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Question

The closed container filled with water is moving with some acceleration as shown in the figure. Then PAPD is :-


A
4×103 N/m2
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B
+4×103 N/m2
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C
+5.6×105 N/m2
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D
5.6×104 N/m2
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Solution

The correct option is D 5.6×104 N/m2
For the given accelerated container the pressure difference in liquid along the horizontal direction is given by,
ΔP=ρax ...(i)


For the accelerated container, the pressure will increase in a direction opposite to that of acceleration of the vessel.
PC>PD
From eq.(i),
PCPD=ρax
PCPD=1000×2×2
(x=2 m)
PCPD=4×103 N/m2 ...(ii)

The pressure difference for liquid due to vertical depth is given as:
ΔP=ρgh
The pressure will increases in vertically downward direction
PCPA=ρgh
where h=6 m is the depth of liquid.
PCPA=1000×10×6
PCPA=60×103 N/m2 ...(iii)
On subtracting (iii) from (ii),
(PCPD)(PCPA)=(4×103)(60×103)
PAPD=56×103
PAPD=5.6×104 N/m2
Thus, difference in pressure (PAPD) is 5.6×104 N/m2

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