An open $U$-tube contains mercury. When $13.6 \,cm$ of water is poured into one of the arms of the tube, then the mercury rise in the other arm from its initial level is ....... $cm$
Medium
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(b)
$P_A+(\rho g h)_{\text {water }}-(\rho g h)_{\text {Mercury }}=P_B$
$(\rho g h)_{\text {water }}=(\rho g h)_{\text {Mercury }}$
$10,00 \times g \times 13.6=13660 \times g \times 2 y$
$y=\frac{1}{2} \Rightarrow y=0.5\,cm$
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