physics : class11 Deriving Relations by Dimensions : Dimensions of Physical Quantities : Physical World and Measurement MCQ Questions and Answers
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Deriving Relations by Dimensions
MCQ questions and answers.
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MCQ Questions
1. The time period T of a simple pendulum is assumed to depend only on its length l and the acceleration due to gravity g, such that T = C l^a g^b, where C is dimensionless. What are the values of a and b obtained by dimensional analysis?
A.
a = 1, b = -1
B.
a = 1/2, b = -1/2
C.
a = -1/2, b = 1/2
D.
a = 1/2, b = 1/2
Correct Answer:
a = 1/2, b = -1/2
Explanation:
Since [T] = [l]^a[g]^b = L^a(LT^-2)^b = L^(a+b)T^(-2b), comparison with T gives a + b = 0 and -2b = 1. Hence b = -1/2 and a = 1/2.
2. A characteristic speed u in a fluid is assumed to depend on pressure p, density rho, and a characteristic length l as u = C p^a rho^b l^c, where C is dimensionless. Which set of exponents is dimensionally possible?
A.
a = 1/2, b = -1/2, c = 0
B.
a = 1, b = -1, c = 1
C.
a = -1/2, b = 1/2, c = 0
D.
a = 1/2, b = 1/2, c = -1
Correct Answer:
a = 1/2, b = -1/2, c = 0
Explanation:
Using [p] = ML^-1T^-2 and [rho] = ML^-3, the dimensions become M^(a+b)L^(-a-3b+c)T^(-2a). Equating these to LT^-1 gives a = 1/2, b = -1/2, and c = 0.
3. The resistive force F acting on a small object moving through a fluid is assumed to depend on fluid density rho, speed v, and cross-sectional area A according to F = C rho^a v^b A^c. What exponents follow from dimensional analysis?
A.
a = 1, b = 1, c = 1
B.
a = 1/2, b = 2, c = 1/2
C.
a = 1, b = 2, c = 1
D.
a = 2, b = 1, c = 1/2
Correct Answer:
a = 1, b = 2, c = 1
Explanation:
The dimensions of the right side are (ML^-3)^a(LT^-1)^b(L^2)^c = M^aL^(-3a+b+2c)T^-b. Equating with [F] = MLT^-2 gives a = 1, b = 2, and c = 1.
4. The total energy E of an oscillator is assumed to depend on its mass m, amplitude A, and angular frequency omega in the form E = C m^a A^b omega^c. Determine the exponents using dimensional analysis.
A.
a = 1, b = 1, c = 2
B.
a = 1/2, b = 2, c = 1
C.
a = 1, b = 2, c = 2
D.
a = 2, b = 1, c = 1
Correct Answer:
a = 1, b = 2, c = 2
Explanation:
Since [E] = ML^2T^-2, while [m^a A^b omega^c] = M^aL^(b)T^-c, comparison gives a = 1, b = 2, and c = 2.
5. The escape speed v from the surface of a spherical body is assumed to depend only on its mass M, radius R, and the gravitational constant G: v = C G^a M^b R^c. What are the required exponents?
A.
a = 1, b = 1, c = -1
B.
a = 1/2, b = 1/2, c = -1/2
C.
a = -1/2, b = 1/2, c = 1/2
D.
a = 1/2, b = -1/2, c = 1/2
Correct Answer:
a = 1/2, b = 1/2, c = -1/2
Explanation:
With [G] = M^-1L^3T^-2, the dimensions of the expression are M^(-a+b)L^(3a+c)T^-2a. Equating to [v] = LT^-1 gives a = 1/2, b = 1/2, and c = -1/2.
6. The capillary rise h in a tube is assumed to depend on surface tension S, liquid density rho, gravitational acceleration g, and tube radius r as h = C S^a rho^b g^c r^d. Which exponent set is dimensionally consistent?
A.
a = 1, b = -1, c = -1, d = -1
B.
a = 1, b = 1, c = -1, d = -1
C.
a = -1, b = 1, c = -1, d = 1
D.
a = 1/2, b = -1/2, c = -1, d = 0
Correct Answer:
a = 1, b = -1, c = -1, d = -1
Explanation:
Surface tension has dimensions ML^0T^-2. Thus the right side has dimensions M^(a+b)L^(-3b+c+d)T^(-2a-2c). Equating with [h] = L gives a = 1, b = -1, c = -1, and d = -1.
7. The viscous force F between two parallel fluid layers is assumed to depend on coefficient of viscosity eta, area A, relative speed v, and separation d as F = C eta^a A^b v^c d^e. Which exponents are obtained dimensionally?
A.
a = 1, b = 1, c = 1, e = -1
B.
a = 1, b = -1, c = 1, e = 1
C.
a = -1, b = 1, c = 1, e = -1
D.
a = 1, b = 1, c = -1, e = 1
Correct Answer:
a = 1, b = 1, c = 1, e = -1
Explanation:
The coefficient of viscosity has dimensions ML^-1T^-1. Therefore the product has dimensions M^aL^(-a+2b+c+e)T^(-a-c). Equating with force dimensions MLT^-2 gives a = 1, b = 1, c = 1, and e = -1.
8. The speed v of a transverse wave on a stretched string is assumed to depend on the tension F and the mass per unit length mu as v = C F^a mu^b. What exponents are required by dimensional consistency?
A.
a = 1, b = -1
B.
a = -1/2, b = 1/2
C.
a = 1/2, b = -1/2
D.
a = 1/2, b = 1/2
Correct Answer:
a = 1/2, b = -1/2
Explanation:
Since [F] = MLT^-2 and [mu] = ML^-1, the expression has dimensions M^(a+b)L^(a-b)T^-2a. Equating to LT^-1 gives a = 1/2 and b = -1/2.
9. A characteristic time t for a system is assumed to depend on mass m, spring constant k, and length l as t = C m^a k^b l^c, where [k] = MT^-2. Which exponent set makes the relation dimensionally valid?
A.
a = 1/2, b = -1/2, c = 0
B.
a = 1, b = -1, c = 1
C.
a = -1/2, b = 1/2, c = 0
D.
a = 1/2, b = 1/2, c = -1
Correct Answer:
a = 1/2, b = -1/2, c = 0
Explanation:
The dimensions are [m^a k^b l^c] = M^(a+b)L^cT^-2b. Comparing with time T gives c = 0, a + b = 0, and -2b = 1. Hence a = 1/2 and b = -1/2.
10. The power P associated with a rotating body is assumed to depend on torque tau and angular speed omega as P = C tau^a omega^b. Which exponents are fixed by dimensional analysis?
A.
a = 1, b = 1
B.
a = 1/2, b = 2
C.
a = 2, b = 1/2
D.
a = -1, b = -1
Correct Answer:
a = 1, b = 1
Explanation:
Torque has dimensions ML^2T^-2 and angular speed has dimensions T^-1. Thus the right side has dimensions M^aL^(2a)T^(-2a-b). Equating with power dimensions ML^2T^-3 gives a = 1 and b = 1.
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