Problem Statement: If a rectangular container has a length of 5m and a height of 2m, and contains water up to a height of 1.5m. If the width of the container given is 3m, find the volume of water spilled when the container is subjected to horizontal acceleration of (i) 4.5g m/s² (ii) 6g m/s²?

Solution:

Given:
Length of container (l) = 5m
Height of container (h) = 2m
Water level (w) = 1.5m
Width of container (b) = 3m

Step 1: Calculate the volume of water in the container
Volume of water (Vw) = l x b x w
= 5 x 3 x 1.5
= 22.5 m³

Step 2: Calculate the maximum horizontal force acting on the water
Maximum horizontal force (Fmax) = ma
= m x a
= ρVg x a
where, ρ = density of water
V = volume of water
g = acceleration due to gravity
a = horizontal acceleration
= 1000 x 22.5 x 9.81 x a
= 220725a N

Step 3: Calculate the force exerted by water on the walls of the container
Force exerted by water (Fw) = ρgh x lb
where, g = acceleration due to gravity
h = height of water level
lb = length x breadth of container
= 1000 x 9.81 x 1.5 x 5 x 3
= 220725 N

Step 4: Calculate the volume of water spilled
(i) When the container is subjected to horizontal acceleration of 4.5g m/s²
Net force acting on the water (Fnet) = Fmax - Fw
= (220725 x 4.5) - 220725
= 674537.5 N
Height of water level on one side (h1) = (Fnet / ρlb) - h
= (674537.5 / (1000 x 5 x 3)) - 1.5
= 0.225 m
Volume of water spilled (Vsp) = lb x h1 x b
= 5 x 0.225 x 3
= 3.375 m³
(ii) When the container is subjected to horizontal acceleration of 6g m/s²
Net force acting on the water (Fnet) = Fmax - Fw
= (220725 x 6) - 220725
= 897450 N
Height of water level on one side (h1) = (Fnet / ρlb) - h
= (897450 / (1000 x 5 x 3)) - 1.5
= 0.525 m
Volume of water spilled (Vsp) = lb x h1 x b
= 5 x 0.525 x

Khud solve karo bhosdi walon