5 Marks Questions

Question 15 Marks

Einstein's mass - energy relation emerging out of his famous theory of relativity relates mass $(\mathrm{m})$ to energy $(\mathrm{E})$ as $\mathrm{E}=$ $\mathrm{mc}^2$, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV , where $1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}$; the masses are measured in unified atomic mass unit ( u ) where $1 \mathrm{u}=1.67 \times 10^{-27} \mathrm{~kg}$.
a. Show that the energy equivalent of 1 u is 931.5 MeV .
b. A student writes the relation as $1 \mathrm{u}=931.5 \mathrm{MeV}$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer

We can apply Einstein’s mass-energy relation in this problem, $E = mc^2$, to calculate the energy equivalent of the given mass.

Here, $1\ \text{amu}=1\text{u}=1.67\times10^{-27}\text{kg}$
Applying $E = mc^2$
Energy $\text{E}=(1.67\times10^{-27})(3\times10^8)\text{J}=1.67\times9\times10^{-11}\text{J}$
$\text{E}=\frac{1.67\times9\times10^{-11}}{1.6\times10^{-13}}\text{MeV}$
$=939.4\text{MeV}\approx931.5\text{MeV}$

$\text{As E}=\text{mc}^2\Rightarrow\text{m}=\frac{\text{E}}{\text{c}^2}$

According to this, $1\text{u}=\frac{931.5\text{MeV}}{\text{c}^2}$
Hence the dimensionally correct relation $1\ \text{amu}\times\text{c}^2=1\text{u}\times\text{c}^2=931.5\text{MeV}$

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Question 25 Marks

The radius of atom is of the order of $\mathring{\text{A}}$ 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?

Answer

According to the question, Radius of atom $1\mathring{\text{A}}=10^{-10}\text{m}$ Radius of nucleus $\approx1\ \text{fermi}=10^{-15}\text{m}$ Volume of atom $\text{V}_\text{Atom}=\frac{4}{3}\pi\text{R}^3_\text{A}$ Volumev of nucleus $\text{V}_\text{nucleus}=\frac{4}{3}\pi\text{R}^3_\text{N}$$\frac{\text{V}_\text{Atom}}{\text{V}_\text{Nucleus}}=\frac{\frac{4}{3}\pi\text{R}^3_\text{A}}{\frac{4}{3}\pi\text{R}^3_\text{N}}=\Big(\frac{\text{R}_\text{A}}{\text{R}_\text{N}}\Big)^3=\Big(\frac{10^{-10}}{10^{-15}}\Big)^3=10^{15}$
Mass of one mole of $_6C^{12}$ atom = 12g Number of atoms in one mole = Avogadro's number = $6.023 \times 10^{23}$
$\therefore$ Mass of one $_6C^{12}$ atom $=\frac{12}{6.023\times10^{23}}\text{g}$
$1\ \text{amu}=\frac{1}{12}\times\text{mass of one}\ _6\text{C}^{12}\ \text{atom}$
$\therefore1\text{ amu}=\Big(\frac{1}{12}\times\frac{12}{6.023\times10^{23}}\Big)\text{g}=1.67\times10^{-24}\text{g}$
$=1.67\times10^{-27}\text{kg}\ \ (\because1\text{g}=10^{-3}\text{kg})$

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Question 35 Marks

Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about $\frac{1}{2}$ A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

Answer

Given that $\frac{\text{D}_\text{mars}}{\text{D}_\text{earth}}=\frac{1}{2}\ \ \ ...(\text{i})$ where D represents diameter. We know that, $\frac{\text{D}_\text{earth}}{\text{D}_\text{sun}}=\frac{1}{100}$$\therefore\ \frac{\text{D}_\text{mars}}{\text{D}_\text{sun}}=\frac{1}{2}\times\frac{1}{100}$ [from Eq. (i)]
At 1AU Sun's diameter $=\Big(\frac{1}{2}\Big)^\circ$$\therefore$ Diameter of Mars $=\frac{1}{2}\times\frac{1}{200}=\Big(\frac{1}{400}\Big)^\circ$
At $\frac{1}{2}$AU, Mars' diameter$=\frac{1}{400}\times2^\circ=\Big(\frac{1}{200}\Big)^\circ$
With 100 magnification, Mars' diameter$=\frac{1}{200}\times100^\circ=\Big(\frac{1}{2}\Big)^\circ=30'$
This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

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Question 45 Marks

A new system of units is proposed in which unit of mass is $\alpha\text{kg},$ unit of length $\beta$ m and unit of time $\gamma\text{ s}.$ How much will 5J measure in this new system?

Answer

Dimension of Energy $=[\text{ML}^2\text{T}^{-2}]$$\text{n}_2\text{u}_2=\text{n}_1\text{u}_1$
$\text{n}_2=\text{n}_1\frac{\text{u}_1}{\text{u}_2}=\text{n}_1\Big[\frac{\text{M}_1}{\text{M}_2}\Big]^1\Big[\frac{\text{L}_1}{\text{L}_2}\Big]^2\Big[\frac{\text{T}_1}{\text{T}_2}\Big]^{-2}$
$n_2$ = New system of unit = ? $n_1$ = S.I system of unit = 5J$\text{M}_2=\alpha\ \text{kg},\ \ \ \text{M}_1=1\text{kg}$
$\text{L}_2=\beta\text{m},\ \ \ \ \ \text{L}_1=1\text{m}$
$\text{T}_2=\gamma\text{s},\ \ \ \ \text{T}_1=1\text{ second}$
$\text{n}_2=5\Big[\frac{1\text{kg}}{\alpha\text{kg}}\Big]^1\Big[\frac{1\text{m}}{\beta\text{m}}\Big]^2\Big[\frac{1\text{ sec}}{\gamma\text{ sec}}\Big]^{-2}$
$\text{n}_2=5\Big[\alpha^{-1}\beta^{-2}\gamma^2\Big]$
New system $=\frac{\gamma^2}{\alpha\beta^2}\ \text{or}\ \Big[\alpha^{-1}\beta^{-2}\gamma^2\Big]$

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Question 55 Marks

Why length, mass and time are chosen as base quantities in mechanics?

Answer

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.

Length, mass and time cannot be derived from one another, that is these quantities are independent.
All other quantities in mechanics can be expressed in terms of length, mass and time.

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Question 65 Marks

A physical quantity X is related to four measurable quantities a, b, c and d as follows:$\text{X}=\text{a}^2\text{b}^3\text{c}^\frac{5}{2}\text{d}^{-2}.$
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

Answer

$\because\frac{\Delta\text{X}}{\text{X}}\times100=\pm\Big[2\frac{\Delta\text{a}}{\text{a}}+3\frac{\Delta\text{b}}{\text{b}}+\frac{5}{2}\frac{\Delta\text{c}}{\text{c}}+2\frac{\Delta\text{d}}{\text{d}}\Big]\times100$$\frac{\Delta\text{X}}{\text{X}}\times100=\pm\Big[\frac{2\times1}{100}+\frac{3\times2}{100}+\frac{5}{2}\times\frac{3}{100}+\frac{2\times4}{100}\Big]\times100$
$=\pm\frac{100}{100}\Big[2+6+\frac{15}{2}+8\Big]$
$\frac{\Delta\text{x}}{\text{x}}\times100=\pm\Big[16+\frac{15}{2}\Big]=\pm\Big[\frac{32+15}{2}\Big]=\pm\frac{47}{2}=\pm23.5\%$
Mean absolute error $=\pm\frac{235}{100}=\pm0.235$
= 0.24 (rounding off in significant figure)
Again rounding off X = 2.763 in two significant figure = 2.8.

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Question 75 Marks

Time for $20$ oscillations of a pendulum is measured as $t_1 = 39.6s; t_2 = 39.9s; t_3 = 39.5s$. What is the precision in the measurements? What is the accuracy of the measurement?

Answer

$t_1 = 39.6s, t_2 = 39.9s$, and $t_3 = 39.5s$ The least count of instrument is 0.1s Hence precision (LC) = 0.1s Mean value of time for 20 oscillations$=\frac{39.6+39.9+39.5}{3}=\frac{119.0}{3}=39.7\text{s}$
Absolute errors in measurement$|\Delta\text{t}_1|=|\bar{\text{t}}-\text{t}_1|=|39.7-39.6|=|0.1|=0.1\text{s}$
$|\Delta\text{t}_2|=|\bar{\text{t}}-\text{t}_2|=|39.7-39.9|=|0.2|=0.2\text{s}$
$|\Delta\text{t}_3|=|\bar{\text{t}}-\text{t}_3|=|39.7-39.5|=|0.2|=0.2\text{s}$
$\therefore$ Mean absolute error $=\frac{0.1+0.2+0.2}{3}=\frac{0.5}{3}\cong0.2\text{s}$
$\therefore$ Accuracy of measurement $=\pm0.2\text{s}.$

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Question 85 Marks

In an experiment to estimate the size of a molecule of oleic acid 1mL of oleic acid is dissolved in 19mL of alcohol. Then 1mL of this solution is diluted to 20mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.
Read the passage carefully and answer the following questions:

Why do we dissolve oleic acid in alcohol?
What is the role of lycopodium powder?
What would be the volume of oleic acid in each mL of solution prepared?
How will you calculate the volume of n drops of this solution of oleic acid?
What will be the volume of oleic acid in one drop of this solution?

Answer

To get a molecular level we have to reduce the concentration of oleic acid by dissolving it in a proper solvent. Oleic acid is an organic compound so cannot be dissolve in ionic solvent water. It can dissolve in organic solvent alcohol.
Lycopodium prevent to mix oleic acid in water, when drop of oleic acid is poured on water. So lycopodium powder spread on water surface first and then thin layer of diluted oleic acid is made on surface of lycopodium spread on water.
The concentration of oleic acid in solution of alcohol in V volume solution is $\frac{1}{20}\times\frac{1}{20}\text{V}=\frac{1}{400}\text{V}$ml if V = 1ml then required concentration in one ml solution $=\frac{1}{400}\text{ml}$ as given in question.
Volume of n drop solution can be calculate by burette, by dropping 1ml, solution drop by drop in beaker and counting its number of drop. If n drops are in 1ml, then volume of 1 drop $=\frac{1}{\text{n}}\text{ml}$
The volume of 1 drop of solution is $\frac{1}{\text{n}}\text{ml}.$ If n drops are measured in one ml in part (d). Then concentration of oleic acid in one drop solution $=\frac{1}{400}\text{V}=\frac{1}{400}.\frac{1}{\text{n}}\text{ml}=\frac{1}{400}\text{ml}$ oleic acid.

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Question 95 Marks

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that $\text{T}=\frac{\text{k}}{\text{R}}\sqrt{\frac{\text{r}^3}{\text{g}}},$ where k is a dimensionless constant and g is acceleration due to gravity.

Answer

According to Kepler;s third law, $\text{T}^2\propto\text{a}^3$ i.e., square of time period $(T^2)$ of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit $(a^3)$. We have to apply Kepler's third law,$\text{T}^2\propto\text{r}^3\Rightarrow\text{T}\propto\text{r}^\frac{3}{2}$
Also, T depends on R and g. Let $\text{T}\propto\text{r}^\frac{3}{2}\text{g}^\text{a}\text{R}^\text{b}$$\Rightarrow\text{T}=\text{kr}^{\frac{3}{2}}\text{R}^\text{a}\text{g}^\text{b}\ \ \ \ \ ...(\text{i})$
Where, k is a dimensionless constant of proportionality. Writing the dimensions of various quantities on both the sides, we get$[\text{M}^0\text{L}^0\text{T}]=[\text{L}]^\frac{3}{2}[\text{LT}^{-2}]^\text{a}[\text{L}]^\text{b}$
$=[\text{M}^0\text{L}^{\text{a}+\text{b}+\frac{3}{2}}\text{T}^{-2\text{a}}]$
On comparing the dimensions of both sides, we get$\text{a}+\text{b}+\frac{3}{2}=0\ \ \ ...(\text{ii})$
$-2\text{a}=1\Rightarrow\text{a}=\frac{-1}{2}\ \ \ ...(\text{iii})$
From Eq. (ii), we get$\text{b}-\frac{1}{2}+\frac{3}{2}=0\Rightarrow\text{b}=-1$
Substituting the values of a and b in Eq. (i), we get$\text{T}=\text{kr}^\frac{3}{2}\text{R}^{-1}\text{g}^{\frac{-1}{2}}\Rightarrow\text{T}=\frac{\text{k}}{\text{R}}\sqrt{\frac{\text{r}^3}{\text{g}}}$

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Question 105 Marks

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Answer

Key point: In geometry, a solid angle (symbol: $\Omega$ or w) is the twodimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

A small object nearby may subtend the solid angle as a larger object farther away. For example, although the Moon is much smaller than the Sun, it is also much closer to Earth. Diagram given below shows that moon almost entirely covers the sphere of the sun. $R_{me}$ = Distance of moon from earth $R_{se}$ = Distance of sun from earth Let the solid angle made by sun and moon is $\text{d}\Omega,$ we can write

$\text{d}\Omega=\frac{\text{A}_\text{sun}}{\text{R}^2_\text{se}}=\frac{\text{A}_\text{moon}}{\text{R}^2_\text{me}}$
Here, $A_{sun}$ = Area of the sun $A_{moon}$ = Area of the moon$\Rightarrow\theta=\frac{\pi\text{R}_\text{s}^2}{\text{R}_\text{se}^2}=\frac{\pi\text{R}_\text{m}^2}{\text{R}_\text{me}^2}$
$\Rightarrow\Big(\frac{\text{R}_\text{s}}{\text{R}_\text{se}}\Big)^2=\Big(\frac{\text{R}_\text{m}}{\text{R}_\text{me}}\Big)^2$
$\Rightarrow\frac{\text{R}_\text{s}}{\text{R}_\text{se}}=\frac{\text{R}_\text{m}}{\text{R}_\text{me}}\ \text{or}\ \frac{\text{R}_\text{s}}{\text{R}_\text{m}}=\frac{\text{R}_\text{se}}{\text{R}_\text{me}}$
(Here, radius of sun and moon represent their sizes respectively)

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