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# Please solve both the questions urgently required answer is known solution not

2 years ago

Khimraj
3007 Points

The container was completely filled with water so the amount of water that will rise in the rear side will go out or will spill out. so. the water level in the rear side will remain constant because it cannot increase more.tan$\Theta$ = h/Lwhere h = fall in water level on the front side.L = length of the containerh = L tan$\Theta$Now volume of the free space above the water level after spilling = (1/2) x h x L x BThis volume is equal to amount of water spilled out.Total volume of container - L x B x HRatio = (h x L x B)/(2 x L x B x H) = h/2H = 1/3h = H - L tan$\Theta$ = H( 1 - tan$\Theta$)h/2H = (1 - tan$\Theta$)/2 = 1/3tan$\Theta$ = 1/3 = a/ga = g/3Let L = length of the containerB = breadth of the containerH = height of the containeras container is a cube.so, L = B = HWe know that angle made by free surface with the horizontal is given by tan$\Theta$ = a/gwhere a = acceleration of the container in the horizontal directionNow we can see that there are sides of the container when seen from the side. These are front side and rear side.When the container is accelerating then the water level on the front side will come down and the water level in the rear side will go up.
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions