${p_C} = {p_a} + {\rho _{water}}g{h_{water'}}$
$where\,{h_{water}} = CE = \left( {65 + 65} \right)mm = 130\,mm$
$pressure\,at\,point\,B,{P_B} = {P_u} + {\rho _{oil}}g{h_{oil}}$
$where\,{h_{oil}} = AB = \left( {65 + 65 + 10} \right)\,mm = 140\,mm$
In liquid, Pressure is same at same liquid level,
${p_B} = {p_c} \Rightarrow {\rho _{oil}}g{h_{oil}} = {\rho _{water}}g\,{h_{water}}$
${\rho _{oil}} = \frac{{130 \times {{10}^3}}}{{140}} = \frac{{13}}{{14}} \times {10^3} = 928.57\,kg\,{m^{ - 3}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $-I$ | List $-II$ | ||
| $A$. | Coefficient of Viscosity | $I$. | $[M L^2T^{–2}]$ |
| $B$. | Surface Tension | $II$. | $[M L^2T^{–1}]$ |
| $C$. | Angular momentum | $III$. | $[M L^{-1}T^{–1}]$ |
| $D$. | Rotational Kimeatic energy | $IV$. | $[M L^0T^{–2}]$ |
