NEET (UG) 2024 — Code S6

Question 1

Match List I with List II.

List I (Spectral Lines of Hydrogen for transitions from):

A. $n_2 = 3$ to $n_1 = 2$

B. $n_2 = 4$ to $n_1 = 2$

C. $n_2 = 5$ to $n_1 = 2$

D. $n_2 = 6$ to $n_1 = 2$

List II (Wavelengths (nm)):

I. 410.2

II. 434.1

III. 656.3

IV. 486.1

Choose the correct answer from t

Accepted Answer

(C) A-III, B-IV, C-II, D-I Energy difference $\Delta E = \dfrac{hc}{\lambda}$ $$ herefore\ \lambda \propto \frac{1}{\Delta E}$$ The further the upper level, the larger the energy gap: $$(\Delta E)_{6-2} > (\Delta E)_{5-2} > (\Delta E)_{4-2} > (\Delta E)_{3-2}$$ so the wavelengths run the other way: $$\lambda_{6-2} < \lambda_{5-2} < \lambda_{4-2} < \lambda_{3-2}$$ Matching the four given wavelengths in increasing order gives A-III, B-IV, C-II, D-I.

Question 2

Match List-I with List-II.

List-I (Material):

A. Diamagnetic

B. Ferromagnetic

C. Paramagnetic

D. Non-magnetic

List-II (Susceptibility ($\chi$)):

I. $\chi = 0$

II. $0 > \chi \geq -1$

III. $\chi \gg 1$

IV. $0 < \chi < \varepsilon$ (a small positive number)

Choose the correct answer from the

Accepted Answer

(B) A-II, B-III, C-IV, D-I

Taking each class of material with its susceptibility:

Diamagnetic — $0 > \chi \geq -1$

Ferromagnetic — $\chi \gg 1$

Paramagnetic — $0 < \chi < \varepsilon$

Non-magnetic — $\chi = 0$

So A-II, B-III, C-IV, D-I.

Question 3

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 	imes 10^{-6}$ kg m$^{2}$. If the magnitude of magnetic moment of the needle is $x 	imes 10^{-5}$ Am$^{2}$, then the value of '$x$' is :

Accepted Answer

(A) $1280\pi^{2}$

Time period of oscillation, $T = 2\pi\sqrt{\dfrac{I}{MB}}$.

Twenty oscillations in 5 s gives $T = \dfrac{5}{20} = \dfrac{1}{4}$ s:

$$\Rightarrow\ \frac{1}{4} = 2\pi\sqrt{\frac{9.8 	imes 10^{-6}}{M 	imes 0.049}}$$

$$\Rightarrow\ \frac{1}{16} = 4\pi^{2} 	imes \frac{9.8 	imes 10^{-6}}{M 	imes 49 	imes 10^{-3}}$$

$$\Rightarrow\ M = \frac{4\pi^{2} 	imes 9.8 	imes 10^{-6}}{49 	imes 10^{-3}} 	imes 16$$

$$= \frac{4\pi^{2} 	imes 9.8 	imes 16 	imes 10^{-3}}{49}$$

$$= 12.8\pi^{2} 	imes 10^{-3} = 1280\pi^{2} 	imes 10^{-5}\ 	ext{Am}^{2}$$

So $x = 1280\pi^{2}$.

Question 4

An unpolarised light beam strikes a glass surface at Brewster's angle. Then

Accepted Answer

(A) The reflected light will be completely polarised but the refracted light will be partially polarised.

At Brewster's angle the reflected ray and the refracted ray are perpendicular to each other, and the reflected light is completely plane polarised with its vibrations perpendicular to the plane of incidence.

The refracted beam still carries both components, though unequally, so it is only partially polarised.

Question 5

Consider the following statements A and B and identify the correct answer.

The graph referred to has current $I$ on the vertical axis and voltage $V$ on the horizontal axis, with the quadrants labelled (I) top right, (II) top left, (III) bottom left and (IV) bottom right.

A. For a solar-cell, the

Accepted Answer

(B) A is correct but B is incorrect

A: The solar cell characteristic runs from the open-circuit voltage $V_{OC}$ on the positive voltage axis down to the short-circuit current $I_{SC}$ on the negative current axis, so the curve lies in the fourth quadrant. Statement A is correct.

B: In a reverse biased $pn$ junction diode, the small current measured in $\mu$A is due to minority charge carriers, not majority ones. Statement B is incorrect.

Question 6

A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:

Accepted Answer

(C) $\dfrac{\sqrt{5}}{2}$

The prism has $A = 90^{\circ}$. The ray emerges along the face $AC$, which means it strikes $AC$ at the critical angle $c$.

In the prism, $r_1 + c = A$:

$$r_1 = 90^{\circ} - c \qquad \ldots(1)$$

$$\sin c = \frac{1}{\mu} \ \Rightarrow\ \cos c = \frac{\sqrt{\mu^{2} - 1}}{\mu}$$

Applying Snell's law on the incidence surface:

$$1 \cdot \sin 30^{\circ} = \mu\sin(r_1) \ \Rightarrow\ 1 	imes \frac{1}{2} = \mu 	imes \sin(90^{\circ} - c)$$

$$\frac{1}{2} = \mu 	imes \frac{\sqrt{\mu^{2} - 1}}{\mu}$$

On squaring, $\dfrac{1}{4} = \mu^{2} - 1$:

$$\Rightarrow\ \mu^{2} = \frac{5}{4} \ \Rightarrow\ \mu = \frac{\sqrt{5}}{2}$$

Question 7

A particle moving with uniform speed in a circular path maintains:

Accepted Answer

(A) Varying velocity and varying acceleration

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration.

It is because the direction of both the velocity as well as the acceleration will change continuously, even though their magnitudes stay the same.

Question 8

The graph which shows the variation of $\left(\dfrac{1}{\lambda^{2}}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

Accepted Answer

(A) A straight line through the origin, rising with $E$

de-Broglie wavelength

$$\lambda = \frac{h}{P} = \frac{h}{mv} = \frac{h}{\sqrt{2mE}} \qquad 	ext{where } E = \frac{1}{2}mv^{2}$$

Squaring both sides,

$$\lambda^{2} = \frac{h^{2}}{2mE}$$

$$\Rightarrow\ \frac{1}{\lambda^{2}} = \left(\frac{2m}{h^{2}}ight)E = (	ext{constant})\,E$$

The graph is therefore a straight line passing through the origin with constant slope.

Question 9

A wire of length '$l$' and resistance 100 $\Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Accepted Answer

(C) 52 $\Omega$

$$R = \frac{ho l}{A}$$

Each of the 10 parts has

$$R' = \frac{ho l}{10A} = \frac{R}{10} = 10\ \Omega$$

Five in series:

$$R_S = 5 	imes \frac{R}{10} = 50\ \Omega$$

Five in parallel:

$$R_P = \frac{R}{50} = 2\ \Omega$$

$$R_{	ext{eq}} = R_S + R_P = 52\ \Omega$$

Question 10

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm$^{2}$. The length of the 400 g rod is nearly:

Accepted Answer

(B) 8.5 cm

Moment of inertia of a rod about a perpendicular axis through its midpoint:

$$I = \frac{m\ell^{2}}{12}$$

$$\Rightarrow\ 2400 = 400\frac{\ell^{2}}{12}$$

$$\Rightarrow\ 72 = \ell^{2}$$

$$\Rightarrow\ \ell = \sqrt{72} = 8.48\ 	ext{cm} \simeq 8.5\ 	ext{cm}$$

Question 11

The output ($Y$) of the given logic gate is similar to the output of an/a

Accepted Answer

(A) AND gate

Input $A$ goes to both terminals of a NAND gate and input $B$ to both terminals of a NOR gate, so

$$Y_1 = \overline{A \cdot A} = \overline{A}$$

$$Y_2 = \overline{B + B} = \overline{B}$$

These two feed a NOR gate:

$$Y = \overline{Y_1 + Y_2} = \overline{\overline{A} + \overline{B}}$$

By De Morgan's law,

$$= \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$$

which is similar to the output of an AND gate.

Question 12

In a vernier callipers, ($N$ + 1) divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:

Accepted Answer

(C) $\dfrac{1}{100(N + 1)}$

$$V.C = 	ext{MSD} - 	ext{VSD} \qquad \ldots(1)$$

Given $(N + 1)\,	ext{VSD} = N\,	ext{MSD}$:

$$	ext{VSD} = \left(\frac{N}{N + 1}ight)	ext{MSD} \qquad \ldots(2)$$

From (1) and (2),

$$V.C = 	ext{MSD} - \frac{N}{N + 1}	ext{MSD} = 	ext{MSD}\left(1 - \frac{N}{N + 1}ight) = \frac{	ext{MSD}}{N + 1}$$

1 MSD = 0.1 mm = 0.01 cm, so

$$= \frac{0.01}{N + 1} = \frac{1}{100(N + 1)}\ 	ext{cm}$$

Question 13

At any instant of time $t$, the displacement of any particle is given by $2t - 1$ (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):

Accepted Answer

(B) 10

$$x = 2t - 1$$

$$v = \frac{dx}{dt} = 2\ 	ext{m s}^{-1}$$

$$P = F \cdot v = 2 	imes 5 = 10\ 	ext{W}$$

Question 14

A thin spherical shell of radius $R = 3$ cm carries a charge $q = 1\ \mu$C. The potential difference between the centre $C$ of the shell and a point $P$ lying on the shell (in V) is:

(Take $\dfrac{1}{4\pi\varepsilon_0} = 9 	imes 10^{9}$ SI units)

Accepted Answer

(A) Zero

For a uniformly charged spherical shell, the potential everywhere inside and on the shell is the same:

$$V = \frac{kq}{R} \qquad (	ext{for } r \leq R)$$

$$	herefore\ V_C = V_P$$

$$V_C - V_P = 	ext{Zero}$$

Question 15

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m$^{-1}$, then the excess force required to take it away from the surface is

Accepted Answer

(B) 19.8 mN

The surface tension acts along the circumference of the disc:

$$	ext{Excess force} = T 	imes 2\pi R$$

$$= \frac{7}{100} 	imes 2 	imes 3.14 	imes \frac{4.5}{100}$$

$$= 197.82 	imes 10^{-4} = 19.8 	imes 10^{-3}\ 	ext{N} = 19.8\ 	ext{mN}$$

Question 16

A logic circuit provides the output $Y$ as per the following truth table :

$A = 0$, $B = 0$ gives $Y = 1$

$A = 0$, $B = 1$ gives $Y = 0$

$A = 1$, $B = 0$ gives $Y = 1$

$A = 1$, $B = 1$ gives $Y = 0$

The expression for the output $Y$ is :

Accepted Answer

(D) $\overline{B}$

Reading the table, whenever $B = 0$ the output is 1, and whenever $B = 1$ the output is 0, no matter what $A$ is.

According to the given truth table, the output is independent of the value of $A$.

$$	herefore\ 	ext{Output } Y = \overline{B}$$

Question 17

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential ($V$) at any axial point, at 2 m distance ($r$) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 	imes 10^{-6}$ C m, is $\pm 9 	imes 10^

Accepted Answer

(D) A is true but R is false.

The potential $V$ at any point at distance $r$ from the centre of a dipole is

$$V = \frac{KP\cos	heta}{r^{2}}$$

At an axial point where $	heta = 0^{\circ}$:

$$V = \frac{KP}{r^{2}} = \frac{9 	imes 10^{9} 	imes 4 	imes 10^{-6}}{2^{2}} = 9 	imes 10^{3}\ 	ext{V}$$

At an axial point where $	heta = 180^{\circ}$:

$$V = \frac{-KP}{r^{2}} = -9 	imes 10^{3}\ 	ext{V}$$

So the Assertion is true. But the correct expression carries no factor of 2 — it is $V = \pm\dfrac{P}{4\pi\varepsilon_0 r^{2}}$ — so the Reason as stated is false.

Question 18

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 $\Omega$, when connected through an external resistance of 4 $\Omega$ as shown in the figure is:

Accepted Answer

(D) 8 V

Current in circuit

$$i = \frac{10}{4 + 1} = 2\ 	ext{A}$$

Terminal voltage

$$= E - iR = 10 - 2 	imes 1 = 8\ 	ext{V}$$

Question 19

In the given diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

Accepted Answer

(B) $AB$ and $DC$

The magnet's north pole is moving away from solenoid-1, so by Lenz's law solenoid-1 opposes the departure by attracting it: the end $B$ of solenoid 1 facing the magnet becomes South.

The magnet's south pole is approaching solenoid-2, so solenoid-2 opposes the approach by repelling it: the end $C$ of solenoid 2 facing the magnet becomes South.

Reading the sense of the winding needed to produce those poles gives the induced current through $AB$ in solenoid-1 and through $DC$ in solenoid-2.

Question 20

In an ideal transformer, the turns ratio is $\dfrac{N_P}{N_S} = \dfrac{1}{2}$. The ratio $V_S : V_P$ is equal to (the symbols carry their usual meaning) :

Accepted Answer

(C) 2 : 1

According to the transformer ratio,

$$\frac{V_S}{V_P} = \frac{N_S}{N_P} = 2 : 1$$

Question 21

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi 	imes 10^{-7}$ SI units):

Accepted Answer

(D) 4.4 mT

The magnitude of magnetic field due to a circular coil of $N$ turns is given by

$$B_C = \frac{\mu_0 iN}{2R}$$

$$= \frac{4\pi 	imes 10^{-7} 	imes 7 	imes 100}{2 	imes 0.1}$$

$$= 4.4 	imes 10^{-3}\ 	ext{T} = 4.4\ 	ext{mT}$$

Question 22

Given below are two statements:

Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.

Statement II: Atoms of each element are stable and emit their characteristic spectrum.

In the light of the above statements, choose the most appropriate answe

Accepted Answer

(D) Statement I is correct but Statement II is incorrect

Statement I is true, as atoms are electrically neutral because they contain equal numbers of positive and negative charges.

Statement II is wrong: atoms of most of the elements are stable and emit a characteristic spectrum, but this statement is not true for every atom — radioactive atoms are not stable.

Question 23

The quantities which have the same dimensions as those of solid angle are:

Accepted Answer

(B) strain and angle

Solid angle $d\Omega = \dfrac{dA}{r^{2}}$ has dimensions $[	ext{M}^{0}	ext{L}^{0}	ext{T}^{0}]$.

Strain $= \dfrac{\Delta l}{l}$ has dimensions $[	ext{M}^{0}	ext{L}^{0}	ext{T}^{0}]$.

Angle measured in radians is also dimensionless, $[	ext{M}^{0}	ext{L}^{0}	ext{T}^{0}]$, since $	heta = \dfrac{l}{r}$.

So strain and angle share the dimensions of solid angle.

Question 24

If $c$ is the velocity of light in free space, the correct statements about photon among the following are:

A. The energy of a photon is $E = h
u$.

B. The velocity of a photon is $c$.

C. The momentum of a photon, $p = \dfrac{h
u}{c}$.

D. In a photon-electron collision, both total energy and to

Accepted Answer

(C) A, B, C and D only

(A) The energy of a photon is $E = h
u$.

(B) The velocity of a photon is equal to the velocity of light, i.e. $c$.

(C) From $\lambda = \dfrac{h}{p}$, $p = \dfrac{h}{\lambda} = \dfrac{h
u}{c}$.

(D) In a photon-electron collision both total energy and total momentum are conserved.

(E) A photon is electrically neutral, so it does not possess a positive charge.

Hence A, B, C and D only.

Question 25

A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2\omega$ while keeping the same radius, the tension in the string becomes:

Accepted Answer

(C) $4T$

The tension supplies the centripetal force:

$$T = m\ell\omega^{2}$$

With the angular speed doubled at the same radius,

$$T' = m\ell(2\omega)^{2} = 4m\ell\omega^{2}$$

$$T' = 4T$$

Question 26

If $x = 5\sin\left(\pi t + \dfrac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

Accepted Answer

(C) 5 m, 2 s

$$x = 5\sin\left(\pi t + \frac{\pi}{3}ight)\ 	ext{m}$$

Comparing with $x = A\sin(\omega t + \phi)$:

Amplitude $= 5$ m

$$\omega = \pi = \frac{2\pi}{T}$$

$$T = \frac{2\pi}{\pi} = 2\ 	ext{s}$$

Question 27

A thermodynamic system is taken through the cycle $abcda$, whose $P$-$V$ diagram has $a$ at (100 cm$^{3}$, 100 kPa), $b$ at (400 cm$^{3}$, 100 kPa), $c$ at (400 cm$^{3}$, 300 kPa) and $d$ at (100 cm$^{3}$, 300 kPa). The work done by the gas along the path $bc$ is:

Accepted Answer

(B) Zero

Along $bc$ the volume stays at 400 cm$^{3}$ while the pressure rises from 100 kPa to 300 kPa, so path $bc$ is an isochoric process.

$$W = \int P\,dV = 0 \qquad (dV = 0)$$

$$	herefore\ 	ext{Work done by gas along path } bc 	ext{ is zero.}$$

Question 28

$$^{290}_{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^{+}\ } Z \xrightarrow{\ \beta^{-}\ } P \xrightarrow{\ e^{-}\ } Q$$

In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are

Accepted Answer

(A) 286, 81

Following each emission in turn:

$$^{290}_{82}X \xrightarrow{\ \alpha\ } {}^{286}_{80}Y \xrightarrow{\ e^{+}\ } {}^{286}_{79}Z \xrightarrow{\ \beta^{-}\ } {}^{286}_{80}P \xrightarrow{\ e^{-}\ } {}^{286}_{81}Q$$

An $\alpha$ emission drops the mass number by 4 and the atomic number by 2; a positron ($e^{+}$) emission drops the atomic number by 1; a $\beta^{-}$ emission raises it by 1; and electron capture written here as $e^{-}$ raises it by 1 again.

$$A \rightarrow 286$$

$$Z = 81$$

Question 29

Two bodies $A$ and $B$ of same mass undergo completely inelastic one dimensional collision. The body $A$ moves with velocity $v_1$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1 : v_2$ is

Accepted Answer

(C) 2 : 1

Before collision, $A$ moves with $v_1$ and $B$ is at rest. It undergoes a completely inelastic collision, so the two move together afterwards.

Using conservation of linear momentum, initial momentum = final momentum:

$$\Rightarrow\ mv_1 = mv_2 + mv_2$$

$$\Rightarrow\ mv_1 = 2mv_2$$

$$\Rightarrow\ \frac{v_1}{v_2} = \frac{2}{1}$$

Question 30

In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is :

The network is a bridge of five 2 $\mu$F capacitors: two in series along the upper arm from $A$ to $B$, two in series along the lower arm from $A$ to $B$, and the fifth bridging the two midpoints.

Accepted Answer

(B) 2 $\mu$F

The given circuit is a balanced Wheatstone bridge: the two arms have equal ratios, so the bridging capacitor carries no charge and can be removed.

What is left is two arms in parallel, each of two 2 $\mu$F capacitors in series:

$$	ext{each arm} = \frac{2 	imes 2}{2 + 2} = 1\ \mu	ext{F}$$

$$C_{AB} = 1 + 1 = 2\ \mu	ext{F}$$

Question 31

A horizontal force 10 N is applied to a block $A$ as shown in figure. The mass of blocks $A$ and $B$ are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is :

Accepted Answer

(D) 6 N

The two blocks move together, so

$$F = (M_1 + M_2)a$$

$$a = \frac{10}{2 + 3} = 2\ 	ext{ms}^{-2}$$

The only horizontal force on $B$ is the contact force from $A$:

$$F' = M_2 a = 3 	imes 2 = 6\ 	ext{N}$$

Question 32

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ($P$ and $Q$ are the highest and lowest points on the wheel, respectively)?

Accepted Answer

(C) Point $P$ moves faster than point $Q$

In the case of pure rolling, the contact point is instantaneously at rest and the speed of any point is proportional to its distance from that contact point.

The topmost point will have velocity $2v$, while point $Q$, i.e. the lowest point, will have zero velocity.

Hence point $P$ moves faster than point $Q$.

Question 33

The mass of a planet is $\dfrac{1}{10}^{	ext{th}}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

Accepted Answer

(A) 3.92 m s$^{-2}$

$$g = \frac{GM}{R^{2}}$$

$$\frac{g_p}{g_e} = \frac{M_p}{M_e}\left(\frac{R_e}{R_p}ight)^{2}$$

$$= \frac{1}{10} 	imes (2)^{2} = \frac{4}{10}$$

$$g_p = 0.4 	imes 9.8 = 3.92\ 	ext{m s}^{-2}$$

Question 34

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are $8 	imes 10^{8}$ N m$^{-2}$ and $2 	imes 10^{11}$ N m$^{-2}$, is:

Accepted Answer

(B) 4 mm

In the case of maximum elongation, stress = elastic limit.

$$\delta_{\max} = \frac{\sigma_{	ext{elastic}} 	imes L}{	ext{Young's modulus}} = \frac{8 	imes 10^{8} 	imes 1}{2 	imes 10^{11}} = 4 	imes 10^{-3}\ 	ext{m}$$

$$= 4\ 	ext{mm}$$

Question 35

If the monochromatic source in Young's double slit experiment is replaced by white light, then

Accepted Answer

(D) There will be a central bright white fringe surrounded by a few coloured fringes

At the central point on the screen, the path difference is zero for all wavelengths. So the central bright fringe is white.

The other fringes depend on wavelength through

$$\beta = \frac{\lambda D}{d}$$

Therefore the other fringes will be coloured.

Question 36

A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:

Accepted Answer

(C) 28

$f_0 = 140$ cm and $f_e = 5$ cm.

For a distant object,

$$m = \frac{f_0}{f_e} = \frac{140}{5} = 28$$

Question 37

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then

A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge a

Accepted Answer

(C) A, C and E only

The capacitor stays connected to the battery, so $V' = V$ = constant.

(i) $C' = \dfrac{\varepsilon_0 A}{d'}$ and $C = \dfrac{\varepsilon_0 A}{d}$ with $d' < d$, so $C' > C$: the final capacitance is greater than the initial capacitance.

(ii) $U' = \dfrac{1}{2}C'V^{2}$ and $U = \dfrac{1}{2}CV^{2}$, so $U' > U$: the final energy is greater than the initial energy, so B is wrong.

(iii) $\dfrac{Q'}{V'} = C'$ and $\dfrac{Q}{V} = C$, and since $C' 
eq C$, the ratio of charge to potential does not remain the same, so D is wrong.

(iv) Product of charge and voltage: $X' = Q'V = C'V^{2}$ and $X = QV = CV^{2}$, so $X' > X$.

Charge $Q = CV$ also rises with $C$, so A is right. Hence A, C and E only.

Question 38

A 10 $\mu$F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ($\pi$ = 3.14):

Accepted Answer

(C) 0.93 A

Capacitive reactance

$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2 	imes 3.14 	imes 50 	imes 10 	imes 10^{-6}} = \frac{1000}{3.14}\ \Omega$$

$V_{	ext{rms}} = 210$ V

$$i_{	ext{rms}} = \frac{V_{	ext{rms}}}{X_C} = \frac{210}{X_C}$$

$$	ext{Peak current} = \sqrt{2}\,i_{	ext{rms}} = \sqrt{2} 	imes \frac{210}{1000} 	imes 3.14 = 0.932 \simeq 0.93\ 	ext{A}$$

Question 39

Choose the correct circuit which can achieve the bridge balance.

Accepted Answer

(B) Circuit (2)

In option (2), the balance condition reads

$$\frac{10}{15} = \frac{10}{5 + R_D}$$

The diode can conduct and have resistance $R_D = 10\ \Omega$, because a diode has a dynamic resistance. In that case the bridge will be balanced.

Question 40

Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:

Accepted Answer

(C) 2 : 9

Power consumed $P = \dfrac{V^{2}}{R}$, so

$$\frac{P_A}{P_B} = \frac{R_B}{R_A} \quad \Rightarrow \quad R_A = 2R_B$$

For the series combination,

$$P_S = \frac{V^{2}}{R_A + R_B} = \frac{V^{2}}{3R_B}$$

For the parallel combination,

$$P_P = \frac{V^{2}}{R_A} + \frac{V^{2}}{R_B} = \frac{V^{2}}{2R_B} + \frac{V^{2}}{R_B} = \frac{3V^{2}}{2R_B}$$

$$\frac{P_S}{P_P} = \frac{\dfrac{1}{3}}{\dfrac{3}{2}} = \frac{2}{9}$$

Question 41

The velocity ($v$) – time ($t$) plot of the motion of a body is a trapezium: the velocity rises linearly from zero, then stays constant, then falls linearly back to zero.

The acceleration ($a$) – time ($t$) graph that best suits this motion is :

Accepted Answer

(D) A positive rectangle, then zero, then a negative rectangle

Initially, the body has zero velocity and zero slope; hence the acceleration would be zero initially.

After that, the slope of the $v$-$t$ curve is constant and positive, so the acceleration is a constant positive value.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of the $v$-$t$ curve is constant and negative, so the acceleration is a constant negative value.

The $a$-$t$ graph is therefore a positive rectangle, followed by zero, followed by a negative rectangle.

Question 42

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\dfrac{x}{2}$ times its original time period. Then the value of $x$ is:

Accepted Answer

(C) $\sqrt{2}$

The time period of a simple pendulum does not depend on the mass of the bob:

$$T' = 2\pi\sqrt{\frac{\ell'}{g}} \qquad 	ext{where } \ell' = \frac{\ell}{2}$$

$$T = 2\pi\sqrt{\frac{\ell}{g}}$$

Given $T' = \dfrac{x}{2}T$:

$$2\pi\sqrt{\frac{\ell}{2g}} = \frac{x}{2} \cdot 2\pi\sqrt{\frac{\ell}{g}}$$

$$\frac{1}{\sqrt{2}} = \frac{x}{2} \ \Rightarrow\ x = \sqrt{2}$$

Question 43

A metallic bar of Young's modulus, $0.5 	imes 10^{11}$ N m$^{-2}$ and coefficient of linear thermal expansion $10^{-5}\ ^{\circ}$C$^{-1}$, length 1 m and area of cross-section $10^{-3}$ m$^{2}$ is heated from 0$^{\circ}$C to 100$^{\circ}$C without expansion or bending. The compressive force develop

Accepted Answer

(C) $50 	imes 10^{3}$ N

Thermal strain = longitudinal strain = $\alpha\Delta T$

$$\Rightarrow\ 	ext{Longitudinal strain}, \delta = 10^{-5} 	imes 10^{2} = 10^{-3}$$

$$\Rightarrow\ 	ext{Compressive stress} = \delta 	imes 	ext{Young's Modulus}$$

$$= 10^{-3} 	imes 0.5 	imes 10^{11} = 0.5 	imes 10^{8}\ 	ext{N m}^{-2}$$

$$\Rightarrow\ 	ext{Compressive force} = 0.5 	imes 10^{8} 	imes 10^{-3} = 0.5 	imes 10^{5}$$

$$= 50 	imes 10^{3}\ 	ext{N}$$

Question 44

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:

A. hold the sheet there if it is magnetic.

B. hold the sheet there if it is non-magnetic.

C. move the sheet away from the pole with uniform velocity if it is conducting.

D. move the sheet away fro

Accepted Answer

(C) A and C only

A. A magnetic pole will repel or attract a magnetic sheet, so a force is needed to hold it there.

B. If the sheet is non-magnetic, no force is needed.

C. If it is conducting, then there will be an eddy current in the sheet which opposes the motion, so a force is needed to move the sheet with uniform speed.

D. The non-conducting and non-polar sheet does not interact with the magnetic field of the magnet, so no force is needed.

Hence A and C only.

Question 45

The property which is not of an electromagnetic wave travelling in free space is that:

Accepted Answer

(A) They originate from charges moving with uniform speed

The EM waves originate from an accelerating charge. A charge moving with uniform velocity produces a steady state magnetic field and radiates no electromagnetic wave, so that statement is not a property of an electromagnetic wave.

Question 46

A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:

Accepted Answer

(C) Displacement current of magnitude equal to $I$ flows in the same direction as $I$

According to modified Ampere's law,

$$\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I_C + I_Dight)$$

For a loop $L_1$ enclosing the wire: $I_C 
eq 0$ and $I_D = 0$.

For a loop $L_2$ enclosing the gap between the plates: $I_C = 0$ and $I_D 
eq 0$.

Since the same $\oint \vec{B} \cdot d\vec{l}$ must result, and by KCL,

$$I_C = I_D$$

so the displacement current has the same magnitude and the same direction as the conduction current.

Question 47

An iron bar of length $L$ has magnetic moment $M$. It is bent at the middle of its length such that the two arms make an angle $60^{\circ}$ with each other. The magnetic moment of this new magnet is :

Accepted Answer

(C) $\dfrac{M}{2}$

The bar is bent at its midpoint, so each arm has length $\dfrac{L}{2}$ and magnetic moment $\dfrac{M}{2}$.

The two arm moments are vectors with $60^{\circ}$ between them, so the resultant is

$$M' = \sqrt{\left(\frac{M}{2}\right)^{2} + \left(\frac{M}{2}\right)^{2} + 2\left(\frac{M}{2}\right)\left(\frac{M}{2}\right)\cos 120^{\circ}}$$

The angle between the moment vectors is $120^{\circ}$ when the arms make $60^{\circ}$, giving

$$M' = \frac{M}{2}$$

Question 48

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$ from the surface of the earth is:

Accepted Answer

(B) $\dfrac{5GmM}{6R}$

The orbit radius is $R + 2R = 3R$. Apply energy conservation,

$$U_i + K_i = U_f + K_f$$

$$\Rightarrow\ -\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2}mv^{2}$$

For a circular orbit of radius $3R$, $v^{2} = \dfrac{GM}{3R}$:

$$\Rightarrow\ -\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2} 	imes m 	imes \frac{GM}{3R}$$

$$\Rightarrow\ K_i = -\frac{1}{6}\frac{GMm}{R} + \frac{GMm}{R}$$

$$K_i = \frac{5}{6}\frac{GMm}{R}$$

Question 49

The following graph represents the $T$-$V$ curves of an ideal gas (where $T$ is the temperature and $V$ the volume) at three pressures $P_1$, $P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. The curve labelled $P_1$ lies furthest to the left, $P_2$ in the middle and

Accepted Answer

(A) $P_1 > P_2 > P_3$

Draw a straight line parallel to the volume axis, i.e. take the three curves at the same temperature.

At the same temperature, the curve with the higher volume corresponds to lower pressure, since $PV = nRT$.

Reading off the graph at fixed $T$,

$$V_3 > V_2 > V_1$$

$$\Rightarrow\ P_1 > P_2 > P_3$$

Question 50

A force defined by $F = \alpha t^{2} + \beta t$ acts on a particle at a given time $t$. The factor which is dimensionless, if $\alpha$ and $\beta$ are constants, is:

Accepted Answer

(C) $\dfrac{\alpha t}{\beta}$

From the principle of homogeneity, every term must have the dimensions of force:

$$[F] = [\alpha t^{2}] = [\beta t]$$

$$[\alpha] = \frac{[F]}{[t^{2}]} \quad 	ext{and} \quad [\beta] = \frac{[F]}{[t]}$$

$$	herefore\ [\alpha][t] = [\beta]$$

$$	herefore\ \frac{\alpha t}{\beta} = 	ext{dimensionless}$$

Question 51

Given below are two statements:

Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.

Statement II : Aniline cannot be prepared through Gabriel synthesis.

In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

(B) Both statement I and Statement II are true

Aniline does not undergo Friedel-Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid which is used as a catalyst.

Aniline (an aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis, because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Both statements are therefore true.

Question 52

Match List I with List II.

List I (Compound):

A. NH$_3$

B. BrF$_5$

C. XeF$_4$

D. SF$_6$

List II (Shape/geometry):

I. Trigonal Pyramidal

II. Square Planar

III. Octahedral

IV. Square Pyramidal

Choose the correct answer from the options given below:

Accepted Answer

(B) A-I, B-IV, C-II, D-III

NH$_3$ is $sp^{3}$ hybridised with 1 lone pair, so the structure will be trigonal pyramidal.

BrF$_5$ is $sp^{3}d^{2}$ hybridised with 1 lone pair, so the structure will be square pyramidal.

XeF$_4$ is $sp^{3}d^{2}$ with two lone pairs, so the structure will be square planar.

SF$_6$ is $sp^{3}d^{2}$ with no lone pair, so the structure will be octahedral.

So A-I, B-IV, C-II, D-III.

Question 53

Match List I with List II.

List I (Molecule):

A. ethane

B. ethene

C. carbon molecule, C$_2$

D. ethyne

List II (Number and types of bond/s between two carbon atoms):

I. one $\sigma$-bond and two $\pi$-bonds

II. two $\pi$-bonds

III. one $\sigma$-bond

IV. one $\sigma$-bond and one $\pi$-bond

Accepted Answer

(D) A-III, B-IV, C-II, D-I

(A) Ethane, H$_3$C–CH$_3$ — one (C–C) $\sigma$ bond

(B) Ethene, H$_2$C=CH$_2$ — one (C–C) $\sigma$ and one (C–C) $\pi$ bond

(C) C$_2$ — two (C–C) $\pi$ bonds, as the molecular orbital picture of dicarbon has no net $\sigma$ bond

(D) Ethyne, H–C$\equiv$C–H — two (C–C) $\pi$ bonds and one (C–C) $\sigma$ bond

So A-III, B-IV, C-II, D-I.

Question 54

For the reaction 2A $ightleftharpoons$ B + C, K$_C$ = $4 	imes 10^{-3}$. At a given time, the composition of reaction mixture is: [A] = [B] = [C] = $2 	imes 10^{-3}$ M.

Then, which of the following is correct?

Accepted Answer

(D) Reaction has a tendency to go in backward direction.

At a given time $t$, $Q_C$ is to be calculated and compared with $K_C$.

$$Q_C = \frac{[B][C]}{[A]^{2}} = \frac{(2 	imes 10^{-3})(2 	imes 10^{-3})}{(2 	imes 10^{-3})^{2}}$$

$$Q_C = 1$$

As $Q_C > K_C$, the reaction has a tendency to move backward.

Question 55

Match List I with List II.

List-I (Process):

A. Isothermal process

B. Isochoric process

C. Isobaric process

D. Adiabatic process

List-II (Conditions):

I. No heat exchange

II. Carried out at constant temperature

III. Carried out at constant volume

IV. Carried out at constant pressure

Choos

Accepted Answer

(A) A-II, B-III, C-IV, D-I

(A) Isothermal process — temperature is constant throughout the process

(B) Isochoric process — volume is constant throughout the process

(C) Isobaric process — pressure is constant throughout the process

(D) Adiabatic process — no exchange of heat ($q$) between system and surroundings

So A-II, B-III, C-IV, D-I.

Question 56

Match List I with List II

List I (Quantum Number):

A. $m_l$

B. $m_s$

C. $l$

D. $n$

List II (Information provided):

I. Shape of orbital

II. Size of orbital

III. Orientation of orbital

IV. Orientation of spin of electron

Choose the correct answer from the options given below :

Accepted Answer

(C) A-III, B-IV, C-I, D-II

Magnetic quantum number $m_l$ informs about the orientation of the orbital.

Spin quantum number $m_s$ informs about the orientation of the spin of the electron.

Azimuthal quantum number ($l$) informs about the shape of the orbital.

Principal quantum number ($n$) informs about the size of the orbital.

So A-III, B-IV, C-I, D-II.

Question 57

Arrange the following elements in increasing order of first ionization enthalpy:

Li, Be, B, C, N

Choose the correct answer from the options given below:

Accepted Answer

(C) Li < B < Be < C < N

The first ionization enthalpies ($\Delta_i$H / kJ mol$^{-1}$) are

Li — 520, Be — 899, B — 801, C — 1086, N — 1402

Boron sits below beryllium because removing boron's 2p electron is easier than removing an electron from beryllium's filled 2s subshell.

So the increasing order of first ionization enthalpy is Li < B < Be < C < N.

Question 58

Among Group 16 elements, which one does NOT show $-2$ oxidation state?

Accepted Answer

(A) Po

Oxygen shows $-2$, $-1$, $+1$ and $+2$ oxidation states.

Selenium shows $-2$, $+2$, $+4$ and $+6$ oxidation states.

Tellurium shows $-2$, $+2$, $+4$ and $+6$ oxidation states.

Polonium shows $+2$ and $+4$ oxidation states only, so it does not show the $-2$ state.

Question 59

In which of the following equilibria, K$_p$ and K$_c$ are NOT equal?

Accepted Answer

(B) $	ext{PCl}_{5(g)} ightleftharpoons 	ext{PCl}_{3(g)} + 	ext{Cl}_{2(g)}$

$$K_p = K_c(RT)^{\Delta n_g}$$

For $K_p 
eq K_c$ we need $\Delta n_g 
eq 0$, where $\Delta n_g = n_p - n_r$.

(1) $\Delta n_g = 2 - 2 = 0$

(2) $\Delta n_g = 2 - 1 = 1$

(3) $\Delta n_g = 2 - 2 = 0$

(4) $\Delta n_g = 2 - 2 = 0$

Only the PCl$_5$ dissociation has $\Delta n_g 
eq 0$.

Question 60

The reagents with which glucose does not react to give the corresponding tests/products are

A. Tollen's reagent

B. Schiff's reagent

C. HCN

D. NH$_2$OH

E. NaHSO$_3$

Choose the correct options from the given below:

Accepted Answer

(D) B and E

Despite having the aldehyde group, glucose does not give Schiff's test and it does not form the hydrogen sulphite addition product with NaHSO$_3$.

These two facts are part of the evidence that the aldehyde group in glucose is not free but is tied up in the cyclic hemiacetal form.

Glucose does react with Tollen's reagent, with HCN to give the cyanohydrin, and with NH$_2$OH to give the oxime.

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