Step 1: The pressure at the closed end of the tube is given by: \[ P = \text{atmospheric pressure} + \text{pressure due to mercury column} \]
Step 2: When the tube is held vertically, the pressure at the closed end is \( 75 \, \text{cm} \) of mercury.
Step 3: When the tube is held horizontally, the length of the air column changes due to the balance of pressures. The total length of the column is \( L = 28 \, \text{cm} \).
A cylindrical vessel, open at the top, contains 15 liters of water. Water drains out through a small opening at the bottom. 5 liters of water comes out in time t1, the next 5 litre in further time t2, and the last 5 litre in further time t3. then:
A cylinder of mass m and material density ρ hanging from a string is lowered into a vessel of cross - sectional area A containing a liquid of density σ (< ρ) until it is fully immersed. The increase in pressure at the bottom of the vessel is:
A big liquid drop splits into 'n' similar small drops under isothermal conditions, then in this process: