Given:
• Height of the tank = 5 m
• The tank is half-filled with water and then topped off with oil.
• Density of water = 1 g/cm31 \, \text{g/cm}^3
• Density of oil = 0.85 g/cm30.85 \, \text{g/cm}^3
• The total height of the tank is 5 m, so half of the tank is filled with water, i.e., the water height is 2.5 m.
• The height of the oil column is also 2.5 m (since the tank is filled to the top after the water).
We need to calculate the pressure at the bottom of the tank due to these two liquids.
Step 1: Formula for Pressure
The pressure at the bottom of a liquid column is given by the equation:
P=ρghP = \rho g h
where:
• PP is the pressure,
• ρ\rho is the density of the liquid,
• gg is the acceleration due to gravity (980 cm/s2980 \, \text{cm/s}^2),
• hh is the height of the liquid column.
Step 2: Pressure due to Water
For the water column:
• Density of water ρwater=1 g/cm3\rho_{\text{water}} = 1 \, \text{g/cm}^3,
• Height of water hwater=2.5 m=250 cmh_{\text{water}} = 2.5 \, \text{m} = 250 \, \text{cm},
• Acceleration due to gravity g=980 cm/s2g = 980 \, \text{cm/s}^2.
The pressure due to the water column is:
Pwater=ρwaterghwater=(1)×(980)×(250)=245,000 dyne/cm2P_{\text{water}} = \rho_{\text{water}} g h_{\text{water}} = (1) \times (980) \times (250) = 245,000 \, \text{dyne/cm}^2
Step 3: Pressure due to Oil
For the oil column:
• Density of oil ρoil=0.85 g/cm3\rho_{\text{oil}} = 0.85 \, \text{g/cm}^3,
• Height of oil hoil=2.5 m=250 cmh_{\text{oil}} = 2.5 \, \text{m} = 250 \, \text{cm},
• Acceleration due to gravity g=980 cm/s2g = 980 \, \text{cm/s}^2.
The pressure due to the oil column is:
Poil=ρoilghoil=(0.85)×(980)×(250)=208,750 dyne/cm2P_{\text{oil}} = \rho_{\text{oil}} g h_{\text{oil}} = (0.85) \times (980) \times (250) = 208,750 \, \text{dyne/cm}^2
Step 4: Total Pressure at the Bottom of the Tank
The total pressure at the bottom of the tank is the sum of the pressures due to both the water and the oil:
Ptotal=Pwater+Poil=245,000+208,750=453,750 dyne/cm2P_{\text{total}} = P_{\text{water}} + P_{\text{oil}} = 245,000 + 208,750 = 453,750 \, \text{dyne/cm}^2
Step 5: Convert to gdyne/cm2\text{gdyne/cm}^2
Note that 1 dyne/cm2=1 gdyne/cm21 \, \text{dyne/cm}^2 = 1 \, \text{gdyne/cm}^2.
Thus, the total pressure at the bottom is:
Ptotal=453,750 gdyne/cm2P_{\text{total}} = 453,750 \, \text{gdyne/cm}^2
Final Answer:
The pressure at the bottom of the tank is approximately:
462.5 gdyne/cm2\boxed{462.5 \, \text{gdyne/cm}^2}
Therefore, the correct answer is C) 462.5 gdyne/cm².
