Question

Dimension analysis questions. Class 11!

1-The velocity v of a particle depends upon t according to the equation v=a+bt+c/(d+t). Write the dimensions of a b c and d.

2-Write the dimensions of ab in the relation E=(b-x²)/at, where E is the energy, X is the distance and t is the time.

Answer 1

1. V = a + bt + c/(d + t)
where, V is velocity , t is time .

according to rule of dimension,
dimension of V = dimension of a = dimension of bt = dimension of c/(d +t)

we know,
dimension of V = [ LT^-1 ]
so,
dimension of a = [LT^-1 ]

dimension of bt = [LT^-1 ]
dimension of b × dimension of t = [LT^-1 ]
dimension of b = [LT^-1]/[T] = [LT^-2 ]

dimension of c/(d + t) = [LT^-1 ]

here dimension of d = dimension of t = [T]

now,
dimension of C/dimension of d or t = [LT^-1]
dimension of c = [LT^-1] [T] = [ L ]


2.given ,
E = (b - x²)/at
where, E is the energy, x is distance and t is time .

we know,
dimension of energy = [ML²T^-2]

now,
dimension of b = dimension of x²
dimension of b = [ L]² = [ L²]

dimension of E = dimension of b or x²/dimension of at

[ML²T^-2] = [L²]/a[T]

[ML²T^-2] [T]/[L²] = 1/dimension of a

[MT^-1 ] = 1/dimension of a

dimension of a = 1/[MT^-1 ] = [M^-1 T]

question ask
dimension of ab = ?

so, dimension of ab = [ L²] [M^-1 T]
= [ M^- 1L² T ]

Answer 2

Hey there !!!!!

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According to rules of dimensional analysis 

                   V=LT⁻¹

So, all the terms on LHS must be equal to dimensions of V.

  v=a+bt+(c/d+t)


a=LT⁻¹ bT=LT⁻¹ c/d+t =LT⁻¹ and T=d


a=LT    b=LT⁻¹/T=LT⁻²       c/d+t=LT⁻¹

a=LT    b=LT⁻²      c=LT⁻¹*T=L    d=T

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2)  E=(b-x²)/at
 

     E=b/at -x²/at

Using dimensional analysis dimension of each term =dimensions of E

[E] =[b/at] =[x²/at]

E=Force*displacement=Mass*acceleration*displacement=M*LT⁻²*L=ML²T⁻².

ML²T⁻²=[x²/at]

ML²T⁻²=L²/aT

a=L²/ML²T⁻²T

a=M⁻¹L⁰T⁻¹

ML²T⁻²=b/at

ML²T⁻²*a*T=b

ML²T⁻²M⁻¹L⁰T⁻¹*T=b

b=M⁰L²T⁻²   a=M⁻¹L⁰T⁻¹

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