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    				| 1. | The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are 2 cm and 8 cm. Atmospheric pressure = 1.01 x 105 N/m2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube. | 
| Answer» (a) The pressure of the gas = Pressure of mercury column (difference) + atmospheric pressure. =13.6*(10⁶/1000)*9.8*{(8-2)/100}+ 1.01 x 10⁵ N/m² =8000+1.01 x 10⁵ N/m² =0.08 x 10⁵+1.01 x 10⁵ N/m² =1.09 x 10⁵ N/m² (b) The pressure of the mercury at the bottom of the U-tube =The pressure of mercury column on the open arm + Atmospheric pressure =13.6(10⁶/1000)*9.8*0.08+1.01 x 10⁵ N/m² =0.107 x10⁵ + 1.01 x 10⁵ N/m² ≈1.12 x 10⁵ N/m² | |