IIT JEE 2001 Physics Question Paper with Answer and Solution

(A) Let the initial speed of the block be $v$ and the height of the track be $h$. By the law of conservation of energy,the speed $v'$ of the block at the highest point of the track is given by:
$\frac{1}{2}mv^2 = \frac{1}{2}m(v')^2 + mgh$
$v' = \sqrt{v^2 - 2gh}$
Since $v$ and $h$ are the same for all tracks,the speed $v'$ at the highest point is the same for all tracks.
At the highest point,the forces acting on the block are the normal reaction $N$ (downwards) and the weight $mg$ (downwards). These provide the necessary centripetal force:
$N + mg = \frac{m(v')^2}{r}$
$N = \frac{m(v')^2}{r} - mg$
where $r$ is the radius of curvature at the highest point.
For $N$ to be maximum,the term $\frac{m(v')^2}{r}$ must be maximum. Since $m$ and $v'$ are constant,$N$ is maximum when the radius of curvature $r$ is minimum.
Comparing the four tracks,the track in option $A$ has the smallest radius of curvature at its highest point. Therefore,the normal reaction is maximum in track $A$.

(C) For the side masses $m$ to be in equilibrium,the tension $T$ in the strings must be equal to the weight of the masses: $T = mg$.
Now,consider the equilibrium of the central mass $\sqrt{2}m$. The forces acting on it are the downward gravitational force $\sqrt{2}mg$ and the upward components of the tension $T$ from the two strings.
Resolving the tension $T$ into vertical components,we have $2T \cos \theta = \sqrt{2}mg$.
Substituting $T = mg$ into the equation:
$2(mg) \cos \theta = \sqrt{2}mg$
$2 \cos \theta = \sqrt{2}$
$\cos \theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Therefore,$\theta = 45^\circ$.

(D) The forces acting on the pulley are:
$1$. The tension $T$ in the horizontal part of the string,where $T = Mg$.
$2$. The tension $T$ in the vertical part of the string,where $T = Mg$.
$3$. The weight of the pulley,$mg$,acting downwards.
$4$. The force exerted by the clamp on the pulley,$F_{pc}$.
For the pulley to be in equilibrium,the vector sum of all forces must be zero. The forces acting on the pulley are the horizontal tension $T$ to the left,the vertical tension $T$ downwards,and the weight $mg$ downwards.
Total horizontal force $F_x = T = Mg$.
Total vertical force $F_y = T + mg = Mg + mg = (M + m)g$.
The magnitude of the force exerted by the clamp on the pulley is the resultant of these two perpendicular forces:
$F_{pc} = \sqrt{F_x^2 + F_y^2}$
$F_{pc} = \sqrt{(Mg)^2 + ((M + m)g)^2}$
$F_{pc} = \sqrt{M^2 + (M + m)^2} g$

(C) The initial momentum of the two-particle system at $t = 0$ is $\vec{P}_i = m_1\vec{v}_1 + m_2\vec{v}_2$.
Collision between the two particles is an internal interaction and does not affect the total momentum of the system.
The only external force acting on the system is gravity,which is $\vec{F}_{ext} = (m_1 + m_2)g$ acting downwards.
According to the impulse-momentum theorem,the change in momentum $\Delta \vec{P}$ is equal to the impulse of the external force: $\Delta \vec{P} = \int_{0}^{2t_0} \vec{F}_{ext} dt$.
Since the force is constant,$\Delta \vec{P} = (m_1 + m_2)g \times (2t_0 - 0) = 2(m_1 + m_2)gt_0$.
Therefore,the magnitude of the change in momentum is $|(m_1\vec{v}_1' + m_2\vec{v}_2') - (m_1\vec{v}_1 + m_2\vec{v}_2)| = 2(m_1 + m_2)gt_0$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{l/g}$.
On the surface of the earth,the acceleration due to gravity is $g$. Thus,$T_1 = 2\pi \sqrt{l/g}$.
At a height $h = R$ above the earth's surface,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Substituting $h = R$,we get $g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{R}{2R} \right)^2 = \frac{g}{4}$.
The time period at height $R$ is $T_2 = 2\pi \sqrt{l/g'} = 2\pi \sqrt{l/(g/4)} = 2 \times 2\pi \sqrt{l/g}$.
Therefore,$T_2/T_1 = 2$.

(A) According to the First Law of Thermodynamics $(FLOT)$,the change in internal energy $dU$ is given by $dU = dQ - dW$.
Given that $dW = 0$ (isochoric process) and $dQ < 0$ (heat is released by the system).
Substituting these values,we get $dU = dQ - 0 = dQ$.
Since $dQ < 0$,it follows that $dU < 0$.
For an ideal gas,internal energy $U$ is a function of temperature $T$ only $(U \propto T)$.
Therefore,a decrease in internal energy $(dU < 0)$ implies a decrease in temperature $(dT < 0)$.

(B) Let the temperature of the junction be $\theta$.
Since rods $B$ and $C$ are connected in parallel to rod $A$,we can calculate the equivalent thermal resistance of the parallel combination of $B$ and $C$.
Let the thermal resistance of each rod be $R$.
Since rods $B$ and $C$ are in parallel,their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$,which implies $R_p = \frac{R}{2}$.
Now,the system acts as two resistances $R$ and $\frac{R}{2}$ in series.
The rate of heat flow $\frac{dQ}{dt}$ through the junction must be conserved.
Thus,the heat flowing from the $90^{\circ}C$ ends through the parallel combination must equal the heat flowing through rod $A$ to the $0^{\circ}C$ end.
Using the formula $\frac{dQ}{dt} = \frac{\Delta T}{R_{eq}}$:
$\frac{90 - \theta}{R/2} = \frac{\theta - 0}{R}$
$2(90 - \theta) = \theta$
$180 - 2\theta = \theta$
$3\theta = 180$
$\theta = 60^{\circ}C$.

(A) Using the equation of motion $x = A \sin(\omega t)$.
For $x = A/2$,we have $\sin(\omega T_1) = 1/2$,which implies $\omega T_1 = \pi/6$,so $T_1 = \frac{\pi}{6\omega}$.
For the particle to reach $x = A$,the total time is $T_1 + T_2$. Thus,$\sin(\omega(T_1 + T_2)) = 1$,which implies $\omega(T_1 + T_2) = \pi/2$,so $T_1 + T_2 = \frac{\pi}{2\omega}$.
Substituting $T_1$,we get $T_2 = \frac{\pi}{2\omega} - \frac{\pi}{6\omega} = \frac{3\pi - \pi}{6\omega} = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$.
Comparing the two,$T_1 = \frac{\pi}{6\omega}$ and $T_2 = \frac{\pi}{3\omega}$,we find that ${T_1} < {T_2}$.
Alternatively,in Simple Harmonic Motion,the speed of the particle is maximum at the mean position $(x = 0)$ and zero at the extreme position $(x = A)$. Since the particle moves faster near the mean position,it covers the distance from $0$ to $A/2$ in less time than it takes to cover the distance from $A/2$ to $A$ where it is slowing down.

(C) The total energy $E$ of a standing wave is given by the formula $E = \frac{1}{4} m \omega^2 A^2$,where $m$ is the mass of the wire,$\omega$ is the angular frequency,and $A$ is the amplitude.
In the first experiment,the frequency is $\omega_1 = \omega$ and the amplitude is $A_1 = A$. Thus,$E_1 = \frac{1}{4} m \omega^2 A^2$.
In the second experiment,the frequency is $\omega_2 = 2\omega$ and the amplitude is $A_2 = A$. Thus,$E_2 = \frac{1}{4} m (2\omega)^2 A^2 = 4 \times (\frac{1}{4} m \omega^2 A^2) = 4E_1$.
Therefore,the energy in the second experiment is four times the energy in the first experiment,i.e.,${E_2} = 4{E_1}$.

(B) The initial distance between the centres of the two pulses is $d = 8 \ cm$.
Each pulse moves towards the other with a speed of $v = 2 \ cm/s$.
The relative speed of the two pulses is $v_{rel} = v + v = 2 + 2 = 4 \ cm/s$.
The time taken for the pulses to meet is $t = d / v_{rel} = 8 \ cm / 4 \ cm/s = 2 \ s$.
After $2 \ s$,the two pulses overlap completely.
Since the pulses have opposite phases (one is a crest and the other is a trough of equal magnitude),their displacements cancel each other out at every point,making the string momentarily straight.
Because the string is straight,there is no deformation,and thus the potential energy is zero.
However,the particles of the string still possess velocity at this instant,so the total energy of the system is purely kinetic.

(A) Let the mass of the complete circular disc be $M_{total}$. Since the sector is one-quarter of the disc,its mass $M = \frac{M_{total}}{4}$,which implies $M_{total} = 4M$.
The moment of inertia $(I)$ of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is given by $I_{total} = \frac{1}{2}M_{total}R^2$.
Substituting $M_{total} = 4M$ into the formula,we get:
$I_{total} = \frac{1}{2}(4M)R^2 = 2MR^2$.
By the principle of symmetry,the moment of inertia of the quarter sector $(I_{sector})$ about the same axis is one-fourth of the moment of inertia of the complete disc:
$I_{sector} = \frac{I_{total}}{4} = \frac{2MR^2}{4} = \frac{1}{2}MR^2$.

(A) Let the mass of the insect be $m$. The forces acting on the insect are its weight $mg$ (downwards),the normal reaction $R$ (radially outwards),and the frictional force $f$ (tangentially upwards along the surface).
At any angle $\alpha$ with the vertical,the components of the weight are $mg \cos \alpha$ (radially inwards) and $mg \sin \alpha$ (tangentially downwards).
For the insect to be in equilibrium without slipping,the forces must be balanced:
$R = mg \cos \alpha$ $(i)$
$f = mg \sin \alpha$ (ii)
For the limiting condition of friction,$f = \mu R$,where $\mu = 1/3$.
Substituting equations $(i)$ and (ii) into the limiting friction condition:
$mg \sin \alpha = \mu (mg \cos \alpha)$
$\tan \alpha = \mu = 1/3$
Since $\tan \alpha = 1/3$,we have $\cot \alpha = 3$.

(D) We know that the capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
Since $A$ has dimensions of $L^2$ and $d$ has dimensions of $L$,the dimensions of $\varepsilon_0 L$ are equivalent to the dimensions of capacitance $C$.
Substituting this into the expression for $X$:
$X = \frac{(\varepsilon_0 L) V}{t} = \frac{C V}{t}$.
Since $Q = CV$,we have $X = \frac{Q}{t}$.
The rate of flow of charge $Q$ with respect to time $t$ is defined as current $I$.
Therefore,the dimensions of $X$ are the same as those of current.

(B) The electric potential decreases in the direction of the electric field.
Since the electric field $\vec{E}$ is in the positive $x$-direction,the potential decreases as $x$ increases.
Point $A$ is at $x = 0$ and point $B$ is at $x = +1 \, cm$. Since $A$ is at a smaller $x$-coordinate than $B$,$V_A > V_B$.
Points $A$ and $C$ lie on the same equipotential line (a line perpendicular to the electric field lines),therefore $V_A = V_C$.
Thus,the correct relationship is $V_A > V_B$ and $V_A = V_C$.

(A) The charges $\pm q$ on the plates of capacitor $A$ are bound charges due to the electric field between them.
Since capacitor $B$ is initially uncharged and is isolated from any external source of charge,closing the switch $S$ connects the inner plate of $A$ to the inner plate of $B$.
However,because the charges on the plates of $A$ are bound by the electrostatic attraction between them,they cannot move to plate $B$.
Therefore,no charge will flow to capacitor $B$,and it will remain uncharged.
Thus,the charge on $B$ is zero.

(C) For the current $I$ to be independent of the resistance $R_6$,the potential difference across $R_6$ must be zero,or no current should flow through $R_6$.
This occurs when the circuit forms a balanced Wheatstone bridge.
In the given circuit,$R_1, R_2, R_3,$ and $R_4$ form the arms of a Wheatstone bridge with $R_6$ as the central galvanometer arm.
The condition for a balanced Wheatstone bridge is that the ratio of resistances in opposite arms must be equal,which is $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.
Cross-multiplying this gives $R_1 R_4 = R_2 R_3$.

(A) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch containing the capacitor $C$.
Let the potential at the right junction be $0$ and at the left junction be $V_L$.
Applying Kirchhoff's loop law to the outer loop (top and bottom branches):
The current $i$ flows through the top branch (resistance $R$) and the bottom branch (resistance $2R$).
Using the loop equation: $V - iR - 2iR + 2V = 0$
$3V = 3iR$
$i = V / R$
Now,find the potential at the right junction $(V_A)$ and left junction $(V_L)$.
Let the potential at the right junction be $0$.
Then the potential at the left junction is $V_L = V - iR = V - (V/R)R = 0$.
Alternatively,$V_L = 2V - i(2R) = 2V - (V/R)(2R) = 0$.
So,the potential at the right junction is $0$ and at the left junction is $0$.
The potential difference across the middle branch is the potential across the capacitor $C$ and the battery $V$.
Since no current flows through the middle branch,the potential drop across the capacitor $V_C$ must balance the battery $V$ in that branch.
Thus,$V_C = V$.

(B) Let $R$ be the resistance and $m$ be the mass of the first wire. The second wire has length $2L$,so its resistance is $2R$ and its mass is $2m$.
Let $E$ be the $EMF$ of each cell and $S$ be the specific heat capacity of the material.
For the first wire,the current is $i_1 = \frac{3E}{R}$. The heat produced is $H_1 = i_1^2 R t = m S \Delta T$.
Substituting $i_1$,we get $(\frac{3E}{R})^2 R t = m S \Delta T \implies \frac{9E^2 t}{R} = m S \Delta T$.
For the second wire,the current is $i_2 = \frac{NE}{2R}$. The heat produced is $H_2 = i_2^2 (2R) t = (2m) S \Delta T$.
Substituting $i_2$,we get $(\frac{NE}{2R})^2 (2R) t = 2m S \Delta T \implies \frac{N^2 E^2 t}{4R^2} \cdot 2R = 2m S \Delta T \implies \frac{N^2 E^2 t}{2R} = 2m S \Delta T$.
Dividing the two heat equations: $\frac{9E^2 t / R}{N^2 E^2 t / 2R} = \frac{m S \Delta T}{2m S \Delta T} \implies \frac{18}{N^2} = \frac{1}{2} \implies N^2 = 36 \implies N = 6$.

(B) When a charged particle moves in a uniform magnetic field perpendicular to its velocity,it follows a circular path.
The radius $r$ of this circular path is given by the formula $r = \frac{mv}{qB}$.
Since the charge $q$ and the magnetic field $B$ are the same for both particles,we have $r \propto mv$.
From the figure,it is clear that the radius of the trajectory of particle $A$ is greater than the radius of the trajectory of particle $B$,i.e.,$r_A > r_B$.
Therefore,$m_A v_A > m_B v_B$.

(C) The number of turns per unit width is $n = \frac{N}{b - a}$.
Consider an elemental ring of radius $x$ and thickness $dx$.
The number of turns in this elemental ring is $dN = n \cdot dx = \frac{N}{b - a} dx$.
The magnetic field at the centre due to this elemental ring is given by $dB = \frac{\mu_0 (dN) I}{2x}$.
Substituting $dN$,we get $dB = \frac{\mu_0 I}{2x} \cdot \frac{N}{b - a} dx = \frac{\mu_0 NI}{2(b - a)} \cdot \frac{dx}{x}$.
To find the total magnetic field $B$ at the centre,we integrate $dB$ from $x = a$ to $x = b$:
$B = \int_a^b \frac{\mu_0 NI}{2(b - a)} \frac{dx}{x} = \frac{\mu_0 NI}{2(b - a)} \int_a^b \frac{dx}{x}$.
$B = \frac{\mu_0 NI}{2(b - a)} [\ln x]_a^b = \frac{\mu_0 NI}{2(b - a)} \ln \frac{b}{a}$.

(D) The magnetic field at point $P(a, 0, a)$ due to the entire loop can be calculated by considering the loop as a combination of two planar loops: $ABCDA$ (in the $xz$-plane) and $AFEBA$ (in the $xy$-plane).
$1$. The magnetic field produced by the loop $ABCDA$ at point $P$ is directed along the positive $x$-axis (i.e.,in the $\hat{i}$ direction) due to the right-hand rule.
$2$. The magnetic field produced by the loop $AFEBA$ at point $P$ is directed along the positive $z$-axis (i.e.,in the $\hat{k}$ direction) due to the right-hand rule.
$3$. Since the geometry of both loops relative to point $P$ is identical,the magnitudes of the magnetic fields produced by both loops are equal.
$4$. Let the magnitude of the magnetic field from each loop be $B_0$. The total magnetic field vector at $P$ is $\vec{B} = B_0\hat{i} + B_0\hat{k}$.
$5$. The direction of the resultant magnetic field is given by the unit vector $\hat{n} = \frac{B_0\hat{i} + B_0\hat{k}}{\sqrt{B_0^2 + B_0^2}} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k})$.

(A) The mutual inductance $M$ between two coils depends on the magnetic flux linkage between them. The flux linkage is maximum when the magnetic field lines produced by one coil pass through the area of the other coil most effectively.
In situation $(A)$,the planes of the two coils are parallel to each other. This orientation allows the maximum number of magnetic field lines produced by the larger coil to pass through the smaller coil,resulting in the highest magnetic flux linkage.
In situation $(B)$,the plane of the smaller coil is perpendicular to the plane of the larger coil. In this case,the magnetic field lines from the larger coil are mostly parallel to the plane of the smaller coil,leading to minimal flux linkage.
In situation $(C)$,the coils are placed side-by-side with their planes perpendicular to each other,which also results in very low flux linkage.
Therefore,the mutual inductance is maximum in situation $(A)$.

(D) When a conductor moves in a magnetic field,a motional electromotive force (emf) is induced across its ends,given by $\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
This motional emf is associated with an induced electric field $\vec{E} = \vec{v} \times \vec{B}$ within the conductor.
In the given figure,the square loop $ABCD$ is moving with velocity $v$ in a uniform magnetic field $B$ perpendicular to the plane of the loop.
For the side $AD$,the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,so an electric field is induced along $AD$.
Similarly,for the side $BC$,the velocity $\vec{v}$ is also perpendicular to the magnetic field $\vec{B}$,so an electric field is also induced along $BC$.
Since both sides $AD$ and $BC$ are moving through the magnetic field,an electric field is induced in both conductors.
Therefore,the correct option is $(d)$.

(A) The wavelength of the characteristic $X$-ray line,such as the ${K_\alpha }$ line $({\lambda _k})$,depends only on the material of the target and is independent of the accelerating voltage $(V)$.
The minimum wavelength of the continuous $X$-ray spectrum $({\lambda _c})$ is given by the Duane-Hunt law: ${\lambda _c} = \frac{hc}{eV}$.
From this relation,it is clear that ${\lambda _c}$ is inversely proportional to the accelerating voltage $(V)$.
As the accelerating voltage $(V)$ is increased,${\lambda _c}$ decreases,while ${\lambda _k}$ remains constant.
Therefore,the difference $({\lambda _K} - {\lambda _C})$ will increase.

(C) During $\beta$-decay,a neutron inside the nucleus transforms into a proton,an electron,and an antineutrino. The process is represented as: $n \rightarrow p + e^- + \bar{\nu}_e$. Since the electron is created during this transformation process within the nucleus,it is emitted as beta radiation.

(D) Let $N_0$ be the initial number of atoms of each species. The number of radioactive nuclei of the first species at time $t$ is $N_1(t) = N_0 e^{-t/\tau}$.
The number of radioactive nuclei of the second species at time $t$ is $N_2(t) = N_0 e^{-t/(5\tau)}$.
The total number of radioactive nuclei is $N(t) = N_1(t) + N_2(t) = N_0(e^{-t/\tau} + e^{-t/(5\tau)})$.
At $t = 0$,$N(0) = 2N_0$. As $t \to \infty$,$N(t) \to 0$.
Since both $e^{-t/\tau}$ and $e^{-t/(5\tau)}$ are monotonically decreasing functions of time,their sum $N(t)$ must also be a monotonically decreasing function of time.
Therefore,the total number of radioactive nuclei will continuously decrease with time,which is represented by the graph in option $(d)$.

(D) According to Snell's law for multiple parallel interfaces,the product of the refractive index and the sine of the angle of incidence at each interface remains constant.
Let $\theta_1$ be the angle of incidence in the first medium and $\theta_4$ be the angle of refraction in the fourth medium.
Applying Snell's law at each interface:
$\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 = \mu_3 \sin \theta_3 = \mu_4 \sin \theta_4$.
Since the emergent ray $CD$ is parallel to the incident ray $AB$,the angle of incidence $\theta_1$ must be equal to the angle of emergence $\theta_4$ (i.e.,$\theta_1 = \theta_4$).
Substituting this into the equation,we get $\mu_1 \sin \theta_1 = \mu_4 \sin \theta_1$.
Therefore,$\mu_1 = \mu_4$.

(B) The prisms $Q$ and $R$ are of the same material and have an identical shape to prism $P$.
When these prisms are arranged as shown in the figure,the combination of prisms $Q$ and $R$ effectively acts as a glass slab with parallel faces.
$A$ light ray passing through a glass slab with parallel faces undergoes lateral displacement but no net angular deviation.
Therefore,the total deviation experienced by the light ray remains the same as the deviation caused by prism $P$ alone,which is the minimum deviation.

(B) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \frac{\pi}{2}$.
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\frac{\pi}{2}) = 5I + 0 = 5I$.
At point $B$,the phase difference $\phi_B = \pi$.
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.
The difference between the resultant intensities at $A$ and $B$ is $|I_A - I_B| = |5I - I| = 4I$.

(B) The width of the segment on the screen is given by $W = n \beta$,where $n$ is the number of fringes and $\beta$ is the fringe width.
Since $\beta = \frac{\lambda D}{d}$,the width of the segment is $W = n \frac{\lambda D}{d}$.
For a fixed segment $W$,the product $n \lambda$ remains constant because $D$ and $d$ are constant.
Therefore,$n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 12$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.
Substituting the values: $12 \times 600 = n_2 \times 400$.
$n_2 = \frac{12 \times 600}{400} = 12 \times 1.5 = 18$.
Thus,$18$ fringes will be observed.

(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2}) \text{ eV}$.
Ultraviolet radiation corresponds to higher energy transitions (Lyman series),while infrared radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).
Given that the transition $n=4 \rightarrow n=3$ results in ultraviolet radiation for this specific atom,it implies that the energy gap $\Delta E_{4 \rightarrow 3}$ is quite large.
To obtain infrared radiation,we need a transition with a significantly smaller energy gap than $\Delta E_{4 \rightarrow 3}$.
Comparing the options:
$(A)$ $2 \rightarrow 1$: This is a transition between lower energy levels,resulting in a larger energy gap than $4 \rightarrow 3$.
$(B)$ $3 \rightarrow 2$: This is also a transition between lower energy levels,resulting in a larger energy gap than $4 \rightarrow 3$.
$(C)$ $4 \rightarrow 2$: This involves a larger jump in quantum numbers,resulting in a larger energy gap.
$(D)$ $5 \rightarrow 4$: This transition occurs between higher energy levels where the energy difference between consecutive levels is much smaller. Thus,$5 \rightarrow 4$ will have the smallest energy gap and will result in infrared radiation.

(C) Electric lines of force due to a positive charge are directed radially outward and are spherically symmetric. Since all three charges are positive and equal in magnitude,they exert a repulsive force on each other. Consequently,the electric field lines originate from each charge and move away from each other. No electric field lines can enter the region between the charges because the fields from the three charges repel each other,creating a neutral point at the centroid of the triangle. The resulting pattern of electric field lines is shown in the provided solution image.