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The pairs of physical quantities that have the same dimensions are : [JEE - 1995'2/100]
In the formula X = 3YZ2, X and Z have dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y in MKSQ system ? [JEE-1995,2/100]
Which of the following pairs have same dimensions : [JEE-1996' 2/100]
The SI unit of inductance, the henry can be written as : [JEE-1998' 2/200]
Let [ε0] denote the dimensional formula of the permittivity of the vaccum, and [μ0] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current : [JEE-1998' 2/200]
The dimensions of 1/2 ε0E2 (ε0 : permittivity of free space, E : electric field ) is : [JEE Sc 2000' 2/200]
A quantity X is given by . where ε0 is the permittivity of free space, L is length, ΔV is potential difference and Δt is time interval. The dimensional formula for X is the same as that of [JEE Sc.2000'3/105]
Pressure depends on distance as, P = α/β exp , where α,β are constants, z is distance, k is Boltzmann's constant and θ is temperature. The dimension of β are [JEE-2004s '3/84]
Which of the following set have different dimensions ?
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency 'ωp' which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.Taking the electronic charge as 'e' and the permittivity as 'ε0 ', use dimensional analysis to determine the correct expression for ωp.