A) \[h{{\rho }_{m}}g\]
B) \[\frac{h({{\rho }_{m+\rho }})gx}{(\ell -x)}\]
C) \[h({{\rho }_{m}}-\rho )g\]
D) \[\frac{h{{\rho }_{m}}gx}{\ell -x}\]
Correct Answer: D
Solution :
Conserving PV before and after the tube is immersed in the liquid, we get \[{{p}_{0}}A\ell =({{p}_{0}}+p)A(\ell -x)\] \[h{{\rho }_{m}}gA\ell =(h{{\rho }_{m}}g+p)A(\ell -x)\] \[p=\frac{h{{\rho }_{m}}gx}{\ell -x}\]You need to login to perform this action.
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