The force of interaction between two atoms is given by `F= alpha beta exp(-(x^2)/(alphakt))` , where x is the distance ,k is the Boltzmann constant and T is temperature and `alpha " and " beta` are two constans. The dimension of `beta` is :
In the relation: P=(alpha)/(beta)e^(-(alphaZ)/(ktheta)),P is pressure Z is distance k is Boltzmann constant and theta is the temperature. The dimensional formula of beta will be
In the relation, P=alpha/beta e^((alphaZ)/(ktheta))P is pressure, Z is distance, k is Boltzmann constant and theta is the temperature. The dimensional formula of beta will be-
In the relation: P=(alpha)/(beta)e^(-(alphaZ)/(ktheta)),P is pressure Z is distance k is Boltzmann constant and theta ils the temperature. The dimensional formula of beta will bes
F=alphabetae^((-(x)/(alphakt))) k = Boltzmann constant t = temperature x = distance The dimensions of beta is
In the equation ((1)/(pbeta))=(y)/(k_(B)T) , where p is the pressure, y is the distance, k_(B) is Boltzmann constant and T is the tempreture. Dimensions of beta are
The relation between time t and displacement x is t = alpha x^2 + beta x, where alpha and beta are constants. The retardation is
Given : force = (alpha)/("density" + beta^3). What are the dimensions of alpha, beta ?
Two waves travelling in a medium in the x-direction are represented by y_(1) = A sin (alpha t - beta x) and y_(2) = A cos (beta x + alpha t - (pi)/(4)) , where y_(1) and y_(2) are the displacements of the particles of the medium t is time and alpha and beta constants. The two have different :-
Energy due to position of a particle is given by, U=(alpha sqrty)/(y+beta) , where alpha and beta are constants, y is distance. The dimensions of (alpha xx beta) are
Force F and density d are related as F=(alpha)/(beta+sqrtd) , Then find the dimensions of alpha and beta
VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
