Question

An L-shaped glass tube is kept inside a bus that is moving with constant acceleration. During the motion, level of the liquid in the left arm is at $12\text{cm}$ whereas in the right arm, it is at $8\text{cm}$ when the orientation of the tube is as shown. Assuming that the diameter of the tube is much smaller than the levels of the liquid and neglecting effects of surface tension, the acceleration of the bus will be $\left({\text{g=10 m/s}}^2\right)$

• A. $1{\text{m/s}}^2$
• B. $2{\text{m/s}}^2$
• C. $4{\text{m/s}}^2$
• D. $5{\text{m/s}}^2$

Answer 1

The correct option is B $2{\text{m/s}}^2$

When the bus starts accelerating, the surface of the liquid in the tube tends to get inclined. The slope of the inclined liquid surface is given by,
$tan\theta =\frac{a}{g}$
where $a$ is the acceleration of the bus.

From geometry, $EF=AC=AB+BC$
Since angle made by tube with horizontal is ${45}^{\circ }$
$BC=\frac{8}{tan{45}^{\circ }}=8cm$
Simillarly $AB=\frac{12}{tan{45}^{\circ }}=12cm$
$\Rightarrow EF=AC=8+12=20cm$
Also $DF=DA-FA=DA-EC=12-8=4cm$
Hence slope of the inclination,
$tan\theta =\frac{DF}{EF}=\frac{4}{20}=\frac{a}{g}$
$\Rightarrow a=2m/s^2$ $(\because g=10m/s^2)$