physics : class11 Dimensional Analysis of G : Gravitational Force : Gravitation MCQ Questions and Answers
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Dimensional Analysis of G
MCQ questions and answers.
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MCQ Questions
1. If the gravitational constant is represented dimensionally by G = F r^x m^y, where F is force, r is length, and m is mass, what are the values of x and y?
A.
x = 2, y = -2
B.
x = -2, y = 2
C.
x = 1, y = -1
D.
x = -1, y = 1
Correct Answer:
x = 2, y = -2
Explanation:
Since [G] = M^-1 L^3 T^-2, [F] = M L T^-2, and [r] = L, the relation [G] = [F][r]^x[m]^y gives M^-1 L^3 T^-2 = M^(1+y)L^(1+x)T^-2. Therefore, 1 + y = -1 and 1 + x = 3, so x = 2 and y = -2.
2. The value of G is 6.67 × 10^-11 m^3 kg^-1 s^-2 in SI units. What is its approximate value in cm^3 g^-1 s^-2?
A.
6.67 × 10^-14
B.
6.67 × 10^-11
C.
6.67 × 10^-8
D.
6.67 × 10^-5
Correct Answer:
6.67 × 10^-8
Explanation:
Using 1 m^3 = 10^6 cm^3 and 1 kg^-1 = 10^-3 g^-1, the conversion factor is 10^6 × 10^-3 = 10^3. Hence, G = 6.67 × 10^-11 × 10^3 = 6.67 × 10^-8 cm^3 g^-1 s^-2.
3. Which of the following is dimensionally equivalent to the gravitational constant G, where ρ denotes mass density and t denotes time?
A.
ρt^2
B.
1/(ρt^2)
C.
ρ/t^2
D.
t^2/ρ^2
Correct Answer:
1/(ρt^2)
Explanation:
The dimensions of density are [ρ] = M L^-3. Therefore, [1/(ρt^2)] = 1/(M L^-3 T^2) = M^-1 L^3 T^-2, which is exactly the dimensional formula of G.
4. Planck's constant h has dimensions M L^2 T^-1. What are the dimensions of the product Gh?
A.
L^5 T^-3
B.
M L^5 T^-3
C.
L T^-1
D.
M^-1 L^5 T^-1
Correct Answer:
L^5 T^-3
Explanation:
The dimensions are [G] = M^-1 L^3 T^-2 and [h] = M L^2 T^-1. Thus, [Gh] = (M^-1 L^3 T^-2)(M L^2 T^-1) = L^5 T^-3.
5. Which combination involving G, mass density ρ, and time t is dimensionless?
A.
Gρt
B.
Gρ/t^2
C.
Gρt^2
D.
G^2ρt^2
Correct Answer:
Gρt^2
Explanation:
Since [G] = M^-1 L^3 T^-2 and [ρ] = M L^-3, their product has dimensions [Gρ] = T^-2. Multiplication by t^2 gives [Gρt^2] = T^-2 T^2 = 1, so the combination is dimensionless.
6. Suppose the gravitational force between two masses is expressed as F = Gm1m2/r^n. Dimensional consistency requires the value of n to be
A.
1
B.
2
C.
3
D.
4
Correct Answer:
2
Explanation:
The dimensions of Gm1m2 are (M^-1 L^3 T^-2)(M)(M) = M L^3 T^-2. Dividing by r^n gives M L^(3-n) T^-2. For this to equal force, M L T^-2, 3 - n = 1, so n = 2.
7. The gravitational constant is 6.67 × 10^-11 m^3 kg^-1 s^-2. What is its numerical value when expressed in km^3 tonne^-1 hour^-2?
A.
8.64 × 10^-12
B.
8.64 × 10^-10
C.
8.64 × 10^-8
D.
8.64 × 10^-6
Correct Answer:
8.64 × 10^-10
Explanation:
Use 1 m^3 = 10^-9 km^3, 1 kg^-1 = 10^3 tonne^-1, and 1 s^-2 = 3600^2 hour^-2. Thus the conversion factor is 10^-9 × 10^3 × 3600^2 = 12.96. Therefore, G = 6.67 × 10^-11 × 12.96 ≈ 8.64 × 10^-10 km^3 tonne^-1 hour^-2.
8. Which expression has the same dimensions as G if a represents acceleration, l represents length, and ρ represents mass density?
A.
aρl
B.
a/(ρl)
C.
aρ/l
D.
a/(ρl^2)
Correct Answer:
a/(ρl)
Explanation:
The dimensions of a/(ρl) are (L T^-2)/[(M L^-3)(L)] = (L T^-2)/(M L^-2) = M^-1 L^3 T^-2, which equals the dimensional formula of G.
9. A new system of units is defined such that the unit of length is 2 m, the unit of mass is 3 kg, and the unit of time is 0.5 s. If the numerical value of G in SI units is G0, what is its numerical value in the new system?
A.
3G0/32
B.
32G0/3
C.
3G0/8
D.
8G0/3
Correct Answer:
3G0/32
Explanation:
Because [G] = L^3 M^-1 T^-2, one new unit of G equals (2)^3(3)^-1(0.5)^-2 = 32/3 SI units. Therefore, the numerical value in the new system is G0 divided by 32/3, giving 3G0/32.
10. If the gravitational force between two spherical bodies is dimensionally represented as F ∝ Gρ^2r^n, where ρ is density and r is length, what must be the value of n?
A.
2
B.
3
C.
4
D.
5
Correct Answer:
4
Explanation:
The dimensions of Gρ^2r^n are (M^-1 L^3 T^-2)(M^2 L^-6)(L^n) = M L^(n-3) T^-2. Equating this to force, M L T^-2, gives n - 3 = 1, so n = 4.
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