The potential energy of a point particle is g | vedclass

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

Question

(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.For the equation to be dimensionally consistent,the d

Vedclass · Accepted Answer

$\frac{\alpha \gamma}{\beta}$

Vedclass · Answer

(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.For the equation to be dimensionally consistent,the dimensions of each term must be the same as the dimensions of potential energy $[V] = [ML^2T^{-2}]$.$1$. For the term $\alpha x$: $[\alpha][x] = [V] \implies [\alpha][L] = [ML^2T^{-2}] \implies [\alpha] = [MLT^{-2}]$.$2$. For the term $\beta \sin(x / \gamma)$: The argument of the sine function must be dimensionless,so $[x / \gamma] = [M^0L^0T^0] \implies [L] / [\gamma] = 1 \implies [\gamma] = [L]$.$3$. Also,$[\beta] = [V] = [ML^2T^{-2}]$.Now,check the dimensions of the given options:For option $(d)$: $\frac{[\alpha][\gamma]}{[\beta]} = \frac{[MLT^{-2}][L]}{[ML^2T^{-2}]} = \frac{[ML^2T^{-2}]}{[ML^2T^{-2}]} = [M^0L^0T^0]$.Since this combination is dimensionless,the correct option is $(d)$.

The potential energy of a point particle is given by the expression $V(x) = -\alpha x + \beta \sin(x / \gamma)$. $A$ dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is

Similar Questions

If the unit of force is $1 \,kN$,the unit of length is $1 \,km$,and the unit of time is $100 \,s$,what will be the unit of mass in $kg$?

If the speed of light $(c)$,acceleration due to gravity $(g)$,and pressure $(p)$ are taken as the fundamental quantities,then the dimension of the gravitational constant $(G)$ is:

In the expression $A=B+\frac{C}{D+E}$,the dimensions of physical quantities $B$ and $C$ are $[L^{1} M^{0} T^{-1}]$ and $[L^{1} M^{0} T^{0}]$ respectively. The dimensions of quantities $A, D$ and $E$ are

What is the dimensional formula of $ab^{-1}$ in the equation $(P+\frac{a}{V^2})(V-b)=RT$,where letters have their usual meaning?

Vedclass Test Series

Exam Paper Generator

Online Exam Module