NCERT Exemplar + Extra MCQs

Hint

(b) length, time and velocity.
Explanation:
A fundamental quantity is a physical quantity that cannot be derived from other quantities. Velocity is defined as the distance traveled over time ( \(\mathrm{V}=\mathrm{L} / \mathrm{T}\) ), meaning it is derived from length and time. Since velocity can be calculated from other fundamental quantities, it cannot be considered a fundamental quantity itself.

Hint

Surface area \(=2 \pi r h+2 \pi r^{2}=2 \pi r(h+r)\) \(=2 \times 22 / 7 \times 2 \times 10(10 \times 10+2 \times 10) \mathrm{mm}^{2}=1.5 \times 10^{4} \mathrm{~mm}^{2}\)

Hint

Density \(=11.3 \mathrm{~g} \mathrm{~cm}^{-3}\)
\(
\begin{aligned}
&=11.3 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\left[1 \mathrm{~kg}=10^{3} \mathrm{~g}, 1 \mathrm{~m}=10^{2} \mathrm{~cm}\right] \\
&=11.3 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-4}
\end{aligned}
\)

Hint

\(1 \mathrm{~m} \quad=\frac{1}{9.46 \times 10^{5}} \mathrm{ly} \approx \frac{1}{10^{16}} \mathrm{ly}=10^{-16} \mathrm{ly}\)

Hint

\(G\)
\(
\begin{aligned}
&=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}=6.67 \times 10^{-11} \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}} \mathrm{~m}^{2} \mathrm{~kg}^{-2} \\
&=6.67 \times 10^{-11} \mathrm{~kg}^{-1} \mathrm{~m}^{3} \mathrm{~s}^{-2} \\
&=6.67 \times 10^{-11} \frac{\mathrm{m}^{3}}{\mathrm{~kg} \mathrm{~s}^{2}}=\frac{6.67 \times 10^{-11} \times\left(10^{2}\right)^{3}}{\left(10^{3}\right)^{2}} \\
&=6.67 \times 10^{-8} \mathrm{~cm}^{-3} \mathrm{~s}^{-2} \mathrm{~g}^{-1} .
\end{aligned}
\)

Hint

Answer: Distance between Sun and Earth
= Speed of light in vacuum x time taken by light to travel from Sun to Earth \(=3 \times 10^{8} \mathrm{~m} / \mathrm{s} \times 8 \mathrm{~min} 20 \mathrm{~s}=3 \times\) \(10^{8} \mathrm{~m} / \mathrm{s} \times 500 \mathrm{~s}=500 \times 3 \times 10^{8} \mathrm{~m}\).
In the new system, the speed of light in vacuum is unity. So, the new unit of length is \(3 \times 10^{8} \mathrm{~m}\).
\(\therefore\) distance between Sun and Earth =500 new units.

Hint

(a) Least count of vernier callipers \(=1 / 20=0.05 \mathrm{~mm}=5 \times 10^{-5} \mathrm{~m}\)
(b) Least count of screw gauge =Pitch/No. of divisions on circular scale \(=1 \times 10^{-3} / 100=1 \times 10^{-5} \mathrm{~m}\)
(c) Least count of optical instrument \(=6000 \mathrm{~A}\) (average wavelength of visible light as \(6000 \mathrm{~A})=6 \times 10^{-7} \mathrm{~m}\). As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.

Hint

As magnification, \(m\) =thickness of image of hair/ real thickness of hair \(=100\) and average width of the image of hair as seen by microscope \(=3.5 \mathrm{~mm}\)
\(\therefore\) Thickness of hair \(=3.5 \mathrm{~mm} / 100=0.035 \mathrm{~mm}\)

Hint

Answer: Here area of the house on slide \(=1.75 \mathrm{~cm}^{2}=1.75 \times 10^{-4} \mathrm{~m}^{2}\) and area of the house of projector-
screen \(=1.55 \mathrm{~m}^{2}\)
\(\therefore\) Areal magnification =Area on screen/Area on slide \(=1.55 \mathrm{~m}^{2} / 1.75 \times 10^{-4} \mathrm{~m}^{2}=8.857 \times 10^{3}\)
\(\therefore\) Linear magnification
\(=\sqrt{\text { Areal magnification }}\)
\(
=\sqrt{(8.857) \times 10^{3}}
\)
\(
=94.1 \text {. }
\)

Hint

(a) \(y=a \sin \frac{2 \pi t}{T}\)
Here, \([L.H.S.]=[y]=[L]\)
and \([R.H.S.]=\left[a \sin \frac{2 \pi t}{T}\right]=\left[L \sin \frac{T}{T}\right]=[L]\)
So, it is correct.
(b) \(y=a \sin v t\)
Here, \([y]=[L]\) and \([a \sin v t]=\left[L \sin \left(L T^{-1} . T\right)\right]=[L \sin L]\) So, the equation is wrong.
(c) \(y=\left(\frac{a}{T}\right) \sin \frac{t}{a}\)
Here, \(\quad[y]=[L]\) and \(\left[\left(\frac{a}{T}\right) \sin \frac{t}{a}\right]=\left[\frac{L}{T} \sin \frac{T}{L}\right]=\left[L T^{-1} \sin T L^{-1}\right]\)
So, the equation is wrong.
(d) \(y=(a \sqrt{2})\left(\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right)\)
Here, \(\quad[y]=[L],[a \sqrt{2}]=[L]\)
and \(\left[\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right]=\left[\sin \frac{T}{T}+\cos \frac{T}{T}\right]=\) dimensionless. So, it is correct.

Hint

Volume of one hydrogen atom = \(\frac{4}{3} \pi r^{3}\) (volume of a sphere)
\(
=4 / 3 \times 3.14 \times\left(0.5 \times 10^{-10}\right) \mathrm{m}^{3}=5.23 \times 10^{-31} \mathrm{~m}^{3}
\)
According to Avagadro’s hypothesis, one mole of hydrogen contains \(6.023 \times 10^{23}\) atoms.
The atomic volume of 1 mole of hydrogen atoms
\(
=6.023 \times 1023 \times 5.23 \times 10^{-31}=3.15 \times 10^{-7} \mathrm{~m}^{3} \text {. }
\)

Hint

As Area \(=(4.234 \times 1.005) \times 2=8.51034=8.5 \mathrm{~m}^{2}\)
\(
\text { Volume }=(4.234 \times 1.005) \times\left(2.01 \times 10^{-2}\right)=8.55289 \times 10^{-2}=0.0855 \mathrm{~m}^{3} \text {. }
\)

Hint

As \(P=\frac{a^{3} b^{2}}{(\sqrt{c} d)}=a^{3} b^{2} c^{-1 / 2} d^{-1}\)
\(\therefore\) Maximum fractional error in the measurement
\(
\frac{\Delta P}{P}=3 \frac{\Delta a}{a}+2 \frac{\Delta b}{b}+\frac{1}{2} \frac{\Delta c}{c}+\frac{\Delta d}{d}
\)
As \(\quad \frac{\Delta a}{a}=1 \%, \quad \frac{\Delta b}{b}=3 \%, \quad \frac{\Delta c}{c}=4 \%\) and \(\frac{\Delta d}{d}=2 \%\)
\(\therefore\) Maximum fractional error in the measurement
\(
\begin{aligned}
\frac{\Delta P}{P} &=3 \times 1 \%+2 \times 3 \%+\frac{1}{2} \times 4 \%+2 \% \\
&=3 \%+6 \%+2 \%+2 \%=13 \%
\end{aligned}
\)
If \(P=3.763\), then \(\Delta P=13 \%\) of \(P\)
\(
=\frac{13 P}{100}=\frac{13 \times 3.763}{100}=0.489
\)
As the error lies in the first decimal place, the answer should be rounded off to the first decimal place. Hence, we shall express the value of \(P\) after rounding it off as \(P=3.8\).

Hint

Answer: Total time \(=100\) years \(=100 \times 365 \times 24 \times 60 \times 60 \mathrm{~s}\)
Error in 1 second \(=0.02 / 100 \times 365 \times 24 \times 60 \times 60\) \(=6.34 \times 10^{-12} \mathrm{~s}\)
\(\therefore\) Accuracy of 1 part in \(10^{11}\) to \(10^{12}\).

Hint

Key concept: 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.
1. All non-zero digits are significant.
2. A zero becomes significant figure if it appears between two non-zero digits.
3. Leading zeros or the zeros placed to the left of the number are never significant.
4. Trailing zeros or the zeros placed to the right of the number are significant.
5. In exponential notation, the numerical portion gives the number of significant figures.
Leading zeros or the zeros placed to the left of the number are never

Hint

(b) 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.

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

Hint

(c) 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
\(
\text { Density }=\frac{4.237 \mathrm{~g}}{2.5 \mathrm{~cm}^3}=1.6948
\)
After rounding off the number, we get density \(=1.7\)

Hint

Key concept: While rounding off measurements, we use the following rules by convention:
1. If the digit to be dropped is less than 5 , then the preceding digit is left unchanged.
2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.
3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one.
4. If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is left unchanged if it is even.
5. If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd.
Units and Measurements
Let us round off 2.745 to 3 significant figures.
Here the digit to be dropped is 5 , then preceding digit is left unchanged, if it is even.
Hence on rounding off 2.745, it would be 2.74.
Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

Hint

Key concept: Error in product of quantities: Suppose \(x=a \times b\)
Let \(\Delta a=\) absolute error in measurement of \(a\),
\(\Delta b=\) absolute error in measurement of \(b\),
\(\Delta x=\) absolute error in calculation of \(x\), i.e. product of \(a\) and \(b\).
The maximum fractional error in \(x\) is \(\frac{\Delta x}{x}= \pm\left(\frac{\Delta a}{a}+\frac{\Delta b}{b}\right)\)
Percentage error in the value of \(x=(\) Percentage error in value of \(a)+\)
(Percentage error in value of \(b\) )
According to the problem, length \(l=(16.2 \pm 0.1) \mathrm{cm}\)
Breadth \(b=(10.1 \pm 0.1) \mathrm{cm}\)
Area \(A=l \times b=(16.2 \mathrm{~cm}) \times(10.1 \mathrm{~cm})=163.62 \mathrm{~cm}^2\)
As per the rule area will have only three significant figures and error will have only one significant figure. Rounding off we get, area \(A=164 \mathrm{~cm}^2\)
If \(\Delta A\) is error in the area, then relative error is calculated as \(\frac{\Delta A}{A}\).
\(
\begin{aligned}
\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b} & =\frac{0.1 \mathrm{~cm}}{16.2 \mathrm{~cm}}+\frac{0.1 \mathrm{~cm}}{10.1 \mathrm{~cm}} \\
& =\frac{1.01+1.62}{16.2 \times 10.1}=\frac{2.63}{163.62} \\
\Rightarrow \Delta A & =A \times \frac{2.63}{163.62} \mathrm{~cm}^2=163.62 \times \frac{2.63}{163.62}=2.63 \mathrm{~cm}^2
\end{aligned}
\)
\(\Delta A=3 \mathrm{~cm}^2\) (By rounding off to one significant figure)
Area, \(A=A \pm \Delta A=(164 \pm 3) \mathrm{cm}^2\)

Hint

\(
\begin{aligned}
& \text { Work }=F \times \Delta x=\left[M L T^{-2}\right][L]=\left[M L^2 T^{-2}\right] \\
& \text { Torque = force } \times \text { distance }=\left[M L^2 T^{-2}\right] \\
& \text { Angular momentum }=m v r=[M]\left[L T^1\right][L]=\left[M L^2 T^{-1}\right] \\
& \text { Planck’s constant }=\frac{E}{V}=\frac{\left[M L^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^2 T^{-1}\right]
\end{aligned}
\)
(b) Angular momentum \(=\operatorname{mvr}=[M]\left[L T^1\right][L]=\left[M L^2 T^{-1}\right]\)
Planck’s constant \(=\frac{E}{V}=\frac{\left[M L^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^2 T^{-1}\right]\)
(c) Tension (force) \(=\left[M L T^{-2}\right]\)
Surface tension \(=\frac{\text { force }}{\text { length }}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right]\)
(d) Impulse \(=F \times \Delta t=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right]\)
Momentum \(=\) mass \(\times\) velocity \(=[M]\left[L T^{-1}\right]=\left[M L T^{-1}\right]\)
So, among the above pairs only tension and surface tension does not have same dimensional formula. They both sound similar but they both have different meaning and different applications.

Hint

(a) According to the problem,
\(
\begin{aligned}
A & =2.5 \mathrm{~ms}^{-1} \pm 0.5 \mathrm{~ms}^{-1}, B=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s} \\
Z & =A B=(2.5)(0.10)=0.25 \mathrm{~m} \\
\frac{\Delta Z}{Z} & =\frac{\Delta A}{A}+\frac{\Delta B}{B}, \Delta Z=Z\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right) \\
& =0.25\left(\frac{0.5}{2.5}+\frac{0.01}{0.10}\right)=0.25(0.2+0.1)=0.075 \\
\Delta Z & =0.075=0.08 \mathrm{~m} \quad \text { (rounding off to two significant figures.) }
\end{aligned}
\)
Thus, measured value of \(A B\), i.e., \(=Z \pm \Delta Z=(0.25 \pm 0.08) \mathrm{m}\)

Hint

(d) According to the problem, \(A=1.0 \mathrm{~m} \pm 0.2 \mathrm{~m}, B=2.0 \mathrm{~m} \pm 0.2 \mathrm{~m}\)
Let, \(Z=\sqrt{A B}=\sqrt{(1.0)(2.0)}=1.414 \mathrm{~m}\)
Rounding off to two significant digits \(Z=1.4 \mathrm{~m}\)
As
\(
\begin{aligned}
\frac{\Delta Z}{Z} & =\frac{1}{2} \frac{\Delta A}{A}+\frac{1}{2} \frac{\Delta B}{B} \\
& =\frac{1}{2}\left(\frac{0.2 \mathrm{~m}}{1 \mathrm{~m}}\right)+\frac{1}{2}\left(\frac{0.2 \mathrm{~m}}{2 \mathrm{~m}}\right)=0.15
\end{aligned}
\)
\(
\Rightarrow \Delta Z=Z(0.15)=1.4 \mathrm{~m}(0.15)=0.212
\)
Rounding off to one significant digit, \(\Delta Z=0.2 \mathrm{~m}\)
The correct value for \(\sqrt{A B}=1.4 \pm 0.2 \mathrm{~m}\).

Hint

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

Hint

Key concept: 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. Example: In target shooting, a high score indicates the nearness to the bull’s eye and is a measure of the shooter’s accuracy.

According to the problem, length \(l=5 \mathrm{~cm}\)
Let us first check the errors in each values by picking options one by one, we get
\(
\begin{array}{ll}
\Delta l_1=5-4.9=0.1 \mathrm{~cm}, & \Delta l_2=5-4.805=0.195 \mathrm{~cm}, \\
\Delta l_3=5.25-5=0.25 \mathrm{~cm} \text { and } & \Delta l_4=5.4-5=0.4 \mathrm{~cm}
\end{array}
\)
Error \(\Delta l_1\) is the least.
Hence \(4.9 \mathrm{~cm}\) is most closer to true value. So, 4.9 is more accurate.

Hint

(c) According to the problem,
Young’s modulus, \(Y=1.9 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
\(1 \mathrm{~N}\) in SI system of units \(=10^5\) dyne in C.G.S system.
Hence, \(\quad Y=1.9 \times 10^{11} \times 10^5\) dyne \(/ \mathrm{m}^2\)
In C.G.S. length is measured in unit ‘ \(\mathrm{cm}\) ‘, so we should also convert \(\mathrm{m}\) into \(\mathrm{cm}\).
\(
\begin{aligned}
\therefore \quad Y & =1.9 \times 10^{11}\left(\frac{10^5 \text { dyne }}{10^4 \mathrm{~cm}^2}\right) \quad[\because 1 \mathrm{~m}=100 \mathrm{~cm}] \\
& =1.9 \times 10^{12} \text { dyne } / \mathrm{cm}^2
\end{aligned}
\)

Hint

(d) According to the problem, fundamental quantities are momentum \((p)\), area \((A)\), and time \((T)\) and we have to express energy in these fundamental quantities.
Let energy \(E\),
\(
E \propto p^a A^A T^c \Rightarrow E=k p^a A^A T^c
\)
where, \(k\) is dimensionless constant of proportionality.
Dimensional formula of energy, \([E]=\left[M L^2 T^{-2}\right]\) and \([p]=\left[M L T^{-1}\right]\)
\(
[A]=\left[L^2\right],[T]=[T] \text { and }[E]=[K][p]^a[A]^b[T]^c
\)
Putting all the dimensions, we get
\(
\begin{aligned}
M L^2 T^{-2} & =\left[M L T^{-1}\right]^a\left[L^2\right]^b[T]^c \\
& =M^a L^{a+2 b} T^{-a+c}
\end{aligned}
\)
According to the principle of homogeneity of dimensions, we get
\(
\begin{aligned}
& a=1 \dots(i) \\
& a+2 b=2 \dots(ii) \\
& -a+c=-2 \dots(iii)
\end{aligned}
\)
By solving these equations (i), (ii) and (iii), we get
\(
a=1, b=\frac{1}{2}, c=-1
\)
Dimensional formula for \(E\) is \(\left[p^1 A^{1 / 2} t^{-1}\right]\).

Hint

(b, 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.
\(
\begin{aligned}
& {[Y]=[L],[a]=[L]} \\
& {\left[\frac{2 \pi t}{T}\right]=\frac{[T]}{[T]}=\left[T^0\right]} \\
& {[v t]=[v][t]=\left[L T^{-1}\right][T]=[L]} \\
& {\left[\frac{a}{T}\right]=\frac{[a]}{[T]}=\frac{[L]}{[T]}=\left[L T^{-1}\right]} \\
& {\left[\frac{t}{a}\right]=\left[L^{-1} T\right]} \\
& {[\text { LHS }] \neq[\text { RHS }]}
\end{aligned}
\)
Hence, (c) is not the correct option. \(\Rightarrow\) LHS \(\neq\) RHS.
So, option (b) is also not correct.

Hint

\((a, e)\)
Key concept: 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) \(P R\) and \(Q^2\) may have same dimensions as those of \(R\).
Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Hint

(b, d) We know that energy of radiation, \(E=h v\). So, we have to compare \(h\) with dimensional formula of each option.
\(
\begin{aligned}
{[h] } & =\frac{[E]}{[v]}=\frac{\text { force } \times \text { displacement }}{\text { frequency }} \\
& =\frac{\left[M L^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^2 T^{-1}\right]
\end{aligned}
\)
(a) Dimension of linear impulse
\(
[I]=[F t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right]
\)
where, \(t\) is the time interval.
(b) Dimension of angular impulse
\(
[J]=[I \omega]=\left[M L^2\right]\left[T^{-1}\right]=\left[M L^2 T^{-1}\right]
\)
(c) Dimensión of linear momentum
\(
[P]=[m v]=[M]\left[L T^{-1}\right]=\left[M L T^{-1}\right]
\)
(d) Dimension of angular momentum
\(
[L]=[m v r]=[M]\left[L T^{-1}\right][L]=\left[M L^2 T^{-1}\right]
\)
Hence, the dimension of angular impulse and angular momentum is the same as Planck’s constant \((h)\).

Hint

(a, b, d) We know that dimension of \(h\)
\(
[h]=\frac{\left[M L^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^2 T^{-1}\right],[G]=\left[M^{-1} L^3 T^{-2}\right],[c]=\left[L T^{-1}\right]
\)
For mass, let \(m=c^a h^b G^c\) where \(a, b\) and \(c\) are exponents of \(c, h\) and \(G\) respectively.
\(
\Rightarrow \quad\left[M L^0 T^0\right]=\left[L T^{-1}\right]^a\left[M L^2 T^{-1}\right]^b\left[M^{-1} L^3 T^{-2}\right]^c
\)
Using the principle of homogeneity of dimensions,
\(
\begin{aligned}
& b-c=1 \dots(i) \\
& a+2 b+3 c=0 \dots(ii) \\
& -a-b-2 c=0 \dots(iii)
\end{aligned}
\)
Adding eqns. (i), (ii) and (iii),
\(
2 b=1 \Rightarrow b=1 / 2
\)
From eqn. (i), \(a=b-1=\frac{1}{2}-1=-\frac{1}{2}\)
From eqn. (iii), \(a=-(b+2 c)=-\left(\frac{1}{2}-1\right)=\frac{1}{2}\)
Hence, \(m=c^{1 / 2} h^{1 / 2} G^{-1 / 2} \Rightarrow m=\sqrt{\frac{c h}{G}}\)
Similarly, for length, \(L=\sqrt{\frac{h G}{c^3}}\) and for time, \(T=\sqrt{\frac{h G}{c^5}}\)
\(\therefore\) Planck’s length \(=\sqrt{\frac{G h}{c^3}} ;\) Planck’s time \(=\sqrt{\frac{G h}{c^5}}\);
and Planck’s mass \(=\sqrt{\frac{c h}{G}}\)
Mass can be expressed by \(m_e\) and \(m_p\).
Hence, (a), (b) or (d) any can be used to express \(L, M\) and \(T\) in terms of three chosen fundamental quantities.

Hint

\((a, b)\) 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
(a) \(\frac{\text { Force }}{\text { Area }}=\frac{\left[M L T^{-2}\right]}{\left[L^2\right]}=\left[M L^{-1} T^{-2}\right]\)
So, this ratio express pressure (In fact this ratio actually represents pressure).
(b) \(\frac{\text { Energy }}{\text { Area }}=\frac{\left[M L^2 T^{-2}\right]}{\left[L^2\right]}=\left[M T^{-2}\right]\)
Dimensions of this ratio are not same as pressure, so this ratio does not express pressure.
(c) \(\frac{\text { Energy }}{\text { Volume }}=\frac{\left[M L^2 T^{-2}\right]}{\left[L^3\right]}=\left[M L^{-1} T^{-2}\right]\)
Dimensions of this ratio is the same as pressure, so this ratio also express pressure.
(d) \(\frac{\text { Force }}{\text { Volume }}=\frac{\left[M L T^{-2}\right]}{\left[L^3\right]}=\left[M L^{-2} T^{-2}\right]\)
Dimensions of this ratio are not same as pressure, so this ratio does not express pressure.

Hint

(b, d) 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.
Important point: 1 light year (distance that light travels in 1 year with speed \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\).) \(=9.46 \times 10^{11} \mathrm{~m}\) And 1 par see \(=3.08 \times 10^{16} \mathrm{~m}\)

Hint

(b) kelvin. In the International System of Units (SI), the standard unit of temperature is the kelvin (K). While Celsius and Fahrenheit are also used to measure temperature, they are not the SI base unit.

Hint

(d) 

\(
\begin{aligned}
& \text { watt }=\text { joule } / \text { second }=\text { ampere } \times \text { volt } \\
& =\text { ampere }^2 \times \text { ohm }
\end{aligned}
\)

Hint

(c) Since, \((m v r)=n \cdot \frac{h}{2 \pi}\) and \(E=h \nu\)
So, unit of \(h=\) joule second = angular momentum

Hint

(c) Thermal capacity and Boltzmann constant.
Explanation:
Thermal capacity: This is measured in units of energy per temperature (J/K).
Boltzmann constant: This is also measured in units of energy per temperature \((\mathrm{J} / \mathrm{K})\).
\(
\begin{aligned}
& \text { Thermal capacity }==\frac{\Delta Q}{\Delta t}=\frac{M L^2 T^{-2}}{K}=M L^2 T^{-2} K^{-1} \\
& \text { Boltzmaan constant }==\frac{P V}{N T}=\frac{M L^2 T^{-2}}{K}=M L^2 T^{-2} K^{-1}
\end{aligned}
\)

Hint

(c) Surface tension is force per unit length: Surface Tension \(=\frac{\text { Force }}{\text { Length }}\).
\(
[\text { Surface Tension }]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]} .
\)
Simplify the expression: \([\) Surface Tension \(]=\left[\mathrm{ML}^{1-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]\).
This simplifies to \(\left[\mathrm{MT}^{-2}\right]\).

Hint

(b) Impulse \(=\) Force \(\times\) Time.
Therefore, the dimensional formula of impulse
\(
\begin{aligned}
& =\text { Dimensional formula of force } \times \text { Dimensional formula of time } \\
& =\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right]
\end{aligned}
\)
and dimensional formula of linear momentum \([p]=M L T^{-1}\).

Hint

(c) The dimensional formula for pressure is \(\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]\).
Stress: is defined as force divided by area: Stress = Force / Area (\(\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]\)).
Bulk modulus \(=\) Bulk Stress \(\times[\text { Bulk strain }]^{-1}\)
(\(
\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]^{-1}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]
\))
\(
\begin{aligned}
& \text { Energy Density }=\text { Energy } \times[\text { Volume }]^{-1} \\
& =\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right] \times\left[\mathrm{M}^0 \mathrm{~L}^3 \mathrm{~T}^0\right]^{-1}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]
\end{aligned}
\)
\(
\text { Thrust }=\text { Mass } \times \text { Acceleration }=[M] \times[L]\left[T^{-2}\right]=\left[M L T^{-2}\right]
\)

Hint

(c)
\(
\begin{aligned}
& F=G \frac{m_1 m_2}{r^2} \\
& G=\frac{F r^2}{m_1 m_2}
\end{aligned}
\)
\(
\begin{aligned}
& G=\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]^2}{[\mathrm{M}][\mathrm{M}]} \\
& G=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}
\end{aligned}
\)
\(
\begin{aligned}
& G=\left[\mathrm{M}^{1-2} \mathrm{~L}^{1+2} \mathrm{~T}^{-2}\right] \\
& G=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

(c)

\(
\text { Force }=\mathrm{MLT}^{-2}, \text { Linear momentum }=\mathrm{MLT}^{-1}, \text { Torque }=\mathrm{ML}^2 \mathrm{~T}^{-2}, \text { Potential energy }=\mathrm{ML}^2 \mathrm{~T}^{-2}
\)

Hint

(d) (d) \(\eta=\frac{F}{6 \pi a v}\)
\(
\text { Substitute the dimensions: }[\eta]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left|\mathrm{LT}^{-1}\right|} \text {. }
\)
\(
[\eta]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] .
\)

Hint

\(
\begin{aligned}
&\text { (a) Since } R=\frac{\rho l}{A} \text {, where } \rho \text { is specific resistance. }\\
&\begin{array}{ll}
\therefore & {[\rho]=\left[\frac{R A}{l}\right], R=\frac{V}{i}, V=\frac{W}{Q}} \\
& {[\rho]=\left[\mathrm{ML}^3 \mathrm{~T}^{-1} \mathrm{Q}^{-2}\right]}
\end{array}
\end{aligned}
\)

Hint

(b) Use the relation \(E=h \nu\).
Rearrange to find \(h=\frac{E}{\nu}\).
Substitute the dimensions: \(h=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}\).
Simplify: \(h=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\).
Use the definition \(L=p r\), where \(p\) is linear momentum and \(r\) is distance.
Substitute the dimensions: \(L=\left[\mathrm{MLT}^{-1}\right] \times[\mathrm{L}]\).
Simplify: \(L=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\).

Hint

(b) The dimensions of electric field \(E\) are \(\left[\mathrm{MLT}^{-3} \mathrm{I}^{-1}\right]\).
The dimensions of permittivity of free space \(\varepsilon_0\) are \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\).
\(
\left[\frac{1}{2} \varepsilon_0 E^2\right]=\left[\varepsilon_0\right]\left[E^2\right]
\)
\(
\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-6} \mathrm{I}^{-2}\right]
\)
\(
\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2} \mathrm{I}^0\right]
\)
\(
\text { The dimensions of } \frac{1}{2} \varepsilon_0 E^2 \text { are }\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \text {. }
\)

Hint

(c) Since, units of length, velocity and force are doubled.
Hence, \([\mathrm{m}]=\frac{[\text { force }][\text { time }]}{[\text { velocity }]},[\) time \(]=\frac{[\text { length }]}{[\text { velocity }]}\)
Hence, unit of mass and time remains same.
Momentum is doubled.

Hint

(d) The equation is \(y=a \cos \left(\frac{t}{p}-q x\right)\). \(t\) represents time. \(x\) represents distance.
The argument of a trigonometric function (like cosine) must be dimensionless.
The term \(\frac{t}{p}\) must be dimensionless.
This implies that the unit of \(t\) must be the same as the unit of \(p\). \(\operatorname{Unit}(t)=\operatorname{Unit}(p)\).
The statement that the unit of \(t\) is the same as that of \(p\) is true.

Hint

(b) Given, \(p=\frac{a-t^2}{b x}\), where \(p\) is the pressure and \(t\) is the time
\(
[p b x]=[a]=\left[t^2\right]
\)
Hence, \([b]=\frac{\left[t^2\right]}{[p x]}\)
Dimensions of \(\frac{a}{b}=[p x]=\left[\mathrm{MT}^{-2}\right]\)

Hint

(c) \(\frac{x}{v}\) have the same dimension as of \(k\).
\(\therefore\) Dimensional formula of \([k]=\frac{[x]}{[v]}=\frac{[\mathrm{L}]}{\left[\mathrm{LT}^{-1}\right]}=[\mathrm{T}]\)

Hint

\(
\text { (a) Muscle } \times \text { Speed }=\text { Power }
\)
\(
\begin{aligned}
\text { Muscle }=\frac{\text { Power }}{\text { Speed }} & =\frac{\text { Work }}{\text { Time } \times \text { Speed }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{T}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{MLT}^{-2}\right] \\
& =\text { Mass } \times \text { Acceleration }=\text { Force }
\end{aligned}
\)
\(
\text { Now, } \begin{aligned}
\frac{\text { SI unit of force }}{\text { CGS unit of force }} & =\frac{\mathrm{kg} \times \mathrm{m} \times \mathrm{s}^{-2}}{\mathrm{~g} \times \mathrm{cm} \times \mathrm{s}^{-2}} \\
& =10^3 \times 10^2=10^5
\end{aligned}
\)

Hint

(b) Since, \(p^x Q^y c^z\) is dimensionless. Therefore,
\(
\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^x\left[\mathrm{MT}^{-3} y^y\left[\mathrm{LT}^{-1}\right]^z=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]\right.
\)
Only option (b) satisfies this expression.
So, \(\quad x=1, y=-1, z=1\)

Hint

(d) \(m \propto v^a \rho^b g^c\). Writing the dimensions on both sides,
\(
\begin{aligned}
{[\mathrm{M}] } & =\left[\mathrm{LT}^{-1}\right]^a\left[\mathrm{ML}^{-3}\right]^b\left[\mathrm{LT}^{-2}\right]^c \\
{[\mathrm{M}] } & =\left[\mathrm{M}^b \mathrm{~L}^{a-3 b+c} \mathrm{~T}^{-a-2 c}\right] \\
\therefore \quad b & =1 \\
a-3 b+c & =0 \\
-a-2 c & =0
\end{aligned}
\)
Solving these, we get
\(
a=6
\)
Hence,
\(
m \propto v^6
\)

Hint

(b) Determine the significant figures by analyzing the digits in the coefficient part of the number.
The coefficient of the number \(0.0310 \times 10^3\) is 0.0310 .
The leading zeros in 0.0310 (the first two zeros) are not significant.
The digits ‘ 3 ‘ and ‘ 1 ‘ are non-zero, so they are significant.
The trailing zero ‘ 0 ‘ after the decimal point and after a non-zero digit is significant.
Therefore, the significant figures are ‘ 3 ‘, ‘ 1 ‘, and ‘ 0 ‘.
The total number of significant figures is 3.

Trick: For Decimal Numbers
Go from 1st non-zero digit to the last. Here first non-zero is 3 and last digit is 0 (So total is 3).

Hint

(c) Coefficient is 11.118. As it is a decimal number, count first non-zero digit to the last digit. Here first non-zero digit is 1 and the last digit is 8. Therefore total count is 5.

Hint

(b) Only in 20.2, all zeros are significant because in this, zero lies between two non-zero digits.

Hint

(c)
\(
\begin{aligned}
&\text { Add the numbers: } 3.20+4.80 \text {. }\\
&3.20+4.80=8.00
\end{aligned}
\)
The number 8.00 has three significant figures.
The trailing zeros after the decimal point are significant.

Hint

(a) \(7.26-0.2=7.06 \mathrm{~J}\)
7.26 has 2 decimal places.
0.2 has 1 decimal place.
The result must be rounded to 1 decimal place, matching the least precise number (0.2).
Round 7.06 to 1 decimal place.
Since the second decimal digit (6) is greater than 5, round up the first decimal digit (0).
The rounded value is 7.1.

Note: 

Addition or subtraction rules:
Suppose in the measured values to be added or subtracted, the least number of significant digits after the decimal is \(n\). Then, in the sum or difference also, the number of significant digits after the decimal should be \(n\). e.g. \(1.2+3.45+6.789=11.439 \approx 11.4\)
Here, the least number of significant digits after the decimal is one. Hence, the result will be 11.4 (when rounded off to smallest number of decimal places).
Similarly, e.g. \(1263-10.2=2.43 \approx 2.4\)

Hint

(d) \(107.88 \times 0.610=65.8068\).

Out of 107.88 and 0.610, the least number of significant digits is 3, so the product must contain 3 significant digits. So, the right answer is 65.8 .

Explanation: 

Multiplication or division
Suppose in the measured values to be multiplied or divided, the least number of significant digits be \(n\), then in the product or quotient, the number of significant digits should also be \(n\).
e.g. \(1.2 \times 36.72=44.064 \approx 44\)
The least number of significant digits in the measured values are two. Hence, the result when rounded off to two significant digits become 44. Therefore, the answer is 44.
Similarly, e.g. \(\frac{1100}{10.2}=107.8431373 \approx 110\)
As 1100 has minimum number of significant figures (i.e. 2), therefore the result should also contain only two significant digits. Hence, the result when rounded off to two significant digits becomes 110.

Hint

(a)

\(
\begin{aligned}
&\text { Volume } V=\text { length } \times \text { breadth } \times \text { thickness. }\\
&\begin{aligned}
& V=4.234 \mathrm{~m} \times 1.005 \mathrm{~m} \times 0.0201 \mathrm{~m} \\
& V=0.085528917 \mathrm{~m}^3
\end{aligned}
\end{aligned}
\)
Length: 4.234 m has 4 significant figures.
Breadth: 1.005 m has 4 significant figures.
Thickness: 2.01 cm (or 0.0201 m ) has 3 significant figures.
The result should be rounded to 3 significant figures, as it is the least number of significant figures among the measurements.
The calculated volume is \(0.085528917 \mathrm{~m}^3\).
Rounding to 3 significant figures gives \(0.0855 \mathrm{~m}^3\).
The volume of the sheet up to the correct significant figures is \(0.0855 \mathrm{~m}^3\).

Hint

(b) Given, \(R=0.16 \mathrm{~mm}\)
Hence, \(A=\pi R^2=\frac{22}{7} \times(0.16)^2=0.080457\)
Since, radius has two significant figures, so answer will also have two significant figures.
Round \(0.0804247 \mathrm{~mm}^2\) to two significant figures.
The first two significant figures are 8 and 0.
The next digit is 4 , which is less than 5 , so we round down.
The area is \(0.080 \mathrm{~mm}^2\).

Hint

(d)

\(
\frac{97.52}{2.54} \approx 38.3937
\)
The number with the least significant figures is 2.54 , which has 3 significant figures.
The final answer must be rounded to 3 significant figures.
Round 38.3937 to 3 significant figures.
The first three significant figures are 3,8 , and 3 .
The digit after the third significant figure is 9 , which is greater than or equal to 5.
Therefore, round up the third significant figure.
The rounded result is 38.4.

Hint

(a) The product must have two significant digit as both the figures being multiplied have two significant digits, i.e. \(25 \times 10^{-14}\). So, the order of magnitude of result is -14.

Hint

(b) Given, number of divisions on circular scale \(=100\)
Pitch \(=0.01 \mathrm{~cm}\)
\(
\begin{aligned}
\therefore \text { Least count }(\mathrm{LC}) & =\frac{\text { Pitch }}{\text { Number of divisions on circular scale }} \\
& =\frac{0.01}{100}=10^{-4} \mathrm{~cm}
\end{aligned}
\)

Hint

(d) Mean = Total/3=\(30.6 \mathrm{~cm} / 3=10.2 \mathrm{~cm}\)

Hint

(c) Since, \(D \propto r\)
where, \(D=\) diameter of a circle
and \(r=\) radius of the circle.
Therefore, there will be no change in error, it will remain \(4 \%\) for radius also.

Hint

(d)

\(
\begin{aligned}
& l=l_1+l_2=11.05+11.05=22.10 \mathrm{~cm} \\
& \Delta l=\Delta_1 l_1+\Delta_2 l_2=0.2+0.2=0.4 \mathrm{~cm} \\
& \text { Hence, }(l \pm \Delta l)=(22.10 \pm 0.4 \mathrm{~cm})
\end{aligned}
\)

Hint

(d) The mean velocity is \(v=\frac{s}{t}\).
The relative error in velocity is \(\frac{\Delta v}{v}=\frac{\Delta s}{s}+\frac{\Delta t}{t}\).
\(
\begin{aligned}
& \frac{\Delta v}{v}=\frac{0.2}{13.8}+\frac{0.3}{4.0} \\
& \frac{\Delta v}{v} \approx 0.01449+0.075 \\
& \frac{\Delta v}{v} \approx 0.08949
\end{aligned}
\)
The absolute error is \(\Delta v=v \times \frac{\Delta v}{v}\).
\(
\begin{aligned}
& \Delta v=3.45 \mathrm{~ms}^{-1} \times 0.08949 \\
& \Delta v \approx 0.3087 \mathrm{~ms}^{-1}
\end{aligned}
\)
Rounding to one significant figure, \(\Delta v \approx 0.3 \mathrm{~ms}^{-1}\).
The velocity of the body within error limits is \((3.45 \pm 0.3) \mathrm{ms}^{-1}\).

Hint

(c) Volume of cuboid,
\(
V=l \times 2 l \times 3 l=6 l^3
\)
\(
\frac{\Delta V}{V} \times 100=3\left(\frac{\Delta l}{l}\right)=3 \% \quad\left(\because \frac{\Delta l}{l} \times 100=1 \%\right)
\)

Hint

\(
\begin{aligned}
& \text { (a) } p=\frac{F}{A}=\frac{F}{L^2}=F L^{-2} \\
& \begin{aligned}
\% \text { error in pressure } & =(\% \text { error in } F)+2(\% \text { error in } L) \\
& =(4 \%)+2(2 \%)=8 \%
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) We know that, } H=i^2 R t\\
&\begin{gathered}
\therefore \% \text { error in } H=2(\% \text { error in } i)+(\% \text { error in } R)+(\% \text { error in } t) \\
=2(2 \%)+1 \%+1 \%=6 \%
\end{gathered}
\end{aligned}
\)

Hint

(c) We know that, \(K=\frac{p^2}{2 m}\)
The error in measurement of momentum is \(+100 \%\).
Therefore, the actual momentum with error, \(p^{\prime}=p+p=2 p\)
\(\because\) Kinetic energy with error,
\(
K^{\prime}=\frac{\left(p^{\prime}\right)^2}{2 m}=\frac{(2 p)^2}{2 m}
\)
\(
\begin{aligned}
K^{\prime} & =\frac{4 p^2}{2 m} \\
K^{\prime} & =4 K
\end{aligned}
\)
So, percentage change in kinetic energy,
\(
\begin{aligned}
\mathrm{KE} & =\left(\frac{K^{\prime}-K}{K}\right) \times 100=\left(\frac{4 K-K}{K}\right) \times 100 \\
& =\left(\frac{4-1}{1}\right) \times 100=300 \%
\end{aligned}
\)

Hint

(a) Radius of ball \(=5.2 \mathrm{~cm}\)
Volume,
\(
\begin{gathered}
V=\frac{4}{3} \pi R^3 \\
\frac{\Delta V}{V}=3\left(\frac{\Delta R}{R}\right)
\end{gathered}
\)
\(
\left(\frac{\Delta V}{V}\right) \times 100=3\left(\frac{0.2}{5.2}\right) \times 100 \simeq 11 \%
\)

Hint

(d)

\(
\begin{aligned}
&\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\\
&\text { Differentiating both sides we get, }\\
&\begin{aligned}
& \Longrightarrow \frac{\Delta R_p}{R_p^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2} \\
& \frac{\Delta R_p}{R_p}=\frac{50}{15}\left(\frac{0.2}{25}+\frac{0.1}{100}\right) \\
& \% \text { error }=\frac{50}{15}\left(\frac{0.8+0.1}{100}\right) \times 100=3 \%
\end{aligned}
\end{aligned}
\)