Test: Dimensions of Physical Quantities - JEE MCQ

The light-year is a unit of length used to express astronomical distances and measures about 9.46 trillion kilometres or 5.88 trillion miles. As defined by the International Astronomical Union, a light-year is the distance that light travels in vacuum in one Julian year. 

The parsec (symbol: pc) is a unit of length used to measure large distances to astronomical objects outside the Solar System. A parsec is defined as the distance at which one astronomical unit subtends an angle of one arcsecond, which corresponds to 648000π astronomical units.

The SI system of units is a modern system and hence involves all the quantities that can't be derived using all the other quantities of the set. While the rest systems are old and local methods and hence are not scientifically accurate and explainable and thus only have three basic quantities while SI have 7.

The incorrect statement is:

d) Unit of surface tension is Newton metre.

The correct unit of surface tension is Newton per metre (N/m), not Newton metre.

Here's a breakdown:

• a) Unit of K.E. is Newton-metre: True, because kinetic energy is measured in joules, and 1 joule equals 1 Newton-metre.
• b) Unit of viscosity is poise: True, poise is indeed a unit of dynamic viscosity.
• c) Work and energy have the same dimensions: True, both work and energy have the same dimensions and are measured in joules.
• d) Unit of surface tension is Newton metre: Incorrect, it should be Newton per metre (N/m).

Unit of angular frequency is rad/sec which can be said as angle/time. As angle is dimensionless and time has dimension T, we get the dimension of angular frequency as T^-1. So it has 0 dimensions in length.

Coefficient of viscosity (η)= Fr/Av   

 F= tangential Force, Area, r= distance between the layers, v= velocity.

Dimensional Formula of Force = M¹L¹T^-2.
Dimensional Formula of Area= M⁰L²T⁰.
Dimensional Formula of distance= M⁰L¹T⁰.
Dimensional Formula of velocity= M⁰L¹T^-1.

Putting these values in above equation we get,

[η]= [M¹L¹T^-2][M⁰L¹T⁰] / [M⁰L²T⁰] [M⁰L¹T^-1] = [M¹L^-1T^-1]

The dimensions of angular momentum are M L²T⁻¹
That of torque is  M L²T⁻²
Also dimension of energy is  M L²T⁻²
Where as same of force is  M LT⁻²
And of power is  M L²T⁻³
Thus we get torque and energy have the same dimensional formulas.

[E] = [F][d]
= [P/T][A]^½ 
[E] = P¹A^½T^-1

Let,


Comparing the coefficients M,L,T, of both sides we get

b−c=1
a + 2b + 3c = 0
−(a + b + 2c) = 0
Solve the Eqs. (ii), (iii) and (iv), we get

So,
M=h^1/2C^1/2G^−1/2

Angular momentum, J = mvr
[J]=[MLT⁻¹L]=[ML²T⁻¹]
This is same as that of Planck's constant.

hence, μ is dimensionless.
Thus, each term on the R.H.S. of given equation should be dimensionless, i.e., B/λ² is dimensionless, i.e., B should have dimension of λ², i.e., cm², i.e., area.

We know that coefficient of viscosity (η)= Fr/Av where F = tangential force, r = distance between the layers , v = velocity and A is the area of the surface.
 Thus we get [η] = MLT^-2.L / L². (L/T)
= M / LT
Thus its unit is kg / m sec

As 'at' and 3 are added in the equation, we get at and 3 have same dimensions i.e.1
Thus a has dimensions same as 1/t.

We know that 10⁵ dyne = 1N
And 10⁴cm²  = 1 m²
Thus we get 10 dyne / cm² = N / m²
Hence 10⁶ dyne / cm² =10⁵ N / m²

The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

The final result should, therefore, be rounded off to one decimal place, i.e. 663.8

We have  1 unit = x meters
So  1 unit² = x² meter²
Hence, we get  1 meter² = 1/x² unit²

G = 6.67 x 10-11 Nm2/kg2

so, when expressed in VGS units we shall convert N to dynes, m to cm and kg to g, thus

G = 6.67 x 10-11 x [(105dynes x (102)2cm2 ) / (103)2 g2]

=  6.67 x 10-11 x [ (105 x 104 ) / 106 ] 

Thus,

in CGS system

G = 6.67 x 10-8 dyne.cm2/g2