MCQ Questions for Class 12 Medical Physics Atoms Quiz 15

According to the Bohr's theory of hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as:

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$\dfrac { 1 }{ n } ,\dfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }$
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$\dfrac { 1 }{ n } ,{ n }^{ 2 },\dfrac { 1 }{ { n }^{ 2 } }$
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${ n }^{ 2 },\dfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }$
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$n,\dfrac { 1 }{ n } ,\dfrac { 1 }{ { n }^{ 2 } }$

In Rutherford experiments on

$\alpha$ - ray scattering the number of particles scattered at

${90}^{o}$ be

$28$ per minute. Then the number of particles scattered per minute by the same foil, but at

${60}^{o}$ is

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$56$
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$112$
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$60$
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$120$

If the radius of Bohr's second orbit of hydrogen atom is

$\alpha$ then the radius of the

$4^{th}$ Bohr orbit in atom with Z=3, is

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$\cfrac {4\alpha}3$
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$\cfrac {16\alpha}3$
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$\cfrac {\alpha}4$
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None of these

The de Broglie wavelength of an electron in

$n^{th}$ Bohr orbit of radius

$a_n(a_0$ being the first orbit radius

$)$ :

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$2\pi a_0$
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$2\pi n^2a_0$
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$2\pi na_0$
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$\dfrac{2\pi a_0}{n}$

Hydrogen atom is excited from ground state to another state with principal quantum number equal to $4$ . Then the number of spectral lines in the emission spectra will be :

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$6$
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$2$
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$3$
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$5$

The radius of the first orbit of hydrogen atom is equal to the radius of the

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first orbit of $He^{ + }$
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second orbit of ${ Be }^{ 3+ }$
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fourth orbit of $He^{ + }$
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third orbit of ${ Be }^{ 3+ }$

A particle of mass

$m$ moves in a circular orbit in a central potential field

$U(r)=\dfrac { 1 }{ 2 } kr^2$ . If Bohr's quantization condition is applied, radii of possible orbitals and energy levels vary with quantum number

$n$ as:

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$r_n \propto n^2, E_n \propto \dfrac { 1}{n^2 }$
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$r_n \propto \sqrt { n } , E_n \propto \dfrac { 1}{n }$
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$r_n \propto n, E_n \propto n$
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$r_n \propto \sqrt { n } , E_n \propto n$

Explanation

$F=\frac { dV}{ dR } =kr=\frac { mv^2}{ r }$

$mvr=\frac { nh}{ 2\pi }$

$r^2\infty n$

$r^2\infty \sqrt {n }$

$E=\frac { 1}{ 2 } kr^2+\frac { 1}{ 2 } mv^2\infty r^2$

Mean free path of gas molecule at constant temperature is inversely proportional to

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$P$
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$V$
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$m$
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$n$ (number density)

If element with principal quantum number

$n > 4$ were not allowed in nature, then the number of possible elements would be

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$60$
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$32$
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$4$
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$64$

The wavelength of second line of Balmer series is 4815

$\mathring A$ . The wavelength of first line Balmer series in the same atom is

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4000 $\mathring A$
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5000 $\mathring A$
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6500 $\mathring A$
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7500 $\mathring A$

The ratio of the largest to shortest wavelength in Lyman series of hydrogen spectra is

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$\dfrac { 25 }{ 9 }$
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$\dfrac { 17 }{ 6 }$
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$\dfrac { 9 }{ 5}$
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$\dfrac { 4 }{ 3 }$

Explanation

For Lyman series

$\dfrac {1}{\lambda_{max}}=R \left[ \dfrac {1}{1^2}-\dfrac {1}{2^2}\right]=\dfrac 34\ R$ and

$\dfrac {1}{\lambda_{min}}=R\left[ \dfrac {1}{1^2}-\dfrac{1}{\infty^2}\right]=\dfrac R1$

$\Rightarrow \dfrac{\lambda_{max}}{\lambda_{min}}=\dfrac {4}{3}$

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength

$980\mathring { A } .$ The radius of the atom in excited state, in terms of Bohr radius a, will be:

$\left( hc=12500\quad eV-\mathring { A } \right)$

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$4{ a }_{ 0 }$
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$16{ a }_{ 0 }$
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$9{ a }_{ 0 }$
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None of these

Explanation

Hint: Energy of photons = $E=\frac{h c}{\lambda}$

Solution:

Step 1: Find the energy of photon.

Given: $h c=12500 \mathrm{eV}-A^{\circ}, \lambda=980 A^{\circ}$

Energy of photons = $E=\frac{h c}{\lambda}$

$\Rightarrow E=\frac{12500}{980}=12.75 \mathrm{eV}$

Step 2: Find the excited state

Now, Energy of hydrogen atom in nth shell = $E_{n}=\frac{-13.6}{n^{2}}$

Energy absorbed by hydrogen atom = energy of photons

$\Rightarrow E_{n}-E_{1}=E$

$\Rightarrow \frac{-13.6}{n^{2}}-\left[\frac{-13.6}{1^{2}}\right]=12.75 \mathrm{eV}$

$\Rightarrow 13.6-\frac{13.6}{n^{2}}=12.75$

$\Rightarrow 1-\frac{1}{n^{2}}=\frac{12.75}{13.6}$

$\Rightarrow 1-\frac{1}{n^{2}}=0.9375$

$\Rightarrow \frac{1}{n^{2}}=1-0.9375=0.0625$

Step 3: Find radius in the excited state.

Radius of hydrogen atom = $r=n^{2} a_{0}$

$\Rightarrow r=(4)^{2} a_{0}$

$\Rightarrow r=16 a_{0}$

So, Radius of hydrogen atom in excited state is $16 a_{0}$

The time taken by the electron in one complete revolution in the

$n^{th}$ Bohr's orbit of the hydrogen atom is

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Inversely proportional to $n^2$
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Directly proportional to $n^3$
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Directly proportional to $\dfrac{h}{2\pi}$
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Inversely proportional to $\dfrac{n}{h}$

When a silver foil

$(Z=47)$ was used in an

$\alpha$ -ray scattering experiment, the number of

$\alpha$ -particles scattered at

$30^o$ was found to be

$200$ per minute. If the silver foil is replaced by aluminium

$(Z=13)$ foil of the same thickness. The number of

$\alpha$ -particles scattered per minute at

$30^o$ is nearly equal to

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$15$
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$30$
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$10$
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$20$

The radius of the orbital of electron in the hydrogen atom is

$0.5 A^0$ . the speed of the electron is

$2 \times 10^5 m/s$ . then, the current in the loop due to the motion of the electron is

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1 mA
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1.5 mA
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2.5 mA
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$1.5 \times 10^{-2} m A$

In hydrogen atom electron is revolving around nucleus in circular orbit with frequency of

$6.57 \times {1^{15}}$ revolutions. If its equivalent magnetic moment is

$9.24 \times {10^{ - 24}}A{m^2}$ , the radius of orbit of electron is

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0.27 A
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0.53 A
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1.27 A
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1.34 A

According to the Bohr theory of the hydrogen atom electrons starting in the

$4^{th}$ energy level and eventually ending in the ground state could produce a total of how many lines in the hydrogen spectra ?

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7
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6
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5
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4

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line H-atom is suitable for this purpose ?

Given:

${ R }_{ H }=1\times { 10 }^{ 5 }cm, h=6.610^{ -34 }Js, C=3\times 10^{ 4}m/{ s }^{ -1 }$

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Paschen $5\rightarrow 3$
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Paschen $\infty \rightarrow 3$
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Lyman, $\infty \rightarrow 1$
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Balmer $\infty \rightarrow 2$

$\alpha-particle$ when accelerated through a potential difference of

$V$ volt has a wavelength

$\lambda$ associated with it. In order to have same

$\lambda$ , by what potential difference a proton must be accelerated?

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$8\ V$
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$6\ V$
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$4\ V$
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$12\ V$

Explanation

We know,

For a charged particle

$q$ of mass

$m$ under potential difference of

$V$ ,

The associated De-broglie wavelength is

$=\dfrac{h}{\sqrt{2mqV}}$

$\textbf{Let,}$

Mass of alpha particle be

$m_{\alpha}$

Mass of proton be

$m_p$

Charge on Alpha particle be

$q_{\alpha}$

Charge on proton be

$q_p$

Voltage associated with proton be

$V'$

$\textbf{Given,}$

Wavelength associated with alpha particle and proton

$=\lambda_{\alpha}=\lambda_p$

Voltage associated with Alpha particle

$=V$

$\textbf{Also,}$

$m_{\alpha}=4\times m_p$

$q_{\alpha}=2\times q_p$

$\dfrac{\lambda_{\alpha}}{\lambda_p}=\sqrt{\dfrac{m_pq_pV'}{m_{\alpha}q_{\alpha}V}}$

$1=\sqrt{\dfrac{0.5V'}{4V}}$

$\textit{On squaring both sides}$

$1=\dfrac{V'}{8V}$

$V'=8V$

Option

$\textbf A$ is the correct answer

In an atom two electrons move round the nucleus in circular orbits of radii R and

$4R$ respectively. The ratio of the time taken by them to complete one revolution is?

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$\dfrac{1}{4}$
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$\dfrac{4}{1}$
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$\dfrac{8}{1}$
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$\dfrac{1}{8}$

Which of following is not a conclusion of Rutherford's atomic model?

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Most of the part inside an atom is empty.
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Almost all mass of an atom is concentrated in the nucleus.
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The size of nucleus is very small in comparison to the size fo atom.
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Electron revolves round the nucleus in definite orbits.

In hydrogen atom, electron revolves around nucleus in circular orbit of radius

$5 \times 10 ^{- 11}\, m$ at a speed

$2.2 \times 10^6 m/s$ . Then the orbital current will be:

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3.2 mA
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2.42 mA
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1.12 mA
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0.02 ma

An electron of energy

$150 eV$ has wavelength of

$10 ^{- 10}m$ . The wavelength of a

$60 keV$ electron is

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$5 pm$
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$2 pm$
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$12 pm$
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$4 pm$

Explanation

we know that wowelength of an electron is given by thy relation

\lambda=\frac{h}{\sqrt{2 m v}}

Keeping

$h$ and

$m$ constant, we have,

$\lambda \propto \frac{1}{\sqrt{V}}$

$\Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{\sqrt{v_{1}}}{\sqrt{v_{2}}}$

$\lambda_{2}=\lambda_{1} \times \sqrt{\frac{150}{600}} \\$

$\lambda_{2}=10^{-10} \times \sqrt{\frac{1}{4}} \\$

$\lambda_{2}=10^{-10} \times \frac{1}{2} \\$

$\lambda_{2}=0.5 \times 10^{-10} \mathrm{~m} \\$

$\lambda_{2}=5 \mathrm{pm}$

\end{array}

$[A]$

$5 \mathrm{pm}$

The ratio of frequencies of the first line of the lyman series and the first line of blamer series is

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$\frac{{27}}{5}$
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$\frac{{27}}{8}$
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$\frac{8}{{27}}$
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$\frac{4}{{27}}$

Hydrogen atoms in a sample are excited to

$n=5$ state and it is found that photons of all possible wavelengths are present in the emission spectra. The minimum number of hydrogen atoms in the sample would be

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$5$
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$6$
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$10$
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$infinite$

Explanation

The wavelengths present in emission spectra are shown in the figure.
Transitions

$a,e,h$ and

$j$ can be performed by a single atom also. This is also true about transitions

$b$ and

$i$ , other transitions require one atom each.
1627842_1763717_ans_f12973bc5086451286e93f94fdceebbc.png

The radius of the shortest orbit in a one-electron system is 18 pm. It may be

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hydrogen
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deuterium
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$He^+$
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$Li^{++}$

The Bohr model for the spectra of a H- atom

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will not be applicable to hydrogen in the molecular from.
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Will not be applicable for a He - atom
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is valid only at room tempreture.
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predicts continuous as well as discrete spectral lines.

Assertion : According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic.
Reason : According to electromagnetic theory an accelerated particle continuously emits radiation.

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If both assertion and reason are true and the reason is the explanation of the assertion.
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If both assertion and reason are true but reason is not the explanation of the assertion.
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If assertion is true but reason is false.
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If the assertion and reason both are false.
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If assertion is false but reason is true.

The figure shows a graph between

$\ln \left|\dfrac {A_n}{A_1}\right|$ and

$1n |n|$ where

$A_n$ is the area enclosed by the

$n^{th}$ orbit in a hydrogen like atom. The correct curve is

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$4$
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$3$
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$2$
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$1$

Explanation

$A_n=\pi r_n^2 \Rightarrow \dfrac {A_n}{A_1} =\left(\dfrac {r_n}{r_1}\right)^2 =\left (\dfrac {n}{1}\right)^4 \quad (\because r_n \propto n^2)$
Taking

$\log_e$ both side

$\log_e \dfrac {A_n}{A_1}=4\log_e(n)$
Comparing it with

$y=mx+c$ , graph

$(4)$ is correct.

CBSE Questions for Class 12 Medical Physics Atoms Quiz 15 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 15 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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