Question

potential energy of an object is given as U=alphax^2-betax where x is position of object. Dimension of alpha/beta is​

Answer 1

Explanation:

Dimension of αx should be the same as dimension of V

So [α]=MLT

−2

dimension of [β] will be same as of V

[β]=ML

2

T

−2

[γ]=L

So we can see that

β

αγ

is dimensionless

Answer 2

The ratio of dimensions of alpha and beta is [L].

Given,

Equation:$U=(alpha)x^{2} -(beta)x.$

x-position of the object.

To find:

the dimension of alpha/beta.

Solution:

• The principle that will be used here is known as "Principle of homogeneity of dimensions".
• It states that the dimensions of all the terms in an equation must be equal.
• Simply, it states that we add or subtract similar physical quantities.

As U here is the potential energy, its dimension will be $[ML^{2} T^{-2} ]$.

Dimensions of $(alpha)x^{2}$=Dimensions of $(beta)x$=Dimensions of U

As x is the position of the object, its dimension will be [L].

The dimensions of beta will be:

Dimensions of $(beta)x$=Dimensions of U

$(beta)x=U\\(beta)[L]=[ML^{2} T^{-2} ]\\(beta)=[ML T^{-2} ].$

The dimensions of alpha will be:

$(alpha)x^{2}=U\\(alpha)[L^{2}]=[ML^{2} T^{-2} ]\\(alpha)=[MT^{-2} ].$

The ratio of dimensions of alpha and beta will be:

$\frac{(alpha)}{(beta)} =\frac{[MLT^{-2}] }{[MT^{-2}] } \\\frac{(alpha)}{(beta)}=[L].$

Hence, dimension of alpha/beta is [L].

#SPJ2