A) \[\frac{200}{3}cm\]
B) \[\frac{50}{3}cm\]
C) \[T=2{{T}_{0}}+\eta {{V}^{2}}\]
D) \[\eta \]
Correct Answer: D
Solution :
We can write As \[\frac{2{{\eta }^{2}}}{g{{a}^{2}}}\] For vessel/P \[\frac{{{\eta }^{2}}}{g{{a}^{2}}}\] For vessel Q; \[\frac{{{\eta }^{2}}}{2g{{a}^{2}}}\] Hence \[\frac{{{\eta }^{2}}}{4g{{a}^{2}}}\] or \[{{t}_{1}}\]You need to login to perform this action.
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