Question

The velocity of a particle v at an instant t is given by v=at +bt²the dimensions of b is

Answer 1

Answer:

Given:

Velocity of a particle is as follows :

$\sf{ \orange{v = at + b {t}^{2}}}$

To find:

Dimensions of "b"

Concept:

Always remember the fact that Quantities with same Dimensions can be added . Otherwise they can't be added.

For example , in this case :

Dimensions of [bt²] will be same as [v] because they are present in addition relationship.

Calculation:

$\sf{ \{v \} = \{b {t}^{2} \}}$

$\sf{ = > \{L {T}^{ - 1} \} = \{b {T}^{2} \}}$

$\sf{ = > \{b \} = \{L {T}^{ - 3} \} }$

So Dimensions of b is as follows :

$\boxed{ \large{ \red{ \bold{ \sf{ \{b \} = \{L {T}^{ - 3} \} }}}}}$

Additional information:

• Only Quantities with same Dimensions can be added.
• Quantities within Trigonometric identities do not have Dimensions.
• Quantities in Exponential or logs do not have Dimensions.

Answer 2

$\huge \star {\underline{\boxed{\red{\sf{Answer :}}}}} \star$

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As we are given,

V = at + bt²

_____________________

Take, V = at

$\bigstar \: \: {\underline{\sf{Dimension \: of \: v \: is \: M^0 LT^{-1}}}}$

$\bigstar \: \: \underline{\sf{Dimenion \: of \: t \: is \: M^0 L^0 T^1}}$

Put Values

$\implies {\sf{[LT^{-1}] \: = \: [T] \: \times \: a}} \\ \\ \implies {\sf{a \: = \: \dfrac{LT^{-1}}{T}}} \\ \\ \implies {\sf{a \: = \: LT^{-2}}} \\ \\ {\underline{\sf{\therefore \: Dimensions \: of \: a \: is \: [LT^{-2}]}}}$

$\rule{200}{2}$

Now Take, V = bt²

$\implies {\sf{[LT^{-1} \: = \: [T^2] \: \times \: b}} \\ \\ \implies {\sf{b \: = \: \dfrac{[LT^{-1}]}{[T^2]}}} \\ \\ \implies {\sf{b \: = \: [LT^{-3}]}} \\ \\ {\underline{\sf{\therefore \: Dimensions \: of \: b \: is \: [LT^{-3}]}}}$