1. Which statement best describes the measurement of a dimensional physical quantity?
ⓐ. It is the assignment of any convenient number to the physical quantity.
ⓑ. It is complete whenever a numerical value is stated, even if no unit is specified.
ⓒ. It is the identification of the unit itself as the physical quantity being measured.
ⓓ. It is a comparison with an accepted reference standard, expressed by a numerical measure and a unit.
Correct Answer: It is a comparison with an accepted reference standard, expressed by a numerical measure and a unit.
Explanation: Measurement involves comparing a physical quantity with an accepted reference standard called a unit.
The result is therefore represented in the form \(Q=n\,u\), where \(n\) is the numerical measure and \(u\) is the chosen unit.
The physical quantity \(Q\) and the unit \(u\) are not the same idea.
For example, length is a physical quantity, whereas metre is one possible unit used to measure length.
A bare number such as \(5\) does not completely specify a dimensional measurement because the corresponding unit is missing.
Writing \(5\,\text{m}\), \(5\,\text{cm}\), or \(5\,\text{kg}\) represents physically different measurements.
Thus measurement is not merely the assignment of a number; the reference unit is an essential part of the statement.
Hence option D correctly describes measurement.
2. A length is written as \(2.4\,\text{m}\). If centimetre is chosen as the unit instead, what is its numerical measure?
ⓐ. \(0.024\)
ⓑ. \(2.4\)
ⓒ. \(240\)
ⓓ. \(24000\)
Correct Answer: \(240\)
Explanation: Given:
\[
L=2.4\,\text{m}.
\]
Required: the numerical measure when the unit is centimetre.
The exact unit relation is
\[
1\,\text{m}=100\,\text{cm}.
\]
Therefore,
\[
2.4\,\text{m}=2.4\times100\,\text{cm}=240\,\text{cm}.
\]
So the physical length has not changed; only its numerical representation has changed.
The centimetre is a smaller unit than the metre, so more centimetres are required to represent the same length.
This illustrates the inverse relation between unit size and numerical measure for a fixed physical quantity.
Hence the required numerical measure is \(240\).
3. Consider the following statements for a fixed physical quantity \(Q=n\,u\).
Statement I: If the chosen unit is made twice as large, the numerical measure becomes half as large.
Statement II: Since the numerical measure changes, the physical quantity itself must also change.
ⓐ. Statement I is correct and Statement II is incorrect.
ⓑ. Both Statement I and Statement II are correct.
ⓒ. Statement I is incorrect and Statement II is correct.
ⓓ. Both Statement I and Statement II are incorrect.
Correct Answer: Statement I is correct and Statement II is incorrect.
Explanation: For a fixed physical quantity,
\[
Q=n_1u_1=n_2u_2.
\]
Suppose the new unit is twice the old unit:
\[
u_2=2u_1.
\]
Then
\[
n_1u_1=n_2(2u_1),
\]
which gives
\[
n_2=\frac{n_1}{2}.
\]
Thus Statement I is correct: increasing the unit size by a factor of \(2\) decreases the numerical measure by the same factor.
However, \(Q\) remains unchanged throughout the conversion because \(n\,u\) represents the same physical quantity.
A change in numerical measure caused solely by a change of unit is not a physical change in the quantity.
Therefore Statement II is incorrect.
4. The same length \(L\) is expressed in two units \(u\) and \(v\) as
\[
L=3.6\,u=0.12\,v.
\]
Which conclusion is correct?
ⓐ. \(v=\dfrac{u}{30}\), so \(v\) is the smaller unit.
ⓑ. \(v=3.48u\), so \(v\) is the larger unit.
ⓒ. \(v=30u\), but \(v\) is the smaller unit because its numerical measure is smaller.
ⓓ. \(v=30u\); \(v\) is the larger unit and therefore gives the smaller numerical measure.
Correct Answer: \(v=30u\); \(v\) is the larger unit and therefore gives the smaller numerical measure.
Explanation: Both expressions represent the same physical length, so
\[
3.6u=0.12v.
\]
Solving for \(v\),
\[
v=\frac{3.6}{0.12}u.
\]
Since
\[
\frac{3.6}{0.12}=30,
\]
we obtain
\[
v=30u.
\]
Thus one unit \(v\) is physically equal to thirty units \(u\), so \(v\) is the larger unit.
The numerical measure in unit \(v\) is only \(0.12\), whereas in the smaller unit \(u\) it is \(3.6\).
This is consistent with the general rule that a larger unit gives a smaller numerical measure for the same physical quantity.
The quantity \(L\) itself is unchanged; only its representation differs.
Hence option D is correct.
5. The following table lists base units for length, mass and time in some systems of units.
| Row | System | Length | Mass | Time |
|---|
| I | CGS | centimetre | gram | second |
| II | FPS | foot | pound | second |
| III | MKS | metre | kilogram | second |
| IV | MKS | metre | gram | second |
Which row is inconsistent with the stated system?
ⓐ. Row I only
ⓑ. Rows II and IV only
ⓒ. Row IV only
ⓓ. Row III only
Correct Answer: Row IV only
Explanation: The CGS system uses centimetre, gram and second as its base units for length, mass and time respectively.
Therefore Row I is correctly matched.
The FPS system uses foot, pound and second, so Row II is also correctly matched.
The MKS system uses metre, kilogram and second.
Hence Row III gives the correct MKS combination.
Row IV differs from MKS in its mass unit because it uses gram instead of kilogram.
Using gram with metre and second would therefore not reproduce the stated MKS base-unit set.
The distinction matters because changing a base unit changes the derived units constructed from the system.
For example, a force unit obtained from mass, length and time would acquire a different physical size if gram replaced kilogram.
Thus only Row IV is inconsistent.
6. A surface area is \(4.0\,\text{cm}^2\), while a volume is \(4.0\,\text{cm}^3\). Which pair gives their respective values in SI units?
ⓐ. \(4.0\times10^{-2}\,\text{m}^2,\;4.0\times10^{-3}\,\text{m}^3\)
ⓑ. \(4.0\times10^{-4}\,\text{m}^2,\;4.0\times10^{-4}\,\text{m}^3\)
ⓒ. \(4.0\times10^{-4}\,\text{m}^2,\;4.0\times10^{-6}\,\text{m}^3\)
ⓓ. \(4.0\times10^{-2}\,\text{m}^2,\;4.0\times10^{-6}\,\text{m}^3\)
Correct Answer: \(4.0\times10^{-4}\,\text{m}^2,\;4.0\times10^{-6}\,\text{m}^3\)
Explanation: The basic length conversion is
\[
1\,\text{cm}=10^{-2}\,\text{m}.
\]
For area, the complete conversion factor must be squared:
\[
1\,\text{cm}^2=(10^{-2}\,\text{m})^2=10^{-4}\,\text{m}^2.
\]
Therefore,
\[
4.0\,\text{cm}^2=4.0\times10^{-4}\,\text{m}^2.
\]
For volume, the complete conversion factor must instead be cubed:
\[
1\,\text{cm}^3=(10^{-2}\,\text{m})^3=10^{-6}\,\text{m}^3.
\]
Hence,
\[
4.0\,\text{cm}^3=4.0\times10^{-6}\,\text{m}^3.
\]
The common error is to apply the linear factor \(10^{-2}\) directly to an area or a volume without raising it to the appropriate power.
Thus option C gives both conversions correctly.
7. Assertion: Plane angle measured in radian and solid angle measured in steradian are dimensionless quantities.
Reason: Plane angle is defined by the ratio of arc length to radius, while solid angle is defined by the ratio of intercepted spherical area to the square of the radius.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: For a plane angle,
\[
\theta=\frac{s}{r},
\]
where both \(s\) and \(r\) are lengths.
Their units therefore cancel:
\[
\frac{\text{m}}{\text{m}}=1.
\]
Thus plane angle has no remaining physical dimension, although its unit radian is used to identify the quantity being described.
Similarly, for a solid angle,
\[
\Omega=\frac{A}{r^2}.
\]
The area \(A\) has unit \(\text{m}^2\), while \(r^2\) also has unit \(\text{m}^2\), so their ratio is dimensionless.
The unit steradian is nevertheless used for solid angle.
Therefore the existence of a named unit does not imply that a quantity must possess non-zero physical dimensions.
The Reason directly explains why both quantities in the Assertion are dimensionless.
Hence option A is correct.
8. Consider the following statements about SI units.
Statement I: Kilogram is an SI base unit.
Statement II: Coulomb is the SI base unit of electric current.
Statement III: Kelvin is an SI base unit of thermodynamic temperature.
Statement IV: Newton is a derived SI unit and can be written as \(\text{kg m s}^{-2}\).
Which combination is correct?
ⓐ. Statements I and II only
ⓑ. Statements II and III only
ⓒ. Statements I and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct because kilogram is the SI base unit of mass.
Statement II is incorrect because the SI base quantity is electric current and its base unit is ampere, \(\text{A}\).
Coulomb is a derived unit because electric charge can be written as current multiplied by time:
\[
1\,\text{C}=1\,\text{A s}.
\]
Statement III is correct because kelvin, \(\text{K}\), is the SI base unit of thermodynamic temperature.
Statement IV is also correct.
From the relation \(F=ma\),
\[
[\text{unit of force}]=\text{kg}\times\text{m s}^{-2}
=\text{kg m s}^{-2}.
\]
This derived SI unit is given the special name newton, \(\text{N}\).
A special name such as newton or coulomb does not make a unit fundamental.
Therefore Statements I, III and IV only are correct.
9. A new system of units is defined such that its unit of mass is \(5\,\text{kg}\), its unit of length is \(2\,\text{m}\), and its unit of time is \(0.5\,\text{s}\). What is the numerical measure of a force of \(120\,\text{N}\) in this new system?
ⓐ. \(120\)
ⓑ. \(40\)
ⓒ. \(3\)
ⓓ. \(0.3\)
Correct Answer: \(3\)
Explanation: Given:
\[
M'=5\,\text{kg},\qquad L'=2\,\text{m},\qquad T'=0.5\,\text{s}.
\]
Required: the numerical measure of \(120\,\text{N}\) in the new force unit.
Force is defined through
\[
F=ma.
\]
Therefore its unit is constructed as
\[
F'=\frac{M'L'}{T'^2}.
\]
Substituting the sizes of the new base units,
\[
F'=\frac{(5\,\text{kg})(2\,\text{m})}{(0.5\,\text{s})^2}.
\]
Since
\[
(0.5)^2=0.25,
\]
we obtain
\[
F'=\frac{10}{0.25}\,\text{kg m s}^{-2}=40\,\text{N}.
\]
Thus one new unit of force is physically equal to \(40\,\text{N}\).
If \(n\) is the required numerical measure,
\[
120\,\text{N}=n(40\,\text{N}),
\]
so
\[
n=3.
\]
The larger force unit therefore gives a numerical measure much smaller than \(120\).
Hence the answer is \(3\).
10. A pressure has numerical value \(6.0\times10^5\) when expressed in the CGS unit \(\text{g cm}^{-1}\text{s}^{-2}\). What is its numerical value in pascal?
ⓐ. \(6.0\times10^3\)
ⓑ. \(6.0\times10^4\)
ⓒ. \(6.0\times10^5\)
ⓓ. \(6.0\times10^6\)
Correct Answer: \(6.0\times10^4\)
Explanation: The CGS unit given is
\[
\text{g cm}^{-1}\text{s}^{-2}.
\]
Convert each base-unit factor to SI.
For mass,
\[
1\,\text{g}=10^{-3}\,\text{kg}.
\]
For the inverse length factor,
\[
1\,\text{cm}^{-1}=(10^{-2}\,\text{m})^{-1}=10^2\,\text{m}^{-1}.
\]
Therefore,
\[
1\,\text{g cm}^{-1}\text{s}^{-2}
=(10^{-3})(10^2)\,\text{kg m}^{-1}\text{s}^{-2}.
\]
Hence,
\[
1\,\text{g cm}^{-1}\text{s}^{-2}=10^{-1}\,\text{kg m}^{-1}\text{s}^{-2}.
\]
But
\[
1\,\text{Pa}=1\,\text{kg m}^{-1}\text{s}^{-2}.
\]
Thus one CGS unit in the stem equals \(0.1\,\text{Pa}\).
The given pressure is therefore
\[
(6.0\times10^5)(10^{-1})\,\text{Pa}
=6.0\times10^4\,\text{Pa}.
\]
The inverse power of centimetre is crucial; treating it like an ordinary positive power would reverse the conversion factor.
Hence option B is correct.
11. A material has density \(7.5\,\text{mg mm}^{-3}\). Student A claims that its numerical density in \(\text{g cm}^{-3}\) is also \(7.5\), whereas Student B claims that it is \(7.5\times10^3\). A sample of the material has volume \(4.0\,\text{cm}^3\). Which option gives the correct student and the corresponding mass of the sample?
ⓐ. Student A; \(30\,\text{g}\)
ⓑ. Student A; \(3.0\,\text{g}\)
ⓒ. Student B; \(30\,\text{g}\)
ⓓ. Student B; \(3.0\times10^4\,\text{g}\)
Correct Answer: Student A; \(30\,\text{g}\)
Explanation: First compare the two density units rather than converting the numerical value blindly.
Since
\[
1\,\text{g}=10^3\,\text{mg},
\]
and
\[
1\,\text{cm}^3=(10\,\text{mm})^3=10^3\,\text{mm}^3,
\]
we have
\[
1\,\text{g cm}^{-3}
=\frac{10^3\,\text{mg}}{10^3\,\text{mm}^3}.
\]
The factors \(10^3\) cancel, giving
\[
1\,\text{g cm}^{-3}=1\,\text{mg mm}^{-3}.
\]
Therefore the numerical density remains \(7.5\), so Student A is correct.
Thus
\[
\rho=7.5\,\text{g cm}^{-3}.
\]
Using
\[
m=\rho V,
\]
we get
\[
m=(7.5\,\text{g cm}^{-3})(4.0\,\text{cm}^3)=30\,\text{g}.
\]
The cubic conversion in the denominator exactly compensates the milligram-to-gram conversion in this particular pair of units.
Hence the correct combination is Student A and \(30\,\text{g}\).
12. A quantity is defined by
\[
X=\left(\frac{F}{A}\right)(Avt),
\]
where \(F\) is force, \(A\) is area, \(v\) is speed and \(t\) is time. If \(F=6\,\text{N}\), \(v=4\,\text{m s}^{-1}\) and \(t=2\,\text{s}\), which option gives the correct SI unit and numerical value of \(X\)?
ⓐ. watt; \(12\)
ⓑ. pascal; \(48\)
ⓒ. joule; \(12\)
ⓓ. joule; \(48\)
Correct Answer: joule; \(48\)
Explanation: The expression appears to contain area, but the same factor \(A\) occurs once in the denominator and once in the numerator.
Therefore,
\[
X=\left(\frac{F}{A}\right)(Avt)=Fvt.
\]
This cancellation also shows why no numerical value of \(A\) is required.
Now examine the units:
\[
[\text{unit of }X]=\text{N}\left(\text{m s}^{-1}\right)\text{s}.
\]
The second cancels with \(\text{s}^{-1}\), giving
\[
[\text{unit of }X]=\text{N m}.
\]
Since
\[
1\,\text{J}=1\,\text{N m},
\]
the SI unit of \(X\) is joule.
For the numerical value,
\[
X=(6)(4)(2)\,\text{N m}.
\]
Hence,
\[
X=48\,\text{N m}=48\,\text{J}.
\]
This question requires both algebraic cancellation and translation of the resulting compound unit into its named SI form.
Therefore option D is correct.
13. In a new system of units, the unit of mass is \(0.20\,\text{kg}\), the unit of length is \(5.0\,\text{m}\), and the unit of time is \(2.0\,\text{s}\). Using \(F=ma\) and \(E=F\ell\), what is the numerical measure of \(250\,\text{J}\) of energy in this system?
ⓐ. \(20\)
ⓑ. \(50\)
ⓒ. \(125\)
ⓓ. \(200\)
Correct Answer: \(200\)
Explanation: Let the new base units be
\[
M'=0.20\,\text{kg},\qquad L'=5.0\,\text{m},\qquad T'=2.0\,\text{s}.
\]
First construct the new unit of force from
\[
F=ma.
\]
Thus,
\[
F'=\frac{M'L'}{T'^2}.
\]
Substitution gives
\[
F'=\frac{(0.20)(5.0)}{(2.0)^2}\,\text{N}.
\]
Therefore,
\[
F'=\frac{1.0}{4.0}\,\text{N}=0.25\,\text{N}.
\]
Now use the supplied relation
\[
E=F\ell.
\]
The new unit of energy is
\[
E'=F'L'=(0.25\,\text{N})(5.0\,\text{m})=1.25\,\text{J}.
\]
Thus one new energy unit equals \(1.25\,\text{J}\).
If \(n\) is the numerical measure of \(250\,\text{J}\),
\[
250\,\text{J}=n(1.25\,\text{J}).
\]
Hence,
\[
n=\frac{250}{1.25}=200.
\]
The calculation illustrates that a derived unit must be reconstructed consistently from all relevant new base units before the numerical measure is found.
Therefore option D is correct.
14. An unfamiliar quantity \(Y\) is defined by
\[
Y=\frac{F}{\rho A},
\]
where \(F\) is force, \(\rho\) is density and \(A\) is area. If its numerical value is \(3.6\) when expressed in its SI base-unit form, which option gives both the correct SI unit and its numerical value when the length unit is changed from metre to centimetre while second remains unchanged?
ⓐ. \(\text{m s}^{-2}\); \(3.6\times10^2\)
ⓑ. \(\text{m}^2\text{s}^{-1}\); \(3.6\times10^4\)
ⓒ. \(\text{m}^2\text{s}^{-2}\); \(3.6\times10^4\)
ⓓ. \(\text{m}^2\text{s}^{-2}\); \(3.6\times10^{-4}\)
Correct Answer: \(\text{m}^2\text{s}^{-2}\); \(3.6\times10^4\)
Explanation: Begin by expressing every quantity in SI base units.
Force has unit
\[
\text{kg m s}^{-2}.
\]
Density has unit
\[
\text{kg m}^{-3},
\]
and area has unit
\[
\text{m}^2.
\]
Therefore the denominator has unit
\[
(\text{kg m}^{-3})(\text{m}^2)=\text{kg m}^{-1}.
\]
Hence,
\[
[\text{unit of }Y]
=\frac{\text{kg m s}^{-2}}{\text{kg m}^{-1}}.
\]
Cancelling kilogram and simplifying the powers of metre gives
\[
[\text{unit of }Y]=\text{m}^2\text{s}^{-2}.
\]
Now change the length unit from metre to centimetre.
Since
\[
1\,\text{m}=10^2\,\text{cm},
\]
we have
\[
1\,\text{m}^2=10^4\,\text{cm}^2.
\]
Thus
\[
3.6\,\text{m}^2\text{s}^{-2}
=3.6\times10^4\,\text{cm}^2\text{s}^{-2}.
\]
The physical quantity is unchanged, but the smaller squared length unit produces a larger numerical measure by \(10^4\).
Therefore option C is correct.
15. A quantity is defined by
\[
Z=\frac{It}{n},
\]
where \(I\) is electric current, \(t\) is time and \(n\) is amount of substance. For \(I=2.0\,\text{A}\), \(t=3.0\,\text{s}\) and \(n=0.50\,\text{mol}\), which statement is correct?
ⓐ. \(Z=12\,\text{A mol}^{-1}\), and its unit is a base SI unit.
ⓑ. \(Z=3.0\,\text{C mol}^{-1}\), and its unit is a base SI unit.
ⓒ. \(Z=12\,\text{mol C}^{-1}\), and its unit is derived.
ⓓ. \(Z=12\,\text{C mol}^{-1}\), and its unit is derived.
Correct Answer: \(Z=12\,\text{C mol}^{-1}\), and its unit is derived.
Explanation: The defining relation is
\[
Z=\frac{It}{n}.
\]
Substituting the numerical data,
\[
Z=\frac{(2.0\,\text{A})(3.0\,\text{s})}{0.50\,\text{mol}}.
\]
The numerator is
\[
(2.0)(3.0)=6.0\,\text{A s}.
\]
Therefore,
\[
Z=\frac{6.0}{0.50}\,\text{A s mol}^{-1}
=12\,\text{A s mol}^{-1}.
\]
Since
\[
1\,\text{C}=1\,\text{A s},
\]
the same unit may be written as
\[
\text{C mol}^{-1}.
\]
Hence
\[
Z=12\,\text{C mol}^{-1}.
\]
Ampere, second and mole are SI base units, but their algebraic combination is not itself a base unit.
Coulomb is a derived unit because it is equivalent to ampere-second.
Thus the complete unit \(\text{C mol}^{-1}\) is derived.
Therefore option D is correct.
16. A quantity \(P\) is defined by
\[
P=\frac{F\ell}{t}.
\]
In an MKS-based measurement its numerical value is \(6\), with \(F\) expressed in newton, \(\ell\) in metre and \(t\) in second. Another observer expresses the same quantity in CGS units. Use \(1\,\text{N}=10^5\,\text{dyn}\) and \(1\,\text{m}=10^2\,\text{cm}\). Which pair gives the CGS numerical measure and the special SI name of the unit of \(P\), respectively?
ⓐ. \(6\times10^7\); watt
ⓑ. \(6\times10^{-7}\); joule
ⓒ. \(6\times10^5\); pascal
ⓓ. \(6\times10^2\); watt
Correct Answer: \(6\times10^7\); watt
Explanation: The defining relation is
\[
P=\frac{F\ell}{t}.
\]
In SI,
\[
[\text{unit of }P]=\frac{\text{N m}}{\text{s}}.
\]
Since
\[
1\,\text{N m}=1\,\text{J},
\]
we obtain
\[
\frac{\text{N m}}{\text{s}}=\frac{\text{J}}{\text{s}}=\text{W}.
\]
Thus the special SI unit of \(P\) is watt.
Now convert one SI unit of \(P\) into the corresponding CGS compound unit.
Using
\[
1\,\text{N}=10^5\,\text{dyn}
\]
and
\[
1\,\text{m}=10^2\,\text{cm},
\]
while the second is unchanged,
\[
1\,\text{N m s}^{-1}
=(10^5)(10^2)\,\text{dyn cm s}^{-1}.
\]
Therefore,
\[
1\,\text{W}=10^7\,\text{dyn cm s}^{-1}.
\]
The physical quantity has numerical value \(6\) in SI units, so
\[
6\,\text{W}=6\times10^7\,\text{dyn cm s}^{-1}.
\]
The CGS unit is smaller, so the numerical measure correspondingly becomes much larger.
Hence the correct pair is \(6\times10^7\) and watt.
17. A derived quantity \(X\) is defined by
\[
X=\frac{Ft}{A},
\]
where \(F\) is force, \(t\) is time and \(A\) is area. Its numerical value is \(2.4\times10^4\) when expressed in the CGS base-unit form \(\text{g cm}^{-1}\text{s}^{-1}\). Which option gives its numerical value and an equivalent SI unit?
ⓐ. \(2.4\times10^5\,\text{Pa s}\)
ⓑ. \(2.4\times10^3\,\text{Pa s}\)
ⓒ. \(2.4\times10^{-3}\,\text{Pa s}\)
ⓓ. \(2.4\times10^4\,\text{Pa s}\)
Correct Answer: \(2.4\times10^3\,\text{Pa s}\)
Explanation: From the defining relation,
\[
X=\frac{Ft}{A}.
\]
In SI, the unit is
\[
\frac{\text{N s}}{\text{m}^2}.
\]
Since \(1\,\text{Pa}=1\,\text{N m}^{-2}\),
\[
\frac{\text{N s}}{\text{m}^2}=\text{Pa s}.
\]
Now convert the supplied CGS base-unit form:
\[
1\,\text{g cm}^{-1}\text{s}^{-1}
=(10^{-3}\,\text{kg})(10^{-2}\,\text{m})^{-1}\text{s}^{-1}.
\]
The inverse centimetre factor is
\[
(10^{-2}\,\text{m})^{-1}=10^2\,\text{m}^{-1}.
\]
Therefore,
\[
1\,\text{g cm}^{-1}\text{s}^{-1}
=10^{-1}\,\text{kg m}^{-1}\text{s}^{-1}.
\]
Also,
\[
\text{Pa s}
=(\text{kg m}^{-1}\text{s}^{-2})\text{s}
=\text{kg m}^{-1}\text{s}^{-1}.
\]
Hence one CGS unit in the stem equals \(0.1\,\text{Pa s}\), so
\[
X=(2.4\times10^4)(10^{-1})\,\text{Pa s}
=2.4\times10^3\,\text{Pa s}.
\]
The negative power of centimetre is the important conversion step; treating it as a positive power would reverse the factor.
Thus option B is correct.
18. Consider the following statements about SI quantities and units.
Statement I: Amount of substance is an SI base quantity and mole is its base unit.
Statement II: Luminous intensity is an SI base quantity and candela is its base unit.
Statement III: Electric charge is an SI base quantity and coulomb is its base unit.
Statement IV: Thermodynamic temperature is an SI base quantity and kelvin is its base unit.
Which combination is correct?
ⓐ. Statements I and II only
ⓑ. Statements II and III only
ⓒ. Statements I, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: The SI framework contains seven base quantities, from which derived quantities and their units are constructed.
Amount of substance is one of these base quantities, and its SI base unit is mole \((\text{mol})\), so Statement I is correct.
Luminous intensity is also a base quantity, with candela \((\text{cd})\) as its SI base unit, so Statement II is correct.
Electric current, not electric charge, is the SI base quantity in the electrical category.
Its base unit is ampere \((\text{A})\).
Electric charge is derived from current and time through
\[
q=It.
\]
Therefore its SI unit is
\[
\text{A s}=\text{C},
\]
so coulomb is a derived unit and Statement III is incorrect.
Thermodynamic temperature is a base quantity with kelvin \((\text{K})\) as its SI base unit, so Statement IV is correct.
Thus Statements I, II and IV are the correct set.
19. Assertion: A standard unit should be reproducible in different laboratories rather than depend on an arbitrary local object available to only one observer.
Reason: SI units are internationally standardized, and modern SI definitions are connected with fixed physical constants and reproducible standards.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: A useful measurement system requires different observers to be able to reproduce the same unit consistently.
If the size of a unit depended on an arbitrary local object, two laboratories could effectively use different standards while attaching the same unit name to them.
That would make quantitative comparison unreliable.
International standardization is intended to prevent such observer-dependent variation.
Modern SI definitions are associated with fixed physical constants and reproducible procedures rather than an independently chosen local reference for every laboratory.
This allows the same unit to have the same physical magnitude wherever a measurement is performed.
The point is not that every student must memorize the numerical value of every defining constant.
The essential idea here is reproducibility and common standardization.
Thus the Reason gives the physical basis for the Assertion.
Therefore both are true and the Reason correctly explains the Assertion.
20. A quantity \(R\) is defined as
\[
R=\frac{E}{At},
\]
where \(E\) is energy, \(A\) is area and \(t\) is time. Initially \(R=0.060\,\text{W cm}^{-2}\). If \(E\) and \(t\) remain unchanged while the area is doubled, what is the new value of \(R\) in \(\text{kW m}^{-2}\)?
ⓐ. \(0.12\,\text{kW m}^{-2}\)
ⓑ. \(0.60\,\text{kW m}^{-2}\)
ⓒ. \(0.30\,\text{kW m}^{-2}\)
ⓓ. \(3.0\,\text{kW m}^{-2}\)
Correct Answer: \(0.30\,\text{kW m}^{-2}\)
Explanation: The defining relation is
\[
R=\frac{E}{At}.
\]
With \(E\) and \(t\) fixed,
\[
R\propto\frac{1}{A}.
\]
Doubling the area therefore halves the physical value of \(R\):
\[
R_{\text{new}}=\frac{0.060}{2}\,\text{W cm}^{-2}
=0.030\,\text{W cm}^{-2}.
\]
Now convert the area unit correctly.
Since
\[
1\,\text{cm}^2=10^{-4}\,\text{m}^2,
\]
we have
\[
1\,\text{W cm}^{-2}
=\frac{1\,\text{W}}{10^{-4}\,\text{m}^2}
=10^4\,\text{W m}^{-2}.
\]
Therefore,
\[
0.030\,\text{W cm}^{-2}
=0.030\times10^4\,\text{W m}^{-2}
=300\,\text{W m}^{-2}.
\]
Using \(1\,\text{kW}=10^3\,\text{W}\),
\[
300\,\text{W m}^{-2}=0.30\,\text{kW m}^{-2}.
\]
The solution requires first applying the physical changed condition and only then converting the compound unit.
Hence option C is correct.
