Photoelectric Effect – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

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Previous Years AIEEE/JEE Mains Questions

Q. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV-nm)

(1) 1.51 eV            (2) 1.68 eV          (3) 3.09 eV              (4) 1.41 eV

[AIEEE – 2009]

Sol. (4)

$\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0} \Rightarrow 1.68=\frac{12400}{4000}-\phi_{0}$

By solving it $\phi_{0}=1.42 \mathrm{eV}$


Q. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\mathrm{max}}$. When the ultraviolet light is replaced by X-rays, both $\mathrm{V}_{0}$ and $\mathrm{K}_{\mathrm{max}}$ increase.

Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light

(1) Statement–1 is true, Statement–2 is false

(2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement– 1

(3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement– 1

(4) Statement–1 is false, Statement–2 is true

[AIEEE – 2010]

Sol. (1)

Speed of emitted electrons is independent of frequency of incident light.


Q. This question has Statememtn-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement–1 : A metallic surface is irradiated by a monochromatic light of frequency $\mathrm{v}>\mathrm{v}_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ respectively. If the frequency incident on the surface is doubled, both the $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ are also boubled.

Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

(1) Statement–1 is true, Statement–2 is true, Statement–2 is not the correct explanationof Statement– 1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true, Statement–2 is the correct explanation of Statement– 1

[AIEEE-2011]

Sol. (2)


Q. The anode voltage of photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows

[AIEEE-2013]

Sol. (4)

For constant intensity as wavelength decreases energy of photons increases and number of photons decreases. So it may seem that current should decrease. But the probability that a photon will be successful in emitting an electron will also increase. So as wavelength decreases current increases


Q. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be :-

[JEE Main-2016]

Sol. (2)

$\mathrm{E}=(\mathrm{KE})_{\max }+\mathrm{f}$

$\left[\frac{\mathrm{hc}}{\lambda}=(\mathrm{KE})_{\max }+\phi\right] \ldots .$ (1)

\frac{4}{3} \frac{\mathrm{hc}}{\lambda}=\left(\frac{4}{3} \mathrm{KE}_{\max }+\frac{\phi}{3}\right)+\phi


Nuclear Physics – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

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Previous Years AIEEE/JEE Mains Questions

Q. The above is a plot of binding energy per nucleon $\mathrm{E}_{\mathrm{b}}$, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions :

(i) $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\varepsilon$

(ii) $\mathrm{C} \rightarrow \mathrm{A}+\mathrm{B}+\varepsilon$

(iii) $\mathrm{D}+\mathrm{E} \rightarrow \mathrm{F}+\varepsilon$

$(\mathrm{iv}) \mathrm{F} \rightarrow \mathrm{D}+\mathrm{E}+\varepsilon$

where $\varepsilon$ is the energy released ? In which reactions is $\varepsilon$ positive ?

(1) (ii) and (iv) (2) (ii) and (iii) (3) (i) and (iv) (4) (i) and (iii)

[AIEEE – 2009]

Sol. (3)


Q. The speed of daughter nuclei is :-

(1) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}}$

(2) $\mathrm{c} \frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}$

(3) $\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$

(4) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}}}$

[AIEEE-2010]

Sol. (3)

Total kinetic energy of products

$=$ Total energy released $\frac{\mathrm{p}^{2}}{2 \mathrm{m}}+\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

$\left.=(\text { mass defect }) \mathrm{c}^{2} \text { (where } \mathrm{m}=\frac{\mathrm{M}}{2} \text { given }\right)$

$\Rightarrow 2\left(\frac{\mathrm{p}^{2}}{2 \mathrm{m}}\right)=\left[(\mathrm{M}+\Delta \mathrm{m})-\left(\frac{\mathrm{M}}{2}+\frac{\mathrm{M}}{2}\right)\right] \times \mathrm{c}^{2}$

$\Rightarrow 2 \times\left[\frac{\mathrm{p}^{2}}{2\left(\frac{\mathrm{M}}{2}\right)}\right]=(\Delta \mathrm{m}) \mathrm{c}^{2}$

$\Rightarrow \frac{2\left(\frac{\mathrm{M}}{2} \mathrm{v}\right)^{2}}{\mathrm{M}}=(\Delta \mathrm{m}) \mathrm{c}^{2} \Rightarrow \mathrm{v}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$


Q. The binding energy per nucleon for the parent nucleus is $E_{1}$ an that for the daughter nuclei is $\mathrm{E}_{2}$. Then:-

(1) $\mathrm{E}_{1}=2 \mathrm{E}_{2}$

$(2) \mathrm{E}_{2}=2 \mathrm{E}_{1}$

(3) $\mathrm{E}_{1}>\mathrm{E}_{2}$

$(4) \mathrm{E}_{2}>\mathrm{E}_{1}$

[AIEEE – 2010]

Sol. (4)

Because energy is releasing $\Rightarrow$ Binding energy per nucleon of product> that of parent $\Rightarrow \mathrm{E}_{2}>$

$\mathrm{E}_{1}$


Q. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be:-

(1) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-2}$

(2) $\frac{\mathrm{A}-\mathrm{Z}-8}{\mathrm{Z}-4}$

(3) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-8}$

(4) $\frac{\mathrm{A}-\mathrm{Z}-12}{\mathrm{Z}-4}$

[AIEEE – 2010]

Sol. (3)


Q. After absorbing a slowly moving neutron of mass $\left.\mathrm{m}_{\mathrm{N}} \text { (momentum } \sim 0\right)$ a nucleus of mass M breaks into two nuclei of masses $\mathrm{m}_{1}$ and $5 \mathrm{m}_{1}\left(6 \mathrm{m}_{1}=\mathrm{M}+\mathrm{m}_{\mathrm{N}}\right)$, respectively. If the de Broglie wavelength of the nucleus with mass $\mathrm{m}_{1}$ is $\lambda$, then de Broglie wavelength of the other nucleus will be:-

(1) $25 \lambda$

(2) $5 \lambda$

(3) $\frac{\lambda}{5}$

( 4)$\lambda$

[AIEEE – 2011]

Sol. (4)


Q. Statement-1: A nucleus having energy $\mathrm{E}_{1}$ decays be $\beta^{-}$ emissionto daughter nucleus having energy $E_{2}$, but the – rays are emitted with a continuous energy spectrum having end point energy $\mathrm{E}_{1}-\mathrm{E}_{2}$.

Statement-1: To conserve energy and momentum in -decay at least three particles must take part in the transformation.

(1) Statement-1 is incorrect, statement-2 is correct

(2) Statement-1 is correct, statement-2 is incorrect

(3) Statement-1 is correct, statement-2 correct; statement-2 is the correct explanation of statement-1

(4) Statement-1 is correct, statement-2 is correct; statement -2 is not the correct explanation of statement-1.

[AIEEE – 2011]

Sol. (3)


Q. Assume that a neutron breaks into a proton and an electron. The energy released during this process is :

(Mass of neutron $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of proton $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of electron $\left.=9 \times 10^{-31} \mathrm{kg}\right)$

(1) 5.4 MeV (2) 0.73 MeV (3) 7.10 MeV (4) 6.30 MeV

[AIEEE – 2012]

Sol. (2)

Released energy

$=\left[1.6747 \times 10^{-27}-1.6725 \times 10^{-27}-9 \times 10^{-31}\right]$

$\times\left(3 \times 10^{8}\right)^{2} \mathrm{J}=0.73 \mathrm{MeV}$


Atomic Structure – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. The transistion from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :-

(1) $4 \rightarrow 2$

(2) $5 \rightarrow 4$

(3) $2 \rightarrow 1$

(4) $3 \rightarrow 2$

[AIEEE – 2009]

Sol. (2)

Use Rydberg equation to find the required transition.


Q. Energy required for the electron excitation in $\mathrm{Li}^{++}$ from the first to the third Bohr orbit is:

(1) 108.8 eV

(2) 122.4 eV

(3) 12.1 eV

(4) 36.3 eV

[AIEEE-2011]

Sol. (1)

Rquired energy $=13.6(Z)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ eV

$=13.6(3)^{2}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=108.8 \mathrm{eV}$


Q. 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 :-

(1) 6 (2) 2 (3) 3 (4) 5

[AIEEE-2012]

Sol. (1)

Number of lines $=^{n} C_{2}=^{4} C_{2}=\frac{(4)(3)}{(2)}=6$


Q. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number $(n-1) .$ If $n>>1$, the frequency of radiation emitted is proportional to :

( 1)$\frac{1}{\mathrm{n}}$

( 2)$\frac{1}{\mathrm{n}^{2}}$

(3) $\frac{1}{\mathrm{n}^{3 / 2}}$

(4) $\frac{1}{n^{3}}$

[JEE Main-2013]

Sol. (4)

Energy of $\mathrm{E}=\mathrm{h} \theta=\mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{1}{(\mathrm{n}-1)^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$

$=\mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{2 \mathrm{n}-1}{\mathrm{n}^{2}(\mathrm{n}-1)^{2}}\right]$

$\mathrm{h} \theta \approx \mathrm{E}_{0} \mathrm{z}^{2}\left[\frac{2 \mathrm{n}}{\mathrm{n}^{4}}\right] \Rightarrow \mathrm{v} \propto \frac{1}{\mathrm{n}^{3}}$


Q. As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion :

(1) kinetic energy decreases, potential energy increases but total energy remains same

(2) kinetic energy and total energy decrease but potential energy increases

(3) its kinetic energy increases but potential energy and total energy decreases

(4) kinetic energy, potential energy and total energy decrease.

[JEE Main-2015]

Sol. (3)

$\mathrm{K}=+13.6 \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}}$ as $\mathrm{n}$ decreases $\mathrm{k}$ increases


Q. Match List-I (Fundament Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list :

(1) A-ii, B-i, C-iii

(2) A-iv, B-iii, C-ii

(3) A-i, B-iv, C-iii

(4) A-ii, B-iv, C-iii

[JEE Main-2015]

Sol. (1)

Self Explanatory/Theory

(A) Franck-Hertz experiment explains disrete energy levels of atom

(B) Photo-electric experiment explain particle nature of light

(C) Davison Germer experiment explain wave nature of electron.


Q. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r=\lambda_{1} / \lambda_{2}$, is given by :

(1) $\mathrm{r}=\frac{3}{4}$

(2) $\mathrm{r}=\frac{1}{3}$

(3) $\mathrm{r}=\frac{4}{3}$

(4) $r=\frac{2}{3}$

[JEE Main-2017]

Sol. (2)

using $\Delta \mathrm{E}=\frac{\mathrm{hC}}{\lambda}$

for $\lambda_{1} \quad-\mathrm{E}-(-2 \mathrm{E})=\frac{\mathrm{hC}}{\lambda_{1}}$

$\lambda_{1}=\frac{\mathrm{hC}}{\mathrm{E}}$

for $\lambda_{2} \quad-\mathrm{E}-\left(-\frac{4 \mathrm{E}}{3}\right)=\frac{\mathrm{hC}}{\lambda_{2}}$

$\lambda_{2}=\frac{3 \mathrm{h} \mathrm{C}}{\mathrm{E}}$

$\frac{\lambda_{1}}{\lambda_{2}}=\mathrm{r}=\frac{1}{3}$


Q. A particle A of mass m and initial velocity v collides with a particle B of mass which is at rest. The collision is head on, and elastic. The ratio of the de–Broglie wavelengths $\lambda_{\mathrm{A}}$ to $\lambda_{\mathrm{B}}$after the collision is :

(1) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{2}{3}$

(2) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}$

(3) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{3}$

(4) $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2$

 

[JEE Main-2017]

Sol. (4)

By conservation of linear momentum

$\mathrm{mv}=\mathrm{mv}_{1}+\frac{\mathrm{m}}{2} \mathrm{v}_{2}$

$2 \mathrm{v}=2 \mathrm{v}_{1}+\mathrm{v}_{2}$ ….(1)

by law of collision

$\mathrm{e}=\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}$

$\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ …(2)

By equation (1) and (2)

Option (4)


Q. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_{\mathrm{n}}, \lambda_{\mathrm{g}}$ be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

$(1) \Lambda_{\mathrm{n}} \approx \mathrm{A}+\mathrm{B} \lambda_{\mathrm{n}}$

$(2) \Lambda_{\mathrm{n}}^{2} \approx \mathrm{A}+\mathrm{B} \lambda_{\mathrm{n}}^{2}$

(3) $\Lambda_{\mathrm{n}}^{2} \approx \lambda$

(4) $\Lambda_{\mathrm{n}} \approx \mathrm{A}+\frac{\mathrm{B}}{\lambda_{\mathrm{n}}^{2}}$

[JEE Main-2018]

Sol. (4)

$\lambda_{\mathrm{n}}=\frac{\mathrm{h}}{\mathrm{mu}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}_{\mathrm{n}}}}$

$\Rightarrow \mathrm{k}_{\mathrm{n}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda_{\mathrm{n}}^{2}} ; \mathrm{k}_{\mathrm{g}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda_{\mathrm{g}}^{2}}$

$\Rightarrow \mathrm{k}_{\mathrm{g}}-\mathrm{k}_{\mathrm{n}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m}}\left[\frac{1}{\lambda_{\mathrm{g}}^{2}}-\frac{1}{\lambda_{\mathrm{n}}^{2}}\right]$

$\mathrm{E}_{\mathrm{n}}=-\mathrm{k}_{\mathrm{n}}$

for emitted photon

$\frac{\mathrm{hc}}{\Lambda_{\mathrm{n}}}=\mathrm{E}_{\mathrm{n}}-\mathrm{E}_{\mathrm{g}}=\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}$

$\frac{1}{\Lambda_{\mathrm{n}}}=\frac{\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}}{\mathrm{hc}}$

$\Lambda_{\mathrm{n}}=\frac{\mathrm{hc}}{\mathrm{K}_{\mathrm{g}}-\mathrm{K}_{\mathrm{n}}} \Rightarrow \Lambda_{\mathrm{n}}=\frac{\mathrm{hc}}{\frac{\mathrm{h}^{2}}{2 \mathrm{m}}\left[\frac{1}{\lambda_{\mathrm{g}}^{2}}-\frac{1}{\lambda_{\mathrm{n}}^{2}}\right]}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc}}{\mathrm{h}\left(\frac{\lambda_{\mathrm{n}}^{2}-\lambda_{\mathrm{g}}^{2}}{\lambda_{\mathrm{g}}^{2} \lambda_{\mathrm{n}}^{2}}\right)}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc} \lambda_{\mathrm{g}}^{2} \lambda_{\mathrm{n}}^{2}}{\mathrm{h}\left(\lambda_{\mathrm{n}}^{2}-\lambda_{\mathrm{g}}^{2}\right)}$

as $\lambda_{\mathrm{n}} \propto \mathrm{n}$

$\lambda_{\mathrm{n}}>>\lambda_{\mathrm{g}}$

$\Lambda_{\mathrm{n}}=\frac{2 \mathrm{mc} \lambda_{\mathrm{g}}^{2}}{\mathrm{h}}\left[1-\left(\frac{\lambda_{\mathrm{g}}}{\lambda_{\mathrm{n}}}\right)^{2}\right]^{-1}$


Q. If the series limit frequency of the Lyman series is $\mathrm{v}_{\mathrm{L}}$, then the series limit frequency of the Pfund series is :

(1) $16 \mathrm{v}_{\mathrm{L}}$

(2) $\mathrm{v}_{\mathrm{L}} / 16$

(3) $\mathrm{v}_{\mathrm{L}} / 25$

(4) $25 \mathrm{v}_{\mathrm{L}}$

[JEE Main-2018]

Sol. (3)


Magnetism – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Previous Years AIEEE/JEE Mains Questions

Q. Due to the presence of the current $\mathrm{I}_{1}$ at the origin:-

(1) The magnitude of the net force on the loop is given by $\frac{\mathrm{I}_{1} \mathrm{I}}{4 \pi} \mu_{0}\left[2(\mathrm{b}-\mathrm{a})+\frac{\pi}{3}(\mathrm{a}+\mathrm{b})\right]$

(2) The magnitude of the net force on the loop is given by $\frac{\mu_{0} \mathrm{I} \mathrm{I}_{1}}{24 \mathrm{ab}}(\mathrm{b}-\mathrm{a})$

(3) The forces on $\mathrm{AB}$ and $\mathrm{DC}$ are zero

(4) The forces on $\mathrm{AD}$ and $\mathrm{BC}$ are zero

[AIEEE – 2009]

Sol. (4)

For AD and BC, $\overrightarrow{\mathrm{B}} \square \overrightarrow{\mathrm{dL}}$. Hence force on AD and BC is zero.


Q. Two short bar magnets of length 1 cm each have magnetic moments $1.20 \mathrm{Am}^{2}$ and $1.00 \mathrm{Am}^{2}$ respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint O of the line joining their centres is close to :-

(Horizontal component of earth’s magnetic induction is $\left.3.6 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}\right)$

(1) $3.6 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}$

(2) $2.56 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

(3) $3.50 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

(4) $5.80 \times 10^{-4} \mathrm{Wb} / \mathrm{m}^{2}$

[JEE(Mains) – 2013]

Sol. (2)


Q. The coercivity of a small magnet where the ferromagnet gets demagnetized is $3 \times 10^{3} \mathrm{A} \mathrm{m}^{-1}$. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is :

(1) 3A          (2) 6 A            (3) 30 mA             (4) 60 mA

[JEE(Mains) – 2014]

Sol. (1)


Q. Hysteresis loops for two magnetic materials A and B are given below :

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use ;

(1) B for electromagnets and transformers.

(2) A for electric generators and transformers.

(3) A for electromagnets and B for electric transformers.

(4) A for transformers and B for electric generators.

[JEE(Mains) – 2016]

Sol. (1)

For electromagnet and transformers, we require the core that can be magnitised and demagnetised quickly when subjected to alternating current. From the given graphs, graph B is suitable.


Q. A magnetic needle of magnetic moment $6.7 \times 10^{-2} \mathrm{Am}^{2}$ and moment of inertia $7.5 \times 10^{-6} \mathrm{kg} \mathrm{m}^{2}$ is performing simple harmonic oscillations in a magnetic field of 0.01 T.

Time taken for 10 complete oscillations is :

(1) 6.98 s

(2) 8.76 s

(3) 6.65 s

(4) 8.89 s

[JEE(Mains) – 2017]

Sol. (3)


Magnetic Effect of Current – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Kinetic Theory of Gases – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. One kg of a diatomic gas is at a pressure of $8 \times 10^{4} \mathrm{N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion ?

(1) $6 \times 10^{4} \mathrm{J}$

(2) $7 \times 10^{4} \mathrm{J}$

(3) $3 \times 10^{4} \mathrm{J}$

(4) $5 \times 10^{4} \mathrm{J}$

Directions : Question number 11, 12 and 13 are based on the following paragraph.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram.

[AIEEE-2009]

Sol. (3 )

$\frac{P M}{R T}=\rho$

$\frac{P}{\rho}=\frac{R T}{M}$

$\mathbf{U}=\frac{3}{2} n R T=\frac{3}{2} m\left(\frac{R T}{M}\right)$


Q. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats $\gamma$. It is moving with speed v and is suddenly broght to rest. Assuming no heat is lost to the surroundings, its temperature increases by :-

(1) $\frac{\gamma \mathrm{Mv}^{2}}{2 \mathrm{R}}$

(2) $\frac{(\gamma-1)}{2 \mathrm{R}} \mathrm{Mv}^{2}$

(3) $\frac{(\gamma-1)}{2(\gamma+1) \mathrm{R}} \mathrm{Mv}^{2}$

(4) $\frac{(\gamma-1)}{2 \gamma \mathrm{R}} \mathrm{Mv}^{2}$

[AIEEE-2011]

Sol. (2)

$\frac{1}{2} m v^{2}=\frac{n R \Delta T}{\gamma-1}$

$\Delta \mathrm{T}=\frac{1}{2} \frac{m}{n} \frac{(\gamma-1) v^{2}}{R}$


Q. Three perfect gases at absolute temperatures $\mathrm{T}_{1}, \mathrm{T}_{2}$ and $\mathrm{T}_{3}$ are mixed. The masses of molecules are m1, m2, and m3 and the number of molecules are $\mathrm{n}_{1}, \mathrm{n}_{2}$ and $\mathrm{n}_{3}$ respectively. Assuming no loss of energy, then final temperature of the mixture is :-

(1) $\frac{n_{1} \mathrm{T}_{1}^{2}+\mathrm{n}_{2} \mathrm{T}_{2}^{2}+\mathrm{n}_{3} \mathrm{T}_{3}^{2}}{\mathrm{n}_{1} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{T}_{3}}$

(2) $\frac{n_{1}^{2} T_{1}^{2}+n_{2}^{2} T_{2}^{2}+n_{3}^{2} T_{3}^{2}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}$

(3) $\frac{\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}}{3}$

(4) $\frac{\mathrm{n}_{1} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$

[AIEEE-2011]

Sol. (4)

$\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{1}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{2}+\mathrm{n}_{3} \mathrm{C}_{\mathrm{V}} \mathrm{T}_{3}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}\right) \mathrm{C}_{\mathrm{v}} \mathrm{T}$


Q. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open of the tube is then closed and sealed and the tube is raised vertically up by addition 46 cm. What will be length of the air column above mercury in the tube now ?

(Atmospheric pressure = 76 cm of Hg)

(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm

[jEE-Mains-2014]

Sol. (3)


Q. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $\mathbf{V} \mathbf{q}$, where V is the volume of the gas. The value of q is :- $\left(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)$

(1) $\frac{\gamma+1}{2}$

( 2)$\frac{\gamma-1}{2}$

(3) $\frac{3 y+5}{6}$

(4) $\frac{3 \gamma-5}{6}$

[jEE-Mains-2015]

Sol. (1)


Q. The temperature of an open room of volume $30 \mathrm{m}^{3}$ increases from $17^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$ due to sunshine. The atmospheric pressure in the room remains $1 \times 10^{5} \mathrm{Pa} .$ If $\mathrm{n}_{1}$ and $\mathrm{n}_{\mathrm{f}}$ are the number of molecules in the room before and after heating, then $\mathrm{n}_{\mathrm{f}}-\mathrm{n}_{\mathrm{i}}$will be :-

(1) $2.5 \times 10^{25}$

(2) $-2.5 \times 10^{25}$

(3) $-1.61 \times 10^{23}$

(4) $1.38 \times 10^{23}$

[jEE-Mains-2017]

Sol. (2)


Q. The mass of a hydrogen molecule is $3.32 \times 10^{-27} \mathrm{kg} .$ If $102^{23}$ hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of $45^{\circ}$ to the normal, and rebound elastically with a speed of $10^{3}$ m/s, then the pressure on the wall is nearly :

(1) $4.70 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}$

(2) $2.35 \times 10^{2} \mathrm{N} / \mathrm{m}^{2}$

(3) $4.70 \times 10^{2} \mathrm{N} / \mathrm{m}^{2}$

(4) $2.35 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}$

[jEE-Mains-2018]

Sol. (4)


Heat Transfer – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length x of the bar from its hot end is best described by which of the following figures ?

[AIEEE – 2009]

Sol. (4)

The heat flow rate is given by

$\frac{d Q}{d t}=\frac{k A\left(\theta_{1}-\theta\right)}{x}$

$\Rightarrow \theta=\theta_{1}-\frac{x}{k A} \frac{d Q}{d t}$

where $\theta_{1}$ is the temp of hot end and $\theta$ is temp at a distance x from hot end.


Q. A liquid in a beaker has temperature $\theta(\mathrm{t})$ at time t and $\theta_{0}$ is temperature of surroundings, then according to Newton’s law of cooling the correct graph between $\log _{\mathrm{e}}\left(\theta-\theta_{0}\right)$ and t is :-

[AIEEE 2012]

Sol. (2)

Newtons law of cooling

$\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)$

$\Rightarrow \frac{d \theta}{\left(\theta-\theta_{0}\right)}=-k d t$

Integrating

$\ln \left(\theta-\theta_{0}\right)=-k t+C$


Q. If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_{0}$ the graph between the temperature T of the metal and time t will be closed to:

[JEE-Main- 2013]

Sol. (3)

According to Newtons law of cooling, The temp goes on decreasing with time non-linearly.


Q. Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = $4 \mathrm{cm}^{2}$. End of copper rod is maintained at $100^{\circ} \mathrm{C}$ where as ends of brass and steel are kept at $0^{\circ} \mathrm{C}$. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :

(1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s

[JEE-Main-2014]

Sol. (1)

Heat flow per unit time through copper rod

$=\frac{(100-40)}{\ell_{\mathrm{c}}}\left(\mathrm{K}_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}\right)$

$=\frac{60}{46} \times 0.92 \times 4$

$=4.8 \mathrm{cal} / \mathrm{s}$


Gravitation – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :-

(1) $\frac{\mathrm{R}}        {2}$ (2) $\sqrt{2} \mathrm{R}$           (3) 2R             (4) $\frac{\mathrm{R}}{\sqrt{2}}$

[AIEEE – 2009]

Sol. (3)

$\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}} \Rightarrow\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}=\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=9$

$\Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3 \Rightarrow \mathrm{h}=2 \mathrm{R}$


Q. Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is :-

$(1)-\frac{6 \mathrm{Gm}}{\mathrm{r}}$

$(2)-\frac{9 \mathrm{Gm}}{\mathrm{r}}$

(3) zero

$(4)-\frac{4 \mathrm{Gm}}{\mathrm{r}}$

[AIEEE – 2011]

Sol. (2)


Q. Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is:-

(1) $\sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$         (2) $\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{R}}}$         (3) $\sqrt{\frac{\mathrm{Gm}}{3 \mathrm{R}}}$           (4) $\sqrt{\frac{\mathrm{Gm}}{2 \mathrm{R}}}$

[AIEEE-2011]

Sol. (2)


Q. The mass of a spaceship is 1000 kg. It is to be launched from the earth’s surface out into free space. The value of ‘g’ and ‘R’ (radius of earth) are 10 m/s2 and 6400 km respectively. The required energy for this work will be :-

(1) $6.4 \times 10^{10}$ Joules

(2) $6.4 \times 10^{11}$ Joules

(3) $6.4 \times 10^{8}$ Joules

(4) $6.4 \times 10^{9}$ Joules

[AIEEE-2012]

Sol. (1)

$\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}}$

$-\mathrm{mgR}+\mathrm{KE}_{\mathrm{i}}=0+0$

$\mathrm{KE}_{\mathrm{i}}=+\mathrm{mgR}=1000 \times 10 \times 6.4 \times 10^{6}$

work done $=6.4 \times 10^{10} \mathrm{J}$


Q. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

(1) $\frac{5 \mathrm{GmM}}{6 \mathrm{R}}$         (2) $\frac{2 \mathrm{GmM}}{3 \mathrm{R}}$        (3) $\frac{\mathrm{GmM}}{2 \mathrm{R}}$        (4) $\frac{\mathrm{GmM}}{3 \mathrm{R}}$

[JEE-Mains 2013]

Sol. (1)

From energy conservation

$\frac{\mathrm{GMm}}{\mathrm{R}}+\mathrm{KE}=\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{1}{2} \mathrm{mV}^{2} \ldots(\mathrm{i})$ …(i)

From force balance at A,

$\frac{\mathrm{GMm}}{(3 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{3 \mathrm{R}} \Rightarrow \mathrm{V}^{2}=\frac{\mathrm{GM}}{3 \mathrm{R}}$ ………..(ii)

from (i) & (ii)

$\mathrm{KE}_{\text {suface }}=\frac{5}{6} \frac{\mathrm{GMm}}{\mathrm{R}}$


Q. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :

(1) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$

(2) $\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$

(3) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

(4) $\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$

[JEE-Mains 2014]

Sol. (2)

Net force on one particle

$\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{1}+2 \mathrm{F}_{2} \cos 45^{\circ}=$ Centripetal force

$\Rightarrow \frac{\mathrm{GM}^{2}}{(2 \mathrm{R})^{2}}+\left[\frac{2 \mathrm{GM}^{2}}{(\sqrt{2} \mathrm{R})^{2}} \cos 45^{\circ}\right]=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$

$\mathrm{V}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$

$\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$

$=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$


Q. From a solid sphere of mass M and radius R, a spherical portion of radius $\frac{\mathrm{R}}{2}$ is removed, as shown in the figure. Taking gravitational potential V = 0 at r = $\infty$, the potential at the centre of the cavity thus formed is : (G = gravitational constant)

(1) $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$

(2) $\frac{-2 \mathrm{GM}}{\mathrm{R}}$

(3) $\frac{-\mathrm{GM}}{2 \mathrm{R}}$

(4) $\frac{-\mathrm{GM}}{\mathrm{R}}$

[JEE-Mains 2015]

Sol. (4)

By principle of superosition

$\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$

$=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$


Q. A satellite is reolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere).

(1) $\sqrt{\mathrm{gR}}(\sqrt{2}-1)$

(2) $\sqrt{2 \mathrm{gR}}$

(3) $\sqrt{\mathrm{gR}}$

(4) $\sqrt{\mathrm{gR} / 2}$

[JEE-Mains 2016]

Sol. (1)

$\mathrm{V}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{\mathrm{gR}}$

$\mathrm{V}_{\mathrm{e}} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{2 \mathrm{gR}}$

$\therefore$ Increase in velocity $=\sqrt{\mathrm{gR}}[\sqrt{2}-1]$


Q. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth’s radius) :-

[JEE-Mains 2017]

Sol. (2)

$\mathrm{g}=\frac{\mathrm{GMx}}{\mathrm{R}^{3}}$ inside the Earth (straight line)

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ outside the Earth

where M is Mass of Earth

option (2)


Ray Optics – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. A transparent solid cyclindrical rod has a refractive index of . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $\theta$ for which the light ray grazes along the wall of the rod is :-

( 1) $\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(2) $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(3) $\sin ^{-1}\left(\frac{1}{2}\right)$

( 4) $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

[AIEEE – 2009]

Sol. (2)


Q. A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is:-

(1) 10 m/s

(2) 15 m/s

(3) $\frac{1}{10} \mathrm{m} / \mathrm{s}$

(4) $\frac{1}{15} \mathrm{m} / \mathrm{s}$

[AIEEE- 2011]

Sol. (4)

$V_{i z f r}=-\left[\frac{f}{f-u}\right]^{2} V_{\text {olrq }}=-\left[\frac{20}{20+280}\right]^{2} \times 15$


Q. When monochromatic red light is used instead of blue light in a convex lens, its focal length will :-

(1) Does not depend on colour of light

(2) Increase

(3) Decrease

(4) Remain same

[AIEEE-2011]

Sol. (2)


Q. A beaker contains water up to a height $\mathrm{h}_{1}$ and kerosene of height $\mathrm{h}_{2}$ above watger so that the total height of (water + kerosene) is $\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$. Refractive index of water is $\mu_{1}$ and that of kerosene is $\mu_{2}$. The apparent shift in the position of the bottom of the beaker when viewed from above is :-

$(1)\left(1-\frac{1}{\mu_{1}}\right) \mathrm{h}_{2}+\left(1-\frac{1}{\mu_{2}}\right) \mathrm{h}_{1}$

(2) $\left(1+\frac{1}{\mu_{1}}\right) \mathrm{h}_{1}-\left(1+\frac{1}{\mu_{2}}\right) \mathrm{h}_{2}$

(3) $\left(1-\frac{1}{\mu_{1}}\right) \mathrm{h}_{1}+\left(1-\frac{1}{\mu_{2}}\right) \mathrm{h}_{2}$

$(4)\left(1+\frac{1}{\mu_{1}}\right) \mathrm{h}_{2}-\left(1+\frac{1}{\mu_{2}}\right) \mathrm{h}_{1}$

[AIEEE- 2011]

Sol. (3)


Q. An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film ?

(1) 5.6 m (2) 7.2 m (3) 2.4 m (4) 3.2 m

[AIEEE- 2012]

Sol. (1)


Q. Diameter of a plano-convex lens is 6cm and thickness at the centre is 3 mm. If speed of light in material of lens is $2 \times 10^{8} \mathrm{m} / \mathrm{s}$, the focal length of the lens is :

(1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm

[JEE-Mains- 2013]

Sol. (3)

$\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\left(\frac{3 \times 10^{8}}{2 \times 10^{8}}-1\right)\left(\frac{1}{15}-\frac{1}{\infty}\right)$

f = 30 cm


Q. The graph between angle of deviation $(\delta)$ and angle of incidence (i) for a triangular prism is represented by :-

[JEE-Mains- 2013]

Sol. (3)

Refer NCERT


Q. A thin convex lens made from crown glass $\left(\mu=\frac{3}{2}\right)$ has focal length ƒ. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$ and , it has the focal length $f_{1}$ and $f_{2}$ respectively. The correct relation between the focal lengths is :

(1) $f_{2}>f$ and $f_{1}$ becomes negative

(2) $f_{1}$ and $f_{2}$ both become negative

(3) $f_{1}=f_{2}<f$

(4) $f_{1}>f$ and $f_{2}$ become negative

[JEE-Mains- 2014]

Sol. (4)


Q. A green light is incident from the water to the air – water interface at the critical angle ($\theta$). Select the correct statement.

(1) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

(2) The entire spectrum of visible light will come out of the water at various angles to the normal

(3) The entire spectrum of visible light will come out of the water at an angle of $90^{\circ}$ to the normal.

(4) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

[JEE-Mains- 2014]

Sol. (4)

Frequency of light () > frequency of green light ($v_{G}$) µ is also greater than $\mu_{\mathrm{G}}$ and critical angle of light is less than green light therefore light will got total internal reflaction and not come out to the air. For frequency of light $(v)<v_{\mathrm{G}}$ ; light will not suffer T.I.R. Therefore light come out to the air


Q. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is $\mu$, a ray, incident at an angle $\theta$, on the face AB would get transmitted through the face AC of the prism provided :

$(1) \theta>\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(2) $\theta<\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(3) $\theta>\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(4) $\theta<\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

[JEE-Mains- 2015]

Sol. (3)

Apply Snell’s law at face AB

$1 \sin \theta=\mu \sin \left(r_{1}\right)$

$\theta=\sin ^{-1}\left(\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right.$

for all light transmitted through $\mathrm{AC}, \mathrm{e}<90^{\circ}$

$\Rightarrow \theta>\sin ^{-1}\left(\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right.$


Q. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :

(1) 20 times nearer

(2) 10 times taller

(3) 10 times nearer

(4) 20 times taller

[JEE-Mains- 2016]

Sol. (4)

Angular magnification is 20.


Q. In an experiment for determination of refractive index of glass of a prism by i – , plot, it was found that a ray incident at angle $35^{\circ}$, suffers a deviation of $40^{\circ}$ and that it emerges at angle $79^{\circ}$. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.8 (2) 1.5 (3) 1.6 (4) 1.7

[JEE-Mains- 2016]

Sol. (2)


Q. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is :

(1) real and at a distance of 40 cm from the divergent lens

(2) real and at a distance of 6 cm from the conver

(3) real and at a distance of 40 cm from convergent lens

(4) virtual and at a distance of 40 cm from convergent lens.

[JEE-Mains- 2017]

Sol. (3)

As parallel beam incident on diverging lens if forms virtual image at $\mathbf{V}_{1}$ = –25 cm from the diverging lense which works as a object for the converging lense (f = 20 cm)

So for converging lens u = –40 cm, f = 20 cm

$\therefore \quad$ Final image

$\frac{1}{\mathrm{V}}-\frac{1}{-40}=\frac{1}{20}$

V = 40 cm from converging lenses.


Alternating Current – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Q. In a series LCR circuit R = $200 \Omega$ and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lages behind the voltage by $30^{\circ}$. On taking out the inductor from the circuit the current leads the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is :

(1) 242 W

(2) 305 W

(3) 210 W

(4) Zero W

[AIEEE – 2010]

Sol. (1)


Q. A fully charged capacitor C with intial charge $\mathrm{q}_{0}$ is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is :-

(1) $2 \pi \sqrt{\mathrm{LC}}$

(2) $\sqrt{1 C}$

(3) $\pi \sqrt{\mathrm{LC}}$

(4) $\frac{\pi}{4} \sqrt{\mathrm{LC}}$

[AIEEE – 2011]

Sol. (4)


Q. In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, $\mathrm{S}_{2}$ kept open, (q is charge on the capacitor and $\tau=\mathrm{RC}$ is Capacitive time constant). Which of the following statement is correct?

(1) Work done by the battery is half of the energy dissipated in the resistor

(2) At t $=\tau, q=C \sqrt{/ 2}$

(3) At $\mathrm{t}=2 \pi, \mathrm{q}=\mathrm{CV}\left(1-\mathrm{e}^{-2}\right)$

(4) At $\mathrm{t}=\frac{\tau}{2}, \mathrm{q}=\mathrm{CV}\left(1-\mathrm{e}^{-1}\right)$

[JEE Main-2013]

Sol. (3)


Q. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to $\mathrm{Q}_{0}$ and then connected to the L and R as shown below. If a student plots graphs of the square of maximum charge on the capacitor with time (t) for two different values $\mathrm{L}_{1}$ and $\mathrm{L}_{2}\left(\mathrm{L}_{1}>\mathrm{L}_{2}\right)$ of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

[JEE Main-2015]

Sol. (3)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

 for


Q. An arc lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :-

(1) 0.065 H

(2) 80 H

(3) 0.08 H

(4) 0.044 H

[JEE Main-2016]

Sol. (1)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

 for


Q. For an RLC circuit driven with voltage of amplitude $\mathrm{v}_{\mathrm{m}}$ and frequency $\mathrm{w}_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$ the current exhibits resonance. The quality factor, Q is given by :

(1) $\frac{\omega_{0} \mathrm{R}}{\mathrm{L}}$

( 2)$\frac{\mathrm{R}}{\left(\omega_{0} \mathrm{C}\right)}$

(3) $\frac{\mathrm{CR}}{\omega_{0}}$

(4) $\frac{\omega_{0} \mathrm{L}}{\mathrm{R}}$

[JEE Main-2018]

Sol. (4)

Quality factor $=\frac{\omega_{0} \mathrm{L}}{\mathrm{R}}$


Q. In an a. c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t i = 20

sin $\left(30 t-\frac{\pi}{4}\right)$

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively.

(1) $\frac{1000}{\sqrt{2}}, 10$

(2) $\frac{50}{\sqrt{2}}, 0$

(3) 50, 0

(4) 50, 10

[JEE Main-2018]

Sol. (1)


Calorimetry – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

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Simulator

 

Previous Years AIEEE/JEE Mains Questions

 

Q. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose $t_{1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_{1} / t_{2}$ will be

(1) 2

(2) 1

(3) 1/2

(4) 1/4

[AIEEE-2010]

Sol. (3)


Q. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g $\mathrm{cm}^{-3}$, the angle remains the same. If density of the material of the sphere is $1.6 \mathrm{g} \mathrm{cm}^{-3}$, the dielectric constant of the liquid is :

(1) 1

(2) 4

(3) 3

(4) 2

[AIEEE – 2010]

Sol. (4)

Therefore K =2


Q. A resistor ‘R’ and $2 \mu \mathrm{F}$ capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5s after the switch has been closed. ($\log _{10}$ 2.5 = 0.4)

(1) $2.7 \times 10^{6} \Omega$

(2) $3.3 \times 10^{7} \Omega$

(3) $1.3 \times 10^{4} \Omega$

(4) $1.7 \times 10^{5} \Omega$

[AIEEE-2011]

Sol. (2)

Neon bulb is filled with gas, so its resistance is infinite, hence no current flows through it.

Now, $\mathrm{V}_{\mathrm{c}}=\mathrm{E}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow 120=200\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=\frac{2}{5} \Rightarrow \mathrm{t}=\mathrm{RCln} 2.5$

$\Rightarrow \mathrm{R}=\frac{\mathrm{t}}{\mathrm{Cln} 2.5}=\frac{\mathrm{t}}{2.303 \mathrm{Clog} 2.5}$

$=2.7 \times 10^{6} \Omega$


Q. Combination of two identical capacitors, a resistor R and a dc voltage source of voltage 6V is used in an experiment on (C–R) circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 second. For series combination the time needed for reducing the voltage of the fully charged series combination by half is :-

(1) 20 second

(2) 10 second

(3) 5 second

(4) 2.5 second

[AIEEE-2011]

Sol. (4)


Q. The figure shows an experimental plot for discharging of a capacitor in an R–C circuit. The time constant t of this circuit lies between:-

(1) 100 sec and 150 sec

(2) 150 sec and 200 sec

(3) 0 and 50 sec

(4) 50 sec and 100 sec

[AIEEE-2012]

Sol. (1)

$\mathrm{t}=0.37 \% \mathrm{of} \mathrm{V}_{0}$

$=0.37 \times 25=9.25$ volt

where is in between 100 and 150 sec.


Q. Two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are charged to 120V and 200V respectively. It is found that by connecting them together the potential on each one can be made zero. Then :

(1) $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$

(2) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$

(3) $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$

(4) $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$

[JEE-Mains-2013]

Sol. (2)

Common voltage $=\frac{\mathrm{C}_{1} \mathrm{v}_{1}-\mathrm{C}_{2} \mathrm{V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

(positive plate of one capacitor is connected with negative plate of second capacitor)

$\Rightarrow 120 \mathrm{C}_{1}=200 \mathrm{C}_{2} \Rightarrow 3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$


Q. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectriic of dielectric constant 2.2 between them. When the electric field in the dielectric field in the dielectric is $3 \times 10^{4}$ V/m, the charge density of the positive plate will be close

to :

(1) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(2) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(3) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

(4) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

[JEE-Mains-2014]

Sol. (3)

$\frac{\sigma}{\mathrm{K} \varepsilon_{0}}=3 \times 10^{4}$

$\frac{\sigma}{2.25 \times 8.86 \times 10^{-12}}=3 \times 10^{4}$

$\sigma=6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$


Q. In the given circuit, charge $\mathrm{Q}_{2}$ on the 2µF capacitor changes as C is varied from $1 \mu \mathrm{F}$ to $3 \mu \mathrm{F} \cdot \mathrm{Q}_{2}$ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale) :-

[JEE-Mains-2015]

Sol. (4)

$\mathrm{C}_{\mathrm{eq}}=\frac{3 \mathrm{C}}{3+\mathrm{C}} \quad \mathrm{Q}=\mathrm{Ceq} \mathrm{E}$

$\mathrm{Q}_{2}=\frac{2 \mathrm{CE}}{(\mathrm{C}+3)}=\frac{3 \mathrm{CE}}{(\mathrm{C}+3)}$

$=2 \mathrm{E}\left[1-\frac{3}{\mathrm{C}+3}\right]$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=1_{\mathrm{HF}}=2 \mathrm{E}\left[\frac{1}{4}\right]=\frac{\mathrm{E}}{2}$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=3_{1 \mathrm{F}}=2 \mathrm{E}\left[\frac{1}{2}\right]=\mathrm{E}$

$\frac{\mathrm{dQ}}{\mathrm{d} \mathrm{C}}=-\frac{2 \mathrm{E} \cdot 3}{(\mathrm{C}+3)}(-1)=\frac{6 \mathrm{E}}{(\mathrm{C}+3)^{2}}>0$

$\frac{\mathrm{d}^{2} \theta}{\mathrm{d} \mathrm{C}^{2}}=\frac{6 \mathrm{E}(-2)}{(\mathrm{C}+3)^{3}}=-\frac{12 \mathrm{E}}{(\mathrm{C}+3)^{2}}<0$


Q. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the $4 \mu \mathrm{F}$ and $9 \mu \mathrm{F}$capacitors), at a point 30 m from it , would equal:

(1) 480 N/C

(2) 240 N/C

(3) 360 N/C

(4) 420 N/C

[JEE-Mains-2016]

Sol. (4)

$\mathrm{Q}=24+18=42 \mu \mathrm{c}$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{E}=\frac{9 \times 10^{9} \times 42 \times 10^{-6}}{(30)^{2}}=420 \mathrm{N} / \mathrm{C}$


Q. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

(1) $\mathrm{CE} \frac{\mathrm{r}_{2}}{\left(\mathrm{r}+\mathrm{r}_{2}\right)}$

(2) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{1}+\mathrm{r}\right)}$

(3) CE

(4) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{2}+\mathrm{r}\right)}$

[JEE-Mains-2017]

Sol. (1)

It steady state, current through AB = 0


Q. A capacitance of 2 $\mu \mathrm{F}$ is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $\mu \mathrm{F}$ capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

(1) 24

(2) 32

(3) 2

(4) 16

[JEE-Mains-2017]

Sol. (2)

To hold 1 KV potential differenceminimum four capacitors are required in series

$\Rightarrow \quad \mathrm{C}_{1}=\frac{1}{4}$ for one series. So for $\mathrm{Ceq}$ to be $2 \mu \mathrm{F}, 8$ parallel combinations are required.

$\Rightarrow$ Minimum no. of capacitors $=8 \times 4=32$


Q. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant $\mathrm{K}=\frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be :-

(1) 0.3 n C

(2) 2.4 n C

(3) 0.9 n C

(4) 1.2 n C

[JEE-Mains-2018]

Sol. (4)

$\mathrm{Q}=(\mathrm{k} \mathrm{C}) \mathrm{V}$

$=\left(\frac{5}{3} \times 90 \mathrm{pF}\right)(20 \mathrm{V})$

= 3000 pC

= 3nC

induced charges on dielectric

$\mathrm{Q}_{\mathrm{ind}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{K}}\right)=3 \mathrm{nC}\left(1-\frac{3}{5}\right)=1.2 \mathrm{nC}$


Capacitor – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Simulator

Previous Years AIEEE/JEE Main Questions

Q. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose $t_{1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_{1} / t_{2}$ will be

(1) 2

(2) 1

(3) 1/2

(4) 1/4

[AIEEE-2010]

Sol. (3)


Q. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g $\mathrm{cm}^{-3}$, the angle remains the same. If density of the material of the sphere is $1.6 \mathrm{g} \mathrm{cm}^{-3}$, the dielectric constant of the liquid is :

(1) 1

(2) 4

(3) 3

(4) 2

[AIEEE – 2010]

Sol. (4)

Therefore K =2


Q. A resistor ‘R’ and $2 \mu \mathrm{F}$ capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5s after the switch has been closed. ($\log _{10}$ 2.5 = 0.4)

(1) $2.7 \times 10^{6} \Omega$

(2) $3.3 \times 10^{7} \Omega$

(3) $1.3 \times 10^{4} \Omega$

(4) $1.7 \times 10^{5} \Omega$

[AIEEE-2011]

Sol. (2)

Neon bulb is filled with gas, so its resistance is infinite, hence no current flows through it.

Now, $\mathrm{V}_{\mathrm{c}}=\mathrm{E}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow 120=200\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=\frac{2}{5} \Rightarrow \mathrm{t}=\mathrm{RCln} 2.5$

$\Rightarrow \mathrm{R}=\frac{\mathrm{t}}{\mathrm{Cln} 2.5}=\frac{\mathrm{t}}{2.303 \mathrm{Clog} 2.5}$

$=2.7 \times 10^{6} \Omega$


Q. Combination of two identical capacitors, a resistor R and a dc voltage source of voltage 6V is used in an experiment on (C–R) circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 second. For series combination the time needed for reducing the voltage of the fully charged series combination by half is :-

(1) 20 second

(2) 10 second

(3) 5 second

(4) 2.5 second

[AIEEE-2011]

Sol. (4)


Q. The figure shows an experimental plot for discharging of a capacitor in an R–C circuit. The time constant t of this circuit lies between:-

(1) 100 sec and 150 sec

(2) 150 sec and 200 sec

(3) 0 and 50 sec

(4) 50 sec and 100 sec

[AIEEE-2012]

Sol. (1)

$\mathrm{t}=0.37 \% \mathrm{of} \mathrm{V}_{0}$

$=0.37 \times 25=9.25$ volt

where is in between 100 and 150 sec.


Q. Two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are charged to 120V and 200V respectively. It is found that by connecting them together the potential on each one can be made zero. Then :

(1) $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$

(2) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$

(3) $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$

(4) $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$

[JEE-Mains-2013]

Sol. (2)

Common voltage $=\frac{\mathrm{C}_{1} \mathrm{v}_{1}-\mathrm{C}_{2} \mathrm{V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

(positive plate of one capacitor is connected with negative plate of second capacitor)

$\Rightarrow 120 \mathrm{C}_{1}=200 \mathrm{C}_{2} \Rightarrow 3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$


Q. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectriic of dielectric constant 2.2 between them. When the electric field in the dielectric field in the dielectric is $3 \times 10^{4}$ V/m, the charge density of the positive plate will be close

to :

(1) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(2) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(3) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

(4) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

[JEE-Mains-2014]

Sol. (3)

$\frac{\sigma}{\mathrm{K} \varepsilon_{0}}=3 \times 10^{4}$

$\frac{\sigma}{2.25 \times 8.86 \times 10^{-12}}=3 \times 10^{4}$

$\sigma=6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$


Q. In the given circuit, charge $\mathrm{Q}_{2}$ on the 2µF capacitor changes as C is varied from $1 \mu \mathrm{F}$ to $3 \mu \mathrm{F} \cdot \mathrm{Q}_{2}$ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale) :-

[JEE-Mains-2015]

Sol. (4)

$\mathrm{C}_{\mathrm{eq}}=\frac{3 \mathrm{C}}{3+\mathrm{C}} \quad \mathrm{Q}=\mathrm{Ceq} \mathrm{E}$

$\mathrm{Q}_{2}=\frac{2 \mathrm{CE}}{(\mathrm{C}+3)}=\frac{3 \mathrm{CE}}{(\mathrm{C}+3)}$

$=2 \mathrm{E}\left[1-\frac{3}{\mathrm{C}+3}\right]$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=1_{\mathrm{HF}}=2 \mathrm{E}\left[\frac{1}{4}\right]=\frac{\mathrm{E}}{2}$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=3_{1 \mathrm{F}}=2 \mathrm{E}\left[\frac{1}{2}\right]=\mathrm{E}$

$\frac{\mathrm{dQ}}{\mathrm{d} \mathrm{C}}=-\frac{2 \mathrm{E} \cdot 3}{(\mathrm{C}+3)}(-1)=\frac{6 \mathrm{E}}{(\mathrm{C}+3)^{2}}>0$

$\frac{\mathrm{d}^{2} \theta}{\mathrm{d} \mathrm{C}^{2}}=\frac{6 \mathrm{E}(-2)}{(\mathrm{C}+3)^{3}}=-\frac{12 \mathrm{E}}{(\mathrm{C}+3)^{2}}<0$


Q. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the $4 \mu \mathrm{F}$ and $9 \mu \mathrm{F}$capacitors), at a point 30 m from it , would equal:

(1) 480 N/C

(2) 240 N/C

(3) 360 N/C

(4) 420 N/C

[JEE-Mains-2016]

Sol. (4)

$\mathrm{Q}=24+18=42 \mu \mathrm{c}$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{E}=\frac{9 \times 10^{9} \times 42 \times 10^{-6}}{(30)^{2}}=420 \mathrm{N} / \mathrm{C}$


Q. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

(1) $\mathrm{CE} \frac{\mathrm{r}_{2}}{\left(\mathrm{r}+\mathrm{r}_{2}\right)}$

(2) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{1}+\mathrm{r}\right)}$

(3) CE

(4) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{2}+\mathrm{r}\right)}$

[JEE-Mains-2017]

Sol. (1)

It steady state, current through AB = 0


Q. A capacitance of 2 $\mu \mathrm{F}$ is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $\mu \mathrm{F}$ capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

(1) 24

(2) 32

(3) 2

(4) 16

[JEE-Mains-2017]

Sol. (2)

To hold 1 KV potential differenceminimum four capacitors are required in series

$\Rightarrow \quad \mathrm{C}_{1}=\frac{1}{4}$ for one series. So for $\mathrm{Ceq}$ to be $2 \mu \mathrm{F}, 8$ parallel combinations are required.

$\Rightarrow$ Minimum no. of capacitors $=8 \times 4=32$


Q. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant $\mathrm{K}=\frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be :-

(1) 0.3 n C

(2) 2.4 n C

(3) 0.9 n C

(4) 1.2 n C

[JEE-Mains-2018]

Sol. (4)

$\mathrm{Q}=(\mathrm{k} \mathrm{C}) \mathrm{V}$

$=\left(\frac{5}{3} \times 90 \mathrm{pF}\right)(20 \mathrm{V})$

= 3000 pC

= 3nC

induced charges on dielectric

$\mathrm{Q}_{\mathrm{ind}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{K}}\right)=3 \mathrm{nC}\left(1-\frac{3}{5}\right)=1.2 \mathrm{nC}$


Center of Mass – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Simulator

 

Previous Years AIEEE/JEE Mains Questions

Q. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be :-

Directions : Question number 9 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best discribes the two statements.

[AIEEE – 2009]

Sol. (1)


Q. Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions.

(1) Statement–1 is true, Statement–2 is false

(2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–1

(3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation

of Statement–1

(4) Statement–1 is false, Statement–2 is true

[AIEEE – 2010]

Sol. (2)


Q. This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements.

Statement – I : A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as $\mathrm{f}\left(\frac{1}{2}\mathrm{mv}^{2}\right)$then$\mathrm{f}=\left(\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}\right)$

Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.

(1) Statement–I is true, Statement–II is true, Statement–II is a correct explanation of Statement–I.

(2) Statement–I is true, Statement–II is true, Statement–II is a not correct explanation of

Statement–I.

(3) Statement–I is true, Statement–II is false.

(4) Statement–I is false, Statement–II is true

[JEE Mains-2013]

Sol. (4)

Energy loss will be maximum when collision will be perfectly inelastic

(By momentum)

Maximum energy loss $=\mathrm{K}_{\mathrm{i}}-\mathrm{K}_{\mathrm{f}}$

$=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \mathrm{v}_{\mathrm{f}}^{2}$

$=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \frac{\mathrm{m}^{2} \mathrm{v}^{2}}{(\mathrm{m}+\mathrm{M})^{2}}$

$=\frac{1}{2} \mathrm{mv}^{2}\left[1-\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}}\right]$

$=\left(\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}\right) \frac{1}{2} \mathrm{mv}^{2}$

statement 1 is false.


Q. A particle of mass m moving in the x direction with speed 2u is hit by another particle of mass 2m moving in the y direction with speed u. If the collisions perfectly inelastic , the percentage loss in the energy during the collision is close to :

(1) 56 % (2) 62% (3) 44% (4) 50%

[JEE Mains-2015]

Sol. (1)

Before collison

Kinetic energy $=\frac{1}{2} \mathrm{m}(2 \mathrm{v})^{2} \times \frac{1}{2} 2 \mathrm{m}(\mathrm{v})^{2}$

$=3 \mathrm{mv}^{2}$

After collison

Applying momentum conservation for inelastic collision

$2 \mathrm{mv} \hat{\mathrm{j}}+\mathrm{m} 2 \mathrm{v} \hat{\mathrm{i}}=3 \mathrm{m} \overrightarrow{\mathrm{v}}_{\mathrm{f}}$

$\left|\overrightarrow{\mathrm{v}}_{\mathrm{f}}\right|=\sqrt{\frac{8}{9}} \mathrm{v}$

$\mathrm{K}_{\mathrm{f}}=\frac{1}{2} \times 3 \mathrm{m} \times\left(\mathrm{v}_{\mathrm{f}}^{2}\right)=\frac{4 \mathrm{mv}^{2}}{3}$

$\% \Delta \mathrm{K}=\frac{\mathrm{K}_{1}-\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{i}}} \times 100=\frac{5 \mathrm{mv}^{2} / 3}{3 \mathrm{mv}^{2}} \times 100=\frac{5}{9} \times 100=56 \%$


Q. Distance of the centre of mass of a solid uniform cone from its vertex is $\mathrm{Z}_{0}$. If the radius of its base is R and its height is h then $\mathrm{Z}_{0}$ is equal to :-

(1) $\frac{5 \mathrm{h}}{8}$

(2) $\frac{3 \mathrm{h}^{2}}{8 \mathrm{R}}$

(3) $\frac{\mathrm{h}^{2}}{4 \mathrm{R}}$

(4) $\frac{3 \mathrm{h}}{4}$

[JEE Mains-2015]

Sol. (4)

for solid cone c.m. is $\frac{\mathrm{h}}{4}$ from base

so $\mathrm{z}_{0}=\mathrm{h}-\frac{\mathrm{h}}{4}=\frac{3 \mathrm{h}}{4}$


Q. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $\mathrm{P}_{\mathrm{d}}$ ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $\mathrm{P}_{\mathrm{c}}$. The values of $\mathrm{P}_{\mathrm{d}}$ and $\mathrm{P}_{\mathrm{c}}$ are respectively :

(1) (.28, .89)

(2) (0, 0)

(3) (0, 1)

(4) (.89, .28)

[JEE Mains-2018]

Sol. (4)

Let initial speed of neutron is $\mathrm{v}_{0}$ and kinetic energy is K.

1st collision :

by momentum conservation

$\mathrm{mv}_{0}=\mathrm{mv}_{1}+2 \mathrm{mv}_{2} \Rightarrow \mathrm{v}_{1}+2 \mathrm{v}_{2}=\mathrm{v}_{0}$

by $\mathrm{e}=1 \quad \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{0}$


Q. In a collinear collision, a particle with an initial speed $\mathrm{V}_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

(1) $\sqrt{2} \mathrm{v}_{0}$

(2)$\frac{\mathrm{v}_{0}}{2}$

(3) $\frac{\mathrm{v}_{0}}{\sqrt{2}}$

(4) $\frac{\mathrm{v}_{0}}{4}$

[JEE Mains-2018]

Sol. (1)

Initial


[/esquestion]

Current Electricity – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Simulator

Previous Years AIEEE/JEE Main Questions

Q. Statement–1 : The temperature dependence of resistance is usually given as $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{t})$. The resistance of a wire changes from $100 \Omega$ to $150 \Omega$ when its temperature is increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$. This implies that  = $2.5 \times 10^{-3} /^{\circ} \mathrm{C}$.

Statement–2 : $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{t})$ is valid only when the change in the temperature $\Delta \mathrm{T}$ is small and $\Delta \mathrm{R}=\left(\mathrm{R}-\mathrm{R}_{0}\right)<<\mathrm{R}_{0}$.

(1) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–1

[AIEEE – 2009]

Sol. (1)


Q. Two conductors have the same resistance at $0^{\circ} \mathrm{C}$ but their temperature coefficients of resistance are $\alpha_{1}$ and $\alpha_{2}$. The respective temperature coefficients of their series and parallel combinations are nearly :

(1) $\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(2) $\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}$

(3) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(4) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$

[AIEEE – 2010]

Sol. (1)

For series combination

$\alpha_{\mathrm{S}}=\frac{\alpha_{1} \mathrm{R}_{01}+\alpha_{2} \mathrm{R}_{02}}{\mathrm{R}_{01}+\mathrm{R}_{02}}$

$\mathrm{R}_{01}=\mathrm{R}_{02}=\mathrm{R}_{0}(\text { given })$

$\alpha_{\mathrm{S}}=\frac{\alpha_{1}+\alpha_{2}}{2}$

For parallel combination

$\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$

$\Rightarrow \frac{1}{\mathrm{R}_{e q}}=\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{1} \mathrm{t}\right)}+\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{2} \mathrm{t}\right)}$

$\frac{1}{\frac{\mathrm{R}_{0}}{2}\left(1+\alpha_{p} \mathrm{t}\right)}=\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{1} \mathrm{t}\right)}+\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{2} \mathrm{t}\right)}$

$2\left(1+\alpha_{p} t\right)^{-1}=\left(1+\alpha_{1} t\right)^{-1}+\left(1+\alpha_{2} t\right)^{-1}$

using binomial expansion

$2-2 \alpha_{\mathrm{p}} \mathrm{t}=1-\alpha_{1} \mathrm{t}+1-\alpha_{2} \mathrm{t} \Rightarrow \alpha_{\mathrm{p}}=\frac{\alpha_{1}+\alpha_{2}}{2}$


Q. If a wire is stretched to make it 0.1 % longer its resistance will :-

(1) decrease by 0.2%

(2) decrease by 0.05%

(3) increase by 0.05%

(4) increase by 0.2%

[AIEEE – 2011]

Sol. (4)

$\mathrm{R}=\rho \frac{\ell}{\mathrm{A}} \Rightarrow \mathrm{R} \alpha \ell^{2}$

$\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{2 \Delta \ell}{\ell}=2[0.1]=0.2 \%$ increase.


Q. If $400 \Omega$ of resistance is made by adding four $100 \Omega$ resistance of tolerance 5%, then the tolerance of the combination is :

(1) 20% (2) 5% (3) 10% (4) 15%

[AIEEE – 2011]

Sol. (2)


Q. The current in the primary circuit of a potentiometer is 0.2 A. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7}$ ohm metre and $8 \times 10^{-7} \mathrm{m}^{2}$ respectively. The potential gradient will be equal to :-

(1) 0.2 V/m (2) 1 V/m (3) 0.5 V/m (4) 0.1 V/m

[AIEEE – 2011]

Sol. (4)


Q. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is :-

(1) 3% (2) 6% (3) zero (4) 1%

[AIEEE – 2012]

Sol. (2)


Q. Two electric bulbs marked 25W-220 V and 100 W-220 V are connected in series to a 440 V supply. Which of the bulbs will fuse ?

(1) Neither (2) Both (3) 100 W (4) 25 W

[AIEEE – 2012]

Sol. (4)


Q. The supply voltage to a room is 120V. The resistance of the lead wires is $6 \Omega$A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

(1) zero Volt (2) 2.9 Volt (3) 13.3 Volt (4) 10.04 Volt

[JEE-Mains 2013]

Sol. (4)


Q. This question has Statement I and Statement II. Of the four choice given after the Statements, choose the one that best describes the two Statements.

Statement-I : Higher the range, greater is the resistance of ammeter.

Statement-II : To increase the range of ammeter, additional shunt needs to be used across it.

(1) Statement-I is true, Statement-II is true, Statement-II is the correct explanation of Statement- I

(2) Statement-I is true, Statement-II is true, Statement-II is not the correct explanation of Statement- I.

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

[JEE-Mains 2013]

Sol. (4)

To increase the range of ammeter, resistance should be decreased (So additional shunt connected in parallel) so total resistance to ammeter decreases.


Q. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

(1) 12 A (2) 14 A (3) 8 A (4) 10 A

[JEE-Mains 2014]

Sol. (1)

All devices are in parallel so total current drawn is gives as


Q. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is $2.5 \times 10^{-4} \mathrm{ms}^{-1}$. If the electron density in the wire is $8 \times 10^{28} \mathrm{m}^{-3}$, the resistivity of the material is close to :-

(1) $1.6 \times 10^{-6} \Omega \mathrm{m}$

(2) $1.6 \times 10^{-5} \Omega \mathrm{m}$

(3) $1.6 \times 10^{-8} \Omega \mathrm{m}$

(4) $1.6 \times 10^{-7} \Omega \mathrm{m}$

[JEE-Mains 2015]

Sol. (2)

$\mathrm{V} \quad=\mathrm{i} \mathrm{R}=\left(\mathrm{neAV}_{\mathrm{d}}\right)\left(\frac{\rho \ell}{\mathrm{A}}\right)$

$\mathrm{V}=\mathrm{neV}_{\mathrm{d}} \ell$

$\rho=\frac{\mathrm{V}}{\mathrm{neV}_{\mathrm{d}} \ell}$

On putting values are got the answer

$=1.6 \times 10^{-5} \Omega \mathrm{m}$


Q. In the circuit shown, the current in the $1 \Omega$ resistor is :-

(1) 0.13 A, from Q to P

(2) 0.13 A, from P to Q

(3) 1.3 A, from P to Q

(4) 0A

[JEE-Mains 2015]

Sol. (1)


Q. A galvanometer having a coil resistance of $100 \Omega$ gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10A, is :-

(1) $3 \Omega$

(2) $0.01 \Omega$

(3) $2 \Omega$

(4) $0.1 \Omega$

[JEE-Mains 2016]

Sol. (2)


Q. Which of the following statements is false ?

(1) A rheostat can be used as a potential divider

(2) Kirchhoff’s second law represents energy conservation

(3) Wheatstone bridge is the most sensitive when all the four resistances are of the same order

of magnitude.

(4) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.

[JEE-Mains 2017]

Sol. (4)

On interchanging Cell & Galvanometer.

On balancing condition

$\frac{\mathrm{R}_{1}}{\mathrm{R}_{3}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{4}}$ ….(1)


Q. When a current of 5 mA is passed through a galvanometer having a coil of resistance $15 \Omega$, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0 – 10 V is :-

(1) $2.535 \times 10^{3} \Omega$

(2) $4.005 \times 10^{3} \Omega$

(3) $1.985 \times 10^{3} \Omega$

(4) $2.045 \times 10^{3} \Omega$

[JEE-Mains 2017]

Sol. (3)

$10=\left(5 \times 10^{-3}\right)(15+\mathrm{R})$

$r=1985 \Omega$


Q. In the above circuit the current in each resistance is :-

(1) 0.5 A (2) 0 A (3) 1 A (4) 0.25 A

[JEE-Mains 2017]

Sol. (2)

Taking voltage of point A as = 0

Then voltage at other points can be written as shown in figure

Hence voltage across all resistance is zero.

Hence current = 0


Q. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of $5 \Omega$, a balance is found when the cell is connected across40 cm of the wire. Find the internal resistance of the cell.

(1) $1.5 \Omega$

(2) $2 \Omega$

(3) $2.5 \Omega$

(4) 1 $\Omega$

[JEE-Mains 2018]

Sol. (1)

without shunting condition :

On balancing

$\mathrm{E}_{\mathrm{s}}-\frac{\mathrm{E}_{\mathrm{s}}}{(\mathrm{r}+\mathrm{R})} \mathrm{r}=40 \times \mathrm{x} \quad \ldots .(2)$

On solving :


Q. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 $\mathrm{k} \Omega$. How much was the resistance on the left slot before interchanging the resistances ?

(1) $505 \mathrm{k\Omega}$

(2) $550 \mathrm{k} \Omega$

(3) $910 \mathrm{k} \Omega$

(4) $990 \mathrm{k} \Omega$

[JEE-Mains 2018]

Sol. (2)

$\mathrm{R}_{1}+\mathrm{R}_{2}=1000 \Rightarrow \mathrm{R}_{2}=1000-\mathrm{R}_{1}$

On balancing condition

$\mathrm{R}_{1}(100-\ell)=\left(1000-\mathrm{R}_{1}\right) \ell \quad \ldots(1)$

On Inter changing resistance

On balancing condition


Q. Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of $10 \Omega$. The internal resistances of the two batteries are $10 \Omega$ and $2 \Omega$respectively. The voltage across the load lies between002E

(1) 11.5 V and 11.6 V

(2) 11.4 V and 11.5 V

(3) 11.7 V and 11.8 V

(4) 11.6 V and 11.7 V

[JEE-Mains 2018]

Sol. (1)


Elasticity – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Previous Years AIEEE/JEE Mains Questions

Q. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of :

(1) 81

( 2)$\frac{1}{81}$

(3) 9

(4) $\frac{1}{9}$

[JEE-Main-2017]

Sol.

Stress $=\frac{\text { Force }}{\text { area }}=\frac{\mathrm{mg}}{\mathrm{A}}=\frac{\text { volume } \times \text { density } \times \mathrm{g}}{\text { Area }}$

Stress $=\frac{\mathrm{L}^{3} \rho \mathrm{g}}{\mathrm{L}^{2}}$

Stress $\propto \mathrm{L}$


Q. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left(\frac{\mathrm{dr}}{\mathrm{r}}\right)$, is :

(1) $\frac{\mathrm{Ka}}{3 \mathrm{mg}}$

(2) $\frac{\mathrm{mg}}{3 \mathrm{Ka}}$

(3) $\frac{\mathrm{mg}}{\mathrm{Ka}}$

(4) $\frac{\mathrm{Ka}}{\mathrm{mg}}$

[JEE-Main-2018]

Sol.

$\left[\text { Bulk Modulus }=\frac{\text { volumetric stress }}{\text { volumetric strain }}\right]$

$\mathrm{K}=\frac{\mathrm{mg}}{\mathrm{a}\left(\frac{\mathrm{dV}}{\mathrm{V}}\right)}$

$\frac{\mathrm{d} \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{mg}}{\mathrm{Ka}} \quad \ldots(\mathrm{i})$

volume of sphere $\rightarrow \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}$

Fractional change in volume $\frac{\mathrm{d} \mathrm{V}}{\mathrm{V}}=\frac{3 \mathrm{dr}}{\mathrm{r}} \ldots .$ (ii)

$\mathrm{U}$ sing eq. (i) $\&(2) \frac{3 \mathrm{dr}}{\mathrm{r}}=\frac{\mathrm{mg}}{\mathrm{Ka}}$

$\frac{\mathrm{dr}}{\mathrm{r}}=\frac{\mathrm{mg}}{3 \mathrm{Ka}}$


Electromagnetic Induction – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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Previous Years AIEEE/JEE Mains Questions

Q. An inductor of inductance L = 400 mH and resistors of resistances $\mathrm{R}_{1}=2 \Omega$ and $\mathrm{R}_{2}$= $2 \Omega$ are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is:-

(1) $6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right) \mathrm{V}$

(2) $12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}$

(3) $6 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}$

( 4)$\frac{12}{\mathrm{t}} e^{-3 \mathrm{t}} \mathrm{V}$

[AIEEE – 2009]

Sol. (2)

$I=I_{0} e^{-\frac{R T}{L}}$


Q. In the circuit show below, the key K is closed at t = 0. The current through the battery is :

(1) $\frac{V\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$ at $t=0$ and $\frac{V}{R_{2}}$ at $t=\infty$

(2) $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=\infty$

(3) $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}{\mathrm{R}_{1} \mathrm{R}_{2}}$ at $\mathrm{t}=\infty$

(4) $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=\infty$

[AIEEE – 2010]

Sol. (3)

At t = 0 inductor behaves as broken wire then

at $\mathrm{t}=\infty$ Inductor behaves as conducting wire


Q. A rectangular loop has a sliding connector PQ of length  and resistance $\mathrm{R} \Omega$ and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $\mathrm{I}_{1}, \mathrm{I}_{2}$ and I are :-

(1) $\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{6 \mathrm{R}}, \mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$

(2) $\mathrm{I}_{1}=-\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$

(3) $\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \quad \mathrm{I}=\frac{2 \mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$

(4) $\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$

[AIEEE – 2010]

Sol. (3)

Circuit can be reduced as


Q. A boat is moving due east in a region where the earth’s magnetic field is 5.0×10–$^{5} \mathrm{NA}^{-1} \mathrm{m}^{-1}$ due north and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is $1.50 \mathrm{ms}^{-1}$, the magnitude of the indueced emf in the wire of aerial is :-

(1) 0.50 mV

(2) 0.15 mV

(3) 1 mV

(4) 0.75 Mv

[AIEEE – 2011]

Sol. (2)

$e=\mathrm{B} \ell \mathrm{v}=\left(5 \times 10^{-5}\right)(2)(1.50)=0.15 \mathrm{mV}$


Q. A horizontal straight wire 20 m long extending from east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field 0.30 $\times 10^{-4}$$\mathrm{Wb} / \mathrm{m}^{2}$. The instantaneous value of the e.m.f. induced in the wire will be :-

(1) 6.0 mV

(2) 3 mV

(3) 4.5 mV

(4) 1.5 mV

[AIEEE – 2011]

Sol. (2)


Q. A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :-

(1) Electromagnetic induction in the aluminium plate giving rise to electromagnetic damping

(2) Development of air current when the plate is placed

(3) Induction of electrical charge on the plate

(4) Shielding of magnetic lines of force as aluminium is a paramagnetic material

[AIEEE – 2012]

Sol. (1)

Due to conducting nature of Al eddy currents are produced


Q. A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is :

(1) $\frac{2 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(2) $\frac{3 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(3) $\frac{4 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(4) $\frac{5 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

[JEE Main-2013]

Sol. (4)

E.M.F. induced $=\frac{1}{2} \mathrm{B} \omega\left[(3 \ell)^{2}-(2 \ell)^{2}\right]$

$=\frac{1}{2} \mathrm{B} \omega\left[5 \ell^{2}\right]$

$=\frac{5}{2} \mathrm{B} \omega \ell^{2}$


Q. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

(1) $9.1 \times 10^{-11}$ weber

(2) $6 \times 10^{-11}$ weber

(3) $3.3 \times 10^{-11}$ weber

(4) $6.6 \times 10^{-9}$ weber

[JEE Main-2013]

Sol. (1)


Q. In the circuit shown here, the point ‘C’ is kept connected to point ‘A’ till the current flowing through the circuit becomes constant. Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to point ‘B’ at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to :

(1) –1

(2) $\frac{1-\mathrm{e}}{\mathrm{e}}$

(3) $\frac{\mathrm{e}}{1-\mathrm{e}}$

(4) 1

[JEE Main-2014]

Sol. (1)

$\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{L}}=0$

$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{L}}}=-1$


Q. An inductor (L = 0.03 H) and a resistor $(\mathrm{R}=0.15 \mathrm{k} \Omega)$ are connected in series to a battery of 15V EMF in a circuit shown below. The key $\mathrm{K}_{1}$ has been kept closed for a long time. Then at t = 0, $\mathbf{K}_{1}$ is opened and key K2 is closed simultaneously. At t = 1ms, the current in the circuit will be ($e^{5}$ 150):-

(1) 6.7 mA

(2) 0.67 mA

(3) 100 mA

(4) 67 mA

[JEE Main-2015]

Sol. (2)

Decay of current

$=6.66 \times 10^{-4}$

$=0.666 \times 10^{-3}$

$=0.67 \mathrm{mA}$


Q. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to $\mathrm{Q}_{0}$ and then connected to the L and R as shown below. If a student plots graphs of the square of maximum charge $\left(\mathrm{Q}_{\mathrm{Max}}^{2}\right)$ on the capacitor with time (t) for two different values $\mathrm{L}_{1}$ and $\mathrm{L}_{2}\left(\mathrm{L}_{1}\right.$ $\left.>\mathrm{L}_{2}\right)$ of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

[JEE Main-2015]

Sol. (3)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

$\therefore$ for


Q. In a coil of resistance $100 \Omega$, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is :-

(1) 250 Wb

(2) 275 Wb

(3) 200 Wb

(4) 225 Wb

[JEE Advance-2017]

Sol. (1)


Electromagnetic Wave – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

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Previous Years AIEEE/JEE Mains Questions

Q. If a source of power $4 \mathrm{kW}$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called

(1) microwaves

(2) y-rays

(3) X-rays

(4) ultraviolet rays

[AIEEEE 2010]

Sol. (3)

Power $=\frac{\text { nhc }}{\lambda} \lambda \approx 1 \mathrm{A} \equiv \mathrm{X}$ Rays


Q. An electromagnetic wave in vacuum has the electric and magnetic fields $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$, which are

always perpendicular to each other. The direction of polarization is given by $\overrightarrow{\mathrm{X}}$ and that ofwave

propagation by $\overrightarrow{\mathrm{k}}$. Then

(1) $\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$

(2) $\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}}$

(3) $\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}}$

(4) $\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$

AIEEE-2012

Sol. (4)


Q. Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists:

[JEE- Mains 2014]

Sol. (2)


Q. During the propagation of electromagnetic waves in a medium:

(1) Electric energy density is equal to the magnetic energy density

(2) Both electric magnetic energy densities are zero

(3) Electric energy density is double of the magnetic energy density

(4) Electric energy density is half of the magnetic energy density.

[JEE- Mains 2014]

Sol. (1)


Q. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 $\mathrm{m}$ from the diode is :-

(1) $5.48 \mathrm{V} / \mathrm{m}$

(2) $7.75 \mathrm{V} / \mathrm{m}$

(3) $1.73 \mathrm{V} / \mathrm{m}$

(4) $2.45 \mathrm{V} / \mathrm{m}$

[JEE- Mains 2015]

Sol. (4)

$\mathrm{I}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{C}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}$

$\mathrm{E}=\sqrt{\frac{2 \mathrm{P}}{4 \pi \mathrm{r}^{2} \varepsilon_{0} \mathrm{c}}}$

On putting value we get

$=2.45 \mathrm{v} / \mathrm{m} .$


Q. Arrange the following electromagnetic radiations per quantum in the order of increasing energy :-

[JEE- Mains 2016]

Sol. (2)

Energy $=\frac{\mathrm{hc}}{\lambda}$

order of wavelength

$x$ ray, VIBGYOR, Radiowaves


Electrostatics – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Simulator

 

Previous Years AIEEE/JEE Mains Questions

Q. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then $\frac{\mathrm{Q}}{\mathrm{q}}$ equals :-

(1) 1

$(2)-\frac{1}{\sqrt{2}}$

(3) $-2 \sqrt{2}$

(4) –1

[AIEEE – 2009]

Sol. (3)


Q. Statement–1 : For a charged particle moving from point P to point Q the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement–2 : The net work done by a conservative force on an object moving along closed loop is zero.

(1) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–

[AIEEE – 2009]

Sol. (4)

Electrostatics field is a conservative field work is independent of path.


Q. Two points P and Q are maintained at the potential of 10V and –4V, respectively. The work done in moving 100 electrons from P to Q is :-

(1) $-2.24 \times 10^{-16} \mathrm{J}$

(2) $2.24 \times 10^{-16} \mathrm{J}$

(3) $-9.60 \times 10^{-17} \mathrm{J}$

(4) $9.60 \times 10^{-17} \mathrm{J}$

[AIEEE – 2009]

Sol. (2)

$\mathrm{q}=-100 \times 1.6 \times 10^{-19} \mathrm{C}$

$\Delta \mathrm{V}=-14 \mathrm{volt}$z

$\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=2.24 \times 10^{-16} \mathrm{J}$


Q. $\operatorname{Let} \mathrm{P}(\mathrm{r})=\frac{\mathrm{Q}}{\pi \mathrm{R}^{4}} \mathrm{r}$ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance $r_{1}$ from the centre of the sphere, the magnitude of electric field is :-

(1) $\frac{\mathrm{Qr}_{1}^{2}}{4 \pi \epsilon_{0} \mathrm{R}^{4}}$

(2) $\frac{\mathrm{Qr}_{1}^{2}}{3 \pi \epsilon_{0} \mathrm{R}^{4}}$

(3) 0

(4) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}}$

[AIEEE – 2009]

Sol. (1)


Q. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field $\overrightarrow{\mathrm{E}}$ at the centre O is :-

(1) $\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

( 2)$\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

$(3)-\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

(4) $-\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

[AIEEE – 2010]

Sol. (4)

$\overrightarrow{\mathrm{E}}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}(-\hat{\mathrm{j}})=\frac{2}{\left(4 \pi \epsilon_{0} \mathrm{r}\right)} \frac{\mathrm{q}}{(\pi \mathrm{r})}(-\hat{\mathrm{j}})$


Q. Let there be a spherically symmetric charge distribution with charge density varying as

(r) = $\rho_{0}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$ upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r < R) from the origion is given by :

( 1)$\frac{\rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(2) $\frac{4 \pi \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(3) $\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(4) $\frac{4 \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

[AIEEE – 2010]

Sol. (3)

Total charge

$Q=\int_{0}^{r} \rho d V=\int_{0}^{r} \rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right) 4 \pi r^{2} d r$

$=4 \pi \rho_{0} \int_{0}^{r}\left(\frac{5 r^{2}}{4}-\frac{r^{3}}{R}\right) d r=4 \pi \rho_{0}\left[\frac{5 r^{3}}{12}-\frac{r^{4}}{4 R}\right]$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \epsilon_{0} \mathrm{r}^{2}} 4 \pi \rho_{0}\left[\frac{5}{12} \mathrm{r}^{3}-\frac{\mathrm{r}^{4}}{4 \mathrm{R}}\right]$

$=\frac{\rho_{0} r}{4 \in_{0}}\left[\frac{5}{3}-\frac{r}{R}\right]$


Q. Two identical charged spheres suspended from a common point by two massless string of length $\ell$are initially a distance d(d << $\ell$) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them :-

(1) $\mathrm{v} \propto \mathrm{x}^{1 / 2}$

(2) $\mathrm{v} \propto \mathrm{x}$

(3) $\mathrm{v} \propto \mathrm{x}^{-1 / 2}$

(4) $\mathrm{v} \propto \mathrm{x}^{-1}$

[AIEEE – 2011]

Sol. (3)

$\tan \theta=\frac{\mathrm{F}}{\mathrm{M}_{9}} \quad(\text { since } \theta \text { small })$


Q. The electrostatic potential inside a charged spherical ball is given by $\phi=\mathrm{ar}^{2}+\mathrm{b}$where r is the distance from the centre; a, b are constant. Then the charge density inside the ball is :-

(1) $-24 \pi \mathrm{a} \in_{0}$

$(2)-6 \mathrm{a} \in_{0}$

(3) $-24 \pi \mathrm{a} \in_{0} \mathrm{r}$

$(4)-6 \mathrm{a} \in_{0} \mathrm{r}$

[AIEEE – 2011]

Sol. (2)


Q. Two positive charges of magnitude ‘q’ are placed at the ends of a side (side 1) of a square of side ‘2a’. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is :-

(1) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{1}{\sqrt{5}}\right)$

(2) zero

(3) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1+\frac{1}{\sqrt{5}}\right)$

(4) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{2}{\sqrt{5}}\right)$

[AIEEE – 2011]

Sol. (1)


Q. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

An insulating solid sphere of radius R has a uniformaly positive charge density . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphre and also at a point out side the sphere. The electric potential at infinity is zero.

Statement-1: When a charge ‘q’ is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{q} \rho}{3 \epsilon_{0}}$

Statement-2 : The electric field at a distance r (r < R) from the centre of the sphere is $\frac{\rho \mathrm{r}}{3 \epsilon_{0}}$

(1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of

Statement-1.

(2) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of

statement-1.

(3) Statement-1 is true, Statement-2 is false

(4) Statement-1 is false, Statement-2 is true

[AIEEE – 2012]

Sol. (4)


Q. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :-

[AIEEE – 2012]

Sol. (4)

For uniformly charged sphere

$\mathrm{E}=\frac{\mathrm{Kqr}}{\mathrm{R}^{3}}(\mathrm{r}<\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{R}^{2}} \quad(\mathrm{r}=\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^{2}} \quad(\mathrm{r}>\mathrm{R})$


Q. Let $\left[\in_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = Length,

T = Time and A = electric current, then :

(1) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{2} \mathrm{A}\right]$

(2) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{4} \mathrm{A}^{2}\right]$

(3) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}^{-2}\right]$

(4) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}\right]$

[AIEEE – 2013]

Sol. (2)


Q. Two charges, each equal to q, are kept at x = –a and x = a on the x-axis. A particle of mass m and charge $\mathrm{q}_{0}=\frac{\mathrm{q}}{2}$ is placed at the origin. If charge $q_{0}$ is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to

(1) y           (2) –y          (3) $\frac{1}{y}$ $          (4)-\frac{1}{\mathrm{y}}$

[AIEEE – 2013]

Sol. (1)


Q. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :-

(1) $\frac{\mathrm{Q}}{8 \pi \epsilon_{0} \mathrm{L}}$

(2) $\frac{3 \mathrm{Q}}{4 \pi \in_{0} \mathrm{L}}$

(3) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{L} \ln 2}$

(4) $\frac{\mathrm{Q} \ln 2}{4 \pi \epsilon_{0} \mathrm{L}}$

[JEE-Main-2013]

Sol. (4)


Q. Assume that an electric field $\overrightarrow{\mathrm{E}}=30 \mathrm{x}^{2} \hat{\mathrm{i}}$ exists in space. Then the potential difference $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{O}}$, where $\mathrm{V}_{\mathrm{O}}$ is the potential at the origin and $\mathrm{V}_{\mathrm{A}}$ the potential at x = 2 m is :-

(1)–80 J

(2) 80 J

(3) 120 J

(4) –120 J

Sol. (1)

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=-10[8-0]=-80 \mathrm{V}$


Q. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density $\rho=\frac{\mathrm{A}}{\mathrm{r}}$, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :-

(1) $\frac{2 Q}{\pi a^{2}}$

(2) $\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}}$

(3) $\frac{\mathrm{Q}}{2 \pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}$

(4) $\frac{2 \mathrm{Q}}{\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}$

[JEE-Main-2016]

Sol. (2)

Gaussian surface at distance r from center


Q. An electric dipole has a fixed dipole moment $\overrightarrow{\mathrm{p}}$, which makes angle  with respect to x-axis. When subjected to an electric field $\overrightarrow{\mathrm{E}}_{1}=\mathrm{E} \hat{\mathrm{i}}$, it experiences a torque $\overrightarrow{\mathrm{T}}_{1}=\tau \hat{\mathrm{k}}$. When subjected to another electric field $\overrightarrow{\mathrm{E}}_{2}=\sqrt{3} \mathrm{E}_{1} \hat{\mathrm{j}}$ it experiences torque . The angle $\theta$ is :

(1) $60^{\circ}$ (2) $90^{\circ}$ (3) $30^{\circ}$ (4) $45^{\circ}$

[JEE-Main-2017]

Sol. (1)

So from $\quad \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}$


Q. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities $+\sigma,-\sigma$ and $+\sigma$ respectively. The potential of shell B is :-

[JEE-Main-2018]

Sol. (1)


Error in Measurement & Instruments – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. In an experiment the angles are required to be measured using an instrument 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $\left(=0.5^{\circ}\right)$, then the least count of the instrument is :-

(1) One degree

(2) Half degree

(3) One minute

(4) Half minute

[AIEEE – 2009]

Sol. (3)

Least count of vernier callipers

L.C. = 1MSD – 1VSD

But here $29 \mathrm{MSD}=30 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{29}{30} \mathrm{MSD}$

$\Rightarrow \mathrm{L.C.}=1 \mathrm{MSD}-\frac{29}{30} \mathrm{MSD}=$

$\frac{1}{30} \mathrm{MSD}=\frac{1}{30} \times 0.5^{\circ}=\left(\frac{1}{60}\right)^{\circ}=1$ minute.


Q. In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $45^{\circ}$ with the x-axis meets the experimental curve at P. The coordinates of P will be :-

(1) (ƒ, ƒ)

(2) (4ƒ, 4ƒ)

(3) (2ƒ, 2ƒ)

(4) $\left(\frac{f}{2}, \frac{f}{2}\right)$

[AIEEE – 2009]

Sol. (3)


Q. The respective number of significant figures for the numbers $23.023,0.0003$ and $2.1 \times 10^{-3}$ are :-

(1) 4, 4, 2          (2) 5, 1, 2           (3) 5, 1, 5             (4) 5, 5, 2

[AIEEE – 2010]

Sol. (2)

5, 1, 2 resp.


Q. A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : 0 mm.

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.

The diameter of wire from the above date is :-

(1) 0.026 cm              (2) 0.005 cm              (3) 0.52 cm                 (4) 0.052 cm

[AIEEE – 2011]

Sol. (4)

Least count $=\frac{1 \mathrm{mm}}{100}=0.01 \mathrm{mm}$

Diameter of the wire = 0 + 52 × 0.01 mm = 0.52 mm = 0.052 cm


Q. A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data :

(1) 59 degree             (2) 58.59 degree             (3) 58.77 degree             (4) 58.65 degree

[AIEEE – 2012]

Sol. (4)

Least count = MSD – VSD

where $\mathrm{MSD}=0.5^{\circ} \& 30 \mathrm{VSD}=29 \mathrm{MSD}$

so least count $=0.5^{\circ}-\left(\frac{29}{30}\right) \times 0.5^{\circ}=\left(\frac{0.5}{30}\right)^{\circ}$

Reading $=58.5^{\circ}+(09) \times\left(\frac{0.5}{30}\right)^{\circ}=58.65^{\circ}$


Q. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is :-

(1) 3%            (2) 6%             (3) zero             (4) 1%

[AIEEE – 2012]

Sol. (2)

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}} \Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}}=\pm\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}+\frac{\Delta \mathrm{I}}{\mathrm{I}}\right)$

$=\pm(3+3) \%=\pm 6 \%$


Q. The current voltage relation of diode is given by $\mathrm{I}=\left(\mathrm{e}^{1000 \mathrm{V} / \mathrm{T}}-1\right) \mathrm{mA}$, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring $\pm 0.01 \mathrm{V}$ while measuring the current of 5 mA at 300 K, what will be error in the value of current in mA ?

(1) 0.5 mA           (2) 0.05 mA             (3) 0.2 mA             (4) 0.02 mA

[JEE-Mains 2014]

Sol. (3)

Given current $\mathrm{I}=\left(e^{1000 \mathrm{V} / \mathrm{T}}-1\right) \mathrm{mA}$

$\Rightarrow \mathrm{I}+1=e^{1000 \mathrm{V} / \mathrm{T}}$

$\mathrm{d} \mathrm{l}=\frac{1000}{\mathrm{T}}\left[e^{\frac{1000 \mathrm{v}}{\mathrm{T}}}\right] \mathrm{d} \mathrm{V}$

$\mathrm{d} \mathrm{l}=\frac{1000}{\mathrm{T}}[1+1] \mathrm{dV}=\frac{1000}{300}[6] \times(0.01)$

dI = 0.2 mA


Q. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ?

(1) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

(2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

(3) A meter scale

(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

[JEE-Mains 2014]

Sol. (4)

Least count of varnier calliper is 0.01 cm

Hence it matches with the reading.


Q. The period of oscillation of a simple pendulum is T = $2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. The accuracy in the determination of g is :

(1)1%           (2) 5%             (3) 2%              (4) 3%

[JEE-Mains 2015]

Sol. (4)

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}$

$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}$

$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{1 \times 10^{-3}}{20 \times 10^{-2}}+2 \times \frac{1}{100 \times \frac{90}{100}}$

$\therefore \frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=2.722 \% \approx 3 \%$


Q. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line ?

(1) 0.50 mm             (2) 0.75 mm             (3) 0.80 mm               (4) 0.70 mm

[JEE-Mains 2016]

Sol. (3)


Q. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :-

(1) $92 \pm 3 \mathrm{s}$

(2) $92 \pm 2 \mathrm{s}$

(3) $92 \pm 5.0 \mathrm{s}$

(4) $92 \pm 1.8 \mathrm{s}$

[JEE-Mains 2016]

Sol. (2)

$\mathrm{T}_{\mathrm{AV}}=92 \mathrm{s}$

$(|\Delta \mathrm{T}|)_{\text {mean }}=1.5 \mathrm{s}$

Least conunt of clock is 1sec. So value that clock can measure should be 1, 2, 3……so on

so reported mean time should be

$92 \pm 2$

Ref : NCERT (XIth) Ex. 2.7, Page. 25


Q. The following observations were taken for determining surface tensiton T of water by capillary method :

Diameter of capilary, $\mathrm{D}=1.25 \times 10^{-2} \mathrm{m}$

rise of water, $h=1.45 \times 10^{-2} \mathrm{m}$

Using $\mathrm{g}=9.80 \mathrm{m} / \mathrm{s}^{2}$ and the simplified relation

$\mathrm{T}=\frac{\mathrm{rhg}}{2} \times 10^{3} \mathrm{N} / \mathrm{m},$ the possible error in surface tension is closest to :

(1) 2.4%                (2) 10%             (3) 0.15%                  (4) 1.5%

[JEE-Mains 2017]

Sol. (4)

$\mathrm{T}=\frac{\mathrm{rhg}}{2} \times 10^{3}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{\Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{h}}{\mathrm{h}}+0$

$100 \times \frac{\Delta \mathrm{T}}{\mathrm{T}}=\left(\frac{10^{-2} \times .01}{1.25 \times 10^{-2}}+\frac{10^{-2} \times .01}{1.45 \times 10^{-2}}\right) 100$

$=(0.8+0.689)$

$=(1.489)$

$100 \times \frac{\Delta \mathrm{T}}{\mathrm{T}}=1.489 \%$

= 1.5%


Q. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is :-

(1) 3.5 %              (2) 4.5 %              (3) 6 %               (4) 2.5 %

[JEE-Mains 2018]

Sol. (2)

Density $=\frac{\text { Mass }}{\text { Volume }}$

$\frac{1 \Delta \mathrm{d}}{\mathrm{d}}=\frac{1 \Delta \mathrm{M}}{\mathrm{M}}+\frac{3 \Delta \mathrm{L}}{\mathrm{L}}$

= 1.5 + 3(1)

= 4.5 %


Fluid Mechanics – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. A ball is made of a material of density r where $\rho_{\text {oil }}<\rho<\rho_{\text {water }}$ with $\rho_{\text {oil }}$ and $\rho_{\text {water }}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position ?

[AIEEE-2010]

Sol. (3)

In equilibrium ball will remain at the interface of water and oil.


Q. Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3}$m. The water velocity as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance $2 \times 10^{-1}$m below the tap is close to :-

(1) $9.6 \times 10^{-3} \mathrm{m}$

(2) $3.6 \times 10^{-3} \mathrm{m}$

(3) $5.0 \times 10^{-3} \mathrm{m}$

(4) $7.5 \times 10^{-3} \mathrm{m}$

[AIEEE-2011]

Sol. (2)

According to equation of continuity

$\mathrm{A}_{1} \mathrm{V}_{1}=\mathrm{A}_{2} \mathrm{V}_{2}$ or $\mathrm{r}_{2}=\sqrt{\frac{\mathrm{r}_{1}^{2} \mathrm{v}_{1}}{\mathrm{v}_{2}}}$

Velocity of stream at 0.2 m below tap.

$\mathrm{V}_{2}^{2}=\mathrm{V}_{1}^{2}+2 \mathrm{as}=0.16+2 \times 10 \times 0.2=4.16 \mathrm{m} / \mathrm{s}$

$\mathrm{r}_{2}=\sqrt{\frac{\mathrm{r}_{1}^{2} \mathrm{v}_{1}}{\mathrm{v}_{2}}}=\sqrt{\frac{16 \times 10^{-6} \times 0.4}{2}}=\sqrt{3.2} \times 10^{-3} \mathrm{m}$

so diameter $=2 \times \sqrt{3.2} \times 10^{-3} \mathrm{m}$

$=2 \times 1.8 \times 10^{-3}=3.6 \times 10^{-3} \mathrm{m}$


Q. Work done in increasing the size of a soap bubble from a radius of 3 cm to 5cm is nearly (Surface tension of soap solution = 0.03 Nm–1) :-

(1) $2 \pi \mathrm{mJ}$

(2) $0.4 \pi \mathrm{mJ}$

(3) $4 \pi \mathrm{mJ}$

(4) $0.2 \pi \mathrm{mJ}$

[AIEEE-2011]

Sol. (2)

$\mathrm{W}=8 \pi \mathrm{T}\left[\left(\mathrm{r}_{2}^{2}\right)-\left(\mathrm{r}_{1}^{2}\right)\right]$

$=8 \times \pi \times 0.03[25-9] \times 10^{-4}$

$=\pi \times 0.24 \times 16 \times 10^{-4}$

$=3.8 \times 10^{-4} \pi$

$=0.384 \pi \mathrm{m} \mathrm{J} \approx 0.4 \pi \mathrm{mJ}$


Q. Two merucry drops (each of radius ‘r’) merge to form a bigger drop. The surface energy of the bigger drop, ifs T is the surface tension, is :

(1) $2^{5 / 3} \pi r^{2} T$

(2) $4 \pi r^{2} \mathrm{T}$

(3) $2 \pi r^{2} T$

(4) $2^{8 / 3} \pi r^{2} \mathrm{T}$

[AIEEE-2011]

Sol. (4)

By volume conservation

$\frac{4}{3} \pi \mathrm{R}^{3}=2\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)$

$\mathrm{R}=2^{1 / 3} \mathrm{r}$

Surface energy E = T (A)

$=\mathrm{T}\left(4 \pi \mathrm{R}^{2}\right)=\mathrm{T}\left(4 \pi 2^{2 / 3} \mathrm{r}^{2}\right)=2^{8 / 3} \pi \mathrm{r}^{2} \mathrm{T}$


Q. If a ball of steel (density $\left.\rho=7.8 \mathrm{g} \mathrm{cm}^{-3}\right)$ attains a terminal velocity of 10 cm s–1 when falling

in a tank of water (coefficient of viscosity $\left.\eta_{\text {water }}=8.5 \times 10^{-4} \mathrm{Pa.s}\right)$ then its terminal velocity in glycerine $\left(\rho=1.2 \mathrm{g} \mathrm{cm}^{-3}, \eta=13.2 \mathrm{Pa.s}\right)$ would be nearly :-

(1) $1.6 \times 10^{-5} \mathrm{cm} \mathrm{s}^{-1}$

(2) $6.25 \times 10^{-4} \mathrm{cm} \mathrm{s}^{-1}$

(3) $6.45 \times 10^{-4} \mathrm{cm} \mathrm{s}^{-1}$

(4) $1.5 \times 10^{-5} \mathrm{cm} \mathrm{s}^{-1}$

[AIEEE-2011]

Sol. (2)

Terminal velocity $\mathrm{V}_{\infty} \frac{\mathrm{d}_{\mathrm{b}}-\mathrm{d}_{\ell}}{\eta}$

$\frac{V_{1}}{V_{2}}=\frac{7.8-1}{8.5 \times 10^{-4}} \times \frac{13.2}{7.8-1.2}$

$\frac{10}{\mathrm{V}_{2}}=1.6 \times 10^{4}$

$\mathrm{V}_{2}=\frac{10}{1.6 \times 10^{4}}=6.25 \times 10^{-4}$


Q. A wooden cube (density of wood ‘d’) of side ‘$\ell$’ floats in a liquid of density ‘r’ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period ‘T’. Then, ‘T’ is equal to :-

(1) $2 \pi \sqrt{\frac{\ell \rho}{(\rho-\mathrm{d}) \mathrm{g}}}$

(2) $2 \pi \sqrt{\frac{\ell \mathrm{d}}{\rho \mathrm{g}}}$

(3) $2 \pi \sqrt{\frac{\ell \rho}{\mathrm{dg}}}$

(4) $2 \pi \sqrt{\frac{\ell \mathrm{d}}{(\rho-\mathrm{d}) \mathrm{g}}}$

[AIEEE-2011]

Sol. (2)

By using $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}$

Where $\mathrm{m}=\ell^{3} \mathrm{d}$ and $\mathrm{A}=\ell^{2}$

$\mathrm{T}=2 \pi \sqrt{\frac{\ell^{3} \mathrm{d}}{\ell^{2} \rho \mathrm{g}}} \Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\ell \mathrm{d}}{\rho \mathrm{g}}}$


Q. A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times 10^{-2} \mathrm{N}$ (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is :-

[AIEEE-2012]

Sol. (2)

weight $=\mathrm{mg}=1.5 \times 10^{-2} \mathrm{N}(\text { given })$

length $=\ell=30 \mathrm{cm}(\text { given })$

= 0.3 m

$2 \mathrm{T} \ell=\mathrm{mg}$

$\mathrm{T}=\frac{\mathrm{mg}}{2 \ell}=\frac{1.5 \times 10^{-2}}{2 \times 0.3}=0.025 \mathrm{N} / \mathrm{m}$


Q. A uniform cylinder of length L and mass M having cross- sectional area A is suspended, with its length vertical, form a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. The extension $\mathrm{X}_{0}$ of the spring when it is in equilibrium is :

(1) $\frac{\mathrm{Mg}}{\mathrm{k}}$

( 2)$\frac{\mathrm{Mg}}{\mathrm{k}}\left(1-\frac{\mathrm{LA\sigma}}{\mathrm{M}}\right)$

(3) $\frac{\mathrm{Mg}}{\mathrm{k}}\left(1-\frac{\mathrm{LA\sigma}}{2 \mathrm{M}}\right)$

(4)$\frac{\mathrm{Mg}}{\mathrm{k}}\left(1+\frac{\mathrm{LA\sigma}}{\mathrm{M}}\right)$

(Here k is spring constant)

[AIEEE-2013]

Sol. (1)

At equilibrium

$\mathrm{Kx}_{0}=\mathrm{Mg}-\mathrm{B}$

$\mathrm{Kx}_{0}=\mathrm{Mg}-\frac{\sigma \mathrm{AL}}{2}$

$\mathrm{X}_{0}=\frac{\left(\mathrm{Mg}-\frac{\sigma \mathrm{AL}}{2} \mathrm{g}\right)}{\mathrm{K}}$

$=\left[1-\frac{\sigma \mathrm{AL}}{2 \mathrm{M}}\right]$

.


Q. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is  and L is its latent heat of vaporization.

(1) $\frac{\rho \mathrm{L}}{\mathrm{T}}$

(2) $\sqrt{\frac{\mathrm{T}}{\rho \mathrm{L}}}$

( 3)$\frac{\mathrm{T}}{\rho \mathrm{L}}$

(4) $\frac{2 \mathrm{T}}{\rho \mathrm{L}}$

[AIEEE-2013]

Sol. (3)

If a layer of thickness dr is evaporates then change in surface energy

= (change in surface area) T

$=\left(\mathrm{d}\left(4 \pi \mathrm{r}^{2}\right)\right) \mathrm{T}=8 \pi \mathrm{rdr} \mathrm{T}$

energy required to evaporate layer of thickness dr $=\left(4 \mathrm{pr}^{2} \mathrm{dr}\right) \rho . \mathrm{L}$

The process of evaporation only starts only if change in surface energy is just sufficient to evaporate the water layer

$\Rightarrow\left(4 \pi r^{2} d r\right) L \rho=(8 \pi r d r) T$

$\Rightarrow \mathrm{r}=\frac{2 \pi}{\rho \mathrm{L}}$


Q. On heating water, bubbles being formed at the bottom of the vessel detatch

and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is:-

(dencity of water is $\rho_{\mathrm{w}}$)

(1) $\mathrm{R}^{2} \sqrt{\frac{\rho_{\mathrm{w}} \mathrm{g}}{\mathrm{T}}}$

(2) $\mathrm{R}^{2} \sqrt{\frac{3 \rho_{\mathrm{w}} \mathrm{g}}{\mathrm{T}}}$

(3) $\mathrm{R}^{2} \sqrt{\frac{\rho_{\mathrm{w}} \mathrm{g}}{3 \mathrm{T}}}$

(4) $\mathrm{R}^{2} \sqrt{\frac{\rho_{\mathrm{w}} \mathrm{g}}{6 \mathrm{T}}}$

[JEE Mains-2014]

Sol. (4)

Force due to Surface Tenstion

$=\mathrm{T}(2 \pi \mathrm{r}) \sin \theta=\mathrm{T}(2 \pi \mathrm{r}) \times \frac{\mathrm{r}}{\mathrm{R}}$

This force will balance the force of Bouyancy

$\mathrm{T}(2 \pi \mathrm{r}) \times \frac{\mathrm{r}}{\mathrm{R}}=\rho_{\mathrm{w}} \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{g}$

$r=R^{2} \sqrt{\frac{2 \rho_{\mathrm{w}} g}{3 T}}$


Q. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ are filled in the tube. Each liquid subtends $90^{\circ}$ angle at centre. Radius joining their interface makes an angle  with vertical. Ratio $\frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}$ is :

[JEE Mains-2014]

Sol. (1)

Let Radius of circular tube is R