Question

A homogeneous aluminum ball of radius $r=0.5cm$ is suspended on a weightless thread from one end of a uniform rod of mass $4.4g$. The rod is placed on the edge of a tumbler with water so that half of the ball is submerged in water when the system is in equilibrium. The densities of aluminum and water are $2.7\times {10}^{3}kg{m}^{-3}$ and $1\times {10}^{-3}$, respectively. The ratio of the two parts of the rod $y/x$ is approximately as shown in figure will be

• A. $1.5$
• B. $2.2$
• C. $1.0$
• D. $3.0$

Answer 1

The correct option is A $1.5$

$ThevolumeofaluminumballisV=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \left({0.5}^3\right)V=0.5266{cm}^{3}Theforceactingontheballis1\left)downwardforceduetoselfweight2\right)upthrustduetobouyantforceBouyantforceBF=weightofwaterdisplaced={\rho }_{w}g\left(0.5266/2\right)=1\times g\times 0.2633=0.2633gSelfweightW={\rho }_{al}g\times 0.5266=2.7\times g\times 0.5266=1.4218gNetweightF=W-BF=1.4218g-.2633gF=1.1585gNow,weightperunitlengthoftherod=\frac{4.4g}{x+y}Takingmomentoneachside1.1585g\times x+\frac{4.4g}{x+y}.x.\frac{x}{2}=\frac{4.4g}{x+y}.y.\frac{y}{2}1.1585g\times x=\frac{4.4g}{2(x+y)}\left(y^2{-x}^2\right)1.1585g\times x=2.2g(y-x)3.3585x=2.2y\frac{y}{x}=1.52\cong 1.5$