The document NCERT Exemplar (Part - 1) - Units and Measurements Notes | EduRev is a part of the NEET Course NEET Revision Notes.

All you need of NEET at this link: NEET

**MULTIPLE CHOICE QUESTIONS **

**Q.1. The number of significant figures in 0.06900 is(a) 5(b) 4(c) 2(d) 3Ans. **(b)

Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.

The following rules are observed in counting the number of significant figures in a given measured quantity.

Leading zeros or the zeros placed to the left of the number are never.

Hence, number of significant figures are four.

(a) 663.821

(b) 664

(c) 663.8

(d) 663.82

The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

436.32

+ 227.2 ← (has only one decimal place)

+ 0.301

------------

663.821 ←(answer should be reported to one decimal place)

------------

The final result should, therefore, be rounded off to one decimal place, i.e. 664.

(a) 1.6048 g cm

(b) 1.69 g cm

(c) 1.7 g cm

(d) 1.695 g cm

Ans.

The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures.

Density = 4.237g/2.5 cm

After rounding off the number, we get density = 1.7

(a) 2.75 and 2.74

(b) 2.74 and 2.73

(c) 2.75 and 2.73

(d) 2.74 and 2.74

While rounding off measurements, we use the following rules by convention:

(a) 164 ± 3 cm

(b) 163.62 ± 2.6 cm

(c) 163.6 ± 2.6 cm

(d) 163.62 ± 3 cm

Ans.

Solution.

Error in product of quantities: Suppose x = a × b

Let Δa = absolute error in measurement of a,

Δb = absolute error in measurement of b,

Δx = absolute error in calculation of x, i.e. product of a and b.

The maximum fractional error in x is Δx/x = ± (Δa/a + Δb/b)

Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b)

According to the problem, length l = (16.2 ± 0.1) cm

Breadth b = (10.1 ± 0.1) cm

Area A = l × b- (16.2 cm) × (10.1 cm) = 163.62 cm

As per the rule area will have only three significant figures and error will have only one significant figure. Rounding off we get, area A = 164 cm

If ΔA is error in the area, then relative error is calculated as ΔA/A.

ΔA/A = Δl/l + Δb/b

= 0.1 cm/16.2 cm + 0.1 cm/10.1 cm

= 1.01 + 1.62/16.2 × 10.1 = 2.63/163.62

⇒ ΔA = A × 2.63/163.62 cm

= 163.62 × 2.63/163.62 = 2.63cm

ΔA = 3 cm

Area, A = A ± ΔA = (164 ± 3) cm

(a) Work and torque.

(b) Angular momentum and Planck’s constant.

(c) Tension and surface tension.

(d) Impulse and linear momentum.

Torque = force × distance = [MLT

Both have same dimensions.

Planck's constant h = F.s/v (∴ E = hv)

= [MLT

Dimensions of h and L are equal.

Surface tension = Force/i = [MLT

Dimensions of both are not same.

Momentum = mv = [MLT

Both have same dimensions.

A = 2.5 m s

B = 0.10 s ± 0.01 s

(b) (0.25 ± 0.5) m

(c) (0.25 ± 0.05) m

(d) (0.25 ± 0.135) m

A = (2.5 ± 0.5)ms

B = (0.10 ± 0.01)s

X = AB = 2.5 × 0.10 = 0.25 m

Δx/x = ΔA/A + ΔB/B

Δx/x = 0.5/2.5 + 0.01./0.10

Δx/x = 0.075/0.25, Δx = 0.007 ≅ 0.08

(Rounding off up to 2 significant figures)

∴ AB = (2.5 ± 0.08) m.

(a) 1.4 m ± 0.4 m

(b) 1.41 m ± 0.15 m

(c) 1.4 m ± 0.3 m

(d) 1.4 m ± 0.2 m

Quantities A and b are measured quantities so, number of significant figures in 1.0 m and 2.0 m are two.

rounding off up to minimum numbers of significant figure in 1.0 and 2.0 result must be in 2 significant figures

Δx = 0.2 m rounding off up to 1 place of decimal.**Q.9. ****Which of the following measurements is most precise?(a) 5.00 mm(b) 5.00 cm(c) 5.00 m(d) 5.00 km**

Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision.

Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct up to two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

(a) 4.9 cm

(b) 4.805 cm

(c) 5.25 cm

(d) 5.4 cm

Accuracy describes the nearness of a measurement to the standard or true value, i.e. a highly accurate measuring device will provide measurements very close to the standard, true or known values.

According to the problem, length l = 5 cm

Let us first check the errors in each values by picking options one by one, we get

Δl

Δl

Δl

Δl

Error Δl

Hence 4.9 cm is most closer to true value. So, 4.9 is more accurate.

According to the problem.

Young's modulus, Y = 1.9 × 10

1 N in SI system of units = 10

Hence, 1.9 × 10

In C.G.S. length is measured in unit ‘cm’, so we should also convert m into cm.

∴ Y = 1.9 × 10

= 1.9 × 10

According to the problem, fundamental quantities arc momentum (p), area (A) and time (T) and we have to express energy in these fundamental quantities.

Let energy E,

E ∝ p

⇒ E = kp

where, k is dimensionless constant of proportionality.

Dimensional formula of energy, [E] = [ML

[A] = [L

Putting all the dimensions, we get

ML

= M

According to the principle of homogeneity of dimensions, we get

a = 1 .... (i)

a + 2b = 2 ....(ii)

- a + c = - 2 ... (iii)

By solving these equations (i), (ii) and (iii), we get

a = 1, b = 1/2, c = -1

Dimensional formula for E is [p

(a) y = a sin 2πt/T

(b) y = a sin vt.

(c)

The argument of trigonometric functions (sin, cos etc.) should be dimensionless. y is displacement and according to the principle of homogeneity of dimensions LHS and RHS.

[Y] = [L], [a] = [L]

[2πt/T] = [T/T]

[vt] = [v] [t] = [LT

[a/T]/[a]/[T] = [L]/[T] = [LT

[t/a] = [LT

[LHS] ≠ [RHS]

Hence, (c) is not the correct option.

⇒ LHS ≠ RHS.

So, option (b) is also not correct.

(a) (P – Q)/R

(b) PQ – R

(c) PQ/R

(d) (PR – Q

(e) (R + Q)/P

Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation.

According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful.

So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q

Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

(a) Linear impulse

(b) Angular impulse

(c) Linear momentum

(d) Angular momentum

∴ E = hv

h = E/v = [ML

Linear impulse = F.t = dp/dt.dt = dp

= mv = [MLT

Angular impulse =τ.dt = dL/dt.dt = dL = mvr

= [M][LT

Linear momentum = mv = [MLT

Angular momentum L = mvr = [ML

So the dimensional formulae of h, Angular impulse and Angular momentum are same.

(a) Mass of electron (m

(b) Universal gravitational constant (G)

(c) Charge of electron (e)

(d) Mass of proton (m

We know that dimension of

Hence, physical quantities (a, b and d) can be used to represent L, M,T in terms of the choosen fundamental quantities.**Q.17. ****Which of the following ratios express pressure?(a) Force/Area(b) Energy/Volume(c) Energy/Area(d) Force/Volume**

Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

We know that pressure

So, this ratio express pressure (In fact this ratio actually represents pressure).

Dimensions of this ratio are not same as pressure, so this ratio does not express pressure.

Dimensions of this ratio is the same as pressure, so this ratio also express pressure.

Dimensions of this ratio are not same as pressure, so this ratio does not express pressure.

(a) Second

(b) Parsec

(c) Year

(d) Light year

Parsec and light year are those practical units which are used to measure large distances. For example, the distance between sun and earth or other celestial bodies. So they are the units of length not time. Here, second and year represent time.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.1. Why do we have different units for the same physical quantity?****Ans. **Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required.**Example:****(a)** Mass ranges from 10^{-30} kg (for an electron) to 10^{53} kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc.**(b)** The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.**Q.2. ****The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?****Ans. **Radius (R) of atom = 1 = 10^{-10} m

Radius (r) of nucleus = 1 fermi =10^{-15 }m

Ratio of volume of atom to nucleus**Q.3.**** Name the device used for measuring the mass of atoms and molecules.****Ans. **A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.**Q.4. Express unified atomic mass unit in kg.****Ans. **The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.**Q.5. A function f (θ) is defined as:****Why is it necessary for f(θ) to be a dimensionless quantity?****Ans. **θ is represented by angle which is equal to

so angle θ is dimensionless physical quantity.First term is 1 which is dimensionless, next term contain only powers of θ, as θ is dimensionless so their powers will also be dimensionless. Hence, each term in R.H.S. expression are dimensionless so left hand side f(θ) must be dimensionless.**Q.6. ****Why length, mass and time are chosen as base quantities in mechanics?****Ans. **Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because**(a)** Length, mass and time cannot be derived from one another, that is these quantities are independent.**(b)** All other quantities in mechanics can be expressed in terms of length, mass and time.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!