Sep 2019 CSAT mock test, Physics I #20

Mass of the object A is $m_A=\dfrac{4}{5}\rho SH$ and of the object B is twice this. Then force acting on A in equilibrium is $m_Ag-h\rho gS$ and on B at that instant is $m_Bg-3h\rho gS$ . Hence using force balance we get $\dfrac{h}{H}=\dfrac{2}{5}$ . So $\dfrac{60h}{H}=\boxed{24}$

For a given height $(h)$ , there are two unknown values: The string tension $(T)$ and the normal reaction force $(N)$ at the edge of the hole upon which block $B$ rests. My process is therefore:

1) Sweep the height $(h)$ of the fluid upwards from zero
2) For each height value, calculate $T$ and $N$ based on force-balance equations
3) Continue sweeping until either $T$ or $N$ becomes negative. When one of them becomes negative, that means that the physical configuration has to change (ex: block B lifting up)

Code below (with extensive comments)

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import math

H = 1.0     # Height of block A
rho = 1.7   # Density of reservoir fluid
S = 0.4     # Base area of block A
g = 10.0    # gravitational acceleration

rhoA = 0.8*rho  # density of block A

mA = S*H*rhoA  # mass of block A
mB = 2.0*mA    # mass of block B

###########################

dh = 10.0**(-6.0)  # infinitesimal change in height "h"

h = dh

N = 0.01  # Junk initialization numbers for normal force and tension
T = 0.01

while (N >= 0.0) and (T >= 0.0):  # run while N and T are both positive

    # mA*g = T + S*h*rho*g       # Force balance for block A

    T = mA*g - S*h*rho*g        # Solve this for the string tension

    # mB*g = T + N + rho*g*h*(3.0*S)  # Force balance for block B

    N = mB*g - T - rho*g*h*(3.0*S)   # Solve this for the normal force

    h = h + dh


###########################

print h
print (60.0*h)/H
#0.400001999996
#24.0001199998