Question

A lead bar of length $50\text{cm}$ and square cross-section of side $5\text{cm}$ is fixed to a vertical wall as shown in figure. A massless tank is suspended from the end of the bar with the help of a massless rope. The tank is slowly filled with mercury at room temperature. Find the level of mercury in the tank so that the end of the bar deviates by $1\text{mm}$.
[Take modulus of rigidity for lead as $5.6\text{Gpa}$ and area of base of cylindrical tank as $0.42{\text{m}}^2;{\rho }_{Hg}=13600{\text{kg/m}}^3\&g=9.8{\text{m/s}}^2$]

• A. $0.21\text{m}$
• B. $1\text{m}$
• C. $0.5\text{m}$
• D. $0.42\text{m}$

Answer 1

The correct option is C $0.5\text{m}$

We know that,
Modulus of rigidity $\left(\eta \right)=\frac{\text{Shear stress}\left({\sigma }_r\right)}{\text{Shear strain}}=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta x}{L}\right)}$
$\Delta x\to$ Deviation at end of bar $=1\text{mm}$
$L\to$ Length of bar $=50\text{cm}$
$A\to$ Cross-sectional area of bar $=5\text{cm}\times 5\text{cm}$

$\therefore \eta \times \frac{\Delta x}{L}=\frac{F}{A}\Rightarrow F=\frac{\eta \times \Delta x\times A}{L}$
From the data given in the question,
$F=\frac{5.6\times {10}^9\times {10}^{-3}\times 25\times {10}^{-4}}{50\times {10}^{-2}}=28\text{kN}$
But, $F=$ weight of mercury
i.e $F=mg={\rho }_{Hg}\times g\times \text{Volume of Hg}$
From the given data,
$28000=13600\times 9.8\times V_{Hg}$
$\Rightarrow V_{Hg}=\frac{28000}{13600\times 9.8}=0.21{\text{m}}^3$
$\therefore$ Level of mercury $=\frac{V_{Hg}}{\text{Base area of tank}}=\frac{V_{Hg}}{0.42}=\frac{0.21}{0.42}=0.5\text{m}$
Thus, option (c) is the correct answer.