Question

The velocity V in cm per second of a particle is given in terms of time t in second by the equation v = at + b/t+c. Find the dimensions of a, b and c.

Answer 1

hello friend...!!

velocity = [L¹T⁻¹]

given that ,

v = at + b/t  + c 

1)  v = at 

[M⁰L¹T⁻¹]  = a[T¹]

implies ,  a = [M⁰L¹T⁻²]-----( 1 answer )

2) v = b/t

[M⁰L¹T⁻¹] = b / [ T¹]

implies, b = [M⁰L¹T⁰] ------------( 2 answer)

3) v = c 

implies,

[M⁰L¹T⁻¹]  = c ----------------------( 3 answer )

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hope it helps...!!!

Answer 2

Hi there

Before solving this lemme give a quick tricks to solve this type of questions

• Try to solve from LHS And proof it equally arrived by RHS

•See over here V = at + b/t + c

you can write it also as

v = at

v = b/t

v = c

Now coming to our Question

$you \: need \: to \: know \: first \: the \: dimensional \: formula \: of \: velocity \: \: \: \\ that \:is \: \: \\ \\ v \: = \frac{L \: }{T} \: \: \: \: \: \: \: \: \: \: \: \: ( since \: \: v \: = s \: = \frac{length}{time \: } ) \\ \\ \: \: \: \: \ = { { {M}^{0} L}^{1} T}^{ - 1} \: \\ \\ see \: if \: \\ \\ v \: = at \\ \\ = > { { {M}^{0} L}^{1} T}^{ - 1} \: \: = a \times \: {T}^{ \: 1} \\ \\ = > a \: = { { {M}^{0} L}^{1} T}^{ - 2} \: \\ \\ simlarly \: \\ \\ v \: = b/t \: \\ then \: \\ \\ { { {M}^{0} L}^{1} T}^{ - 1} \: = b/t \\ \: \\ b \: = { { {M}^{0} L}^{1} T}^{ 0 } \\ \\ and \: last \: \: one \: \\ v \: = c \\ \\ c \: = { { {M}^{0} L}^{1} T}^{ - 1} \:$

Hope this helps you