question_answer

An upright U-tube manometer with its limbs \[0.6\text{ }m\]high and spaced \[0.3\text{ }m\]apart contains a liquid to a height of \[0.4\text{ }m\]on each limb. If the U-tube is rotated at 10 radians/second about a vertical axis at \[0.1\text{ }m\]from one limb. \[{{z}_{1}}\] and \[{{z}_{2}}\] are heights of liquid columns on rotation. Choose the correct option. (Use\[g=10\text{ }m/{{s}^{2}}\])

A)  \[{{z}_{2}}-{{z}_{1}}=30\,cm\]

B)        \[{{z}_{2}}-{{z}_{1}}=10\,cm\]

C)  \[{{z}_{2}}-{{z}_{1}}=20\,cm\]          

D) \[{{z}_{2}}-{{z}_{1}}=15\,cm\]

Correct Answer: D

Solution :

[d]  \[{{z}_{1}}=0.4-x\] & \[{{z}_{2}}=0.4+x\] So,  \[{{z}_{2}}-{{z}_{1}}=2x\] \[{{y}_{2}}-{{y}_{1}}=\frac{{{\omega }^{2}}{{\left( 0.2 \right)}^{2}}}{2g}-\frac{{{\omega }^{2}}{{\left( 0.1 \right)}^{2}}}{2g}\] \[=\frac{{{10}^{2}}{{\left( 0.2 \right)}^{2}}}{2\times 10}-\frac{{{10}^{2}}{{\left( 0.1 \right)}^{2}}}{2\times 10}=\frac{15}{100}m\] \[{{y}_{2}}-{{y}_{1}}={{z}_{2}}-{{z}_{1}}=2x=15cm\]