A) \[\frac{La}{g}\]
B) \[\frac{Lg}{a}\]
C) \[L\left( 1+\frac{a}{g} \right)\]
D) \[L\left( 1-\frac{a}{g} \right)\]
Correct Answer: A
Solution :
Let \[{{P}_{1}}\]and\[{{P}_{2}}\] be the pressures at the bottom of limbs A and B, then \[{{P}_{1}}-{{P}_{2}}=h\rho g\] (\[\rho =\]density of liquid) If A is the cross-sectional area of the U-tube, the force at the horizontal part of the tube is \[F=({{P}_{1}}-{{P}_{2}})A=h\rho \,g\,A\] (1) Now F = ma, where\[m=\rho AL\]is the mass of the liquid in length L. Hence \[F=\rho \,AL\,a\] (2) Equating (1) and (2), we get \[h=\frac{La}{g}.\]So the correct choice is (a). Hence, the correction option is (a).You need to login to perform this action.
You will be redirected in
3 sec