Advertisements
Advertisements
प्रश्न
The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as `v = π/8 (pr^4)/(ηl)` where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula ML–1 T–1. Check whether the equation is dimensionally correct.
उत्तर
If the dimensions of LHS of an equation are equal to the dimensions of RHS, then the equation is said to be dimensionally correct.
According to the problem, the volume of a liquid flowing out per second of a pipe is given by V = `pi/8 (pr^4)/(ηl)`
(where V = rate of volume of liquid per unit time)
Dimension of given physical quantities,
`[V] = "Dimension of Volume"/"Dimension of time" = ([L^3])/([T]) = [L^3T^-1], [p] = [ML^-1T^-2]`
`[η] = [ML^-1T^-1], [l] = [L], [r] = [L]`
LHS = `[V] = ([L^3])/([T]) = [L^3T^-1]`
RHS = `([ML^-1T^-2] xx [L^4])/([ML^-1 T^-1] xx [L]) = [L^3T^-1]`
Dimensionally, L.H.S = R.H.S.
Therefore, the equation is correct dimensionally.
APPEARS IN
संबंधित प्रश्न
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
(a) y = a sin `(2pit)/T`
(b) y = a sin vt
(c) y = `(a/T) sin t/a`
d) y = `(a/sqrt2) (sin 2πt / T + cos 2πt / T )`
(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by `Å: 1Å = 10^(-10)m`. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?
If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
- (P – Q)/R
- PQ – R
- PQ/R
- (PR – Q2)/R
- (R + Q)/P
A function f(θ) is defined as: `f(θ) = 1 - θ + θ^2/(2!) - θ^3/(3!) + θ^4/(4!)` Why is it necessary for q to be a dimensionless quantity?
Why length, mass and time are chosen as base quantities in mechanics?
Give an example of a physical quantity which has a unit but no dimensions.
If velocity of light c, Planck’s constant h and gravitational contant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.
An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that `T = k/R sqrt(r^3/g)`. where k is a dimensionless constant and g is acceleration due to gravity.
Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m ) to energy (E ) as E = mc2, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV= 1.6 × 10–13 J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 × 10–27 kg.
- Show that the energy equivalent of 1 u is 931.5 MeV.
- A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
The entropy of any system is given by `S = alpha^2betaIn[(mukR)/(Jbeta^2) + 3]` Where α and β are the constants µ J, k, and R are no. of moles, the mechanical equivalent of heat, Boltzmann constant, and gas constant respectively. `["take S" = (dQ)/T]`
Choose the incorrect option from the following.
