The closed container filled with water is moving with some acceleration as shown in the figure. Then PA−PD is :-
A
−4×103N/m2
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B
+4×103N/m2
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C
+5.6×105N/m2
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D
−5.6×104N/m2
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Solution
The correct option is D−5.6×104N/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), PC−PD=ρax ⇒PC−PD=1000×2×2 (∵x=2m) ⇒PC−PD=4×103N/m2...(ii)
The pressure difference for liquid due to vertical depth is given as: ΔP=ρgh ∵ The pressure will increases in vertically downward direction PC−PA=ρgh
where h=6m is the depth of liquid. ⇒PC−PA=1000×10×6 ⇒PC−PA=60×103N/m2...(iii)
On subtracting (iii) from (ii), ⇒(PC−PD)−(PC−PA)=(4×103)−(60×103) ⇒PA−PD=−56×103 ∴PA−PD=−5.6×104N/m2
Thus, difference in pressure (PA−PD) is −5.6×104N/m2