Question

The velocity V of a particle depends on time 't' as V = At2
+ Bt. Find the dimensions and units of A and B.

Answer 1

Explanation:

We have,

$\sf{V = At^{2}+Bt}$

Dimensions of LHS = Dimensions of RHS

$\sf{[V] = [At^{2}+Bt]}$

Ac. to the principle of homogeneity,,

Dimensions of At² = Dimensions of Bt

$\sf{\implies{[At^{2}]=[Bt]}}$

$\sf{\implies{[A][T^{2}]=[B][T]}}$

$\sf{\implies{\frac{[A][T^{2}]}{[T]}=[B]}}$

$\sf{\implies{[B]=[A][T] ........(i)}}$

Therefore,,

$\sf{[V] = [At^{2}]}$

$\sf{\implies{[M^{0}L^{1}T^{-1}] = [A][T^{2}]}}$

$\sf{\implies{\frac{[M^{0}L^{1}T^{-1}]}{T^{2}} = [A]}}$

$\sf{\implies{\boxed{\sf{[A]=[M^{0}L^{1}T^{-3}]}}}}$

Substituting value of [A] in eqn. (i),

We get,

$\sf{[B]=[M^{0}L^{1}T^{-3}][T] }$

$\sf{\implies{\boxed{\sf{[B]=[M^{0}L^{1}T^{-2}]}} }}$

Hence, Dimensions of A,

$\sf{\boxed{\to{\sf{[A]=[M^{0}L^{1}T^{-3}]}}} }$

And Dimensions of B,

$\sf{\boxed{\to{\sf{[B]=[M^{0}L^{1}T^{-2}]}}} }$

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Answer 2

Given:

v = At² + Bt

To find:

Unit and dimensions of

• A and B

Calculation:

The dimensions of LHS and RHS of any equation have to be equal.

$\rm[v] = [A {t}^{2} ]$

$\rm \implies [L{T}^{ - 1} ] = [A {T}^{2} ]$

$\rm \implies [A] = [{M }^{0} L{T}^{ - 3} ]$

• So, unit of A is metre/sec³.

$\rm[v] = [B t ]$

$\rm \implies [L{T}^{ - 1} ] = [BT ]$

$\rm \implies[B] = [{M}^{0} L{T}^{ - 2} ]$

• So, unit of B is metre/sec².