Question

A uniform rod $\text{AB},4\text{m}$ long and weight $12\text{kg}$, is supported at end A, with a $6\text{kg}$ lead weight at B. The rod floats as shown in figure with one-half of its length submerged. The buoyant force on the lead mass is negligible as it is of negligible volume. Find the tension in the cord and the total volume of the rod.

• A. $10\text{N},32\times {10}^{-3}{\text{m}}^3$
• B. $20\text{N},32\times {10}^{-3}{\text{m}}^3$
• C. $20\text{N},16\times {10}^{-3}{\text{m}}^3$
• D. $10\text{N},16\times {10}^{-3}{\text{m}}^3$

Answer 1

The correct option is B $20\text{N},32\times {10}^{-3}{\text{m}}^3$

Balancing forces
$T+U=60+W.....\left(i\right)$


$\sum$ (Moments) about = 0
$\therefore U\left(\frac{l}{4}\right)+T\left(l\right)=W\left(\frac{l}{2}\right)$
or $U+4T=2W....\left(ii\right)$
$W=120\text{N}....\left(iii\right)$
Solving these three equations we get $T=20\text{N}$ and $U=160\text{N}$
Now, $160=\left(\frac{V}{2}\right){\rho }_{w}g=\left(\frac{V}{2}\right)\left(1000\right)\left(10\right)$
$\therefore V=32\times {10}^{-3}{\text{m}}^3$