JEE Main 2026 Jan 24 shift 2

We need to find the intensity on the screen in front of one of the slits in Young's double slit experiment. The intensity from each slit is $I_0$, the slit separation is d = 2 mm = 2 \times 10⁻³ m, the screen distance is D = 10 m, and the wavelength is $\lambda = 6000 \text{ Å} = 6 \times 10^{-7}$ m. The fringe width is given by $\beta = \frac{\lambda D}{d} = \frac{6 \times 10^{-7} \times 10}{2 \times 10^{-3}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}$. The point \in front of one slit lies at y = d/2 = 1 mm from the central maximum, so the path difference is $\Delta = \frac{yd}{D} = \frac{1 \times 10^{-3} \times 2 \times 10^{-3}}{10} = 2 \times 10^{-7}$ m. The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \times \Delta = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$. The resultant intensity is $I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\cos\phi = 2I_0 + 2I_0\cos\frac{2\pi}{3},$ which simplifies to $= 2I_0 + 2I_0\left(-\frac{1}{2}\right) = 2I_0 - I_0 = I_0.$ Therefore, the intensity is Option 3 : $I_0$.

We need to find the wavelength of the first light given photoelectric stopping potentials. For light of wavelength $\lambda$ the stopping potential is $V_1 = 3.2$ V, and for light of wavelength $2\lambda$ the stopping potential is $V_2 = 0.7$ V. According to Einstein's photoelectric equation, $eV = \frac{hc}{\lambda} - \phi.$ Applying this to the first light gives $eV_1 = \frac{hc}{\lambda} - \phi \quad \cdots (1)$ and to the second light gives $eV_2 = \frac{hc}{2\lambda} - \phi \quad \cdots (2).$ Subtracting equation (2) from (1) yields $e(V_1 - V_2) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda},$ which leads to $\lambda = \frac{hc}{2e(V_1 - V_2)}.$ Substituting the numerical values, we get $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 1.6 \times 10^{-19} \times (3.2 - 0.7)}$ $= \frac{19.89 \times 10^{-26}}{2 \times 1.6 \times 10^{-19} \times 2.5}$ $= \frac{19.89 \times 10^{-26}}{8 \times 10^{-19}} = 2.486 \times 10^{-7} \text{ m} \approx 2.5 \times 10^{-7} \text{ m}.$ Therefore, the wavelength is Option 2: $2.5 \times 10^{-7}$ m.

We need to find the height of the submerged part of a floating cubical block. The density of the block is $\rho_b = 600$ kg/m³, and the density of the liquid is $\rho_l = 900$ kg/m³. The block has height H = 8.0 cm. By the principle of floatation, the weight of the block equals the weight of the liquid displaced, so $\rho_b \times V_{block} \times g = \rho_l \times V_{submerged} \times g$ Because the block is cubical with base area A, its total volume is A \times H and its submerged volume is A \times h, giving $\rho_b \times A \times H = \rho_l \times A \times h$ Solving for h yields $h = \frac{\rho_b}{\rho_l} \times H = \frac{600}{900} \times 8.0 = \frac{2}{3} \times 8.0 = 5.33 \text{ cm} \approx 5.3 \text{ cm}$ Therefore, the height of the submerged part is Option 4 : 5.3 cm.

The stability of a nucleus is measured by its total binding energy: larger binding energy ⇒ greater stability. Let the binding energies of ${}_{2}He^{3},\; {}_{2}He^{4}$ and ${}_{2}He^{5}$ be $B_3,\; B_4$ and $B_5$ (\in MeV). Write the given reactions \in the form ${}_{2}He^{3} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{4} + 20\;{\rm MeV}$ ${}_{2}He^{4} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{5} - 0.9\;{\rm MeV}$ For any nuclear reaction, energy released $=\;(\text{binding energy of products})-(\text{binding energy of reactants})$. Case 1: Products' binding energy $=\;B_4$, reactants' binding energy $=\;B_3 + 0$ (a free neutron has zero binding energy). Hence $B_4 - (B_3 + 0) = 20$ $-(1)$ $\Rightarrow\; B_4 = B_3 + 20$ Case 2: Products' binding energy $=\;B_5$, reactants' binding energy $=\;B_4 + 0$. Here the reaction absorbs 0.9 MeV, so energy released = $-0.9$ MeV: $B_5 - (B_4 + 0) = -0.9$ $-(2)$ $\Rightarrow\; B_5 = B_4 - 0.9$ Substitute $B_4$ from $(1)$ into $(2)$: $B_5 = (B_3 + 20) - 0.9 = B_3 + 19.1$ Thus $B_4 = B_3 + 20$ $B_5 = B_3 + 19.1$ Since $20 \gt 19.1 \gt 0$, we get the order $B_4 \gt B_5 \gt B_3$. Greater binding energy means higher stability, so $X_4 \gt X_5 \gt X_3$. Therefore the correct option is Option A .

The truth table corresponds to the behavior of the logic circuit, which has two inputs, A and B. The correct operation indicates that the output Y is true (1) when either input is false (0) and the other is true (1), except when both inputs are true (1). This is characteristic of the exclusive OR (XOR) function, which matches the truth table labeled B. Therefore, B is the correct answer.

Label the six vertices of the regular hexagon in order as $A,\,B,\,C,\,D,\,E,\,F$ and let the centre be $O$. Each side and each spoke (vertex-centre wire) has resistance $r$. The current enters at vertex $A$ and leaves at the opposite vertex $D$. Because the geometrical arrangement is symmetric about the line $AD$, the following potentials will be equal: \bullet $V_B = V_F = x$ (call this common potential $x$) \bullet $V_C = V_E = y$ (call this $y$) \bullet $V_O = z$ (centre potential) Let the potential at $A$ be $V_A = V$ (unknown for now) and at $D$ be $V_D = 0$. We shall write Kirchhoff's Current Law (KCL) at the three distinct internal potentials $x,\,y,\,z$. (All currents are taken as leaving the node, and each resistance is $r$.) Case 1: Node $B$ (or $F$) at potential $x$ Connections: to $A$, $C$, $O$. $\frac{x - V}{r} + \frac{x - y}{r} + \frac{x - z}{r} = 0$ Multiplying by $r$ gives: $3x - V - y - z = 0 \quad -(1)$ Case 2: Node $C$ (or $E$) at potential $y$ Connections: to $B$, $D$, $O$. $\frac{y - x}{r} + \frac{y - 0}{r} + \frac{y - z}{r} = 0$ Multiplying by $r$ gives: $3y - x - z = 0 \quad -(2)$ Case 3: Centre $O$ at potential $z$ Connections: to all six vertices. $\frac{z - V}{r} + \frac{z - x}{r} + \frac{z - y}{r} + \frac{z - 0}{r} + \frac{z - y}{r} + \frac{z - x}{r} = 0$ Grouping like terms and multiplying by $r$: $6z - V - 2x - 2y = 0 \quad -(3)$ Solving the simultaneous equations From $-(1)$: $x = \frac{V + y + z}{3} \quad -(1a)$ From $-(2)$: $y = \frac{x + z}{3} \quad -(2a)$ Substitute $x$ from $(1a)$ into $(2a)$: $y = \frac{\dfrac{V + y + z}{3} + z}{3} = \frac{V + y + 4z}{9}$ $\Rightarrow 9y = V + y + 4z \Longrightarrow 8y = V + 4z$ $\therefore \; y = \frac{V + 4z}{8} \quad -(4)$ Put $y$ from $(4)$ into $(1a)$: $x = \frac{V + \dfrac{V + 4z}{8} + z}{3} = \frac{9V + 12z}{24} = \frac{3V + 4z}{8} \quad -(5)$ Now use equation $-(3)$: $6z - V - 2x - 2y = 0$ Insert $x$ and $y$ from $-(5)$ and $-(4)$: $6z - V - 2\!\left(\frac{3V + 4z}{8}\right) - 2\!\left(\frac{V + 4z}{8}\right)=0$ $\Longrightarrow 6z - V - \frac{3V + 4z}{4} - \frac{V + 4z}{4}=0$ $\Longrightarrow 6z - V - (V + 2z)=0$ $\Longrightarrow 4z - 2V=0 \;\; \Rightarrow\;\; z = \frac{V}{2}$ Back-substitute $z$: $y = \frac{V + 4\!\left(\dfrac{V}{2}\right)}{8} = \frac{3V}{8}, \qquad x = \frac{3V + 4\!\left(\dfrac{V}{2}\right)}{8} = \frac{5V}{8}$ Total current entering at $A$ Three resistors leave $A$: to $B$, $O$, $F$. Hence $I = \frac{V - x}{r} + \frac{V - z}{r} + \frac{V - x}{r} = \frac{2(V - x) + (V - z)}{r}$ Substitute $x = \dfrac{5V}{8}, \; z = \dfrac{V}{2}$: $I = \frac{2\!\left(V - \dfrac{5V}{8}\right) + \left(V - \dfrac{V}{2}\right)}{r} = \frac{2\!\left(\dfrac{3V}{8}\right) + \dfrac{V}{2}}{r} = \frac{\dfrac{6V}{8} + \dfrac{4V}{8}}{r} = \frac{\dfrac{10V}{8}}{r} = \frac{5V}{4r}$ Equivalent resistance $R_{\text{eq}} = \frac{V}{I} = \frac{V}{\dfrac{5V}{4r}} = \frac{4r}{5}$ Therefore the equivalent resistance between the two opposite corners is $\dfrac{4}{5}\,r$. Option A is correct.

after removing capacitor branch:(as in steady state capacitor is an open circuit) from middle node → one 8Ω to bottom from middle → 4Ω → right node right node → two 8Ω in parallel to bottom step 1: right side $8||8=4\Omega$ step 2: path via right: $4+4=8\Omega$ so from middle node to bottom we have: $8\Omega\text{ (direct)}\parallel8\Omega\text{ (via right)}$ R=4\Omega step 3: total resistance: $R_{total}=1+4=5\Omega$ step 4: current: $I=\frac{10}{5}=2A$

From the PV diagram, the two curved paths are adiabatic (their shape indicates $PV^{\gamma}=\text{constant}$, and the process from $P_1$ to $P_2$​ at $V=1m^3$ is vertical, i.e. an isochoric process. Since volume is constant, work done is zero: W=0 So heat supplied equals change \in internal energy: $Q=nC_v(T_2-T_1)$ Using ideal gas law, $T_1=\frac{P_1V}{nR}=\frac{21.7\times1}{10\times8.3}$ $T_2=\frac{P_2V}{nR}=\frac{30\times1}{10\times8.3}$ Thus, $T_2−T_1=\frac{\left(30−21.7\right)}{83}=\frac{8.3}{83}=0.1K$ Therefore, $Q=10\times21\times0.1$ $Q=21J$ So, $\alpha =21$

let $C_0=\frac{\varepsilon_0A}{d}$ case (A) top layer (d/2, full area, $k_1$​): $C_t=\frac{k_1\varepsilon_0A}{d/2}=\frac{2k_1\varepsilon_0A}{d}$ bottom layer (d/2 split): $C_{b1}=\frac{k_1\varepsilon_0(A/2)}{d/2}=\frac{k_1\varepsilon_0A}{d}$ $C_{b2}=\frac{k_2\varepsilon_0(A/2)}{d/2}=\frac{k_2\varepsilon_0A}{d}$ parallel: $C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$ series: $\frac{1}{C_A}=\frac{1}{C_t}+\frac{1}{C_b}=\frac{d}{2k_1\varepsilon_0A}+\frac{d}{(k_1+k_2)\varepsilon_0A}$ $C_A=\frac{\varepsilon_0A}{d}\cdot\frac{2k_1(k_1+k_2)}{3k_1+k_2}$ case (B) top: $k_2$ $C_t=\frac{2k_2\varepsilon_0A}{d}$ bottom same as before: $C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$ $C_B=\frac{\varepsilon_0A}{d}\cdot\frac{2k_2(k_1+k_2)}{k_1+3k_2}$ case (C) top split: $C_t=\frac{(k_1+k_2)\varepsilon_0A}{d}$ bottom split: $C_b=\frac{(k_1+k_2)\varepsilon_0A}{d}$ series of equal: $C_C=\frac{1}{2}\cdot\frac{(k_1+k_2)\varepsilon_0A}{d}$ now compare (given $k_1>k_2$): $C_A>C_C>C_B$

Moment of Inertia about the Pivot ($I_P$): The rod is pivoted at a distance of $L/3$ from the bottom. The center of mass (CM) of the rod is at its midpoint, $L/2$ from the bottom. The distance ($d$) between the pivot and the center of mass is: $d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$ Using the Parallel Axis Theorem: $I_P = I_{cm} + Md^2$ $I_P = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2$ $I_P = \frac{1}{12}ML^2 + \frac{1}{36}ML^2 = \frac{1}{9}ML^2$ Change \in Potential Energy ($\Delta PE$): When the rod falls from the vertical position to the horizontal position, the center of mass descends by a vertical height equal to its initial distance from the pivot. $\Delta h = d = \frac{L}{6}$ $\Delta PE = Mg\Delta h = Mg\left(\frac{L}{6}\right)$ Conservation of Energy: $\Delta PE = \Delta KE_{rot}$ $Mg\left(\frac{L}{6}\right) = \frac{1}{2} I_P \omega^2$ $Mg\frac{L}{6} = \frac{1}{2} \left(\frac{1}{9}ML^2\right) \omega^2$ $\omega = \sqrt{\frac{3g}{L}}$

The key concept is the superposition of magnetic fields from two circular loops carrying currents. Each loop produces a magnetic field at point O, with loop P generating a field directed towards P and loop Q towards Q. Since the currents in both loops are equal (J in P and 41 in Q) and both are in the same clockwise direction, the magnetic field contributions at O add constructively towards Q, yielding a net magnetic field of $\frac{3\mu_{0}I}{4\sqrt{2}r}$ towards Q. Thus, the correct answer is B.

For a point source, Intensity varies as $I\propto\frac{1}{r^2}$ If volume of spherical detector increases 8 \times , $V'=8V$ For sphere, $V=\frac{4}{3}\pi r^3$ so $r'^3=8r^3$ $r'=2r$ Now intensity becomes $I'=\frac{1}{(2r)^2}$ $=\frac{1}{4}\cdot\frac{1}{r^2}$ So $I'=\frac{I}{4}$ Hence intensity becomes one-fourth of original.

Take the chain and split it into two equal halves. Because both supports are at the same level, the situation is symmetric. Now focus on only one half of the chain. For this half: its weight acts downward and equals $\frac{mg}{2}$​ the tension at the support is $T_s$ making an angle $30^{\circ}$ with the horizontal at the lowest point, the tension is horizontal (no vertical component there) Now apply equilibrium. Vertical direction: Only the support tension provides an upward component, so it must balance the weight of half the chain: $T_s\sin30^{\circ}=\frac{mg}{2}$ $Since\ \sin30^{\circ}=\frac{1}{2}$ $T_s\cdot\frac{1}{2}=\frac{mg}{2}$ $\Rightarrow T_s=mg$ Now look at horizontal direction. The horizontal component of tension is the same everywhere \in the chain. That means the tension at the lowest point is equal to the horizontal component of TsT_sTs​: $T_{lowest}=T\cos30^{\circ\ }$ Substitute $T_s=mg\ and\cos30^{\circ}=\frac{\sqrt{3}}{2}$ $T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$ So the tension at the lowest point is $T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$

We need to find the least count of the vernier callipers. 50 vernier scale divisions (VSD) = 48 main scale divisions (MSD) 1 MSD = 0.05 mm Find the value of 1 VSD: $1 \text{ VSD} = \frac{48}{50} \text{ MSD} = \frac{48}{50} \times 0.05 = 0.048 \text{ mm}$ Calculate the least count: $\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 - 0.048 = 0.002 \text{ mm}$ Therefore, the least count is Option 3 : 0.002 mm.

We need to find x where the ratio of lengths of closed pipe to open pipe is 5/x. Key formulas: For a closed organ pipe, the nth harmonic frequency (only odd harmonics): $f_n = \frac{nv}{4L_c}$ where n = 1, 3, 5, ... For an open organ pipe, the nth harmonic: $f_n = \frac{nv}{2L_o}$ where n = 1, 2, 3, ... Fifth harmonic of closed pipe = First harmonic of open pipe. Fifth harmonic of closed pipe: $f = \frac{5v}{4L_c}$ First harmonic of open pipe: $f = \frac{v}{2L_o}$ Setting them equal: $\frac{5v}{4L_c} = \frac{v}{2L_o}$ $\frac{5}{4L_c} = \frac{1}{2L_o}$ $\frac{L_c}{L_o} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$ So $\frac{L_c}{L_o} = \frac{5}{2}$, which means x = 2. Therefore, the value of x is Option 3 : 2.

In vertical circular motion, the tension in the thread becomes zero when the centripetal force required to maintain the circular path is provided entirely by gravity. At an angle of $30^\circ$, the gravitational component acting towards the center is $mg \cos(30^\circ) = mg \frac{\sqrt{3}}{2}$. The centripetal force required at the top of the circular path is given by $\frac{mv^2}{r}$. Setting the gravitational force equal to the centripetal force, we solve for $v$ and find $v = \sqrt{4gr}$ at point A, making the correct option $C$ as $\sqrt{\frac{7}{2}gr}$.

The relationship between velocity ($v$) and distance ($x$) shows a linear decrease, indicating that acceleration ($a$) is constant. Since acceleration is the rate of change of velocity, when velocity decreases uniformly with distance, the acceleration remains constant over that interval. Thus, graph D, which depicts a constant acceleration, correctly represents this scenario. In contrast, other graphs either show varying acceleration or no acceleration.

We need to find the focal length of a concave mirror that produces a 3x magnified real image with object-image distance of 40 cm. Magnification: m = -3 (real image is inverted, so negative for a mirror) Distance between object and image = 40 cm Set up equations: For a mirror: $m = -\frac{v}{u}$ $-3 = -\frac{v}{u} \implies v = 3u$ Using sign convention (both u and v are negative for concave mirror with real image): Let u = -x (object distance), then v = -3x (image distance, real image). Distance between object and image: |v - u| = |-3x - (-x)| = |-2x| = 2x = 40 cm So x = 20 cm, u = -20 cm, v = -60 cm. Find focal length using mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-20} = -\frac{1}{60} - \frac{1}{20} = -\frac{1+3}{60} = -\frac{4}{60} = -\frac{1}{15}$ $f = -15$ cm Therefore, the focal length is Option 2 : -15 cm.

The galvanometer behaves like a resistor $R_g = 100\Omega$ that can carry a maximum (full-scale) current of $I_g = 1\text{ mA} = 0.001\text{ A}$. To measure larger currents we connect a small resistance $R_s$ \in parallel with the galvanometer. This combination is called a shunt-type ammeter. Let the required full-scale current of the ammeter be $I = 5\text{ mA} = 0.005\text{ A}$. When this current enters the parallel combination, part $I_g$ flows through the galvanometer and the remainder $I_s$ flows through the shunt: $I = I_g + I_s$ Because the galvanometer and the shunt are \in parallel, the voltage across each branch is the same: $I_g\,R_g = I_s\,R_s$ $-(1)$ From the current relation, the shunt current is $I_s = I - I_g = 0.005 - 0.001 = 0.004\text{ A}$ Substitute $I_g$, $I_s$ and $R_g$ into equation $-(1)$ to obtain $R_s$: $R_s = \frac{I_g\,R_g}{I_s} = \frac{0.001 \times 100}{0.004} = \frac{0.1}{0.004} = 25\Omega$ Therefore, the resistance that must be connected \in parallel with the galvanometer to convert it into an ammeter of full-scale 5 mA is $25\Omega$. Option C is correct.

Convex lens with power +5D (focal length f = 1/5 m = 20 cm). Check readings of (u, v) pairs. Apply lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for each pair: With $f = 20$ cm, $u$ is negative (real object), $v$ is positive (real image). For $P_1$: $(u, v) = (-25, 96)$. $\frac{1}{96} + \frac{1}{25} = 0.0104 + 0.04 = 0.0504 \approx 0.05 = 1/20$. Correct. For $P_2$: $(u, v) = (-30, 62)$. $\frac{1}{62} + \frac{1}{30} = 0.0161 + 0.0333 = 0.0495 \approx 0.05$. Approximately correct. For $P_3$: $(u, v) = (-35, 37)$. $\frac{1}{37} + \frac{1}{35} = 0.0270 + 0.0286 = 0.0556 \neq 0.05$. INCORRECT. (Expected v: $\frac{1}{v} = 1/20 - 1/35 = 3/140$, $v = 46.67$). For $P_4$: $(u, v) = (-45, 35)$. $\frac{1}{35} + \frac{1}{45} = 0.0286 + 0.0222 = 0.0508 \approx 0.05$. Approximately correct. For $P_5$: $(u, v) = (-50, 32)$. $\frac{1}{32} + \frac{1}{50} = 0.03125 + 0.02 = 0.05125 \approx 0.05$. Approximately correct. Only $P_3$'s reading is significantly off. The correct answer is Option D : Readings recorded by $P_3$ person are incorrect.

For the ideal gas, $C_v = C_p - R = \dfrac{7}{2}R - R = \dfrac{5}{2}R$. At constant volume, $Q = nC_v\Delta T$. With $Q = 300$ J, $\Delta T = 50 - 20 = 30$ K, and $R = 8.314$ J/mol K: $300 = n \times \dfrac{5}{2} \times 8.314 \times 30 \implies n = \dfrac{300}{623.55} \approx 0.481$ mol. The answer is $\boxed{481}$.

We need to find the ratio $I_1/I_2$ of moments of inertia of a solid cylinder and a smaller co-centric cylinder carved from it. Moment of inertia of a solid cylinder about its axis: $I = \frac{1}{2}MR^2$ For the large cylinder: Mass: $M_1 = \rho \pi R^2 L$ $I_1 = \frac{1}{2}M_1 R^2 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{1}{2}\rho \pi R^4 L$ For the small cylinder: Radius = R/3, Length = L/2 Mass: $M_2 = \rho \pi (R/3)^2 (L/2) = \rho \pi R^2 L / 18$ $I_2 = \frac{1}{2}M_2 (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}$ Ratio: $\frac{I_1}{I_2} = \frac{\frac{1}{2}\rho \pi R^4 L}{\frac{\rho \pi R^4 L}{324}} = \frac{324}{2} = 162$ Therefore, $I_1/I_2 =$ 162 .

We begin by noting that we need to find x where the work done in blowing a soap bubble from diameter 7 cm to 14 cm is (15000 - x) μJ. We recall that a soap bubble has two surfaces, so the formula for the work done is: $W = T \times \Delta A \times 2 = 2T \times 4\pi(R_2^2 - R_1^2)$ $W = 8\pi T(R_2^2 - R_1^2)$ Here, T = 0.04 N/m. Also, $R_1 = 3.5$ cm = 0.035 m, $R_2 = 7$ cm = 0.07 m. We take $\pi = 22/7$. Substituting into the formula gives: $W = 8 \times \frac{22}{7} \times 0.04 \times (0.07^2 - 0.035^2)$ $= 8 \times \frac{22}{7} \times 0.04 \times (0.0049 - 0.001225)$ $= 8 \times \frac{22}{7} \times 0.04 \times 0.003675$ $= 8 \times \frac{22}{7} \times 0.000147$ $= 8 \times 22 \times \frac{0.000147}{7}$ $= 8 \times 22 \times 0.000021$ $= 8 \times 0.000462$ $= 0.003696 \text{ J} = 3696 \text{ \mu J}$ Thus, $15000 - x = 3696$ and $x = 15000 - 3696 = 11304$ Therefore, x = 11304 .

Wire length L=10 cm=0.1mCharge on wire $Q=24\mu C=24\times10^{-6}C$ Point charge $q=1\mu C=10^{-6}C$ Distance of charge from nearer end: a=2 cm=0.02m Far end distance: $a+L=0.12m$ Linear charge density $\lambda=\frac{Q}{L}=\frac{24\times10^{-6}}{0.1}=2.4\times10^{-4}C/m$ For small element dx, $dF=\frac{1}{4\pi\epsilon_0}\frac{q\lambda dx}{x^2}$ Integrate from x=0.02 to 0.12: $F=9\times10^9(10^{-6})(2.4\times10^{-4})\int_{0.02}^{0.12}\frac{dx}{x^2}=9\times10^9(2.4\times10^{-10})\left[-\frac{1}{x}\right]_{0.02}^{0.12}$ $=2.16\left(\frac{1}{0.02}-\frac{1}{0.12}\right)$ $=2.16(50-8.33)$ $=2.16(41.67)$ =90

We need to find the unknown resistance x connected in parallel with 3Ω in a meter bridge. Initially, 2Ω is on the left and 3Ω is on the right, and the null point is at a distance l from the left. $\frac{2}{3} = \frac{l}{100-l} \implies l = 40 \text{ cm}$ When x is connected \in parallel with the 3\Omega resistor, the effective resistance on the right becomes $R_{eff} = \frac{3x}{3+x}$. The null point shifts 10 cm to the right, so the new null point is at $l' = 50$ cm. $\frac{2}{\frac{3x}{3+x}} = \frac{50}{50} = 1$ $\frac{2(3+x)}{3x} = 1$ $6 + 2x = 3x$ $x = 6 \, \Omega$ Therefore, x = 6 Ω.

We need to find the wavelength of a spectral line in $Li^{2+}$ for a transition where the \sum of levels is 4 and difference is 2. First, we find the energy levels by solving $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$, which gives $n_2 = 3, n_1 = 1$. Next, we use the Rydberg formula for $Li^{2+}$ (Z = 3): $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \times 9 \times \left(1 - \frac{1}{9}\right) = 9R \times \frac{8}{9} = 8R$ Here, $R = 1.097 \times 10^5 \text{ cm}^{-1}$. Substituting this value gives $\frac{1}{\lambda} = 8 \times 1.097 \times 10^5 = 8.776 \times 10^5 \text{ cm}^{-1}$, and hence $\lambda = \frac{1}{8.776 \times 10^5} = 1.14 \times 10^{-6} \text{ cm}$. Therefore, the wavelength is Option 3: $1.14 \times 10^{-6}$ cm.

We need to identify which statements are NOT true about organic chemistry concepts. Statement A: "Resonating structures with more number of covalent bonds and lesser charge separation are more stable." This is TRUE . More covalent bonds and less charge separation contribute to stability of resonance structures. Statement B: "In electromeric effect, an unsaturated system shows +E effect with nucleophile and -E effect with electrophile." This is NOT TRUE . The electromeric effect is the opposite: +E effect occurs with electrophiles (electron shift toward the electrophile) and -E effect occurs with nucleophiles (electron shift away from the nucleophile). The statement has it reversed. Statement C: "Inductive effect is responsible for high melting point, boiling point and dipole moment of polar compounds." This is TRUE . Inductive effect contributes to polarity of molecules, affecting physical properties. Statement D: "The greater the number of alkyl groups attached to the doubly bonded carbon atoms, higher is the heat of hydrogenation." This is NOT TRUE . More alkyl groups stabilize the double bond (hyperconjugation), resulting in LOWER heat of hydrogenation. Statement E: "Stability of carbanion increases with the increase in s-character of the carbon carrying the negative charge." This is TRUE . Greater s-character means the electrons are held closer to the nucleus, stabilizing the carbanion (sp > sp² > sp³). The statements that are NOT true are B and D only. Therefore, the answer is Option 4 : B & D only.

We need to identify the oxoanion X of the lightest element of group 7, with metal in +6 oxidation state. First, the lightest element of group 7 (Group VIIB / Mn group) is Manganese (Mn). Next, Mn in +6 oxidation state forms the manganate ion: $MnO_4^{2-}$. Check: Mn + 4(-2) = -2, so Mn = +6. ✓ Then, potassium manganate $K_2MnO_4$ has a green color (this is different from $KMnO_4$ which is purple, and $K_2Cr_2O_7$ which is orange). Therefore, the color is Option 4: green.