4 Marks Question

Question 14 Marks

Read the passage given below and answer the following questions from (i) to (v).
All engineering phenomena deal with definite and measured quantities and so depend on the making of the measurement. We must be
clear and precise in making these measurements. To make a measurement, magnitude of the physical quantity (unknown) is compared.
The record of a measurement consists of three parts, i.e. the dimension of the quantity, the unit which represents a standard quantity and a number which is the ratio of the measured quantity to the standard quantity.

A device which is used for measurement of length to an accuracy of about 10^-5m, is:

Screw gauge
Spherometer
Vernier callipers
Either (a) or (b)

Which of the technique is not used for measuring time intervals?

Electrical oscillator
Atomic clock
Spring oscillator
Decay of elementary particles

The mean length of an object is 5cm. Which of the following measurements is most accurate?

4.9cm
4.805cm
5.25cm
5.4cm

If the length of rectangle l = 10.5cm, breadth b = 2.1cm and minimum possible measurement by scale = 0.1cm, then the area is:

22.0cm²
21.0cm²
22.5cm²
21.5cm²

Age of the universe is about 10¹⁰ yr, whereas the mankind has existed for 10⁶yr. For how many seconds would the man have existed, if age of universe were 1 day?

9.2s
10.2s
8.6s
10.5s

Answer

(d) Either (a) or (b)

Explanation:

A screw gauge and a spherometer can be used to measure length accurately as less as 10^-5m

Explanation:

Spring oscillator cannot be used to measure time intervals.

(a) 4.9cm

Explanation:

Given, length, l = 5cm

Now, checking the errors with each options one - by - one, we get

$\triangle\text{l}_1=5-4.9=0.1\text{cm}$

$\triangle\text{l}_2=5-4.805=0.195\text{cm}$

$\triangle\text{l}_3=5.25-5=0.25\text{cm}$

$\triangle\text{l}_4=5.4-5=0.4\text{cm}$

Error $\triangle\text{l}_1$ is least.

Hence, 4.9cm is most precise or accurate.

(a) 22.0cm²

Explanation:

Area of rectangle, A = Length × Breadth

So, A = lb = 10.5 × 21 = 2205cm²

Minimum possible measurement of scale = 0.1cm.

So, area measured by scale = 22.0cm²

Explnation:

Magnification in time $=\frac{\text{Age of mankind}}{\text{Age of universe}}$

$=\frac{10^6}{10^{10}}=10^{-4}$

Apparent age of mankind = 10^-4 × 1 day

= 10^-4 × 86400s

= 8.64s = 8.6s

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Question 24 Marks

Read the passage given below and answer the following questions from 1 to 5.
In general, the errors in measurement can be broadly classified as (a) systematic errors and (b) random errors.
Systematic errors: The systematic errors are those errors that tend to be in one direction, either positive or negative. Some of the sources of systematic errors are:
(a) Instrumental errors that arise from the errors due to imperfect design or calibration of the measuring instrument, zero error in the instrument, etc. For example, the temperature graduations of a thermometer may be inadequately calibrated (it may read 104 °C at the boiling point of water at STP whereas it should read 100 °C); in a vernier calipers the zero mark of vernier scale may not coincide with the zero mark of the main scale, or simply an ordinary metre scale may be worn off at one end.
(b) Imperfection in experimental technique or procedure to determine the temperature of a human body, a thermometer placed under the armpit will always give a temperature lowers than the actual value of the body temperature.
(c) Personal errors that arise due to an individual’s bias, lack of proper setting of the apparatus or individual’s carelessness in taking observations without observing proper precautions, etc. For example, if you, by habit, always hold your head a bit too far to the right while reading the position of a needle on the scale, you will introduce an error due to parallax.Systematic errors can be minimized by improving experimental techniques, selecting better instruments and removing personal bias as far as possible. For a given set-up, these errors may be estimated to a certain extent and the necessary corrections may be applied to the readings.
Random errors:The random errors are those errors, which occur irregularly and hence are random with respect to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions (e.g. unpredictable fluctuations in temperature, voltage), personal (unbiased) errors by the observer taking readings, etc. For example, when the same person repeats the same observation, it is very likely that he may get different readings every time.
Least count error: The smallest value that can be measured by the measuring instrument is called its least count. All the readings or measured values are good only up to this value. The least count error is the error associated with the resolution of the instrument.

The errors due to imperfect design or calibration of the measuring instrument:

Instrumental error
Random error
Least count error
None of the above

The errors which occur irregularly

Instrumental error
Personal error
Random error
None of these

Write a note on least count error

Write a note on random error

Write a note on systematic error

Answer

(a) Instrumental error
(c) Random error
The least count is the smallest possible value which can be measured with the help of the measuring instrument. All the readings or measured values are read. only up to this value. The least count error is the error related to the resolution of the instrument.
The random errors are irregular errors and comes in measurement randomly hence called as random errors. The causes are unpredictable fluctuations in experimental conditions like temperature, voltage, personal errors by the observer taking readings, etc.
Systematic errors: The systematic errors are those errors that tend to be in one direction, either positive or negative. Some of the sources of systematic errors are:

Instrumental errors that arise from the errors due to imperfect design or calibration of the measuring instrument, zero error in the instrument.

Imperfection in experimental technique or procedure.

Personal errors.

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Question 34 Marks

Read the passage given below and answer the following questions from (i) to (v).
Maximum absolute error in the sum or difference of two quantities is equal to sum of the absolute error in the individual quantities, i.e.
Z = A + B, then, $\pm\triangle\text{Z}=\pm\triangle\text{A}\pm\text{B}$
Maximum fractional error in a product or division of quantities is equal to the sum of fractional errors in the individual quantities i.e.
AB or $\frac{\text{A}}{\text{B}},$ then, $\frac{\triangle\text{Z}}{\text{Z}}=\pm\frac{\triangle\text{A}}{\text{A}}+\frac{\triangle\text{B}}{\text{B}}$
Two resistors of resistances $\text{R}_1=100\pm3\Omega$ are connected (a) in series and (b) in parallel.

The percentage error in the value of R₁ is:

3%
4%
6%
0.3%

The fractional error in the value of R₂ is:

$\frac{1}{40}$
$\frac{1}{50}$
$\frac{1}{100}$
$\frac{1}{200}$

Find the equivalent resistance of the series combination.

$(250\pm7)\Omega$
$(320\pm6)\Omega$
$(300\pm7)\Omega$
$(300\pm1)\Omega$

The percentage error in equivalent resistance in series combination is:

2%
2.3%
2.5
3%

Find the equivalent resistance of the parallel combination having error of $1.8\Omega.$

$(66\pm1)\Omega$
$(66.7\pm1.18)\Omega$
$(66.3\pm2)\Omega$
$(67\pm3)\Omega$

Answer

(a) 3%

Explanation:

Given, $\text{R}_1=100\pm3\Omega$

$\therefore\frac{\triangle\text{R}_1}{\text{R}_2}\times100=\frac{3}{100}\times100=3\text{%}$

$\frac{1}{50}$

Explanation:

Given, $\text{R}_2=(200\pm4)\Omega$

$\therefore\frac{\triangle\text{R}_2}{\text{R}_2}=\frac{4}{200}=\frac{1}{50}$

Explanation:

The equivalent resistance of series combination, i.e.

$\text{R}_{\text{s}}=\text{R}_1+\text{R}_2=(100\pm3)\Omega+(200\pm4)\Omega=(300\pm7)\Omega$

(b) 2.3%

Explanation:

As, $\therefore\frac{\triangle\text{R}_{\text{s}}}{\text{R}_{\text{s}}}\times100=\frac{7}{300}\times100=2.3\text{%}$

(b) $(66.7\pm1.18)\Omega$

Explanation:

The equivalent resistance of parallel
combination,

$\text{R}'=\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}$

$=\frac{200}{3}=66.7\Omega$

Given, $\triangle\text{R}'=1.18\Omega$

$\therefore\text{R}'=(66.7\pm1.18)\Omega$

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Question 44 Marks

Read the passage given below and answer the following questions from 1 to 5.
The rules for determining the uncertainty or error in the measured quantity in arithmetic operations can be understood from the following examples.
a.) If the length and breadth of a thin rectangular sheet are measured, using a metre scale as 16.2cm and, 10.1 cm respectively, there are three significant figures in each measurement. It means that the length L may be written as L = 16.2 ± 0.1cm = 16.2cm ± 0.6%.
Similarly, the breadth b may be written as b = 10.1 ± 0.1 cm = 10.1 cm ± 1%
Then, the error of the product of two (or more) experimental values, using the combination of errors rule, will be L*b = 163.62cm² + 1.6% = 163.62 + 2.6cm²
This leads us to quote the final result as L*b = 164 + 3cm². Here 3cm² is the uncertainty or error in the estimation of area of rectangular sheet.
b) If a set of experimental data is specified to n significant figures a result obtained by combining the data will also be valid to n significant figures.However, if data are subtracted, the number of significant figures can be reduced.For example, 12.9g – 7.06g, both specified to three significant figures, cannot properly be evaluated as 5.84g but only as 5.8g, as uncertainties in subtraction or addition combine in a different fashion (smallest number of decimal places rather than the number of significant figures in any of the number added or subtracted).
c.) The relative error of a value of number specified to significant figures depends not only on n but also on the number itself. For example, the accuracy in measurement of mass 1.02g is ± 0.01g whereas another measurement 9.89g is also accurate to ± 0.01g. The relative error in 1.02g is:
= (± 0.01/1.02) × 100% = ± 1%
Similarly, the relative error in 9.89 g is = (± 0.01/9.89) × 100% = ± 0.1%
Finally, remember that intermediate results in a multi-step computation should be calculated to one more significant figure in every measurement than the number of digits in the least precise measurement.
d.) The nature of a physical quantity is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental or base quantities. We shall call these base quantities as the seven dimensions of the physical world, which are denoted with square brackets [ ]. Thus, length has the dimension [L], mass [M], time [T], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol]. The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. Note that using the square brackets [ ] round a quantity means that we are dealing with ‘the dimensions of’ the quantity. In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] and [T]. For example, the volume occupied by an object is expressed as the product of length, breadth and height, or three lengths. Hence the dimensions of volume are [L] × [L] × [L] = [L³].

Dimensions of area is:

[L²]
[L³]
[M²]
None of these

dimensions of volume are:

[L²]
[L]
[L³]
None of these

What is uncertainty in physics? Explain with one example:

define dimensions of a physical quantity:

Give list for 7 base quantities with dimensions:

Answer

(a) [L²]
(c) [L³]
Uncertainty means the range of possible values within which true values of the measurement lies. For example. If the length and breadth of a thin rectangular sheet are measured, using a metre scale as 16.2 cm and, 10.1 cm respectively, there are three significant figures in each measurement. It means that the length L may be written as L = 16.2 ± 0.1cm = 16.2 cm ± 0.6%.

Similarly, the breadth b may be written as b = 10.1 ± 0.1cm = 10.1cm ± 1% Then, the error of the product of two (or more) experimental values, using the combination of errors rule, will be L*b = 163.62cm² + 1.6% = 163.62 + 2.6cm²

This leads us to quote the final result as L*b = 164 + 3cm². Here 3cm² is the uncertainty or error in the estimation of area of rectangular sheet.

The dimensions of a physical quantity are defined as the powers to which the base quantities must be raised to represent that quantity.

All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental or base quantities. These quantities are
Length has the dimension [L]

Mass [M]
Time [T]
Electric current [A]
Thermodynamic temperature [K]
Luminous intensity [cd]
Amount of substance [mol].

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Question 54 Marks

Read the passage given below and answer the following questions from (i) to (v).
Measurement of Physical Quantity All engineering phenomena deal with definite and measured quantities and so depend on the making of the measurement. We must be clear and precise in making these measurements. To make a measurement, magnitude of the physical quantity (unknown) is compared. The record of a measurement consists of three parts, i.e. the dimension of the quantity, the unit which represents a standard quantity and a number which is the ratio of the measured quantity to the standard quantity.

A device which is used for measurement of length to an accuracy of about 10”5m, is:

screw gauge
spherometer
vernier callipers
Either (a) or (b)

Which of the technique is not used for measuring time intervals?

Electrical oscillator
Atomic clock
Spring oscillator
Decay of elementary particles

The mean length of an object is 5cm. Which of the following measurements is most accurate?

4.9cm
4.805cm
5.25 cm
5.4 cm 63.

If the length of rectangle I = 105 cm, breadth b = 2.1 cm and minimum possible measurement by scale = 0.1 cm, then the area is

22.0cm²
21.0cm²
22.5cm²
21.5cm

Age of the universe is about 10¹⁰ yr, whereas the mankind has existed for 10⁶ yr. For how many seconds would the man have existed, if age of universe were 1day?

9.2s
10.2s
8.6s
10.5s

Answer

(a) screw gauge

Explanation:

A screw gauge and a spherometer can be used to measure length accurately as less as 10⁵m.

Explanation:

Spring oscillator cannot be used to measure time intervals.

(a) 4.9cm

Explanation:

Given, length, 1 = 5 cm

Now, checking the errors with each options one-by-one, we get

AZ, =5 - 4.9 = 0.1 cm

A Z2 = 5 - 4.805 = 0.195 cm

A l 3 = 5.25 - 5 = 0.25 cm

A Z4 = 5.4 - 5 = 0.4 cm

Error I, is least.

Hence, 4.9 cm is most precise or accurate

(a) 22.0cm²

Explanation:

Area of rectangle, A = Length x Breadth

So, A = u) = 10.5 x 2.1 = 22.05cm²

Minimum possible measurement of scale = 01cm.

So, area measured by scale = 220cm²

Explanation:

Magnihcation in time $=\frac{\text{Age of mankind}}{\text{Age of universe}}=\frac{10^6}{10^{10}}=10^{-4}$

Apparent age of mankind =10^-4 x 1day

= 10^-4 x 86400s

= 8.64s = 8.6s

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Question 64 Marks

Read the passage given below and answer the following questions from 1 to 5.
Every measurement involves errors. Thus, the result of measurement should be reported in a way that indicates the precision of measurement. Normally, the reported result of measurement is a number that includes all digits in the number that are known reliably plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as significant digits or significant figures. If we say the period of oscillation of a simple pendulum is 1.62 s, the digits 1 and 6 are reliable and certain, while the digit 2 is uncertain. Thus, the measured value has three significant figures.A choice of change of different units does not change the number of significant digits or figures in a measurement. This important remark makes most of the following observations clear,

All the non-zero digits are significant.
All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.
If the number is less than 1, the zero(s) on the right of decimal point but to the left of the first non-zero digit are not significant.
The terminal or trailing zero(s) in a number without a decimal point are not significant.[Thus 123 m = 12300 cm = 123000 mm has three significant figures, the trailing zero(s) being not significant.
The trailing zero(s) in a number with a decimal point are significant. [The numbers 3.500 or 0.06900 have four significant figures each]
For a number greater than 1, without any decimal, the trailing zero(s) are not significant.
For a number with a decimal, the trailing zero(s) are significant

(b) The digit 0 conventionally put on the left of a decimal for a number less than 1 (like 0.1250) is never significant. However, the zeroes at the end of such number are significant in a measurement.
(c) The multiplying or dividing factors which are neither rounded numbers nor numbers representing measured values are exact and have infinite number of significant digits.
(d) In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures.In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal
places. For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic addition, is 663.821 g. But the least precise measurement (227.2 g) is correct to only one decimal place. The final result should, therefore, be rounded off to 663.8 g.

Significant figures in 12300 cm are:

5
4
3
None of these

All the non-zero digits are:

Significant
Non significant
None of these

Give rules for significant figures

Give rules for addition and subtraction operations with significant figure

Give rules for multiplication and division operations with significant figure

Answer

(c) 3
(a) Significant
All the non-zero digits are significant.

All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.
If the number is less than 1, the zero on the right of decimal point but to the left of the first non-zero digit are not significant.
The terminal or trailing zero(s) in a number without a decimal point are not significant.
The trailing zero(s) in a number with a decimal point are significant.
For a number greater than 1, without any decimal, the trailing zero(s) are not significant.
For a number with a decimal, the trailing zero(s) are significant.
The digit 0 conventionally put on the left of a decimal for a number less than 1 (like 0.1250) is never significant. However, the zeroes at the end of such number are significant in a measurement.

In addition or subtraction, the final result after the operation should have as many decimal places as are there in the number with the least decimal places.
In multiplication or division, the final result after operation should have as many significant figures as are there in the original number with the least significant figures. if the speed of light is given as 3 × 10⁸ m s^-1 (one significant figure) and one year (1y = 365.25 d) has 3.1557 × 10⁷ s (five significant figures), the light year is 9.47 × 1015m (three significant figures.

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Question 74 Marks

Read the passage given below and answer the following questions from (i) to (v).

Dimensional analysis and its applications The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity. The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or
subtracted which have the same dimensions. A thorough understanding of dimensional analysis helps us in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions. When magnitudes of two or more physical quantities are multiplied, their units should be treated in the same manner as ordinary algebraic symbols. We can cancel identical units in the numerator and denominator. The same is true for dimensions of a physical quantity. Similarly, physical quantities represented by symbols on both sides of a mathematical equation must have the same dimensions.

Statement I The method of dimensions analysis cannot validate the exact relationship between physical quantities in any equation.

Statement II It does not distinguish between the physical quantities having same dimensions.

Which of the following statement(s) is/are correct?

Only I
I and II
Only II
None of these

The quantity having same dimension as that of Planck’s constant is

Work
Linear momentum
Angular momentum
Impulse

If speed v, acceleration A and force F, are considered as fundamental units, the dimension of Young’s modulus will be:

[v^-4 A^-2 F]
[v^-2 A² F²]
[v^-2 A²F^-2]
[v^-4 A² F¹]

Given that, the amplitude of the scattered light is:

Directly proportional to amplitude of incident light
Directly proportional to the volume of the scattering dust particle
Inversely proportional to its distance from the scattering particle and
Dependent upon the wavelength l of the light.

Then, the relation of intensity of scattered light with the wavelength is:

$\frac{1}{\lambda^2}$
$\frac{1}{\lambda^4}$
$\frac{1}{\lambda^6}$
$\frac{1}{\lambda^7}$

Find the value of power of 60 J/min on a system that has 100g, 100cm and 1min as the base units.

2.16 × 10⁴ units
2.16 × 10⁶ units
3 × 10⁴ units
4 × 10⁷ units

Answer

(b) I and II

Explanation:

The method of dimensions can only test the dimensional validity but not the exact relationship between physical quantities in any equation.

This is because, it does not distinguish between the physical quantities having same dimensions.

Explanation:

Planck’s constant, $\text{h}=\frac{\text{E}}{\text{v}}$

So, dimensions of $\text{h}=\Big[\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}}\Big]=[\text{ML}^2\text{T}^{-1}]$

Angular momentum, L = mvr

Dimensions of L = [M] [LT^-1] [L] = [ML²T^-1]

Work, W = Force × Displacement

$\therefore$ Dimenstion of W = [MLT²] × [L] = [ML²T^-2]

Linear momentum, p = Mass × Velocity

Dimensions of P = [M] [LT^-1] = [MLT^-1]

Impulse, $\text{I}=\frac{[\text{MLT}^{-2}]}{[\text{T}]}=[\text{MLT}^{-3}]$

Hence, only angular momentum has same dimensions as that of Planck’s constant.

(d) [v^-4 A² F¹]

Explanation:

Dimensions of speed [v] = [LT^-1]

Dimensions of acceleration [A] = [LT^-2]

Dimensions of force [F] = [MLT^-2]

Dimensions of Young modulus [Y] = [ML^-1T^-2]

Let dimensions of Young’s modulus is expressed in terms of speed, acceleration and force as

$[\text{Y}]=[\text{v}]^{\alpha}[A]^{\beta}[\text{F}]^{\curlyvee} ....(\text{i})$

Then substituting dimensions in terms of M, L and T, we get

$[\text{ML}^{-1}\text{T}^{-2}]=[\text{LT}^{-1}]^{\alpha}[\text{LT}^{-2}]^{\beta}[\text{MLT}^{-2}]^{\curlyvee}$

$=[\text{M}\curlyvee\text{L}^{\alpha+\beta+\curlyvee}\text{T}^{-\alpha-2\beta-2\curlyvee}]$

Now comparing powers of basic quantities on both sides, we get

$\curlyvee=1$

$\alpha+\beta+\curlyvee=-1$

and $-\alpha-2\beta-2\curlyvee=-2$

Solving these, we get

$\alpha=-4,\beta=2,\curlyvee=1$

Substituting the values of a b, and g in Eq. (i), we get

[Y] [v^-4 A² F¹]

(b) $\frac{1}{\lambda^4}$

Explanation:

According to the question, the expression for the scattered amplitude of light (A_s) in terms of amplitude of incident light ( A_i), volume
(V), distance from scattering particle (x) and wavelength $(\lambda)$ l can be given as

$\therefore\text{A}_{s}=\text{kA}_{\text{i}}^{1}\text{V}^1\text{X}^{-1}\lambda^{\text{d}}$

where, k is the constant of proportionality. Writing the dimensions on both sides of the above equation, we get

[L] = [L] [L³] [L^-1] [L^d] = [L^3+d]

Comparing the powers of L on both sides, we get,

or 1 = 3 + d

or d = -2

i.e., $\text{A}_\text{s}\propto\frac{1}{\lambda^2}$

But, intensity $(\text{I}_\text{s})\propto[\text{amplitude}(\text{A}_\text{s})]^2$

$\therefore\text{I}_\text{s}\propto\frac{1}{\lambda^4}$

(b) 2.16 × 10⁶ units

Explanation:

Given, power, $\text{P}_1=\frac{\text{Work done}}{\text{Time taken}}$

$=\frac{60\text{J}}{1\text{min}}=\frac{60\text{J}}{60\text{s}}=1\text{W}$ or Kg-m²s^-3

which is the SI unit of power.

Given, p₁ = 1 w, m₁ = 1kg = 1000g

L₁ = 1m = 100cm, T₁ = 1s

In new system, P₂ = ?, M₂ = 100g, L₂ = 100cm, T₂ = 1min =60s

$\therefore$ Conversion of 60J per min or 1W in a new system, i.e.

$\text{P}_2=\text{P}_1\Big[\frac{\text{M}_1}{\text{M}_2}\Big]^\text{a}\Big[\frac{\text{L}_1}{\text{L}_2}\Big]^\text{b}\Big[\frac{\text{T}_1}{\text{T}_2}\Big]^\text{c}$

Now, [power] = [ML²T^-3]

So, a = 1, b = 2 and c = -3

$\Rightarrow\text{P}_2=1\Big[\frac{1000}{100}\Big]^1\Big[\frac{100}{100}\Big]^2\Big[\frac{1}{60}\Big]^{-3}$

= 2.16 × 10⁶ new units of power

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Physics 4 Marks Question

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