Question

A uniform rod $AB,12\text{m}$ long and weighing $24\text{kg}$, is supported at end $B$ by a flexible light string and a lead mass (of very small size) of $12\text{kg}$ is attached at end $A$. The rod floats in water with half of its length submerged. For the situation described, choose the correct statement(s). [Take $g=10{\text{m/s}}^2$, density of water $=1000{\text{kg/m}}^3$]

• A. The tension in the string is $40\text{N}$.
• B. The tension in the string is $120\text{N}$.
• C. The volume of the rod is $6.4\times {10}^{-2}{\text{m}}^3$.
• D. The point of action of the buoyant force is $C$ (centre of mass of rod).

Answer 1

Answer: (A), (C)

The correct options are
A The tension in the string is $40\text{N}$.
C The volume of the rod is $6.4\times {10}^{-2}{\text{m}}^3$.

The length $CP=OC=\frac{l}{2}$
Bouyant force or upthrust $\left(U\right)$ will act at the mid point$\left(M\right)$ of the length of the rod submerged in water.
i.e option (d) is wrong.
$OM=MC=\frac{l}{4}$


From the FBD of rod, applying equilibrium condition in vertical direction,
$T+U=W_1+W_2=360\text{N}...\left(i\right)$
Upthrust will be given by
$U=V^{\prime }\rho g$, where $V^{\prime }=$ volume of immersed rod, i.e half of total volume.
$\Rightarrow U=\frac{V}{2}{\rho }_{w}g=\left(V/2\right)\left({10}^3\right)\left(10\right)$
$\Rightarrow U=0.5\times {10}^{4}V...\left(ii\right)$
For rotational equilibrium, balancing the moments of the forces on the rod about point $M$:
$\text{Net anticlockwise moment}=\text{Net clockwise moment}$
$\Rightarrow 120\left(\frac{l}{4}\cos \theta \right)+T\left(\frac{3l}{4}\cos \theta \right)=240\left(\frac{l}{4}\cos \theta \right)$
$\Rightarrow T=\frac{(60-30)\times 4}{3}$
$\therefore T=40\text{N}$
i.e Option (a) is correct and (b) is wrong.
Now from eqs. $\left(i\right)$ and $\left(ii\right)$ we get,
$U=360-T=360-40=320\text{N}$
$\Rightarrow 320=0.5V\times {10}^4$
$\therefore V=6.4\times {10}^{-2}{\text{m}}^3$
where $V$ is the volume of rod.
Option (c) is also correct.