Explanation
Pressure on the piston=\dfrac{F}{A}
Force F=m×a
=2000×9.8
Area of cross section A=2.25×10^{-2}m^2
Therefore the pressure P=\dfrac{2000×9.8}{2.25×10^(-2)} P=0.8711×10^6\dfrac{N}{m^2}
Compression of the streamlines means that the stream tube above the winghas a smaller cross-sectional area than that in front of the plane and from the continuity equation , the velocity v' of the air must therefore be greater above the wing.Area {A}'= \dfrac{8}{10}A \Rightarrow \dfrac{A}{{A}'}=\dfrac{10}{8}From continuity equation,Av={A}'{v}'\therefore {v}' =\dfrac{A}{{A}'}v= \dfrac{10}{8} \times 200= 250 \: ms^{-1}The greater velocity (v',as obtained above) implies lower pressure than the normal pressure of the air in front of the plane. Given that the flow-lines under the wing are not compressed at all. The pressure under the wing is just the normal pressure of the air in front of the wing.
From Bernoulli's equation , with both points 1 (for p1) and 2 (for p2) at effectively same elevation.P_a + \dfrac{1}{2}\rho_av_1^2 = P_b + \dfrac{1}{2}\rho_bv_2^2Given:v_2=200\: ms^{-1}\therefore P_2-P_1 = \dfrac{1}{2}\rho_a( v_1^2-v_2^2)= \dfrac{1}{2}\times 1.3( 250^2-200^2)= 1.46 \times 10^4\: Pa
According to to Torricelli's theorem, velocity of efflux at point A ( shown in the figure ) isv_A=\sqrt{(2gh)}After emerging from the orifice the water adopts parabolic path. if it takes time t secs in falling a vertical height (H=h) and covers a horizontal range R_1, then H-h = \dfrac{1}{2}gt^2 \Rightarrow t= \sqrt{\dfrac{2(H-h)}{g}}R_1= v_At= \sqrt{(2gh)} * \sqrt{\dfrac{2(H-h)}{g}}= 2\sqrt{h(H-h)} ....(1) where R_1 is the range.The above results show that Range is same whether the hole is made at distance h from the top or at (H-h) from the top. \therefore h= H-h \Rightarrow h=\dfrac{1}{2}H ....(2)If h is changing at the rate \dfrac{dh}{dt}=Rdifferentiating eqn (1) wrt t, we get \dfrac{dR_1}{dt}= (h(H-h))^{-\dfrac{1}{2}}(\dfrac{dh}{dt}(H-h) + h( -\dfrac{dh}{dt}))=(h(H-h))^{-\dfrac{1}{2}} (R(H-h) -Rh) ....(3)Substituting h=\dfrac{H}{2} from eqn(2) in eqn(3) we get \dfrac {dR_1}{dt} = (\dfrac{H}{2})^{-1}R(H-2h)= (\dfrac{H}{2})^{-1}R(H-H)=0
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