Create the summary notes of the provided text. PHYSICS-IC dp If a = 0 and a, =

Create the summary notes of the provided text.

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PHYSICS-IC dp If a = 0 and a, = 0 then =0 and That is, pressure does not vary along the x-axis if a, = 0 thus pressure at all the PHYSIC dx dy points on plane are small when a, 0 pressure increases in the opposite direction of acceleration. The angle of inclination of the free surface is obtained by When a contai dp dy accele pressan the same in the horizo dx tane = dx a dp g+a, dy Pressure difference in an accelerated frame When a liquid is filled in a container, generally its free surface remains hori- zontal as shown in figure (a) as for its eguilibrium its free surface must be normal to gravity i.e., horizontal. Now, the container is accelerated. Due to the acceleration of the container, liquid filled in it experiences a pseudo force relative to the container and due to this the free surface of liquid which remians normal to the gravity now is filled as shown in the figure and normal to the direction effective gravity. Fig. 2.12 The inclination angle e of the free surface of liquid from the horizontal is given by a a Tan e = 8 = Tan -1 The pressure at the point A is given by PA = Po + hp/g² +a? .2 %3D Where h is the depth of the point A below the free surface of liquid along effectively gravity and P, is the atmospheric pressure acting on this free surface of the liquid. The pressure at point A can also be expressed as PA = Po +4pg, PA = Po +4pa, Pa = Po + hp/g² +a? .2 %3D %3D 2. From figure we have, cosece =2, L2 = hcosece, cosece = Va²+g² a h/a² +g² L2 2 *E, PA = P, + hp/g² +a² %3D %3D a L2 L2 PA = Po + P.Vg²+a², PA =Po + p.Vg²+a² g*+a' %3D cosec 0 Vg +a? a PA = Po+L2pa, similarly PA = Po +Lpg Pressure distribution in a closed accelerated container 不 L- P. A a Fig. 2.13 82 ELITE SERIES for Sri Chaltanya Jr. ICON Students 不 %3D %3D PHYSICS-IC dp If a = 0 and a, = 0 then =0 and That is, pressure does not vary along the x-axis if a, = 0 thus pressure at all the PHYSIC dx dy points on plane are small when a, 0 pressure increases in the opposite direction of acceleration. The angle of inclination of the free surface is obtained by When a contai dp dy accele pressan the same in the horizo dx tane = dx a dp g+a, dy Pressure difference in an accelerated frame When a liquid is filled in a container, generally its free surface remains hori- zontal as shown in figure (a) as for its eguilibrium its free surface must be normal to gravity i.e., horizontal. Now, the container is accelerated. Due to the acceleration of the container, liquid filled in it experiences a pseudo force relative to the container and due to this the free surface of liquid which remians normal to the gravity now is filled as shown in the figure and normal to the direction effective gravity. Fig. 2.12 The inclination angle e of the free surface of liquid from the horizontal is given by a a Tan e = 8 = Tan -1 The pressure at the point A is given by PA = Po + hp/g² +a? .2 %3D Where h is the depth of the point A below the free surface of liquid along effectively gravity and P, is the atmospheric pressure acting on this free surface of the liquid. The pressure at point A can also be expressed as PA = Po +4pg, PA = Po +4pa, Pa = Po + hp/g² +a? .2 %3D %3D 2. From figure we have, cosece =2, L2 = hcosece, cosece = Va²+g² a h/a² +g² L2 2 *E, PA = P, + hp/g² +a² %3D %3D a L2 L2 PA = Po + P.Vg²+a², PA =Po + p.Vg²+a² g*+a' %3D cosec 0 Vg +a? a PA = Po+L2pa, similarly PA = Po +Lpg Pressure distribution in a closed accelerated container 不 L- P. A a Fig. 2.13 82 ELITE SERIES for Sri Chaltanya Jr. ICON Students 不 %3D %3D