A) 1 cm of Hg
B) 5 cm of Hg
C) 13 cm of Hg
D) 15 cm of Hg
Correct Answer: A
Solution :
Pressure at the bottom \[P={{h}_{1}}{{d}_{1}}g+{{h}_{2}}{{d}_{2}}g\] \[=(3\times {{10}^{-2}}\times 1.2\times {{10}^{3}}\times g)\] \[+(10\times {{10}^{-2}}\times {{10}^{3}}g)\] \[=13.6\times {{10}^{-2}}\times {{10}^{3}}g=136g\] \[=136\frac{g}{13.6\times {{10}^{3}}g}={{10}^{-2}}m\] \[=1cm\] of \[Hg\]
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