Question

The magnitude of force experienced by an object moving
with speed is given by F= kv2. Find dimensions of k.

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Answer 1

AnswEr :

Given Expression,

$\sf F = Kv^2$

Here,

• "F" is the Force experienced by the object

• "v" is the Velocity of the object

Now,

$\sf{Dimensions} \begin{cases} \sf{F \longrightarrow MLT^{-2} } \\ \sf{v \longrightarrow LT^{-1} } \end{cases}$

Thus,

$\sf [MLT^{-2}] = K[LT^{-1} ] \\ \\ \dashrightarrow \sf K = \dfrac{M\cancel{L}T^{-2}}{\cancel{L}T^{-1}} \\ \\ \dashrightarrow \sf K = MT^{-2 + 1} \\ \\ \dashrightarrow \boxed{\boxed{\sf K = MT^{-1} }}$

Dimension of 'K' is $\sf MT^{-1}$

Answer 2

Given :

• Equation is F = kv²

To Find :

• Dimensions of k

Solution :

We are given F = kv²

Where,

• F is Force
• V is velocity

As we know that, Dimensional formula of Force is $\sf{[MLT^{-2}]}$ And dimension formula of Velocity is $\sf{[LT^{-1}]}$

putting these values

$\implies \sf{[MLT^{-2}] \: = \: k [LT^{-1}]} \\ \\ \implies \sf{k \: = \: \dfrac{[MLT^{-2}]}{[LT^{-1}]}} \\ \\ \implies \sf{k \: = \: \dfrac{MT^{-2}]}{[T^{-1}]}} \\ \\ \implies \sf{k \: = \: [MT^{-1}]}$