Question

Dimension analysis question! Class 11!

1-Write the dimension of a/b in the relation p=(a-t²)/bx, where p is the pressure, x is the distance and t is the time.

2-An apprentice engineer found that the volume V of water which passes any point of canal during time t is connected with the cross section a of the canal and the velocity of v of water by the relation, V=kavt. Verify the correctness of the relation.

3-The height h to which the liquid of surface tension T and density p rises in a capillary of radium r is given by the relation h=2tcos/r² pg, where Ф is the angle of contact and g is the acceleration due to gravity. Verify the correctness of the statement.

Answer 1

The 3rd bit is not clear.
the angle of the cosine function is not specified

Answer 2

Hey there !!!!!

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1)P=(a-t²)/bx

  P=a/bx -t²/bx

Dimensions of a/bx and t²/bx = dimensions of Pressure


Pressure = Force/Area = Mass*acceleration/area=MLT⁻²/L²=ML⁻¹T⁻².

 ML⁻¹T⁻²=a/bx

ML⁻¹T⁻²*L=a/b

Dimensions of a/b=ML⁻¹⁺¹T⁻²=ML⁰T⁻²

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2)  Volume = Area*Height = L²*L=L³(dimensions of volume)

According to observation by engineer

  V=kAvt

k= dimensionless constant    A= area    v=Velocity    t=Time

[V]= L²*LT⁻¹*T¹= L³

So observations made by engineer are correct and [kAvt] is dimensionally equal to Volume.

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3)  h = 2TcosФ/r²рg

If the above statement is dimensionally correct then dimensions of 

  2TcosФ/r²рg must be equal to "L".

T = tension=force= mass*acceleration=MLT⁻²

r= radius =L

р=density=Mass/volume=M/L³=ML⁻³

g=gravity=acceleration=LT⁻² cosФ is dimensionless

Now,

  2TcosФ/r²рg = MLT⁻²/L²ML⁻³LT⁻²= M*M⁻¹L*L⁻³*L³*T⁻²*T²=L

So dimensions of 2TcosФ/r²рg equals to dimensions of "h" so given relation  of 2TcosФ/r²рg=h is dimensionally correct.

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Hope this helped you..............