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"""give me complete correct appropriate detailed answer in 4 steps step 1 about introduction or overview of question step 2 & 3 about answer with proper calculation and its formula , tables if required also mention step titles and last only final answer as per question asked format and requirements " provide ALL THE ANSWER WITH FUIL CALVCULATION WITHOUT AI VOICES STRICTLYCO aaa EEE Item 6 EEE Te Besides ca di res TL So eR EE Tn [ee Ena ne es Ee fi toe 0 dW iB 500 kg/m and p,, = 1000 kg/m? (Figure 1) [1] RE Fu, = | Value Units [Ep Ft ef AB ri or - “rome Fu, = | Value Units Wy Answers Gre [Figre 1 v]oft Partc am Tmo = ~1 I 1 “rome ‘ul d= | Value Units [3 [5m JET——y

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"""give me complete correct appropriate detailed answer in 4 steps step 1 about introduction or overview of question step 2 & 3 about answer with proper calculation and its formula , tables if required also mention step titles and last only final answer as per question asked format and requirements " provide ALL THE ANSWER WITH FUIL CALVCULATION WITHOUT AI VOICES STRICTLYUploaded ImageCO aaa EEE Item 6 EEE Te Besides ca di res TL So eR EE Tn [ee Ena ne es Ee fi toe 0 dW iB 500 kg/m and p,, = 1000 kg/m? (Figure 1) [1] RE Fu, = | Value Units [Ep Ft ef AB ri or - “rome Fu, = | Value Units Wy Answers Gre [Figre 1 v]oft Partc am Tmo = ~1 I 1 “rome ‘ul d= | Value Units [3 [5m JET——y

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Asked by: Yash Suthar
Created at: 2025-06-22 16:09:37

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Let's solve the problem step by step as you requested: --- ### **Step 1: Introduction / Overview** The tank is filled with water to a depth \( d = 2 \) m. We are to determine the resultant force exerted by the water on side **A** and side **B** of the tank. The tank sides have the following dimensions: - Side **A**: width = 3 m, height = d = 2 m - Side **B**: width = 2 m, height = d = 2 m We then need to find the depth to which oil (with density \( \rho_o = 900 \) kg/m³) must be filled so that it exerts the same resultant force as water (\( \rho_w = 1000 \) kg/m³) at 2 m depth. --- ### **Step 2: Calculating Resultant Forces for Water** The resultant force due to a fluid on a vertical wall is given by: \[ F = \rho g A \frac{h}{2} \] where: - \( \rho \) = density of fluid - \( g \) = acceleration due to gravity (9.81 m/s²) - \( A \) = area of the wall (\( \text{width} \times \text{height} \)) - \( h \) = depth of fluid #### **Force on Side A:** - Width = 3 m, Height = 2 m, Area \( A_A = 3 \times 2 = 6 \) m² \[ F_{R_A} = \rho_w \cdot g \cdot A_A \cdot \frac{d}{2} \] \[ F_{R_A} = 1000 \cdot 9.81 \cdot 6 \cdot \frac{2}{2} \] \[ F_{R_A} = 1000 \cdot 9.81 \cdot 6 \cdot 1 \] \[ F_{R_A} = 58,860 \ \text{N} \] #### **Force on Side B:** - Width = 2 m, Height = 2 m, Area \( A_B = 2 \times 2 = 4 \) m² \[ F_{R_B} = \rho_w \cdot g \cdot A_B \cdot \frac{d}{2} \] \[ F_{R_B} = 1000 \cdot 9.81 \cdot 4 \cdot \frac{2}{2} \] \[ F_{R_B} = 1000 \cdot 9.81 \cdot 4 \cdot 1 \] \[ F_{R_B} = 39,240 \ \text{N} \] --- ### **Step 3: Calculating Required Oil Depth** Set the oil force on side A equal to that of water: \[ F_{R_A, \, oil} = F_{R_A, \, water} \] \[ \rho_o g A_A \frac{d_{oil}}{2} = \rho_w g A_A \frac{d_{water}}{2} \] \[ \rho_o d_{oil} = \rho_w d_{water} \] \[ d_{oil} = \frac{\rho_w}{\rho_o} d_{water} \] \[ d_{oil} = \frac{1000}{900} \times 2 \] \[ d_{oil} = 2.222 \ \text{m} \] --- ### **Step 4: Final Answers** **Part A:** \[ F_{R_A} = 58,860 \ \text{N} \] **Part B:** \[ F_{R_B} = 39,240 \ \text{N} \] **Part C:** \[ d = 2.22 \ \text{m} \]

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