Question

Dimensional Analysis Questions Class 11

Q.31 to 35

Answer 1

Hey there !!!!

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Gas bubble oscillates such that T is proportional to PᵃρᵇEˣ.

⇒⇒⇒⇒ T=[PᵃρᵇEˣ]


           M⁰L⁰T¹=(ML⁻¹T⁻²)ᵃ(ML⁻³)ᵇ(ML²T⁻²)ˣ

           M⁰L⁰T¹=MᵃL⁻ᵃT⁻²ᵃMᵇL⁻³ᵇMˣL²ˣT⁻²ˣ
   
           M⁰L⁰T¹=Mᵃ⁺ᵇ⁺ˣ L⁻ᵃ⁻³ᵇ⁺²ˣ T⁻²ᵃ⁻²ˣ
 M,L,T bases  are equal ,So, equating the powers 

 0=a+b+x ----Equation 1

0=-a-3b+2x----Equation 2

1=-2a-2x--------Equation 3

From equation 3 a+x=-1/2 substituting in equation 1

0=a+b+x

0=-1/2+b

b=1/2

From equation 2  0=-a-3b+2x

0=2x-a-3/2

3/2 =2x-a------- equation 4

 from equation 3    -2a-2x=-1

Solving equation 3&4 we get  a=-5/6 x=1/3 .

So, a=-5/6    b=1/2    x=1/3

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32) Dimensions of Force= MLT⁻²

      Dimensions of velocity=LT⁻¹
 
      Dimensions of  area=L²

     Dimensions of   density=ML⁻³.

Let us consider area,density,velocity are proportional to force.


   F∞vᵃAᵇρˣ------Equation 1


  MLT⁻²=(LT⁻¹)ᵃ(L²)ᵇ(ML⁻³)ˣ


  MLT⁻²=LᵃT⁻ᵃL²ᵇMˣL⁻³ˣ


  MLT⁻²=Mˣ Lᵃ⁺²ᵇ⁻³ˣ T⁻ᵃ


Comparing powers of M,L,T

1=x----Equation 2

1=a+2b-3x----Equation 3

-2=-a-----Equation 4

From equations 2&4 we get x=1 and a=2 

Substituting x=1 and a=2 in equation 3

a+2b-3x=1

2+2b-3=1

2b=2

b=1 a=2 x=1 

 F∞vᵃAᵇρˣ

 F∞vA²ρ

removing ∞ by introducing proportionality const "k"

F=kvA²ρ.


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33)

  Dimensions of "d"=L

  Dimensions of Elasticity(E)=ML⁻¹T⁻²

  Dimensions of Energy(K)=ML²T⁻²

Let us consider "d" is proportional to E & K


    d∞EᵃKᵇ


   M⁰L¹T⁰=(ML⁻¹T⁻²)ᵃ(ML²T⁻²)ᵇ


   M⁰L¹T⁰=MᵃL⁻ᵃT⁻²ᵃMᵇL⁻²ᵇT⁻²ᵇ

   M⁰L¹T⁰=Mᵃ⁺ᵇL⁻ᵃ⁺²ᵇT⁻²ᵃ⁻²ᵇ

Comparing powers of M,L,T

 0=a+b----Equation 1

 1=2b-a-----Equation 2

 0=-2a-2b----Equation 3

From equation 1 and 3 we get a+b=0⇒⇒⇒ a=-b

Substituting a=-b in equation 2

1=2b-a

1=2b-(-b)

 1=3b

  b=1/3 and a = -1/3


  d∞EᵃKᵇ

  d∞E⁻¹/³K¹/³

 ⇒⇒⇒⇒ d∞(K/E)¹/³

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34)  Let us consider Surface tension (S) is proportional to                                                                                  
                            S=k*Mass*Pressure*radius


                    S=k(mᵃPᵇrˣ)

Dimensions of Surface tension=ML⁰T⁻²

Dimensions of Pressure=ML⁻¹T⁻²

Dimensions of mass=M

Dimensions of radius=L

 and k=1/2

 S=k(mᵃPᵇrˣ)

  ML⁰T⁻²=(Mᵃ(ML⁻¹T⁻²)ᵇ(L)ˣ)/2

 ML⁰T⁻²=Mᵃ⁺ᵇL⁻ᵇ⁺ˣT⁻²ᵇ/2

Comparing powers of M,L,T

1=a+b----Equation 1

0=x-b------Equation 2

-2=-2b----Equation 3

From equation 3    b=1 substituting b=1 in equations 1&2 

we get a=0 and x=1  

Now a=0 b=1 x=1 k=1/2

S=k(mᵃPᵇrˣ)

S=Pr/2

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35) Let us consider Time period "T" is proportional to 

                    T=radius*density*gravitational constant

Dimensions of time period =T

Dimensions of radius=L

Dimensions of density=ML⁻³

Dimensions of gravitational constant(G) = M⁻¹L³T⁻².

  
                   T∞rᵃρᵇGˣ

  
                   T=Lᵃ(ML⁻³)ᵇ(M⁻¹L³T⁻²)ˣ

                   M⁰L⁰T¹=Mᵇ⁻ˣLᵃ⁻³ᵇ⁻³ˣT⁻²ˣ

Comparing powers of M,L,T

 b-x=0  ----Equation 1

a-3b-3x=0 -----Equation 2

1=-2x----Equation 3

From equation 3  x=-1/2  and from equation 1  x=b 

a-3b-3x=0

But b=x

a-3b-3b=0

a=0 

So, a=0 b=-1/2 x=-1/2

 T∞rᵃρᵇGˣ

  T=k*r⁰ρ⁻¹/²G⁻¹/²

  T=k/√ρG.


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