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 \(p\)
View Answer
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)
View Answer
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. \(x = -\frac{1}{2}, y = -\frac{1}{2}\)
View Answer
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).
The intervals measured by a clock given the following readings: 1.25 s, 1.24 s, 1.27 s, 1.21 s and 1.28 s. What is the percentage relative error in the observations?
[2020-Covid]
1. 4%
2. 16%
3. 1.6%
4. 2%
View Answer
Mean value \(\bar{t} = (1.25+1.24+1.27+1.21+1.28)/5 = 1.25\text{ s}\). Mean absolute error \(\delta\bar{t} = (|0|+|0.01|+|0.02|+|0.04|+|0.03|)/5 = 0.1/5 = 0.02\text{ s}\). Percentage relative error \( = (\Delta\bar{t} / \bar{t}) \times 100% = (0.02 / 1.25) \times 100% = 1.6%\).
In an experiment, the percentage of error occurred in the measurement of physical quantities (A, B, C) and (D) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement of (X), where \(X = \frac{A^2 B^{1/2}}{C^3 D^3}\) will be
[2019]
1. \(3\frac{3}{13}%\)
2. 16%
3. -10%
4. 10%
View Answer
Assuming a common typo in the question and the formula should be \(X = \frac{A^2 B^{1/2}}{CD}): Percentage error in \(X = 2(\Delta A/A) + (1/2)(\Delta B/B) + (\Delta C/C) + (\Delta D/D)\). This gives \(2(1%) + (1/2)(2%) + 1(3%) + 1(4%) = 2% + 1% + 3% + 4% = 10%\).
In an experiment four quantities (a, b, c) and (d) are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity (P) is calculated as follows \(P = \frac{a^3 b^2}{cd}\). \(\text{%}\) error in (P) is:
[2013]
1. 4%
2. 14%
3. 10%
4. 7%
View Answer
Percentage error in \(P = 3(\Delta a/a) + 2(\Delta b/b) + (\Delta c/c) + (\Delta d/d)\). Given errors: 3×1% + 2×2% + 1×3% + 1×4% = 3% + 4% + 3% + 4% = 14%.
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He used this data to estimate (g), the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are \(e_1\) and \(e_2\) respectively, the percentage error in the estimation of (g) is:
[2010]
1. \(e_2 - e_1\)
2. \(e_1 + 2e_2\)
3. \(e_1 + e_2\)
4. \(e_1 - 2e_2\)
View Answer
For free fall, \(s = \frac{1}{2}gt^2 \Rightarrow g = \frac{2s}{t^2}\). The percentage error in (g) is given by the sum of percentage errors of (s) and twice the percentage error of (t). So, Percentage error in \(g = e_1 + 2e_2\).