Question

An object of specific gravity $\rho$ is hung from a massless string. The tension of a string is T. The object is immersed in water so that one half of its volume is submerged. The new tension in the string is

• A. $\left(\frac{2\rho +1}{2\rho }\right)r$
• B. $\left(\frac{2\rho -1}{2\rho }\right)r$
• C. $\left(\frac{\rho -1}{\rho }\right)r$
• D. $\left(\frac{\rho +1}{\rho }\right)r$

Answer 1

The correct option is B $\left(\frac{2\rho -1}{2\rho }\right)r$

Hence, option $\left(B\right)$ is correct answer.

V-volume of body

T- Tension in study initial

T'- New tension in string

F- Buoyant force /Thrust force

$\rho$- R.D of body

We know that is air

$T=\rho vg$.........(1)

After body is immersed in liquid

weight of body-Buoyant force =T'

$\rho vg-1\times \frac{V}{2}g=T^{\prime }$

$vg[\rho -\frac{1}{2}]=T^{\prime }$

$\rho vg\left[\frac{2\rho -1}{2l}\right]=T^{\prime }$

$T\left[\frac{2l-1}{2l}\right]=T^{\prime }$ [From (1)]

Hence (2) option is the right answer