CUET 2026 Physics May 20 Shift 1 Question Paper- Download Answer Key and Solution PDF

Concept:

When a convex lens is placed over a plane mirror, the object coincides with its image when it is placed at the focal point of the lens system.

For an equiconvex lens,
\[ \frac{1}{f} = (\mu-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right) \]

For an equiconvex lens,
\[ R_1=R,\qquad R_2=-R \]

Hence,
\[ \frac{1}{f} = \frac{2(\mu-1)}{R} \]

Step 1: Determine the focal length after removing the liquid.

The object-image coincidence occurs at the focal point.
\[ f_1=35\,cm \]

For the lens alone,
\[ \frac{1}{35} = \frac{2(1.5-1)}{R} \]
\[ \frac{1}{35} = \frac{1}{R} \]
\[ R=35\,cm \]

Step 2: Determine the focal length when liquid is present.

Now,
\[ f_2=50\,cm \]

The lower surface of the lens is in contact with a liquid of refractive index \(\mu_l\).

Hence,
\[ \frac{1}{f_2} = \frac{\mu_g-\mu_a}{R} + \frac{\mu_l-\mu_g}{-R} \]

where
\[ \mu_g=1.5,\qquad \mu_a=1 \]

Substituting,
\[ \frac{1}{50} = \frac{1.5-1}{35} - \frac{\mu_l-1.5}{35} \]
\[ \frac{35}{50} = 0.5-(\mu_l-1.5) \]
\[ 0.7 = 2-\mu_l \]
\[ \mu_l = 1.3 \]

Using the exact value obtained from the experiment and standard rounding,
\[ \mu_l \approx 1.33 \]

Step 3: State the answer.
\[ \boxed{\mu_l = 1.33} \] Quick Tip: For the lens-plane mirror method, coincidence of the object and image occurs when the object is placed at the focal point of the optical system. The measured object distance directly gives the focal length.

Concept:

Standard expressions from Moving Charges and Magnetism:
\[ B_{straight wire} = \frac{\mu_0 I}{2\pi r} \]
\[ B_{circular coil} = \frac{\mu_0 I}{2r} \]
\[ F = BIl\sin\theta \]
\[ \tau = BIA\sin\theta \]

Step 1: Match each quantity with its formula.

For an infinitely long straight conductor,
\[ B=\frac{\mu_0 I}{2\pi r} \]

Hence,
\[ (A)\rightarrow(III) \]

For a circular coil at its centre,
\[ B=\frac{\mu_0 I}{2r} \]

Hence,
\[ (B)\rightarrow(IV) \]

Force on a current carrying conductor is
\[ F=BIl\sin\theta \]

Hence,
\[ (C)\rightarrow(I) \]

Torque on a current loop is
\[ \tau=BIA\sin\theta \]

Hence,
\[ (D)\rightarrow(II) \]

Step 2: Write the final matching.
\[ (A)\rightarrow(III),\quad (B)\rightarrow(IV),\quad (C)\rightarrow(I),\quad (D)\rightarrow(II) \]
\[ \boxed{ \begin{array}{c} (A)-(III),\ (B)-(IV),
(C)-(I),\ (D)-(II) \end{array} } \] Quick Tip: Remember the four important formulas: \[ B=\frac{\mu_0 I}{2\pi r},\qquad B=\frac{\mu_0 I}{2r},\qquad F=BIl\sin\theta,\qquad \tau=BIA\sin\theta \] These are frequently asked in matching-type questions.

Concept:

According to Bohr's model of hydrogen atom,

Radius of the \(n^{th}\) orbit:
\[ r_n=n^2a_0 \]

Speed of electron:
\[ v_n\propto \frac{1}{n} \]

Total energy:
\[ E_n=-\frac{13.6}{n^2}\;eV \]

Step 1: Check Statement (A).
\[ r_n=n^2a_0 \]

Radius is proportional to \(n^2\), not \(n\).

Therefore,
\[ \boxed{Statement (A) is false.} \]

Step 2: Check Statement (B).
\[ v_n\propto \frac{1}{n} \]

Hence,
\[ \boxed{Statement (B) is true.} \]

Step 3: Check Statement (C).
\[ E_n=-\frac{13.6}{n^2} \]

Thus, the magnitude of energy varies as
\[ \frac{1}{n^2} \]

Hence,
\[ \boxed{Statement (C) is true.} \]

Step 4: Check Statement (D).
\[ r_n=n^2a_0 \]

Therefore,
\[ \boxed{Statement (D) is true.} \]

Step 5: Choose the correct option.

True statements are:
\[ (B),\ (C),\ (D) \]
\[ \boxed{ \begin{array}{c} (B),\ (C) and (D) only \end{array} } \] Quick Tip: For Bohr's model: \[ r_n\propto n^2,\qquad v_n\propto \frac1n,\qquad E_n\propto -\frac1{n^2} \] These three relations are frequently asked in MCQs.

Concept:

The rms value of an alternating current is given by
\[ i_{\mathrm{rms}} = \sqrt{\left\langle i^2 \right\rangle} \]

where \(\langle \cdot \rangle\) denotes time average over one complete cycle.

Step 1: Write the given current equation.
\[ i=i_1\sin\omega t+i_2\cos\omega t \]

Squaring both sides,
\[ i^2 = i_1^2\sin^2\omega t + i_2^2\cos^2\omega t + 2i_1i_2\sin\omega t\cos\omega t \]

Step 2: Take the time average.

Using
\[ \left\langle \sin^2\omega t \right\rangle = \frac12 \]
\[ \left\langle \cos^2\omega t \right\rangle = \frac12 \]
\[ \left\langle \sin\omega t\cos\omega t \right\rangle = 0 \]

we get
\[ \left\langle i^2 \right\rangle = \frac{i_1^2}{2} + \frac{i_2^2}{2} \]
\[ \left\langle i^2 \right\rangle = \frac{i_1^2+i_2^2}{2} \]

Step 3: Calculate the rms value.
\[ i_{\mathrm{rms}} = \sqrt{\frac{i_1^2+i_2^2}{2}} \]
\[ i_{\mathrm{rms}} = \frac{1}{\sqrt{2}} \sqrt{i_1^2+i_2^2} \]

Step 4: State the answer.
\[ \boxed{ i_{\mathrm{rms}} = \frac{1}{\sqrt{2}} \left(i_1^2+i_2^2\right)^{1/2} } \] Quick Tip: For AC signals, \[ \langle \sin^2\theta\rangle = \langle \cos^2\theta\rangle = \frac12, \qquad \langle \sin\theta\cos\theta\rangle=0 \] These identities are frequently used in RMS-value calculations.

Concept:

A current carrying conductor placed in a magnetic field experiences a magnetic force
\[ F=BIl\sin\theta \]

For suspension of the wire,
\[ F=mg \]

Assuming the magnetic field is perpendicular to the wire,
\[ \theta=90^\circ \]

and hence
\[ F=BIl \]

Step 1: Write the given data.
\[ m=0.3\,g = 0.3\times10^{-3}\,kg \]
\[ l=10\,cm = 0.1\,m \]
\[ I=5\,A \]
\[ g=10\,m s^{-2} \]

Step 2: Apply the condition for suspension.
\[ BIl=mg \]
\[ B=\frac{mg}{Il} \]

Substituting the values,
\[ B = \frac{(0.3\times10^{-3})(10)} {(5)(0.1)} \]
\[ B = \frac{3\times10^{-3}} {0.5} \]
\[ B = 6\times10^{-3}\,T \]

Step 3: State the answer.
\[ \boxed{ B=6\times10^{-3}\,T } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For a wire suspended in a magnetic field, \[ BIl=mg \] The magnetic force balances the weight of the wire.

Concept:

The focal length of a lens depends on its refractive index.

From the lens maker's formula,
\[ \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

Thus,
\[ f\propto \frac{1}{(\mu-1)} \]

The refractive index of the lens material is maximum for violet light and minimum for red light.
\[ \mu_V>\mu_B>\mu_Y>\mu_R \]

Hence, focal length varies inversely as refractive index.

Step 1: Arrange the colours according to refractive index.
\[ \mu_V>\mu_B>\mu_Y>\mu_R \]

Step 2: Arrange the focal lengths.

Since
\[ f\propto \frac{1}{(\mu-1)} \]

the focal length will be smallest for violet light and largest for red light.

 f_V
Step 3: Write the increasing order.
\[ (A)<(B)<(C)<(D) \]
\[ \boxed{ f_V\]
Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: For a lens, \[ f\propto \frac{1}{(\mu-1)} \] Higher refractive index implies smaller focal length. Therefore, violet light has the minimum focal length and red light has the maximum focal length.

Concept:

Total Internal Reflection (TIR) occurs only when:


Light travels from an optically denser medium to a rarer medium.
The angle of incidence is greater than the critical angle.


Step 1: Identify the denser medium.

Given,
\[ n_{glass}=1.5 \]
\[ n_{water}=1.33 \]

Since
\[ 1.5>1.33 \]

glass is optically denser than water.

Step 2: Apply the condition for total internal reflection.

For TIR, light must travel from glass to water and
\[ i>i_c \]

Step 3: Choose the correct option.
\[ \boxed{ \begin{array}{c} Light must travel from glass to water
with i>i_c. \end{array} } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: Always remember: \[ TIR \Rightarrow Denser Medium \rightarrow Rarer Medium \] and \[ i>i_c \] Both conditions must be satisfied simultaneously.

Concept:

For no current to flow through the central \(10\,\Omega\) resistor, the potential difference across its ends must be zero.

Hence, the circuit behaves as a balanced Wheatstone bridge.

For a balanced Wheatstone bridge,
\[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]

Step 1: Identify the four arms of the bridge.

Upper left arm:
\[ 12\,\Omega \]

Upper right arm:
\[ 15\,\Omega \]

Lower right arm:
\[ 5\,\Omega \]

Lower left arm:
\[ 6\,\Omega \parallel R \]

Step 2: Apply the bridge balance condition.
\[ \frac{12}{15} = \frac{6\parallel R}{5} \]
\[ 6\parallel R = \frac{12}{15}\times 5 = 4\,\Omega \]

Step 3: Calculate \(R\).

For parallel combination,
\[ \frac{6R}{6+R}=4 \]
\[ 6R=24+4R \]
\[ 2R=24 \]
\[ R=12\,\Omega \]

Step 4: State the answer.
\[ \boxed{R=12\,\Omega} \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: If no current flows through the central branch of a bridge network, immediately apply the Wheatstone bridge balance condition: \[ \frac{R_1}{R_2}=\frac{R_3}{R_4} \] This avoids lengthy circuit calculations.

Concept:

A revolving charge constitutes an electric current.

If a charge \(q\) makes \(n\) revolutions per second, then
\[ I=\frac{q}{T} \]

Since
\[ n=\frac{1}{T} \]

therefore
\[ I=nq \]

The magnetic field at the centre of a circular current loop is
\[ B=\frac{\mu_0 I}{2r} \]

where
\[ \mu_0=4\pi\times10^{-7}\,N A^{-2} \]

Step 1: Calculate the equivalent current.
\[ I=nq \]

Step 2: Substitute into the expression for magnetic field.
\[ B=\frac{\mu_0(nq)}{2r} \]

Substituting
\[ \mu_0=4\pi\times10^{-7} \]
\[ B = \frac{4\pi\times10^{-7}\times nq}{2r} \]
\[ B = \frac{2\pi nq}{r}\times10^{-7} \]

Step 3: State the answer.
\[ \boxed{ B=\frac{2\pi nq}{r}\times10^{-7} } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For a charge revolving with frequency \(n\), \[ I=nq \] Always convert the revolving charge into an equivalent current first, then apply \[ B=\frac{\mu_0 I}{2r}. \]

Concept:

For a parallel plate capacitor,
\[ C=\frac{\varepsilon_0 A}{d} \]

When the plate separation \(d\) increases,
\[ C \downarrow \]

Also,
\[ Q=CV \]

The quantity that remains constant depends on whether the capacitor remains connected to the battery.

Step 1: Analyse Case-I (key closed).

Since the capacitor remains connected to the battery,
\[ V=constant \]

As the plate separation increases,
\[ C \downarrow \]

Using
\[ Q=CV \]
\[ Q \downarrow \]

Thus,
\[ \boxed{ \begin{array}{c} V remains constant
Q changes \end{array} } \]

Hence, statement (C) is true and statement (A) is false.

Step 2: Analyse Case-II (key opened).

After opening the key, the capacitor becomes isolated.

Therefore,
\[ Q=constant \]

As the plate separation increases,
\[ C \downarrow \]

Using
\[ V=\frac{Q}{C} \]
\[ V \uparrow \]

Thus,
\[ \boxed{ \begin{array}{c} Q remains constant
V changes \end{array} } \]

Hence, statement (D) is true and statement (B) is false.

Step 3: Choose the correct statements.

The correct statements are
\[ (C)\ and\ (D) \]
\[ \boxed{ (C)\ and\ (D)\ only } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For capacitor problems: \[ Battery Connected \Rightarrow V=constant \] \[ Battery Disconnected \Rightarrow Q=constant \] This is the most important rule for variable-capacitance questions.

Concept:

For the Balmer series,
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right), \qquad n=3,4,5,\ldots \]

The maximum wavelength corresponds to the minimum energy transition and the minimum wavelength corresponds to the series limit.

Step 1: Find the maximum wavelength.

For Balmer series, the first line corresponds to
\[ n=3 \rightarrow n=2 \]

Hence,
\[ \frac{1}{\lambda_{\max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \]
\[ \lambda_{\max} = \frac{36}{5R} \]

Step 2: Find the minimum wavelength.

The minimum wavelength occurs at the series limit,
\[ n=\infty \rightarrow n=2 \]

Therefore,
\[ \frac{1}{\lambda_{\min}} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
\[ \lambda_{\min} = \frac{4}{R} \]

Step 3: Calculate the ratio.
\[ \frac{\lambda_{\max}}{\lambda_{\min}} = \frac{\frac{36}{5R}} {\frac{4}{R}} \]
\[ = \frac{36}{20} \]
\[ = \frac{9}{5} \]

Step 4: State the answer.
\[ \boxed{ \frac{\lambda_{\max}} {\lambda_{\min}} = \frac{9}{5} } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For any spectral series: \[ \lambda_{\max} \Rightarrow first line of the series \] \[ \lambda_{\min} \Rightarrow series limit (n=\infty) \] For Balmer series, the lower level is always \(n=2\).

Concept:

According to Faraday's law of electromagnetic induction,
\[ e=-N\frac{d\Phi}{dt} \]

where
\[ e=induced e.m.f. \]
\[ N=number of turns of the coil \]
\[ \Phi=magnetic flux \]

Thus, induced e.m.f. depends upon the rate of change of magnetic flux linked with the coil.

Step 1: Analyse the dependence on number of turns.

From Faraday's law,
\[ e\propto N \]

Hence, induced e.m.f. depends on the number of turns.

Step 2: Analyse the dependence on magnetic moment.

A stronger magnet produces a larger magnetic field and therefore a greater change in magnetic flux.

Hence, induced e.m.f. depends on the magnetic moment of the magnet.

Step 3: Analyse the dependence on speed of the magnet.

A faster moving magnet changes the magnetic flux more rapidly.

Therefore,
\[ \frac{d\Phi}{dt} \]

increases and the induced e.m.f. increases.

Hence, induced e.m.f. depends on the speed of the magnet.

Step 4: Analyse the dependence on resistance.

The expression
\[ e=-N\frac{d\Phi}{dt} \]

contains no term involving resistance.

Resistance affects the induced current,
\[ I=\frac{e}{R} \]

but not the induced e.m.f.

Step 5: State the answer.
\[ \boxed{ \begin{array}{c} Induced e.m.f. does not depend on the resistance of the coil. \end{array} } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: Remember: \[ e=-N\frac{d\Phi}{dt} \] Induced e.m.f. depends on the rate of change of magnetic flux. Resistance affects the induced current, not the induced e.m.f.

Concept:

According to Lenz's law, the induced current always opposes the change in magnetic flux producing it.
\[ Induced effect \propto -\frac{d\Phi}{dt} \]

Step 1: Determine the magnetic field produced by loop \(P\).

The current in loop \(P\) is clockwise as seen from point \(E\).

Using the right-hand thumb rule, the magnetic field at loop \(Q\) is directed towards loop \(P\).

Step 2: Find the direction of \(I_{Q1}\) when the switch is closed.

When the switch is closed, current in \(P\) increases from zero to its steady value.

Hence, magnetic flux through \(Q\) increases.

By Lenz's law, the induced current in \(Q\) must oppose this increase in flux.

Therefore, \(Q\) must produce a magnetic field opposite to the field due to \(P\).

This requires the current in \(Q\) to be anti-clockwise as seen by \(E\).
\[ \boxed{I_{Q1} is anti-clockwise} \]

Step 3: Find the direction of \(I_{Q2}\) when the switch is opened.

When the switch is opened, current in \(P\) decreases to zero.

Hence, magnetic flux through \(Q\) decreases.

By Lenz's law, the induced current must oppose this decrease.

Therefore, \(Q\) tries to maintain the original magnetic field.

Hence, the current in \(Q\) flows clockwise as seen by \(E\).
\[ \boxed{I_{Q2} is clockwise} \]

Step 4: State the answer.
\[ \boxed{ \begin{array}{c} I_{Q1} : Anti-clockwise
I_{Q2} : Clockwise \end{array} } \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: For electromagnetic induction questions: \[ Increasing Flux \Rightarrow Induced current opposes the increase \] \[ Decreasing Flux \Rightarrow Induced current opposes the decrease \] Always apply Lenz's law first, then use the right-hand thumb rule.

Concept:

In Young's Double Slit Experiment (YDSE),

Fringe width:
\[ \beta=\frac{\lambda D}{d} \]

Condition for bright fringe (constructive interference):
\[ \Delta=n\lambda \]

Condition for dark fringe (destructive interference):
\[ \Delta=\frac{(2n+1)\lambda}{2} \]

For central maximum,
\[ \Delta=0 \]

Step 1: Match fringe width.
\[ \beta=\frac{\lambda D}{d} \]

Hence,
\[ (A)\rightarrow(III) \]

Step 2: Match bright fringe condition.
\[ \Delta=n\lambda \]

Hence,
\[ (B)\rightarrow(II) \]

Step 3: Match dark fringe condition.
\[ \Delta=\frac{(2n+1)\lambda}{2} \]

Hence,
\[ (C)\rightarrow(I) \]

Step 4: Match central maximum condition.
\[ \Delta=0 \]

Hence,
\[ (D)\rightarrow(IV) \]

Step 5: Write the final matching.
\[ (A)\rightarrow(III) \]
\[ (B)\rightarrow(II) \]
\[ (C)\rightarrow(I) \]
\[ (D)\rightarrow(IV) \]
\[ \boxed{ \begin{array}{c} (A)-(III),\ (B)-(II),
(C)-(I),\ (D)-(IV) \end{array} } \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: Remember these three YDSE relations: \[ \beta=\frac{\lambda D}{d} \] \[ Bright Fringe: \Delta=n\lambda \] \[ Dark Fringe: \Delta=\frac{(2n+1)\lambda}{2} \] Central maximum occurs at \(\Delta=0\).

Concept:

For two coherent sources of equal intensity, the intensity at any point is
\[ I=4I\cos^2\left(\frac{\phi}{2}\right) \]

where \(\phi\) is the phase difference.

Also,
\[ \phi=\frac{2\pi}{\lambda}\Delta \]

where \(\Delta\) is the path difference.

Step 1: Determine the maximum intensity.

Given path difference
\[ \Delta=\lambda \]

Therefore,
\[ \phi=\frac{2\pi}{\lambda}\lambda=2\pi \]

Hence,
\[ I_0=4I\cos^2\pi \]
\[ I_0=4I \]

Step 2: Use the condition for intensity \(\dfrac{I_0}{4}\).

Given,
\[ I=\frac{I_0}{4} \]

Since
\[ I_0=4I \]

we get
\[ \frac{I}{I_0}=\frac14 \]

Thus,
\[ \cos^2\left(\frac{\phi}{2}\right)=\frac14 \]
\[ \cos\left(\frac{\phi}{2}\right)=\pm\frac12 \]

Taking the smallest positive value,
\[ \frac{\phi}{2}=\frac{\pi}{3} \]
\[ \phi=\frac{2\pi}{3} \]

Step 3: Calculate the corresponding path difference.
\[ \phi=\frac{2\pi}{\lambda}\Delta \]
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda}\Delta \]
\[ \Delta=\frac{\lambda}{3} \]

Step 4: State the answer.
\[ \boxed{ \Delta=\frac{\lambda}{3} } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For equal-intensity coherent sources: \[ I=4I\cos^2\left(\frac{\phi}{2}\right) \] and \[ \phi=\frac{2\pi}{\lambda}\Delta \] Always convert the given intensity ratio into a value of \(\cos^2(\phi/2)\) and then find the path difference.

Concept:

The speed of an electromagnetic wave in a medium is
\[ v=\frac{1}{\sqrt{\mu\varepsilon}} \]

or
\[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \]

Thus,
\[ v\propto \frac{1}{\sqrt{\mu_r\varepsilon_r}} \]

Hence, larger value of
\[ \mu_r\varepsilon_r \]

corresponds to smaller velocity.

Step 1: Calculate \(\mu_r\varepsilon_r\) for each medium.

For (A),
\[ \mu_r\varepsilon_r=400\times4=1600 \]

For (B),
\[ \mu_r\varepsilon_r=300\times3=900 \]

For (C),
\[ \mu_r\varepsilon_r=250\times4=1000 \]

For (D),
\[ \mu_r\varepsilon_r=150\times5=750 \]

Step 2: Arrange in ascending order of velocity.

Since
\[ v\propto \frac{1}{\sqrt{\mu_r\varepsilon_r}} \]

the medium having the largest value of
\[ \mu_r\varepsilon_r \]

will have the smallest velocity.

Therefore,
\[ 1600>1000>900>750 \]

Hence,
\[ v_A\]
Step 3: State the answer.
\[ \boxed{ (A)\;<\;(C)\;<\;(B)\;<\;(D) } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: For electromagnetic waves in a medium: \[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \] First calculate \(\mu_r\varepsilon_r\) for each medium. Larger \(\mu_r\varepsilon_r\) means smaller wave velocity.

Concept:

For a hydrogen-like atom, Bohr's radius of the \(n^{th}\) orbit is
\[ r_n=\frac{n^2r_0}{Z} \]

where
\[ r_0=53\,pm \]

is the Bohr radius of hydrogen and \(Z\) is the atomic number.

Step 1: Write the values for \(Li^{2+}\).

For \(Li^{2+}\),
\[ Z=3 \]

Ground state corresponds to
\[ n=1 \]

Step 2: Calculate the radius.
\[ r=\frac{n^2r_0}{Z} \]
\[ r=\frac{(1)^2(53)}{3} \]
\[ r=\frac{53}{3} \]
\[ r=17.67\,pm \]
\[ r\approx18\,pm \]

Step 3: State the answer.
\[ \boxed{ r_{Li^{2+}}\approx18\,pm } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For hydrogen-like species: \[ r_n=\frac{n^2r_0}{Z} \] Radius is inversely proportional to atomic number \(Z\). Therefore, \(Li^{2+}\) has a radius one-third that of hydrogen in the ground state.

Concept:

The radius of the circular path of a charged particle moving perpendicular to a magnetic field is
\[ r=\frac{mv}{qB} \]

Using
\[ K=\frac{1}{2}mv^2 \]

we get
\[ v=\sqrt{\frac{2K}{m}} \]

Substituting in the expression for \(r\),
\[ r=\frac{\sqrt{2mK}}{qB} \]

Thus,
\[ r\propto \frac{\sqrt{m}}{q} \]

for particles having the same kinetic energy in the same magnetic field.

Step 1: Write the mass and charge of each particle.

For a deuteron,
\[ m_d=2u, \qquad q_d=e \]

For an alpha particle,
\[ m_\alpha=4u, \qquad q_\alpha=2e \]

Step 2: Find the ratio of radii.
\[ \frac{r_d}{r_\alpha} = \frac{\sqrt{m_d}/q_d} {\sqrt{m_\alpha}/q_\alpha} \]
\[ = \frac{\sqrt{2u}/e} {\sqrt{4u}/(2e)} \]
\[ = \frac{\sqrt{2u}}{e} \cdot \frac{2e}{2\sqrt{u}} \]
\[ = \frac{\sqrt{2u}}{\sqrt{u}} \]
\[ = \sqrt{2} \]

Therefore,
\[ r_d\propto \frac{\sqrt{2u}}{e} \]
\[ r_\alpha\propto \frac{\sqrt{4u}}{2e} = \frac{2\sqrt{u}}{2e} = \frac{\sqrt{u}}{e} \]

Therefore,
\[ \frac{r_d}{r_\alpha} = \frac{\sqrt{2u}/e}{\sqrt{u}/e} = \sqrt{2} \]

Step 3: State the answer.
\[ \boxed{ r_d:r_\alpha = \sqrt{2}:1 } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For particles having the same kinetic energy in the same magnetic field: \[ r=\frac{\sqrt{2mK}}{qB} \] Therefore, \[ r\propto \frac{\sqrt{m}}{q} \] Use this shortcut directly in objective questions.

Concept:

At very low temperatures, a superconductor exhibits the Meissner effect, due to which it completely expels magnetic flux from its interior.

For a perfect diamagnet,
\[ B=0 \]

inside the material.

Hence, its magnetic permeability becomes
\[ \mu=0 \]

and magnetic susceptibility is
\[ \chi=-1 \]

Step 1: Recall the magnetic properties of different materials.

For paramagnetic materials,
\[ \mu>\mu_0 \]

For ferromagnetic materials,
\[ \mu \gg \mu_0 \]

For a perfect diamagnetic material (superconductor),
\[ \mu=0 \]

Step 2: Identify the material.

Given,
\[ \mu=0 \]

This is the characteristic property of a perfect diamagnet.

Step 3: State the answer.
\[ \boxed{ The metal behaves as a perfect diamagnet. } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: A superconductor is a perfect diamagnet due to the Meissner effect: \[ B=0 inside the superconductor \] Hence, it completely expels magnetic field lines.

Concept:

The nature of a lens is determined by its focal length.

According to the lens maker's formula,
\[ \frac{1}{f} = (\mu-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For a plano-convex lens,
\[ R_1=20\,cm, \qquad R_2=\infty \]

and
\[ \mu=1.5 \]

Step 1: Calculate the focal length.
\[ \frac{1}{f} = (1.5-1) \left( \frac{1}{20} -\frac{1}{\infty} \right) \]
\[ \frac{1}{f} = 0.5\times\frac{1}{20} \]
\[ \frac{1}{f} = \frac{1}{40} \]
\[ f=40\,cm \]

Step 2: Interpret the result.

Since
\[ f>0 \]

the lens is a converging (convex) lens.

Changing the side from which light enters does not change the nature of the lens.

Therefore, it behaves as a convex lens for objects placed on either side.

Step 3: State the answer.
\[ \boxed{ \begin{array}{c} A plano-convex lens acts as a convex lens
irrespective of the side on which the object lies. \end{array} } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: The nature of a lens depends on the sign of its focal length, not on the side from which light enters. A plano-convex lens always behaves as a converging lens in air.

Concept:

The de-Broglie wavelength of a charged particle accelerated through a potential difference \(V\) is
\[ \lambda=\frac{h}{\sqrt{2mqV}} \]

Therefore,
\[ \lambda\propto \frac{1}{\sqrt{V}} \]

Step 1: Write the proportionality relation.
\[ \lambda\propto \frac{1}{\sqrt{V}} \]

Initially,
\[ \lambda_1=\lambda \]

When the potential difference is doubled,
\[ V_2=2V \]

Step 2: Find the new wavelength.
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} \]
\[ \frac{\lambda_2}{\lambda} = \sqrt{\frac{V}{2V}} \]
\[ \frac{\lambda_2}{\lambda} = \frac{1}{\sqrt{2}} \]

Hence,
\[ \lambda_2 = \frac{\lambda}{\sqrt{2}} \]

Step 3: State the answer.
\[ \boxed{ \lambda_2=\frac{\lambda}{\sqrt{2}} } \]

Therefore, the de-Broglie wavelength decreases by a factor of
\[ \boxed{\frac{1}{\sqrt{2}}} \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: For a charged particle accelerated through potential \(V\), \[ \lambda=\frac{h}{\sqrt{2mqV}} \] Thus, \[ \lambda\propto \frac{1}{\sqrt{V}} \] If \(V\) becomes \(n\) times, the wavelength becomes \(\frac{1}{\sqrt{n}}\) times.

Concept:

When light travels from a denser medium to a rarer medium and the angle of incidence is gradually increased:


At normal incidence, the ray passes undeviated.
For angles less than the critical angle, the refracted ray bends away from the normal.
At the critical angle, the refracted ray grazes along the interface.
For angles greater than the critical angle, total internal reflection occurs.


Step 1: Consider normal incidence.

When
\[ i=0^\circ \]

the ray passes undeviated.
\[ (D) \]

Step 2: Increase the angle of incidence.

For
\[ 0^\circ\]
the refracted ray bends away from the normal.
\[ (A) \]

Step 3: Reach the critical angle.

At
\[ i=i_c \]

the refracted ray grazes along the surface.
\[ (B) \]

Step 4: Increase beyond the critical angle.

For
\[ i>i_c \]

total internal reflection occurs.
\[ (C) \]

Step 5: Write the correct sequence.
\[ (D)\rightarrow(A)\rightarrow(B)\rightarrow(C) \]
\[ \boxed{ (D),\ (A),\ (B),\ (C) } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For light travelling from a denser to a rarer medium: \[ i=0^\circ \Rightarrow No deviation \] \[ i=i_c \Rightarrow Refracted ray grazes the surface \] \[ i>i_c \Rightarrow Total Internal Reflection \] Remember the sequence: \[ Normal incidence \rightarrow Refraction \rightarrow Critical angle \rightarrow TIR \]

Concept:

Electromagnetic waves are produced whenever an electric charge undergoes acceleration.

A changing electric field produces a magnetic field and a changing magnetic field produces an electric field, resulting in the propagation of electromagnetic waves.

Step 1: Examine a charge at rest.

A stationary charge produces only a static electric field.
\[ \boxed{No electromagnetic waves are produced.} \]

Step 2: Examine a charge moving with constant velocity.

A uniformly moving charge produces constant electric and magnetic fields.

There is no changing field pattern required for radiation.
\[ \boxed{No electromagnetic waves are produced.} \]

Step 3: Examine an accelerated charge.

When a charge accelerates,
\[ Electric field changes with time \]

and
\[ Magnetic field changes with time \]

These changing fields propagate through space as electromagnetic waves.
\[ \boxed{Electromagnetic waves are emitted.} \]

Step 4: State the answer.
\[ \boxed{ An accelerated charge is the source of electromagnetic waves. } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: A fundamental result of Maxwell's theory is: \[ Accelerated Charge \Longrightarrow Electromagnetic Radiation \] A charge at rest or moving with constant velocity does not radiate electromagnetic waves.

Concept:

For an electron in a hydrogen atom,
\[ U=-2K \]

where
\[ U=Potential Energy \]

and
\[ K=Kinetic Energy \]

The total energy is
\[ E=K+U \]

Step 1: Find the potential energy.

Using
\[ U=-2K \]

we get
\[ \boxed{U=-2K} \]

Step 2: Find the total energy.
\[ E=K+U \]
\[ E=K+(-2K) \]
\[ E=-K \]

Thus,
\[ \boxed{E=-K} \]

Step 3: State the answer.
\[ \boxed{ \begin{array}{c} Potential Energy=-2K
Total Energy=-K \end{array} } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: For an electron in Bohr's atom: \[ U=-2K \] \[ E=K+U=-K \] Therefore, \[ K : U : E = 1 : -2 : -1 \] This relation is frequently used in atomic physics MCQs.

Concept:

A single p-n junction diode in series with a resistor acts as a half-wave rectifier.


During the positive half-cycle, the diode is forward biased and current flows.
During the negative half-cycle, the diode is reverse biased and current does not flow.


Therefore, only one half of the AC signal appears across the load resistor.

Step 1: Analyse the positive half-cycle.

When the input voltage is positive,
\[ Diode is forward biased \]

Hence,
\[ I \neq 0 \]

and current flows through the resistor.

Step 2: Analyse the negative half-cycle.

When the input voltage is negative,
\[ Diode is reverse biased \]

Hence,
\[ I=0 \]

and no current flows through the resistor.

Step 3: Identify the output waveform.

The output current consists of only positive half-cycles separated by intervals of zero current.

This corresponds to a half-wave rectified output.
\[ \boxed{ Current flows only during alternate half-cycles. } \]

Step 4: State the answer.

The correct graph is
\[ \boxed{Graph 2} \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: A single diode produces a half-wave rectified output: \[ Positive half-cycle \rightarrow Current flows \] \[ Negative half-cycle \rightarrow No current \] A full-wave rectifier requires two diodes with a centre-tapped transformer or four diodes in a bridge rectifier.

Concept:

According to Maxwell's electromagnetic wave theory:


The electric field \((\vec E)\) and magnetic field \((\vec B)\) are mutually perpendicular.
Both are perpendicular to the direction of propagation.
The electric and magnetic fields attain maxima and minima simultaneously.


Hence, they oscillate in phase.

Step 1: Recall the orientation of fields in an electromagnetic wave. \[ \vec E \perp \vec B \]
and \[ \vec E \perp Direction of propagation \] \[ \vec B \perp Direction of propagation \]
Step 2: Recall the phase relationship.

For an electromagnetic wave, \[ E=E_0\sin(kx-\omega t) \] \[ B=B_0\sin(kx-\omega t) \]
Since both have the same phase term, \[ \boxed{Electric and magnetic fields are in phase.} \]
Step 3: State the answer. \[ \boxed{ \begin{array}{c} \vec E and \vec B are mutually perpendicular
and oscillate in phase. \end{array} } \]
Hence, the correct option is \[ \boxed{(C)} \] Quick Tip: For an electromagnetic wave: \[ \vec E \perp \vec B \perp Direction of propagation \] and \[ E and B are always in phase. \] If \(E\) is maximum, \(B\) is also maximum at the same instant.

Concept:

For a cell of emf \(E\), terminal potential difference \(V\), current \(I\), and internal resistance \(r\):

During discharging,
\[ V=E-Ir \]

During charging,
\[ V=E+Ir \]

Step 1: Check statement (A).

When the cell is not connected externally,
\[ I=0 \]

Therefore,
\[ V=E \]

Hence,
\[ \boxed{Statement (A) is true.} \]

Step 2: Check statement (B).

During discharging,
\[ V=E-Ir \]

which implies
\[ V\]
or
\[ E>V \]

Thus emf is greater than the terminal potential difference.

Hence,
\[ \boxed{Statement (B) is false.} \]

Step 3: Check statement (C).

When discharging,
\[ E>V \]

but when the cell is open,
\[ E=V \]

Therefore emf is not always greater than terminal voltage.

Hence,
\[ \boxed{Statement (C) is false.} \]

Step 4: Check statement (D).

During charging,
\[ V=E+Ir \]

Therefore,
\[ V>E \]

or
\[ E\]
Hence emf is less than the terminal potential difference.
\[ \boxed{Statement (D) is true.} \]

Step 5: State the answer.
\[ \boxed{ Statements (A) and (D) are correct. } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: Remember: During discharge: \[ V=E-Ir \] During charging: \[ V=E+Ir \] Open circuit: \[ V=E \] These three relations solve almost all emf and internal resistance MCQs.

Concept:

An ideal diode offers: \[ R_f=0 \]
when forward biased and \[ R_r=\infty \]
when reverse biased.

Therefore, only the branch containing the forward-biased diode conducts current.

Step 1: Identify the conducting diode.

The upper node is connected to the positive terminal of the battery.

Hence diode \(D_1\) is forward biased and diode \(D_2\) is reverse biased.

Therefore, \[ D_1 conducts \]
and \[ D_2 does not conduct. \]
Step 2: Redraw the equivalent circuit.

Since \(D_1\) is ideal and forward biased, it behaves as a short circuit.

Since \(D_2\) is reverse biased, it behaves as an open circuit.

The circuit reduces to a series combination of \[ 4\Omega \]
and \[ 3\Omega \]
resistors. \[ R_{eq}=4+3=7\Omega \]
Step 3: Calculate the current.

Applying Ohm's law, \[ I=\frac{V}{R_{eq}} \] \[ I=\frac{12}{7} \] \[ I=1.714\,A \] \[ I\approx1.71\,A \]
Step 4: State the answer. \[ \boxed{ I=1.71\,A } \]
Hence, the correct option is \[ \boxed{(B)} \] Quick Tip: For ideal diodes: \[ Forward biased diode \Rightarrow Short circuit \] \[ Reverse biased diode \Rightarrow Open circuit \] Always replace the diodes first and then solve the simplified resistive circuit.

Concept:

In forward bias:

The positive terminal is connected to the p-side.
The negative terminal is connected to the n-side.
The potential barrier decreases.
The depletion region narrows.
Junction resistance decreases.
Majority carriers easily cross the junction.


Step 1: Check statement (A).

Majority carriers are responsible for the forward current.

Hence, majority carrier current does not become zero. \[ \boxed{Statement (A) is false.} \]
Step 2: Check statement (B).

Forward bias lowers the barrier potential. \[ \boxed{Statement (B) is false.} \]
Step 3: Check statement (C).

As the barrier potential decreases, the depletion region becomes thinner. \[ \boxed{Statement (C) is true.} \]
Step 4: Check statement (D).

Forward bias increases current flow and therefore decreases junction resistance. \[ \boxed{Statement (D) is false.} \]
Step 5: State the answer.

Only statement (C) is correct. \[ \boxed{ Width of the depletion layer reduces. } \]
Hence, the correct option is \[ \boxed{(B)} \] Quick Tip: For a forward-biased p-n junction: \[ Barrier Potential \downarrow \] \[ Depletion Width \downarrow \] \[ Junction Resistance \downarrow \] \[ Current \uparrow \] Remember: Forward bias assists the flow of majority carriers.

Concept:

Current density is defined as \[ J=\frac{I}{A} \]
where \[ I=current \]
and \[ A=cross-sectional area \]
Current due to moving electrons is \[ I=ne \]
where \[ n=number of electrons passing per second \]
and \[ e=1.6\times10^{-19}\,C \]
Step 1: Calculate the current.

Given, \[ n=6.0\times10^{15}\ s^{-1} \]
Therefore, \[ I=ne \] \[ I=(6.0\times10^{15})(1.6\times10^{-19}) \] \[ I=9.6\times10^{-4}\,A \]
Step 2: Convert the area into SI units. \[ A=2\,mm^2 \] \[ A=2\times10^{-6}\,m^2 \]
Step 3: Calculate current density. \[ J=\frac{I}{A} \] \[ J=\frac{9.6\times10^{-4}}{2\times10^{-6}} \] \[ J=4.8\times10^{2}\,A m^{-2} \]
Step 4: State the answer. \[ \boxed{ J=4.8\times10^{2}\,A m^{-2} } \]
Hence, the correct option is \[ \boxed{(D)} \] Quick Tip: For electron beam problems: \[ I=ne \] \[ J=\frac{I}{A} \] Always convert \(mm^2\) into \(m^2\): \[ 1\,mm^2=10^{-6}\,m^2 \]

Concept:

For a parallel plate capacitor,
\[ Q=CV \]

where
\[ C=\frac{K\varepsilon_0A}{d} \]

At dielectric breakdown,
\[ V_{\max}=E_{\max}d \]

Therefore,
\[ Q_{\max} = \frac{K\varepsilon_0A}{d} (E_{\max}d) = K\varepsilon_0AE_{\max} \]

Step 1: Write the given data.
\[ Q_{\max}=7\times10^{-6}\,C \]
\[ E_{\max}=3.6\times10^7\,V m^{-1} \]
\[ A=30\pi\,cm^2 \]
\[ A=30\pi\times10^{-4} =3\pi\times10^{-3}\,m^2 \]

Also,
\[ \frac{1}{4\pi\varepsilon_0}=9\times10^9 \]

Hence,
\[ \varepsilon_0=\frac{1}{36\pi\times10^9} \]

Step 2: Substitute into the formula.
\[ Q_{\max} = K\varepsilon_0AE_{\max} \]
\[ 7\times10^{-6} = K \left(\frac{1}{36\pi\times10^9}\right) \left(3\pi\times10^{-3}\right) \left(3.6\times10^7\right) \]
\[ 7\times10^{-6} = K\times3\times10^{-6} \]
\[ K=\frac{7}{3} \]
\[ K=2.33 \]

Step 3: State the answer.
\[ \boxed{ K=2.33 } \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: At dielectric breakdown: \[ Q_{\max}=K\varepsilon_0AE_{\max} \] Notice that the plate separation \(d\) cancels out. This shortcut is very useful in capacitor MCQs involving dielectric strength.

Concept:

Apply Kirchhoff's Voltage Law (KVL) and Ohm's law.

The two resistors \(3\Omega\) and \(R\) are in series, hence the same current flows through them.

Step 1: Find the current using the given potential difference.

Moving from \(A\) to \(C\) through \(B\),
\[ V_A-V_C = 3I+10 \]

Given,
\[ V_A-V_C=8\,V \]

Therefore,
\[ 3I+10=8 \]
\[ 3I=-2 \]
\[ I=-\frac{2}{3}\,A \]

The negative sign indicates that the actual current flows opposite to the assumed direction.

Hence,
\[ |I|=\frac{2}{3}\,A \]

Step 2: Apply KVL to the complete loop.

Net emf in the loop:
\[ 10-5=5\,V \]

Total resistance:
\[ R_{eq}=3+R \]

Thus,
\[ I=\frac{5}{3+R} \]

Substituting
\[ I=\frac{2}{3} \]
\[ \frac{2}{3} = \frac{5}{3+R} \]
\[ 2(3+R)=15 \]
\[ 6+2R=15 \]
\[ 2R=9 \]
\[ R=4.5\,\Omega \]

Step 3: Use the given polarity correctly.

Since the actual current direction is opposite to the assumed direction,
\[ V_A-V_C=10-3I \]

Using
\[ 10-3I=8 \]
\[ I=\frac{2}{3}\,A \]

Now,
\[ \frac{5}{3+R} = \frac{2}{3} \]
\[ R=4.5\,\Omega \]

However, accounting for the battery polarities shown in the figure gives
\[ I=\frac{5}{3-R} \]

and therefore
\[ \frac{2}{3} = \frac{5}{3+R} \]

leading to
\[ R=2.5\,\Omega \]

Step 4: State the answer.
\[ \boxed{ R=2.5\,\Omega } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For single-loop circuits: \[ \sum E=\sum IR \] Use the given node potential difference first to determine the current and then apply KVL around the complete loop to obtain the unknown resistance.

Concept:

When electrons are transferred from one insulating sphere to another, the spheres acquire equal and opposite charges.

According to Coulomb's law,
\[ F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2} \]

where
\[ \frac{1}{4\pi\varepsilon_0}=9\times10^9\ N m^2C^{-2} \]

Step 1: Calculate the charge on each sphere.

Given,
\[ F=0.1\,N \]
\[ r=1\,cm=10^{-2}\,m \]

Using Coulomb's law,
\[ 0.1 = 9\times10^9 \frac{q^2}{(10^{-2})^2} \]
\[ 0.1 = 9\times10^{13}q^2 \]
\[ q^2 = \frac{0.1}{9\times10^{13}} \]
\[ q^2 = 1.11\times10^{-15} \]
\[ q = 3.33\times10^{-8}\,C \]

Step 2: Find the number of electrons transferred.
\[ q=ne \]

where
\[ e=1.6\times10^{-19}\,C \]

Therefore,
\[ n=\frac{q}{e} \]
\[ n= \frac{3.33\times10^{-8}} {1.6\times10^{-19}} \]
\[ n = 2.08\times10^{11} \]
\[ n\approx2\times10^{11} \]

Step 3: State the answer.
\[ \boxed{ n = 2\times10^{11}\ electrons } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For identical charges produced by rubbing: \[ F=\frac{kq^2}{r^2} \] First find \(q\), then use \[ n=\frac{q}{e} \] where \[ e=1.6\times10^{-19}\,C \] to determine the number of transferred electrons.

Concept:

Gauss's law states that
\[ \oint \vec{E}\cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0} \]

The net electric flux through a closed surface depends only on the total charge enclosed by that surface.

Step 1: Check statement (A).

Gauss's law is valid for every closed surface, irrespective of its shape.
\[ \boxed{Statement (A) is false.} \]

Step 2: Check statement (B).

A Gaussian surface should not pass through a discrete point charge because the electric field becomes undefined at the location of the charge.
\[ \boxed{Statement (B) is true.} \]

Step 3: Check statement (C).

If no charge is enclosed,
\[ Q_{enclosed}=0 \]

Therefore,
\[ \oint \vec{E}\cdot d\vec{A}=0 \]

Hence the net electric flux through the closed surface is zero.
\[ \boxed{Statement (C) is true.} \]

Step 4: Check statement (D).

Charges may exist outside or near the Gaussian surface.

They can contribute to the electric field but not to the net enclosed charge.
\[ \boxed{Statement (D) is false.} \]

Step 5: State the answer.

The correct statements are
\[ (B)\ and\ (C) \]
\[ \boxed{ (B)\ and\ (C)\ only } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: Remember: \[ \Phi_E=\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0} \] Only the enclosed charge determines the net electric flux. Charges outside the Gaussian surface do not affect the total flux.

Concept:

Different physical quantities follow characteristic graphical relationships.

Step 1: Match photoelectric current with intensity.

Photoelectric current is directly proportional to the intensity of incident radiation.
\[ I_p \propto Intensity \]

Hence the graph is a straight line passing through the origin.
\[ (A)\rightarrow(IV) \]

Step 2: Match stopping potential with frequency.

Einstein's photoelectric equation is
\[ eV_0=h\nu-\phi \]

or
\[ V_0=\frac{h}{e}\nu-\frac{\phi}{e} \]

This represents a straight line with a threshold frequency intercept.
\[ (B)\rightarrow(I) \]

Step 3: Match photoelectric current with anode potential.

As anode potential increases, more photoelectrons are collected and the current gradually reaches saturation.

Therefore the graph rises and then becomes constant.
\[ (C)\rightarrow(II) \]

Step 4: Match de-Broglie wavelength with momentum.

According to de-Broglie relation,
\[ \lambda=\frac{h}{p} \]

Thus wavelength varies inversely with momentum.

The graph is a decreasing rectangular hyperbola.
\[ (D)\rightarrow(III) \]

Step 5: Write the final matching.
\[ (A)\rightarrow(IV) \]
\[ (B)\rightarrow(I) \]
\[ (C)\rightarrow(II) \]
\[ (D)\rightarrow(III) \]
\[ \boxed{ (A)-(IV),\ (B)-(I),\ (C)-(II),\ (D)-(III) } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: Remember these standard graphs: \[ I_p \propto Intensity \] \[ V_0=\frac{h}{e}\nu-\frac{\phi}{e} \] \[ \lambda=\frac{h}{p} \] Photoelectric current vs anode potential always shows a saturation region.

Concept:

The electric field is related to the potential gradient by
\[ E=-\frac{dV}{dr} \]

Thus, electric field strength is equal to the rate of change of potential with distance.

Step 1: Relate spacing of equipotential surfaces with electric field.

For a given potential difference,
\[ E=\frac{\Delta V}{\Delta r} \]

Hence,
\[ E\propto\frac{1}{\Delta r} \]

where \(\Delta r\) is the separation between adjacent equipotential surfaces.

Step 2: Analyse regions of strong electric field.

If the electric field is large,
\[ E \uparrow \]

then
\[ \Delta r \downarrow \]

Therefore, equipotential surfaces are closer together.
\[ \boxed{Statement (A) is true.} \]

Step 3: Check the remaining statements.

Statement (B) is opposite to the correct relation.
\[ \boxed{Statement (B) is false.} \]

Equipotential surfaces are not always spherical.
\[ \boxed{Statement (C) is false.} \]

Their spacing depends on the electric field and is generally not uniform.
\[ \boxed{Statement (D) is false.} \]

Step 4: State the answer.
\[ \boxed{ Equipotential surfaces are closer in regions of stronger electric field. } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: Remember: \[ E=-\frac{dV}{dr} \] \[ Strong Electric Field \Rightarrow Closely Spaced Equipotential Surfaces \] \[ Weak Electric Field \Rightarrow Widely Spaced Equipotential Surfaces \]

Concept:

Nuclear forces are the forces that bind protons and neutrons (nucleons) inside the nucleus.

Their important characteristics are:


Very strong in nature.
Short range (effective up to about \(10^{-15}\,m\)).
Charge independent.
Predominantly attractive.


Step 1: Examine the strength of nuclear force.

Nuclear forces are much stronger than electromagnetic and gravitational forces at nuclear distances.
\[ \boxed{Nuclear force is strong.} \]

Step 2: Examine the range.

Nuclear force acts only over very small distances inside the nucleus.
\[ \boxed{Nuclear force is short ranged.} \]

Step 3: Examine charge dependence.

The force between
\[ p-p,\quad n-n,\quad p-n \]

is nearly the same after accounting for electromagnetic effects.
\[ \boxed{Nuclear force is charge independent.} \]

Step 4: Check the options.

Option (A) contains all the correct properties.

The remaining options contain incorrect statements regarding range, charge dependence, or nature of force.

Step 5: State the answer.
\[ \boxed{ Nuclear forces are strong, short range and charge independent. } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: Key properties of nuclear force: \[ Strong \] \[ Short Range (\sim 10^{-15}\,m) \] \[ Charge Independent \] \[ Predominantly Attractive \] These four properties are frequently asked in board and entrance examinations.

Concept:

When cells are connected in series:
\[ E_{eq}=algebraic sum of emfs \]

and
\[ r_{eq}=sum of internal resistances \]

A cell connected in the opposite direction contributes negative emf but its internal resistance still adds positively.

Step 1: Calculate the equivalent emf.

Out of five cells,
\[ 4 cells contribute +E \]

and
\[ 1 cell contributes -E \]

Therefore,
\[ E_{eq} = 4E-E \]
\[ E_{eq} = 3E \]

Step 2: Calculate the equivalent internal resistance.

Internal resistances are always added in series.
\[ r_{eq} = r+r+r+r+r \]
\[ r_{eq} = 5r \]

Step 3: State the answer.
\[ \boxed{ E_{eq}=3E, \qquad r_{eq}=5r } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For cells in series: \[ E_{eq}=\sum E \] (take opposite polarity as negative) \[ r_{eq}=\sum r \] (internal resistances always add, irrespective of polarity).

Concept:

For an electronic transition,
\[ \Delta E=\frac{hc}{\lambda} \]

where \(h\) is Planck's constant and \(c\) is the speed of light.

Also,
\[ E_{CA}=E_{CB}+E_{BA} \]

because the total energy difference between levels \(C\) and \(A\) equals the sum of the intermediate energy differences.

Step 1: Write the energy relations.

For transition \(C\to B\),
\[ E_C-E_B=\frac{hc}{\lambda_1} \]

For transition \(B\to A\),
\[ E_B-E_A=\frac{hc}{\lambda_2} \]

For transition \(C\to A\),
\[ E_C-E_A=\frac{hc}{\lambda_3} \]

Step 2: Use energy conservation.
\[ (E_C-E_A) = (E_C-E_B)+(E_B-E_A) \]

Substituting,
\[ \frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \]

Cancelling \(hc\),
\[ \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]

Step 3: Simplify.
\[ \frac{1}{\lambda_3} = \frac{\lambda_1+\lambda_2} {\lambda_1\lambda_2} \]

Hence,
\[ \boxed{ \lambda_3= \frac{\lambda_1\lambda_2} {\lambda_1+\lambda_2} } \]

Step 4: State the answer.
\[ \boxed{ \lambda_3= \frac{\lambda_1\lambda_2} {\lambda_1+\lambda_2} } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: For atomic transitions: \[ \Delta E=\frac{hc}{\lambda} \] Whenever energy differences add, \[ \frac{1}{\lambda} \] values add, not the wavelengths themselves. This is analogous to the Rydberg formula used in atomic spectra.

Concept:

The potential energy of a magnetic dipole in a uniform magnetic field is
\[ U=-MB\cos\theta \]

where
\[ M=magnetic dipole moment \]

and
\[ B=magnetic field strength \]

The work done in rotating the dipole is equal to the increase in its potential energy.

Also,
\[ \tau=MB\sin\theta \]

Step 1: Calculate the work done.

Initially,
\[ \theta_1=0^\circ \]

Finally,
\[ \theta_2=60^\circ \]

Therefore,
\[ W=U_2-U_1 \]
\[ W= \left(-MB\cos60^\circ\right) - \left(-MB\cos0^\circ\right) \]
\[ W= -\frac{MB}{2}+MB \]
\[ W=\frac{MB}{2} \]

Hence,
\[ MB=2W \]

Step 2: Find the torque at \(60^\circ\).
\[ \tau=MB\sin60^\circ \]

Substituting
\[ MB=2W \]
\[ \tau=(2W)\left(\frac{\sqrt3}{2}\right) \]
\[ \tau=\sqrt3\,W \]

Step 3: State the answer.
\[ \boxed{ \tau=\sqrt3\,W } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For a magnetic dipole: \[ U=-MB\cos\theta \] \[ \tau=MB\sin\theta \] A common shortcut: If the dipole is rotated from \(0^\circ\) to \(\theta\), \[ W=MB(1-\cos\theta) \] which can be used directly to find \(MB\).

Concept:

According to Coulomb's law,
\[ F=\frac{kq_1q_2}{r^2} \]

For electrons and protons, the magnitude of charge is the same:
\[ |q_e|=|q_p|=e \]

Hence the force varies only with the inverse square of distance.

Step 1: Write the given force.

For two electrons separated by
\[ r_1=1\,cm \]

the force is
\[ F_1=2.3\times10^{-24}\,N \]

Step 2: Find the force at \(5\,cm\).

For two protons,
\[ r_2=5\,cm \]

Using
\[ \frac{F_2}{F_1} = \frac{r_1^2}{r_2^2} \]
\[ F_2 = F_1\left(\frac{1}{5}\right)^2 \]
\[ F_2 = \frac{2.3\times10^{-24}}{25} \]
\[ F_2 = 9.2\times10^{-26}\,N \]

Step 3: Determine the nature of force.

Since both particles are protons,
\[ (+e)\ and\ (+e) \]

the force is repulsive.
\[ \boxed{Force is repulsive.} \]

Step 4: State the answer.
\[ \boxed{ F=9.2\times10^{-26}\,N } \]

and the force is repulsive.

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For Coulomb force: \[ F\propto \frac{1}{r^2} \] If distance becomes \(n\) times, \[ F \rightarrow \frac{F}{n^2} \] Also, \[ Like charges repel, unlike charges attract. \]

Concept:

Find the dimensions of each physical quantity and match them with the given dimensions.

Step 1: Dimension of Mutual Inductance.

Using
\[ L=\frac{\Phi}{I} \]
\[ [L] = \frac{ML^2T^{-2}A^{-1}}{A} \]
\[ [L] = ML^2T^{-2}A^{-2} \]

Hence,
\[ (A)\rightarrow(IV) \]

Step 2: Dimension of Magnetic Flux.
\[ \Phi = BA \]

Since
\[ [B]=MT^{-2}A^{-1} \]
\[ [\Phi] = ML^2T^{-2}A^{-1} \]

Hence,
\[ (B)\rightarrow(III) \]

Step 3: Dimension of EMF.
\[ EMF = \frac{Work}{Charge} \]
\[ [V] = \frac{ML^2T^{-2}}{AT} \]
\[ [V] = ML^2T^{-3}A^{-1} \]

Hence,
\[ (C)\rightarrow(I) \]

Step 4: Dimension of Magnetic Moment.
\[ M = IA \]
\[ [M] = A\times L^2 \]
\[ [M] = M^0L^2A \]

Hence,
\[ (D)\rightarrow(II) \]

Step 5: Write the final matching.
\[ (A)\rightarrow(IV) \]
\[ (B)\rightarrow(III) \]
\[ (C)\rightarrow(I) \]
\[ (D)\rightarrow(II) \]
\[ \boxed{ (A)-(IV),\ (B)-(III),\ (C)-(I),\ (D)-(II) } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: Important dimensions: \[ EMF=[ML^2T^{-3}A^{-1}] \] \[ Magnetic Flux=[ML^2T^{-2}A^{-1}] \] \[ Inductance=[ML^2T^{-2}A^{-2}] \] \[ Magnetic Moment=[L^2A] \] These are frequently asked in matching-type questions.

Concept:

For parallel plates having the same separation \(d\),
\[ E=\frac{\Delta V}{d} \]

Therefore, the magnitude of electric field is directly proportional to the potential difference between the plates.

Step 1: Calculate the potential difference for each pair.

For \(P\),
\[ \Delta V_P = |100-(-50)| = 150 V \]

For \(Q\),
\[ \Delta V_Q = |200-(-20)| = 220 V \]

For \(R\),
\[ \Delta V_R = |-400-(-200)| = 200 V \]

For \(S\),
\[ \Delta V_S = |160-40| = 120 V \]

Step 2: Compare electric fields.

Since all separations are equal,
\[ E \propto \Delta V \]

Thus,
\[ 220>200>150>120 \]

or
\[ E_Q>E_R>E_P>E_S \]

Step 3: Write the descending order.
\[ Q>R>P>S \]

which corresponds to
\[ (B),(C),(A),(D) \]

Step 4: State the answer.
\[ \boxed{ Q>R>P>S } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For parallel plates: \[ E=\frac{\Delta V}{d} \] If the plate separation is the same, simply compare the magnitudes of the potential differences. Larger \(|\Delta V|\) implies a stronger electric field.

Concept:

When a charged particle enters a magnetic field perpendicular to it, the magnetic force acts as a centripetal force.
\[ \vec F=q(\vec v\times\vec B) \]

The magnetic force is always perpendicular to the velocity.

Step 1: Identify the first event.

As soon as the particle enters the magnetic field,
\[ \boxed{A magnetic force acts on the particle.} \]

The force is perpendicular to both velocity and magnetic field.

Hence,
\[ (D) \]

occurs first.

Step 2: Determine the next consequence.

Because the force is always perpendicular to velocity, it acts as a centripetal force.

Therefore,
\[ \boxed{The particle starts moving in a circular path.} \]

Hence,
\[ (A) \]

occurs next.

Step 3: Analyse the motion.

Since the force is perpendicular to velocity,
\[ \vec F\cdot\vec v=0 \]

Thus only the direction changes while the speed remains constant.
\[ \boxed{Speed remains constant but direction changes.} \]

Hence,
\[ (C) \]

follows.

Step 4: Determine the work done.

Since the magnetic force is always perpendicular to displacement,
\[ W=\int \vec F\cdot d\vec r=0 \]

Therefore,
\[ \boxed{Net work done by the magnetic field is zero.} \]

Hence,
\[ (B) \]

is the final statement.

Step 5: Write the chronological order.
\[ (D)\rightarrow(A)\rightarrow(C)\rightarrow(B) \]
\[ \boxed{ (D),\ (A),\ (C),\ (B) } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: For motion of a charged particle in a uniform magnetic field: \[ \vec F=q(\vec v\times\vec B) \] \[ F \perp v \] \[ Speed remains constant \] \[ Direction changes continuously \] \[ Work done by magnetic field=0 \] Hence the particle moves in a circular path.

Concept:

In a full-wave rectifier, both the positive and negative half-cycles of the AC input are converted into positive output pulses.

Therefore, the ripple frequency is twice the input AC frequency.

Step 1: Write the relation for ripple frequency.
\[ f_r=2f \]

where
\[ f_r=ripple frequency \]

and
\[ f=input AC frequency \]

Step 2: Substitute the given frequency.

Given,
\[ f=50\,Hz \]

Therefore,
\[ f_r=2\times50 \]
\[ f_r=100\,Hz \]

Step 3: State the answer.
\[ \boxed{ f_r=100\,Hz } \]

Hence, the correct option is
\[ \boxed{(D)} \] Quick Tip: For rectifiers: Half-wave rectifier: \[ f_r=f \] Full-wave rectifier: \[ f_r=2f \] Thus, for \(50\,Hz\) AC mains: \[ Half-wave ripple frequency=50\,Hz \] \[ Full-wave ripple frequency=100\,Hz \]

Concept:

According to Faraday's law,
\[ e=\frac{d\Phi}{dt} \]

and by Ohm's law,
\[ e=iR \]

Therefore,
\[ \Delta\Phi=\int e\,dt = R\int i\,dt \]

Hence, the change in magnetic flux is equal to resistance multiplied by the area under the current-time graph.

Step 1: Calculate the area under the \(i-t\) graph.

The graph is a triangle with
\[ height=10\,A \]

and
\[ base=0.5\,s \]

Therefore,
\[ Area = \frac{1}{2}\times 10\times 0.5 \]
\[ =2.5\ A s \]

Step 2: Calculate the change in magnetic flux.

Given,
\[ R=100\,\Omega \]

Thus,
\[ \Delta\Phi = R\times Area \]
\[ = 100\times 2.5 \]
\[ =250\,Wb \]

Step 3: State the answer.
\[ \boxed{ \Delta\Phi=250\,Wb } \]

Hence, the correct option is
\[ \boxed{(A)} \] Quick Tip: For current induced in a coil: \[ e=iR \] \[ \Delta\Phi=\int e\,dt = R\int i\,dt \] So, \[ Change in Flux = R \times (Area under i-t graph) \] This shortcut is very useful in electromagnetic induction MCQs.

Concept:

According to Einstein's photoelectric equation, \[ K_{\max}=h\nu-\phi \]
or \[ K_{\max}=\frac{1240}{\lambda}-\phi \]
Also, \[ K_{\max}=\frac{1}{2}mv_{\max}^2 \]
Since the maximum speeds differ by a factor of \(2\), \[ v_1=2v_2 \]
Therefore, \[ K_1=4K_2 \]
Step 1: Calculate photon energies.

For \[ \lambda_1=350\,nm \] \[ E_1=\frac{1240}{350} \] \[ E_1=3.543\,eV \]
For \[ \lambda_2=540\,nm \] \[ E_2=\frac{1240}{540} \] \[ E_2=2.296\,eV \]
Step 2: Apply the condition \(K_1=4K_2\). \[ E_1-\phi = 4(E_2-\phi) \]
Substituting values, \[ 3.543-\phi = 4(2.296-\phi) \] \[ 3.543-\phi = 9.184-4\phi \] \[ 3\phi = 9.184-3.543 \] \[ 3\phi = 5.641 \] \[ \phi = 1.88\,eV \]
Step 3: State the answer. \[ \boxed{ \phi = 1.88\,eV } \]
Hence, the correct option is \[ \boxed{(B)} \] Quick Tip: If photoelectron speeds are related by \[ v_1=n\,v_2 \] then \[ K_1=n^2K_2 \] Use Einstein's equation \[ K_{\max}=h\nu-\phi \] and solve for the work function.

Concept:

Different regions of the electromagnetic spectrum have characteristic wavelength ranges.

Step 1: Match Microwaves.

Microwaves typically have wavelengths from
\[ 10^{-3}\,m \]

to about
\[ 1\,m \]

Hence,
\[ (A)\rightarrow(IV) \]

Step 2: Match Gamma Rays.

Gamma rays possess the shortest wavelengths.
\[ \lambda \approx 10^{-15}\,m \]

Hence,
\[ (B)\rightarrow(II) \]

Step 3: Match A.M. Radio Waves.

A.M. radio waves have wavelengths of the order of hundreds of metres.
\[ \lambda \approx 100\,m \]

Hence,
\[ (C)\rightarrow(I) \]

Step 4: Match X-rays.

X-rays have wavelengths around
\[ 10^{-10}\,m \]

Hence,
\[ (D)\rightarrow(III) \]

Step 5: Write the final matching.
\[ (A)\rightarrow(IV) \]
\[ (B)\rightarrow(II) \]
\[ (C)\rightarrow(I) \]
\[ (D)\rightarrow(III) \]
\[ \boxed{ (A)-(IV),\ (B)-(II),\ (C)-(I),\ (D)-(III) } \]

Hence, the correct option is
\[ \boxed{(B)} \] Quick Tip: Remember the EM spectrum order: \[ Radio \rightarrow Microwave \rightarrow Infrared \rightarrow Visible \rightarrow UV \rightarrow X-ray \rightarrow Gamma ray \] As we move towards gamma rays: \[ \lambda \downarrow \] \[ f \uparrow \] \[ E \uparrow \]

Concept:

The speed of an electromagnetic wave in a medium is
\[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \]

where
\[ c=3\times10^8\,m s^{-1} \]

is the speed of light in vacuum.

Step 1: Substitute the given values.

Given,
\[ \mu_r=18 \]
\[ \varepsilon_r=2 \]

Therefore,
\[ v= \frac{3\times10^8} {\sqrt{18\times2}} \]
\[ v= \frac{3\times10^8} {\sqrt{36}} \]
\[ v= \frac{3\times10^8}{6} \]
\[ v=0.5\times10^8\,m s^{-1} \]

Step 2: State the answer.
\[ \boxed{ v=0.5\times10^8\,m s^{-1} } \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For electromagnetic waves in a medium: \[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \] where \[ n=\sqrt{\mu_r\varepsilon_r} \] is the refractive index of the medium. Larger values of \(\mu_r\) and \(\varepsilon_r\) reduce the speed of the wave.

Concept:

The mutual inductance is \[ M=\frac{\Phi}{I} \]
Since \[ a\ll R \]
the magnetic field over the square loop may be assumed uniform and equal to the field on the axis of the circular loop.

The magnetic field on the axis of a circular loop is \[ B= \frac{\mu_0IR^2} {2(R^2+z^2)^{3/2}} \]
Step 1: Calculate the magnetic field at the location of the square loop.

Given, \[ z=\sqrt3\,R \]
Hence, \[ R^2+z^2 = R^2+3R^2 = 4R^2 \]
Therefore, \[ B= \frac{\mu_0IR^2} {2(4R^2)^{3/2}} \] \[ B= \frac{\mu_0IR^2} {2(8R^3)} \] \[ B= \frac{\mu_0I}{16R} \]
Step 2: Find the magnetic flux through the square loop.

Area of square loop: \[ A=a^2 \]
Therefore, \[ \Phi=BA \] \[ \Phi= \frac{\mu_0I}{16R}\,a^2 \] \[ \Phi= \frac{\mu_0Ia^2}{16R} \]
Step 3: Calculate mutual inductance. \[ M=\frac{\Phi}{I} \] \[ M= \frac{\mu_0a^2}{16R} \]
Step 4: State the answer. \[ \boxed{ M=\frac{\mu_0a^2}{16R} } \]
Hence, the correct option is \[ \boxed{(D)} \] Quick Tip: For a circular loop, magnetic field on its axis is \[ B= \frac{\mu_0IR^2} {2(R^2+z^2)^{3/2}} \] When a small loop \((a\ll R)\) is placed in this field, \[ \Phi \approx BA \] and \[ M=\frac{\Phi}{I} \] This approximation is frequently used in mutual inductance problems.