Question

A particle experiences a force F=kr^2r. Where r is the unit vector along position vector r. The dimensional formula of 'k' is :​

Answer 1

Given:

$\vec{F} =( k {r}^{2} ) \hat{r}$

To find:

Dimensional formula of 'k' ?

Calculation:

Let's represent the force magnitude:

$| \vec{F}| = k {r}^{2}$

Now, we know that, dimension of LHS should be equal to that of RHS, for a dimensionally correct equation:

$\implies \: \bigg[ML{T}^{ - 2} \bigg] = \bigg[k \bigg] \times \bigg[{L}^{2} \bigg]$

$\implies \: \bigg[k\bigg] = \bigg[M{L}^{ (1 - 2)} {T}^{ - 2} \bigg]$

$\implies \: \bigg[k\bigg] = \bigg[M{L}^{ - 1} {T}^{ - 2} \bigg]$

So, dimensional form of [k] is :

$\boxed{ \bf \: \bigg[k\bigg] = \bigg[M{L}^{ - 1} {T}^{ - 2} \bigg] }$

Answer 2

The dimensional formula of 'k' is [ML⁻¹T⁻²]

Explanation:

Given:

The expression of the force

$\vec F=k (\vec r)^2\hat r$

To find out:

The dimensional formula of k

Solution:

In dimensional analysis we know that

For a given expression, the

LHS dimension = RHS dimension

Dimensions of force = $[MLT^{-2}]$

The dimensions of a position vector $\vec r$ will be equal to the dimension of displacement i.e. $[L]$

Since unit vector is found out by dividing a vector with its magnitude, both of which will have the same unit

Therefore, the unit vector will be dimensionless quantity

Therefore,

The dimensions of k = Dimensions of $\vec F$/(Dimension of $\vec r$)²

or, $[k]=\frac{[MLT^{-2}]}{[L]^2}$

or, $[k]=[ML^{-1}T^{-2}]$

Hope this answer is helpful.

Know More:

Q: the dimensional formula of physical quantities Z is M^a L^b T^-c the percentage error in measurement of mass length and time are Alpha percent beta percent and gamma percent respectively . the percentage error in Z is​:

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