Q#15
Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powers of mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.Answer:
(a) E ∝ macb
E = k macb
Where taking k as a dimensionless constant. Writing the dimensions on both sides,
[ML²T⁻²] = [M]a[LT⁻¹]b = [MaLbT⁻b]
Equating the dimensions on both sides we have,
a = 1 and b = 2
Hence the relation among the quantities may be,
E = kmc²
Q#16
Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm's law, the product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for R and V are ML²I⁻²T⁻³ and ML²T⁻³I⁻¹ respectively.
Answer:
(a) The dimensions of the current, I = [I]
Since the product of two of these given quantities equals the third, we need to know whether the product of V and R gives [I] or their ratio.
It is clear that the product will not give [I]. If we divide V by R we get,
[ML²T⁻³I⁻¹]/[ML²I⁻²T⁻³] = [I]
So we get the desired relation,
V/R = I
V = IR
Hence it is the Ohm's law.
Q#17
The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and it's mass per unit length m. Guess the expression for its frequency from dimensional analysis.
Answer:
(a) From the given condition,
frequency, 𝜈 ∝ LaFbmc
𝜈 = kLa Fbmc
{Where k is a dimensionless constant}
Putting the dimensions on both sides,
[T⁻¹] = [L]a[MLT⁻²]b[ML⁻¹]c
[T⁻¹] = [Mb+c La+b-c T⁻²b ]
Equating the powers of M, L and T on bothe sides,
b + c = 0, a + b – c = 0 and -2b = -1
It gives, b = 1/2, c = – 1/2 and a + b – c = 0
a + ½ – (–1/2) =0
a = – 1
Now the expression for the frequency becomes,
𝜈 = k L⁻¹ √F(1/√m)
𝜈 = (k/L)√(F/m)
Q#18
Test if the following equations are dimensionally correct:
(a) h = 2S Cosθ/ρrg
(b) v = √(P/ρ)
(c) V = πPr⁴t/8ηl
(d) 𝞶 = (1/2π)√(mgl/I)
where h = height, S = surface tension, ρ = density, P = pressure, V = volume, η = coefficient of viscosity, 𝞶 = frequency and I = moment of inertia.
Answer:
(a) Left side:
The dimension of h = [L]
Right side:
Dimensions of S = Force/Length = [MLT⁻²]/[L] = [MT⁻²]
Cosθ is a ratio hence dimensionless.
The dimensions of ρ = [ML⁻³]
The dimensions of r = [L]
The dimensions of g = [LT⁻²]
Hence the dimensions of 2S cosθ/ρrg
= [MT⁻²]/[ML⁻³][L][LT⁻²]
= [MT⁻²]/[ML⁻¹T⁻²]
= [L]
Hence the dimensions of both sides are the same. So, it is dimensionally correct.
(b) v = √(P/ρ)
Left side:
Dimensions of velocity, v = [LT⁻¹]
Right side:
Dimensions of pressure P = Force/area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²]
Dimensions of ρ = [ML⁻³]
Hence the dimensions of √(P/ρ)
= √[ML⁻¹T⁻²]/[ML⁻³]
= √[L²T⁻²]
= [LT⁻¹]
Same dimensions on both sides. Hence it is dimensionally correct.
(c) V = πPr⁴t/8ηl
Left side:
Dimensions of volume, V = [L³]
Right side:
π is dimensionless. Dimensions of P = [ML⁻¹T⁻²]
Dimensions of r = [L]
Dimensions of t = [T]
Dimensions of l = [L]
Dimensions of η = Force/area = [ML⁻¹T⁻¹]
Hence the dimensions of right side
= [ML⁻¹T⁻²][L]⁴[T]/[ML⁻¹T⁻¹][L]
= [ML³T⁻¹]/[MT⁻¹]
= [L³]
Dimensions of both sides are same. Hence it is dimensionally correct.
(d) 𝞶 = (1/2π)√(mgl/I)
Left side:
Dimensions of frequency, 𝜈 = [T⁻¹]
Right side:
π is dimensionless.
Dimensions of m = [M]
Dimensions of g = [LT⁻²]
Dimensions of l = [L]
Dimensions of I = [ML²]
Dimensions of the right side,
= √{[M][LT⁻²][L]}/√{[ML²]}
= √[T⁻²]
= [T⁻¹]
Dimensions of both sides are the same, hence it is dimensionally correct.
So, all are dimensionally correct.
Q#19
Let x and a stand for distance. Is ∫dx/√(a² – x²) = (1/a)sin⁻¹(a/x) dimensionally correct?
Answer:
(a) Since both x and a are distances their dimensions are = [L]. dx is infinitesimally small distance hence its dimension is also = [L]. Inverse sine is an angle hence it is also dimensionless. So, the dimensions of the left side,
= [L]/√[L²] = [L]/[L] = L⁰ = Dimensionless
The dimensions of the right side = 1/[L] = [L⁻¹]
We see that the dimensions of both sides are not the same. Hence the equation is not dimensionally correct.
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