\rho_w$, the buoyant force acting on the submerged portion of the rod will be $Al\rho_wg$

This will be equal to the weight of the rod

Thus,

$mLg=Al\rho_wg$

$\implies AL\rho g=Al\rho_wg$

$\implies l=\frac{\rho}{\rho_w}L$

But $\frac{\rho}{\rho_w}$ is specific gravity which is given to be 0.5

Therefore,

$l=0.5L$

$\implies l=\frac{1}{2}L$

Thus, the length that should be submerged = 1/2

Hope this answer is helpful.

Know More:

Q: A uniform rod of length2m a uniform rod of 2 metres and mass 5 kg is lying on a horizontal surface the work done in raising one end of the road with the other m into contact with the surface until the road makes an angle 30 degree with the horizontal is :

Click Here: https://brainly.in/question/1418920