If Force \((F)\), Velocity \((V)\), and Time \((T)\), are taken as fundamental units, then the dimensions of mass are:
[2014]
1. \(FVT^{-1}\)
2. \(FVT^{-2}\)
3. \(FV^{-1}T^{-1}\)
4. \(FV^{-1}T\)
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Given fundamental units: Force \(F = [MLT^{-2}]\), Velocity \(V = [LT^{-1}]\), Time \(T = [T]\). We want to find dimensions of Mass \(M = F^x V^y T^z\). Equating dimensions: \([M] = [MLT^{-2}]^x [LT^{-1}]^y [T]^z = [M^x L^{x+y} T^{-2x-y+z}]\). Comparing powers: \(x=1\), \(x+y=0 ⇒ 1+y=0 ⇒ y=-1\), \(-2x-y+z=0 ⇒ -2(1)-(-1)+z=0⇒ -2+1+z=0 ⇒ z=1\). Thus, mass dimensions are \(FV^{-1}T\).
Dimensions of resistance in an electrical circuit, in terms of dimension of mass (M), of length (L), of time (T) and of current (I), would be
[2007]
1. \(ML^2T^{-2}I^{-2}\)
2. \(ML^2T^{-1}I^{-1}\)
3. \(ML^2T^{-3}I^{-2}\)
4. \(ML^2T^{-3}I^{-1}\)
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Resistance \(R = V/I = W/(QI)\). (W) is work done, (Q) is charge. \(W = [ML^2T^{-2}]\), \(Q = [IT]\). So, \(R = [ML^2T^{-2}] / ([IT][I]) = [ML^2T^{-3}I^{-2}]\).
The velocity (v) of a particle at time (t) is given by \[v = at + \frac{b}{t+c}\] where (a), (b) and (c) are constants. The dimensions of (a), (b) and (c) are respectively:
[2006]
1. \((LT^{-2}), (L) and (T)\)
2. \((L), (T) and (LT^2)\)
3. \((L^2T^{-2}), (LT) and (L)\)
4. \((L), (LT) and (T^2)\)
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From dimensional homogeneity: ([c] = [t] = [T]). ([at] = [v]) so ([a] = [v]/[t] = [LT^{-1}]/[T] = [LT^{-2}]). ([b/(t+c)] = [v]) so ([b] = [v][t] = [LT^{-1}][T] = [L]).
An equation is given here \[\left(P + \frac{a}{V^2}\right) = b\frac{\theta}{V}\] where P = Pressure, V = Volume and \(\theta =\) Absolute temperature. If (a) and (b) are constants, the dimensions of (a) will be:
1. \(ML^{-5}T^{-1}\)
2. \(ML^5T^{-1}\)
3. \(ML^5T^{-2}\)
4. \(M^{-1}L^5T^2\)
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From dimensional homogeneity, \([a/V^2] = [P]\). \([a] = [P][V^2]\). Pressure \(P = [ML^{-1}T^{-2}]\), Volume \(V = [L^3]\). So, \([a] = [ML^{-1}T^{-2}][L^3]^2 = [ML^{-1}T^{-2}L^6] = [ML^5T^{-2}]\).
A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible, error in the measurement of mass and volume are 0.01 g and 0.1 cc. Then maximum error in the density will be:
[1991]
1. 22%
2. 2%
3. 0.2%
4. 0.02%
View Answer
Density \(\rho = M/V\). Fractional error \(\Delta\rho/rho = (\Delta M/M) + (\Delta V/V)\). Given \(M=22.42\text{ g}\), \(\Delta M=0.01\text{ g}\). \(V=4.7\text{ cc}\), \(\Delta V=0.1\text{ cc}\). So, \(Delta M/M = 0.01/22.42 \approx 0.000446\). \(\Delta V/V = 0.1/4.7 \approx 0.02127\). Total fractional error \(\approx 0.021716\). Percentage error \(\approx 2.17%\), which is closest to 2%.
The time dependence of a physical quantity (p) is given by \(p = p_0 \text{exp } (-\alpha t^2)\), where (alpha) is constant and (t) is the time. The constant \(\alpha\):
[1993]
1. Is dimensionless
2. Has dimensions \(T^{-2}\)
3. Has dimensions \(T^2\)
4. Has dimensions of \(p\)
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For \(\text{exp }(-\alpha t^2)\) to be dimensionless, \(\alpha t^2\) must be dimensionless. \([\alpha][t^2] = [M^0L^0T^0]\). Since \([t] = [T]\), \([\alpha][T^2] = [1]\). Thus, \([\alpha] = [T^{-2}]\).
(P) represents radiation pressure, (c) represents speed of light and (S) represents radiation energy striking per unit area per sec. The non-zero integers (x, y, z) such that \(P^x S^y c^z\) is dimensionless are:
[1992]
1. (x = 1, y = 1, z = 1)
2. (x = -1, y = 1, z = 1)
3. (x = 1, y = -1, z = 1)
4. (x = 1, y = -1, z = -1)
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Dimensions: \(P = [ML^{-1}T^{-2}]\), \(c = [LT^{-1}]\), \(S = [MT^{-3}]\). For \(P^x S^y c^z\) to be dimensionless, powers of M, L, T must be zero. \(M: x+y=0\). \(L: -x+z=0\). \(T: -2x-3y-z=0\). Solving gives \(y=-x\) and \(z=x\). Taking \(x=1\) yields \(y=-1\), \(z=1\).
The frequency of vibration (f) of a mass (m) suspended from a spring of spring constant (k) is given by a relation \(f = a.m^x k^y\), where (a) is a dimensionless constant. The values of (x) and (y) are:
[1990]
1. \(x = \frac{1}{2}, y = \frac{1}{2}\)
2. \(x = -\frac{1}{2}, y = \frac{1}{2}\)
3. \(x = \frac{1}{2}, y = -\frac{1}{2}\)
4. \(x = -\frac{1}{2}, y = -\frac{1}{2}\)
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Frequency \(f = [T^{-1}]\). Mass (m = [M]). Spring constant \(k = [MT^{-2}]\). Comparing dimensions of \(f = m^x k^y\): \([T^{-1}] = [M]^x [MT^{-2}]^y = [M^{x+y} T^{-2y}]\). Solving (x+y=0) and (-2y=-1) gives (y = 1/2) and (x = -1/2).