1. Differentiate between accuracy and precision of measurement. (2075 ‘A’, 2074 ‘B’, 2067)
Ans: Accurate measurement is the degree of agreement between experimental value and the true value. Good accuracy means the reading or the mean of a set of readings is very close to the true value and it associated with small systematic uncertainties.
The limit or resolution to which a physical quantity is measured by the instrument is known as its precision.
e.g., If one student finds acceleration due to gravity as 7.91 m/s2, 7.92 m/s2, 7.91 m/s2, 7.71 m/s2 in several experiments, then the measurement is precise, since all the values are very close to their mean.
However, if one finds 9.81 m/s2, 9.80 m/s2, 9.79 m/s2, 9.80 m/s2 then measurement is accurate and precise, which are very close to standard value 9.8 m/s2.
2. In one of the printed documents, the unit of universal gravitational constant is given as NmKg-2. Check its correctness from dimensional analysis. (2075 ‘B’)
Ans: Gravitational constant (G) = [latex]\frac{Fd^2}{M_1M_2}[/latex]
Or, G = [latex][\frac{MLT^2L^2}{MM}][/latex]
= [latex][M^{-1}L^3T^{-2}][/latex]
= [latex][Nm^2Kg^{-2}][/latex]
Hence, the given unit is dimensionally incorrect.
3. Is dimensionally correct equation necessarily be a correct physical relation? What about dimensionally wrong equation? (2074 ‘S’, 2063)
Ans: No, dimensionally correct equation may not be a correct physical relation. Consider the following relation,
v = u + 10at ……………… (i)
Dimensional formula of v is LT-1 and that of (u + 10at) is also LT-1. Hence, the relation (i) is dimensionally correct. But, it is established fact that the relation (i) isn’t correct physical relation.
Dimensionally wrong equation can’t be correct physical relation. For example, v = u + at2 is dimensionally wrong equation and it is also incorrect physical relation.
4. A student writes an expression for the momentum (p) of a body of mass (m) with total energy (E) and considering the duration of time (t) as p = [latex]\sqrt{\frac{2mE}{t}}[/latex]. Check its correctness on dimensional analysis. (2073 ‘S’)
Ans: p = [latex]\sqrt{\frac{2mE}{t}}[/latex]
Dimension of p = [MLT-1]
Dimension of m = [M]
Dimension of E = [mL2T-2]
Dimension of t = [T]
Dimension of right hand side = [latex]\sqrt{\frac{MML^2T^{-2}}{T}}[/latex] [∵ 2 is dimensionless]
= [latex]\frac{\sqrt{M^2L^2T^{-2}}}{\sqrt{T}}[/latex]
= [latex]\frac{MLT^{-1}}{T^{1/2}} = MLT^{-1-1/2}[/latex]
= MLT-3/2
Since, dimension of RHS [latex]\neq[/latex] Dimension of LHS, the given expression is dimensionally incorrect.
5. Check dimensionally correctness of the Stoke’s formula, F = [latex]6\pi \eta rv[/latex] where symbols have their usual meanings. (2073 ‘C’)
Ans: F = [latex]6\pi \eta rv[/latex]
Dimension of F = [MLT-2]
Dimension of = [ML-1T-1]
Dimension of r = [L]
Dimension of v = [LT-1]
∴ [MLT-2] = [latex]6\pi [ML^{-1}T^{-1}][L][LT^{-1}] = 6\pi [ML^{-1+1+1}T^{-1-1}][/latex]
[latex][MLT^{-2}] = 6\pi [MLT^{-2}][/latex]
Here, [latex]6\pi[/latex] is constant and unitless. Since dimension of LHS is equal to dimension of RHS. Hence, the relation is correct.
6. Check the correctness of the formula, PV = RT. (2072 ‘S’)
Ans: Here, PV = RT
Dimension of P = [ML-1T-2]
Dimension of V = [L3]
Dimension of R = [ML2T-2mol-1K-1]
Dimension of T = [K]
∴ [latex][ML^{-1}T^{-2}L^3] = [ML^2T^{-2}mol^{-1}K^{-1}K][/latex]
Or, [ML2T-2] = [ML2T-2mol-1]
But, for 1 mole of gas,
[ML2T-2] = [ML2T-2]
Hence, for 1 mole of gas, the equation is correct.
7. The diameter of a steel rod is given as 56.47 [latex]\pm[/latex] 0.02 mm. What does it mean? (2072 ‘C’)
Ans: The precision of a number is generally indicated by following the number with the symbol [latex]\pm[/latex]; and the second number indicating the maximum likely error.
If the diameter of a steel rod is given as 56.47 [latex]\pm[/latex] 0.02 mm then the true diameter lies between 56.45 mm and 56.49 mm. The precession can also be expressed in terms of maximum likely percentage error.
8. The length of rod is exactly 1 cm. An observer records the readings as 1.0 cm, 1.00 cm and 1.000 cm, which is the most accurate measurement? (2072 ‘D’)
Ans: The digits whose values are accurately known in a particular measurement are called its significant figures. In 1.0 cm, there are 2 significant figures, in 1.00 cm, there are 3 and in 1.000 cm there are 4 significant figures. Greater the no. of significant figures obtained when making a measurement, more accurate the measurement is. Hence, in the given three observations, 1.000 cm is more accurate.
9. What do you mean by significant figure? (2071 ‘S’)
Ans: The number of meaningful digits in a number is called the number of significant figures. The significant figures in a measured physical quantity indicate the number of digits for which we have more confidence.
For example, if an internal radius of a hollow cylinder is 2.71 cm by using a vernier callipers having the least count 0.01 cm, the significant figures in the above example are 3. In the significant figures, the right most digit must be rounded to the figures of the least count of the instrument.
10. A student writes an expression of the force causing a body of mass (m) to move in a circular motion with a velocity (v) as F = mv2. Use the dimensional method to check its correctness. (2071 ‘D’)
Ans: Given, F = mv2
Dimension of F = [MLT-2]
Dimension of m = [M]
Dimension of v = [LT-1]
Dimension of RHS = [MLT-1]
Since, dimension of LHS is not equal to dimension of RHS, the given equation is dimensionally incorrect.
11. Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension? Explain. (2070 ‘C’)
Ans: 1st Part: The two physical quantities which have same dimension are work and torque.
Work = Force x displacement = [MLT-2] x [L] = [ML2T-2]
Torque = Force x perpendicular distance = [MLT-2] x [L] = [ML2T-2]
2nd Part: Yes, a quantity may have unit but no dimension. Eg. Angle. The angle has no dimension but its S.I. unit is radian.
Angle = [latex]\frac{arc length}{radius} = \frac{m}{m}[/latex] = dimensionless
12. If y = a + bt + ct2, where y is the distance and t is time. What is the dimension and unit of c? (2070 ‘D’)
Ans: Here, y = a + bt + ct2
From the principle of homogeneity, the dimension of physical quantity of left side must be equal to dimension of physical quantity on right side.
dimension of y = dimension of ct2
Or, [L] = c[T2]
∴ C = [latex][LT^{-2}][/latex]
Unit: ms-2
13. Check the correctness of the formula t = [latex]2\pi \sqrt{\frac{l}{g}}[/latex] using dimensional analysis, t is the time period of simple pendulum, [latex]l[/latex] is the length of simple pendulum and g is the acceleration due to gravity. (2069 ‘B’)
Ans: Given, formula is
t = [latex]2\pi \sqrt{\frac{l}{g}}[/latex]
Dimensional formula of L.H.S. = [T]
Dimensional formula of R.H.S. = [latex]\frac{L}{LT^{-2}} = \sqrt{T^{-2}} = [T][/latex]
[∵ constant [latex]2\pi[/latex] can be neglected in dimensional analysis]
Since, L.H.S. = R.H.S., the given formula is dimensionally correct.
14. Write dimensional formula of gravitational constant and latent heat. (2067 ‘S’, 2056)
Ans: (i) We have, F = G [latex]\frac{m_1m_2}{d^2}[/latex]
Or, G = [latex]\frac{Fd^2}{m_1m_2}[/latex] ………………. (1)
Where, F = force, d = distance, m1 and m2 = masses of two bodies
∴ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}][/latex]
(ii) Dimensional formula of latent heat,
Latent heat of a substance is given by
L = [latex]\frac{Q}{m}[/latex]
Now, dimension of Q = [ML2T-2]
Dimension of m = [M]
∴ Dimension of L = [latex]\frac{ML^2T^{-2}}{M} = [M^oL^2T^{-2}][/latex]
15. Check the correctness of formula t = [latex]2\pi \sqrt{\frac{m}{k}}[/latex] where t be the time period, m is the mass and k is the force per unit displacement. (2064)
Ans: Given, t = [latex]2\pi \sqrt{\frac{m}{k}}[/latex] ………………….. (i)
Where, t = time period, m = mass, k = force/displacement
In dimensional form, the equation (i) becomes
Or, T = [latex]\sqrt{\frac{M}{MLT^{-2}}\times L}[/latex] [∵ Constant, [latex]2\pi[/latex] can be neglected in dimensional analysis]
Or, T = [latex]\sqrt{T^2}[/latex]
Or, T = T which is true.
Hence, the equation (i) is dimensionally correct. Notice that the equation (i) remains dimensionally correct if 2[latex]\pi[/latex] is replaced by other non-dimensional constants like 3[latex]\pi , 4\pi , 100, etc.[/latex]
16. Derive the dimensional equation of power. (2063 ‘S’)
Ans: Power = [latex]\frac{Work done}{Time} = \frac{Force\times Displacement}{Time}[/latex]
= [latex]\frac{(Mass\times Acceleration)\times Displacement}{Time}[/latex]
∴ [latex][P] = \frac{[M]\times [LT^{-2}]\times [L]}{[T]} = [ML^2T^{-3}][/latex]
Hence, the dimensional equation of power is [P] = [ML2T-3]
Note: Power has 1 dimension in mass, 2 dimensions in length and -3 dimensions in time. [ML2T-3] is the dimensional formula of power and [P] = [ML2T-3] is the dimensional equation of power. Carefully notice the difference among dimension, dimensional formula and dimensional equation.
17. A student writes [latex]\sqrt{\frac{R}{2GM}}[/latex] for escape velocity. Check the correctness of the formula by using dimensional analysis. (2062)
Ans: Given, v = [latex]\sqrt{\frac{R}{2GM}}[/latex] ………………… (i)
Where, v is the escape velocity, R is the radius, G is universal gravitational constant and M is the mass.
In dimensional form, the equation (i) becomes,
[v] = [latex][\sqrt{\frac{R}{GM}}][/latex] ………………….. (ii) [∵ numerical constant 2 can be neglected]
Here,
Dimensional formula of [v] = LT-1
Dimensional formula of [R] = L
Dimensional formula of [G] = M-1L3T-2 [∵ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}[/latex]]]
Dimensional formula of [M] = M
Now, the equation (ii) becomes,
LT-1 = [latex]\sqrt{\frac{L}{M^{-1}L^3T^{-2}\times M}}[/latex]
Or, LT-1 = [latex]\sqrt{L^{-2}T^2}[/latex]
Or, LT-1 = L-1T which is inconsistent.
Hence, the equation (i) is incorrect dimensionally.
18. Convert 10 ergs in joules. (2061)
Ans: We can convert 10 ergs in joules by the method of dimensions. Since ergs and joules are both units of energy and dimension formula for energy is MLT-2. Let, 10 ergs = n2 joules
∴ [latex]n_1[M_1L_1^2T_1^{-2}] = n_2[M_2L_2^2T_2^{-2}][/latex] ……………………… (i)
Here,
n1 = 10 n2 =?
M1 = 1 g M2 = 1 kg
L1 = 1 cm L2 = 1 m
T1 = 1 s T2 = 1 s
From (i), we have,
10 x 1 g x (1 cm)2 x (1s)-2 = n2 x 1 kg x (1 m)2 x (1s)-2
Or, 10 x 1 g x (1 cm2) = n2 x (100g) x (100 cm)2
Or, 10 = n2 x 1000 x 10000
Or, n2 = 10-6 joules
Hence, 10 ergs equal 10-6 joules.
19. Check the correctness of the formula v2 = u2 + 2as using dimensional analysis.
Ans: Given formula is
v2 = u2 + 2as …………… (i)
Dimensional formula of L.H.S. = [LT-1]2 = [L2T-2]
Dimensional formula of R.H.S. = [LT-1]2 + [LT-2L]= [L2T-2]
(Notice that numerical constant can be neglected in dimensional analysis)
Since, L.H.S. = R.H.S., therefore, the given formula (i) is dimensionally correct.
20. Taking force, length and time to be fundamental quantities, find the dimensional formula for the density.
Ans: Force, length and time are taken as fundamental quantities. Let, F, L and T be the dimensional formula of force, length and time respectively.
Since, Density = [latex]\frac{Mass}{Volume}[/latex]
= [latex]\frac{Force}{Acceleration\times Volume}[/latex] (∵ [latex]Mass = \frac{Force}{Acceleration}[/latex])
= [latex]\frac{Force}{\frac{Velocity}{time}\times Volume}[/latex] (∵ [latex]Acceleration = \frac{Velocity}{Time}[/latex])
= [latex]\frac{Force}{\frac{Displacement}{(time)^2}\times (length)^3}[/latex] (∵ [latex]Velocity = \frac{Displacement}{Time}[/latex])
∴ [Density] = [latex]\frac{F}{LT^{-2}\times L^3} = [FL^{-4}T^2][/latex]
Hence, dimensional formula of density is [FL-4T2].
21. The escape velocity of a body is v = [latex]\sqrt{\frac{2GM}{R}}[/latex]. Check the correctness of the formula using dimension.
Ans: v = [latex]\sqrt{\frac{2GM}{R}}[/latex] ……………… (i) where, v is the escape velocity, G is the universal gravitational constant, M is for mass of the earth and R is the radius of the earth.
L.H.S. = [V] = [LT-1]
R.H.S. = [latex]\sqrt{\frac{2[G][M]}{[R]}}[/latex]
Since, 2 is numerical constant, which can be neglected in dimensional form.
= [latex]\sqrt{\frac{M^{-1}L^3T^{-2}M}{L}} = \sqrt{L^2T^{-2}}=[LT^{-1}][/latex] [∵ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]}] = [M^{-1}L^3T^{-2}[/latex]]
L.H.S. = R.H.S., so the equation (i) is dimensionally correct.
22. The density of gold is 19.3 g/cc. Express its value in SI unit.
Ans: 19.3 g/cc = [latex]\frac{19.3\times 10^{-3}kg}{10^{-6}m^3}[/latex] {∵ 1000 g = 1 kg & 1 cc = [latex]1 cm^3 = 10^{-6} m^3[/latex]}
= 19.3 x 103 kgm-3 = 19300 kgm-3
23. Obtain dimensions of specific heat capacity and gravitational constant.
Ans: Dimensional formula of specific heat capacity:
Specific heat capacity of a substance is defined by the relation:
Q = [latex]ms\Delta \theta[/latex]
Or, s = [latex]\frac{Q}{\Delta \theta m}[/latex]
Where, m = mass, [latex]\Delta \theta = [/latex] change in temperature,
Q = Heat energy supplied
∴ Dimension of Q = [latex][ML^2T^{-2}][/latex]
Dimension of m = [M]
Dimension of [latex]\Delta \theta = [/latex] [K]
Dimension of s = [latex]\frac{[ML^2T^{-2}]}{[MK]} = [M^oL^2T^{-2}K^{-1}][/latex]
Dimensional formula of gravitational constant:
We have, F = G [latex]\frac{m_1m_2}{d^2}[/latex]
Or, G = [latex]\frac{Fd^2}{m_1m_2}[/latex] ………………. (1)
Where, F = force, d = distance, m1 and m2 = masses of two bodies
∴ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}][/latex]