physics : class11 Unknown Exponents Method : Dimensions of Physical Quantities : Physical World and Measurement MCQ Questions and Answers
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Unknown Exponents Method
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MCQ Questions
1. The coefficient of viscosity η of a fluid is assumed to depend on its density ρ, characteristic speed v, and characteristic length l as η = Cρ^a v^b l^c, where C is dimensionless. Using the unknown exponents method, determine (a, b, c).
A.
(1, 1, 1)
B.
(1, -1, 1)
C.
(1, 1, -1)
D.
(-1, 1, 1)
Correct Answer:
(1, 1, 1)
Explanation:
Since [η] = ML⁻¹T⁻¹, [ρ^a v^b l^c] = M^aL⁻³ᵃ⁺ᵇ⁺ᶜT⁻ᵇ. Equating powers gives a = 1, b = 1, and c = 1.
2. Surface tension S is represented as S = Cρ^a g^b l^c, where ρ is density, g is acceleration due to gravity, l is length, and C is dimensionless. What are the values of a, b, and c?
A.
(1, 1, 3)
B.
(1, -1, 3)
C.
(-1, 1, 3)
D.
(1, 1, -3)
Correct Answer:
(1, 1, 3)
Explanation:
The dimensions of surface tension are [S] = MT⁻². The assumed expression has dimensions M^aL⁻³ᵃ⁺²ᵇ⁺ᶜT⁻²ᵇ. Comparing powers gives a = 1, b = 1, and c = 3.
3. Suppose the gravitational constant is expressed as G = CF^aρ^b l^c, where F is force, ρ is density, l is length, and C is dimensionless. Determine the exponents a, b, and c.
A.
(1, -2, -4)
B.
(-1, 2, 4)
C.
(1, 2, -4)
D.
(1, -2, 4)
Correct Answer:
(1, -2, -4)
Explanation:
Using [G] = M⁻¹L³T⁻², [F] = MLT⁻², and [ρ] = ML⁻³, the assumed dimensions are Mᵃ⁺ᵇLᵃ⁻³ᵇ⁺ᶜT⁻²ᵃ. Equating powers gives a = 1, b = −2, and c = −4.
4. Planck's constant h is assumed to have the form h = Cρ^a v^b l^c, where ρ is mass density, v is speed, l is length, and C is dimensionless. Find the values of the unknown exponents.
A.
(1, 1, 4)
B.
(1, -1, 4)
C.
(-1, 1, 4)
D.
(1, 1, -4)
Correct Answer:
(1, 1, 4)
Explanation:
Since [h] = ML²T⁻¹, while [ρ^a v^b l^c] = M^aL⁻³ᵃ⁺ᵇ⁺ᶜT⁻ᵇ, comparison gives a = 1, b = 1, and c = 4.
5. A length scale r is related to the gravitational constant G, mass m, and speed v by r = CG^a m^b v^c. If C is dimensionless, which set of exponents is obtained by the unknown exponents method?
A.
(1, 1, -2)
B.
(1, -1, 2)
C.
(1, 1, 2)
D.
(-1, 1, -2)
Correct Answer:
(1, 1, -2)
Explanation:
The dimensions are [r] = L, [G] = M⁻¹L³T⁻², [m] = M, and [v] = LT⁻¹. Equating powers in G^a m^b v^c gives −a + b = 0, 3a + c = 1, and −2a − c = 0. Hence a = 1, b = 1, c = −2.
6. The time period T of a phenomenon is assumed to depend only on G and density ρ according to T = CG^aρ^b. Determine the exponents a and b using dimensional analysis.
A.
(1/2, 1/2)
B.
(-1/2, 1/2)
C.
(1/2, -1/2)
D.
(-1/2, -1/2)
Correct Answer:
(1/2, -1/2)
Explanation:
Because [T] = T, [G] = M⁻¹L³T⁻², and [ρ] = ML⁻³, the mass and length equations give −a + b = 0. The time equation gives −2a = 1, so a = −1/2 and b = −1/2. Therefore the expression is proportional to (Gρ)⁻¹ᐟ².
7. Energy density u is represented by u = CG^aρ^b v^c, where G is the gravitational constant, ρ is density, v is speed, and C is dimensionless. Find (a, b, c).
A.
(0, 1, 2)
B.
(1, 1, 2)
C.
(0, -1, 2)
D.
(0, 1, -2)
Correct Answer:
(0, 1, 2)
Explanation:
Energy density has dimensions ML⁻¹T⁻². Equating dimensions in G^aρ^b v^c gives −a + b = 1, 3a − 3b + c = −1, and −2a − c = −2. Solving yields a = 0, b = 1, c = 2.
8. The angular momentum L of a particle is assumed to be L = Cm^a v^b r^c, where m is mass, v is speed, r is distance, and C is dimensionless. Determine the exponents a, b, and c.
A.
(1, 1, 1)
B.
(1, -1, 1)
C.
(-1, 1, 1)
D.
(1, 1, -1)
Correct Answer:
(1, 1, 1)
Explanation:
Angular momentum has dimensions ML²T⁻¹. Since [m^a v^b r^c] = M^aLᵇ⁺ᶜT⁻ᵇ, comparison gives a = 1, b = 1, and c = 1.
9. The gravitational potential energy U of two masses is assumed to be U = CG^a m^b r^c, where r is their separation. Determine the exponents a, b, and c.
A.
(1, 2, -1)
B.
(-1, 2, 1)
C.
(1, -2, -1)
D.
(2, 1, -1)
Correct Answer:
(1, 2, -1)
Explanation:
Potential energy has dimensions ML²T⁻². The dimensions of G^a m^b r^c are M⁻ᵃ⁺ᵇL³ᵃ⁺ᶜT⁻²ᵃ. Equating powers gives a = 1, b = 2, and c = −1.
10. A characteristic speed v is assumed to depend on the gravitational constant G, a mass M, and a distance R as v = CG^aM^bR^c. Using the unknown exponents method, determine (a, b, c).
A.
(1/2, 1/2, -1/2)
B.
(1/2, -1/2, 1/2)
C.
(-1/2, 1/2, -1/2)
D.
(1, 1, -1)
Correct Answer:
(1/2, 1/2, -1/2)
Explanation:
Since [v] = LT⁻¹ and [G^aM^bR^c] = M⁻ᵃ⁺ᵇL³ᵃ⁺ᶜT⁻²ᵃ, equating powers gives −a + b = 0, 3a + c = 1, and −2a = −1. Thus a = 1/2, b = 1/2, and c = −1/2.
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