The potential energy of a particle varies wit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The potential energy $U$ is given by $U = \frac{A \sqrt{x}}{x^2 + B}$.According to the principle of homogeneity of dimensions,the dimensions of te

Vedclass · Accepted Answer

$[ML^{11/2} T^{-2}]$

Vedclass · Answer

(A) The potential energy $U$ is given by $U = \frac{A \sqrt{x}}{x^2 + B}$.According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same. Therefore,the dimensions of $B$ must be equal to the dimensions of $x^2$.Since $x$ is distance,$[x] = [L]$,so $[B] = [L^2]$.The dimensions of potential energy $U$ are $[ML^2 T^{-2}]$.From the equation,$A = \frac{U(x^2 + B)}{\sqrt{x}}$.Substituting the dimensions:$[A] = \frac{[ML^2 T^{-2}] [L^2]}{[L^{1/2}]} = [ML^2 T^{-2} L^2 L^{-1/2}] = [ML^{7/2} T^{-2}]$.Now,we need the dimensions of $AB$:$[AB] = [A][B] = [ML^{7/2} T^{-2}] [L^2] = [ML^{11/2} T^{-2}]$.Thus,the correct option is $A$.

The potential energy of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x^2 + B}$,where $A$ and $B$ are dimensional constants. Then,the dimensional formula for $AB$ is:

Similar Questions

Which of the following relations can be derived using dimensional analysis?

In the equation $[X+\frac{a}{Y^2}][Y-b]= RT$,$X$ is pressure,$Y$ is volume,$R$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is

If $z = xP + G$,where $P$ is pressure and $G$ is the universal gravitational constant; then the dimensional formulas for $x$ and $z$ respectively are (here,$G = \frac{Fr^2}{m_1 m_2}$,$P = \frac{\text{Thrust}}{\text{Area}}$).

The speed of light $(c)$,gravitational constant $(G)$,and Planck's constant $(h)$ are taken as fundamental units in a system. The dimensions of time in this new system should be

Vedclass Test Series

Exam Paper Generator

Online Exam Module