Question

1-Test by method of dimensions the correctness of the equation =mgl²/4bd³Y

Where is the depression produced at the centre of a bar of length l breadth and depth d, placed symmetrically on two knife edges near its ends and loaded in the middle by mass m, and Y is the Young's modulus off the metical of the bar
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2-Speed (v) of sound in the medium of elasticity E and density p is given by v=√E/p. Check the accuracy of the above statement.

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3-Check the dimensional accuracy of the relation 1/2 mv²=mgh, where the letters have their usual meanings.

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4-Check by the method of dimensions the validity of expression Q=πPr^4/8nl

Where Q is the volume of liquid of viscosity n flowing per second through a capillary tube of length l and radium r and pressure different b/w the ends of the tube is P.

Answer 1

Hey there !!!!!

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1) Angle of depression =∅     cos∅= mgl²/4bd³Y

cos∅  is dimensionless as generally it is ratio of sides i.e, L/L =L⁰ which is dimensionless.

⇒⇒mgl²/4bd³Y must be dimensionless.

Y represents "young's modulous"   = Stress/strain =[dimensions of pressure]

Y=[Force/area]=[MLT⁻²/L²]=[ML⁻¹T⁻²]

Dimensions of mgl²/4bd³Y = MLT⁻²*L²/L*L³*ML⁻¹T⁻² =ML³T⁻²/ML³T⁻²
                        
                                                                                  =M⁰L⁰T⁰

So, mgl²/4bd³Y is dimensionless.

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2)  v=(√Elasticity/density)

Dimensions of velocity =LT⁻²  

Dimensions of elasticity = [Dimensions of pressure] 

As, it is defined as the ratio of tangential stress to the shearing strain and strain is dimensionless and stress has the dimensions of pressure.

[Elasticity] = [Force/area]=ML⁻¹T⁻²

[Density]=Mas/volume=ML⁻³

So,
      
       √E/p = √ML⁻¹T⁻²/ML⁻³ =√L²T⁻²=LT⁻¹

LT⁻¹ is the dimension of velocity so given relation is dimensionally correct.

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3) mv²/2=mgh

Both of these formulas represent diff forms of energies.

mv²/2 = Kinetic energy and mgh=Potential Energy

Dimensions of mv²/2 = M*(LT⁻¹)²=ML²T⁻²---- Equation 1

Dimensions of mgh = M*LT⁻²*L=ML²T⁻²------- Equation 2

From equations 1 & 2 given relation is dimensionally correct.

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4) Q=πPr⁴/8ηl  

Q=Volume/time =L³T⁻¹  

Pressure=Force/area = ML⁻¹T⁻²

r=radius=L    η=ML⁻¹T⁻¹ length=L

Q=πPr⁴/8ηl  

   = ML⁻¹T⁻²*L⁴/ML⁻¹T⁻¹L

    = ML³T⁻²/ML⁰T⁻¹  = L³T⁻¹ which is equal to dimensions of Q=L³T⁻¹

So, given relation is dimensionally correct.

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