(a) The pressure must be equal in both arms:
"\\rho_kgh_k=\\rho_mg\\Delta h_m,\\\\\\space\\\\\nh_k=\\Delta h_m\\frac{\\rho_m}{\\rho_k}=25.5\\text{ cm}."(b) When the levels of mercury are the same, we can ignore its presence and weights forces created by the kerosene and water only:
"\\rho_kgh_k=\\rho gh,\\\\\\space\\\\\nh=h_k\\frac{\\rho_k}{\\rho}=20.4\\text{ cm}."
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