A) \[\frac{aL}{2g}\]
B) \[\frac{gL}{2a}\]
C) \[\frac{gL}{a}\]
D) \[\frac{aL}{g}\]
Correct Answer: D
Solution :
Let \[{{P}_{1}}\]and \[{{P}_{2}}\]be the pressures at the bottom of the left and right end of the tube respectively. Then \[F=({{P}_{1}}-{{P}_{2}})A=\rho ghA\] Where A is the cross-section of the tube. The mass of the liquid in the horizontal position is \[m=\rho LA\] Now,\[F=ma,\]so \[\rho gh\,A=\rho LAa,\]\[\therefore \]\[h=\frac{aL}{g}\] Hence, the correction option is [d].You need to login to perform this action.
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