Question

An open rectangular tank of dimensions $5m\times 4m\times 3m$ containing water upto a height of $2m$ is accelerated horizontally along the longer side. (Take $g=10m/s^2$)

• A. The maximum acceleration that can be given without spilling the water is $4m/s^2$
• B. If the maximum acceleration is increased by $20\%$, the percentage of water spilt over is $10\%$.
• C. If initially, the tank is closed at the top and is accelerated horizontally by $9m/s^2$, the gauge pressure at the bottom of the rear wall of the tank will be $4.5\times {10}^{4}Pa$
• D. If initially, the tank is closed at the top and is accelerated horizontally by $9m/s^2$, the gauge pressure at the bottom of the rear wall of the tank will be $9\times {10}^{4}Pa$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The maximum acceleration that can be given without spilling the water is $4m/s^2$
B If the maximum acceleration is increased by $20\%$, the percentage of water spilt over is $10\%$.
C If initially, the tank is closed at the top and is accelerated horizontally by $9m/s^2$, the gauge pressure at the bottom of the rear wall of the tank will be $4.5\times {10}^{4}Pa$


(a) Volume of water inside that tank remains constant

$\left(\frac{3+y_0}{2}\right)5\times 4=5\times 2\times 4$

or $y_0=1m$

$\therefore \tan {\theta }_0=\frac{3-1}{5}=0.4$

Since, $\tan {\theta }_0=\frac{a_0}{g}$, therefore $a_0=0.4g=4m/s^2$


(b) When acceleration is increased by $20\%$

$a=1.2$ $a_0=0.48g$

$\therefore tan\theta =\frac{a}{g}=0.48$

Now, $y=3-5tan\theta =3-5\left(0.48\right)=0.6m$

Fraction of water spilt over

$=\frac{4\times 2\times 5-\frac{(3+0.6)}{2}\times 5\times 4}{2\times 5\times 4}=0.1$

Percentage of water spilt over $=10\%$


(c) $a^{\prime }=0.9g$

$tan{\theta }_0=\frac{a^{\prime }}{g}=0.9$

volume of air remains constant

$4\times \frac{1}{2}yx=\left(5\right)\left(1\right)\times 4$

Since $y=xtan{\theta }_0=5$

or $x=3.33m;y=3.0m$

Gauge pressure at the bottom of the rear wall:
$P_r=x{\rho }_w{a}^{\prime }+3{\rho }_{w}g=3.33\times {10}^3\times 9+3\times {10}^3\times 10=4.5\times {10}^{4}Pa$