4 Apr 2024 Shift 2

Use dimensional analysis. Assume the relation: $T\propto G^aM^br^c$ Write dimensions: $[T]=T$ $[G]=M^{-1}L^3T^{-2}$ $[M]=M$ $[r]=L$ Substitute: $T=(M^{-1}L^3T^{-2})^a\cdot M^b\cdot L^c$ $=M^{-a+b}\cdot L^{3a+c}\cdot T^{-2a}$ Now compare powers with LHS$[T^1]$: For mass: $-a+b=0$ For length: $3a+c=0$ For time: $-2a=1\Rightarrow a=-\frac{1}{2}$ Substitute aaa: $b=a=-\frac{1}{2}$ $3a+c=0\Rightarrow c=\frac{3}{2}$ Final relation: $T\propto G^{-1/2}M^{-1/2}r^{3/2}$ $T\propto\sqrt{\frac{r^3}{GM}}$ Correct expression: $T=k\sqrt{\frac{r^3}{GM}}$ $T^2 = \frac{k r^3}{GM}$

given: radius of circle = 2 km start point = P (top of circle) end point = S (left side of circle) step 1: what is displacement? displacement = shortest straight-line distance between start and end NOT the curved path along the circle so we just need straight line PS step 2: visualize positions center = O P is directly above O → distance 2 S is directly left of O → distance 2 so OP = 2 km (vertical) OS = 2 km (horizontal) and OP ⟂ OS (90° angle) step 3: triangle OPS you now have a right triangle: OP = 2 OS = 2 PS = hypotenuse apply Pythagoras: PS² = OP² + OS² PS² = 2² + 2² PS² = 4 + 4 = 8 PS = √8 km step 4: final understanding even though cyclist moved along a curved path, displacement ignores path it only depends on start and end points so straight-line distance = √8 km

Given that the block begins to slide so $f_l=Force\ acting\ down\ the\ incline$ We know that $f_l=\mu\ _s\times \ N$ , Where N is the normal force acting on the block Here the force acting down the incline is $Mg\sin\theta\$ so $f_l=Mg\sin\theta\$ Now $f_l=Mg\sin\theta\ =\mu\ _s\times \ N$ N is normal force which is, $N=Mg\cos\theta\$ $f_l=Mg\sin\theta\ =\mu\ _s\times \ Mg\cos\theta\$ $\mu\ _s=\tan\ \theta\$ Its given that $\theta\ =45^{\circ\ }$ so $\mu\ _s=1$

Potential Energy at point A, $PE_A=mgH_A\ =mgR\left(1+\sin45^{\circ\ }\right)\ =m\times \ 10\times \ \left(1+\frac{1}{\sqrt{\ 2}}\right)\ =\ 240m\ J$ Potential Energy at point B = 0 From conservation of energy, $PE_A+KE_A=PE_B+KE_B$ \Rightarrow $KE_B=PE_A+KE_A-PE_B\ =240m+0-0=240m$ \Rightarrow $\frac{1}{2}mv_B^2=240m\ \Rightarrow \ v_B^2\ =480$ $\therefore\ v_B=\sqrt{\ 480}=21.91\$ m/s

A 90 kg body is placed at a height $2R$ above Earth's surface and the gravitational pull on it is to be determined. The acceleration due to gravity at a height $h$ above Earth's surface is given by $g' = g\left(\frac{R}{R + h}\right)^2$. This follows from Newton's law of gravitation, since $g' = \frac{GM}{(R+h)^2}$ and $g = \frac{GM}{R^2}$. For $h = 2R$, the distance from Earth's center is $R + h = 3R$, so substituting into the formula yields $g' = g\left(\frac{R}{3R}\right)^2 = g\times \frac{1}{9} = \frac{10}{9}\,\text{m/s}^2$. The gravitational force on the body is then $F = m\,g' = 90 \times \frac{10}{9} = 100\,\text{N}$. The correct answer is Option A: 100 N.

Derive the correct formula for the height of a satellite above Earth's surface. For a satellite in circular orbit at distance $r$ from Earth's center, the gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2 r$ where $\omega = \frac{2\pi}{T}$. $\frac{GM}{r^2} = \omega^2 r = \frac{4\pi^2}{T^2}r$ $GM = \frac{4\pi^2 r^3}{T^2}$ Since $g = \frac{GM}{R^2}$, we have $GM = gR^2$. Substituting: $gR^2 = \frac{4\pi^2 r^3}{T^2}$ $r^3 = \frac{gR^2T^2}{4\pi^2} = \frac{T^2R^2g}{4\pi^2}$ $r = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3}$ Since $r = R + h$: $h = r - R = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$ The correct answer is Option B: $\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$.

Evaluate the two statements about contact angle and capillary rise. Statement I: "The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well." This is TRUE . The contact angle depends on the interfacial tensions between three phases: solid-liquid ($\gamma_{SL}$), solid-vapor ($\gamma_{SV}$), and liquid-vapor ($\gamma_{LV}$). These tensions are determined by the molecular properties of both the solid and the liquid. For example, water on glass has a different contact angle than water on wax, because the solid-liquid interaction differs. Similarly, mercury on glass has a different contact angle than water on glass, because the liquid is different. Statement II: "The rise of a liquid \in a capillary tube does not depend on the inner radius of the tube." This is FALSE . The capillary rise formula is: $h = \frac{2T\cos\theta}{\rho g r}$ where $T$ is the surface tension, $\theta$ is the contact angle, $\rho$ is the liquid density, $g$ is gravitational acceleration, and $r$ is the inner radius of the tube. Clearly, $h \propto \frac{1}{r}$, so the capillary rise does depend on the inner radius. A narrower tube gives a greater rise. The correct answer is Option A: Statement I is true but Statement II is false.

Find the work done when a gas at temperature $T$ is adiabatically expanded to double its volume with $\gamma = 3/2$ and $\mu = 1$ mole. For an adiabatic process with an ideal gas, the relation $TV^{\gamma-1} = \text{constant}$ implies $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$ With $V_f = 2V_i$ and $\gamma - 1 = 3/2 - 1 = 1/2$, we have $T \cdot V_i^{1/2} = T_f \cdot (2V_i)^{1/2}$ $T_f = T \cdot \frac{V_i^{1/2}}{(2V_i)^{1/2}} = T \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}}$ For an adiabatic process, $Q = 0$, so the work done by the gas is $W = -\Delta U$. Equivalently, integrating $W = \int P\,dV$ with $PV^\gamma = \text{const}$ yields $W = \int P\,dV = \frac{P_iV_i - P_fV_f}{\gamma-1} = \frac{nRT_i - nRT_f}{\gamma-1} = \frac{nR(T_i-T_f)}{\gamma-1}$ Substituting $n = 1$, $T_i = T$, $T_f = T/\sqrt{2}$, and $\gamma - 1 = 1/2$ gives $W = \frac{1 \times R \times (T - T/\sqrt{2})}{1/2} = 2R \cdot T\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = RT(2 - \sqrt{2})$ The correct answer is Option C: $RT[2 - \sqrt{2}]$.

Methane CH4 is a non-linear polyatomic molecule (tetrahedral structure). For any molecule: Translational degrees of freedom are always $f_t=3$ because it can move along x,y,z directions. For rotational degrees of freedom: Linear molecules have $f_r=2$ Non-linear molecules have $f_r=3$ Since $\text{CH}_4$ is non-linear, $f_r=3$

For SHM, total energy is $E=\frac{1}{2}kA^2$ where k = spring constant A = amplitude Notice mass does not appear \in this expression. If mass is doubled, $m\to2m$, angular frequency changes: $\omega=\sqrt{\frac{k}{m}}$ so it becomes smaller, but energy depends only on k and A. Since amplitude is same, energy remains unchanged:

Charge q is at the center of one face of the cube. Imagine attaching an identical cube to this face so the charge lies on the common face center, which becomes inside a rectangular box made of two cubes. Now the charge is at the center of this combined box. By Gauss law, total flux through the combined closed surface is $\Phi_{\text{total}}=\frac{q}{\varepsilon_0}$ The two cubes are identical and symmetric, so flux splits equally between them. Hence flux linked with one cube is $\Phi_{\text{total}}=\frac{q}{2\varepsilon_0}$

We need to find the power dissipated by a 50 W - 200 V bulb when connected to a 100 V supply. The filament resistance can be found from the rated power and voltage using $P = \frac{V^2}{R}$. From this, $R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \; \Omega$. When the bulb is connected to 100 V, the resistance remains $800 \; \Omega$, so the power dissipated is $P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \; \text{W}$. Alternatively, since $P \propto V^2$ at constant $R$, one can write $\frac{P}{P_{\text{rated}}} = \left(\frac{V}{V_{\text{rated}}}\right)^2 = \left(\frac{100}{200}\right)^2 = \frac{1}{4} \implies P = \frac{50}{4} = 12.5 \text{ W}$. The correct answer is Option (2): 12.5 W.

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This question relates to identifying the relationship between pairs of carbohydrate molecules.
(P) α-Glucose is an aldose, while α-Fructose is a ketose. Since they have the same molecular formula (C₆H₁₂O₆) but different functional groups, they are functional group isomers.
(Q) α-Glucose and α-Mannose are stereoisomers that differ in configuration only at the C-2 carbon atom. Such pairs are called epimers.
(R) α-Glucose and β-Glucose differ in the orientation of the -OH group at the anomeric carbon (C-1). These isomers are called anomers.
(S) The PDF solution pairs α-Glucose and Ribose as homologous, though they are a hexose and a pentose, respectively.

Electromagnetic waves are arranged according to frequency (or energy) and, inversely, according to wavelength $\lambda$ using the relation $c = \nu \lambda$, where $c$ is the speed of light and $\nu$ is the frequency. Higher frequency $\Rightarrow$ lower wavelength, and lower frequency $\Rightarrow$ higher wavelength. Typical order from the highest frequency (smallest $\lambda$) to the lowest frequency (largest $\lambda$) is: $\text{Gamma rays} \;(\gamma) \; \lt\; \text{X-rays} \;\lt\; \text{Ultraviolet} \;\lt\; \text{Visible} \;\lt\; \text{Infrared} \;\lt\; \text{Microwaves} \;\lt\; \text{Radio waves}$ Using only the given categories: $\lambda_1\;(\text{Gamma rays}) \; \lt \; \lambda_2\;(\text{X-rays}) \; \lt \; \lambda_3\;(\text{Infrared}) \; \lt \; \lambda_4\;(\text{Microwaves})$ Hence, the ascending (increasing) order of wavelength is $\lambda_1 \lt \lambda_2 \lt \lambda_3 \lt \lambda_4$ This corresponds to Option C.

In Young's double slit experiment, when one slit has width 4 times that of the other, the amplitudes are related to the slit widths. The intensity is proportional to the slit width (since wider slits allow more light through). So if one slit has width $w$ and the other has width $4w$: $I_1 : I_2 = 1 : 4$ Since amplitude is proportional to the square root of intensity: $a_1 : a_2 = 1 : 2$ The maximum intensity occurs when the waves interfere constructively: $I_{max} = (a_1 + a_2)^2 = (1 + 2)^2 = 9$ The minimum intensity occurs when the waves interfere destructively: $I_{min} = (a_1 - a_2)^2 = (1 - 2)^2 = 1$ Therefore, the ratio of maximum to minimum intensity is: $\frac{I_{max}}{I_{min}} = \frac{9}{1} = 9 : 1$ The answer is Option C: $9 : 1$.

We need to evaluate the Assertion and Reason about photons and the photoelectric effect. Assertion A: "Number of photons increases with increase in frequency of light." The energy of each photon is $E = h\nu$. For a light source of constant power $P$, the number of photons per second is $N = \frac{P}{h\nu}$. As $\nu$ increases, each photon carries more energy, so fewer photons are emitted for the same power. Therefore, A is incorrect . Reason R: "Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation." By Einstein's photoelectric equation: $KE_{\max} = h\nu - \phi$. This shows $KE_{\max}$ increases linearly with $\nu$. Therefore, R is correct . The correct answer is Option (3) : A is not correct but R is correct.

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To measure the dynamic resistance ($r_d = \frac{\Delta V}{\Delta I}$) of a p-n junction diode, it must be operated \in forward bias. This requires connecting the p-side to the positive terminal and the n-side to the negative terminal of a variable DC power supply. A voltmeter is connected \in parallel across the diode to measure the voltage ($V$), and a milliammeter is connected \in series to measure the current ($I$). The dynamic resistance is then found by slightly changing the voltage and measuring the corresponding change in current around a specific operating point.

v=72 km/h=20 m/s. t=4s. a=-20/4=-5 m/s². s=ut+½at²=20(4)+½(-5)(16)=80-40=40 m. The answer is $\boxed{40}$.

We need to find how far $m_2$ must move to keep the centre of mass at its original position. The position of the centre of mass is $x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$, and for $x_{\text{cm}}$ to remain unchanged the net displacement must satisfy $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. Here $m_1 = 3$ kg moves towards the centre of mass by $\Delta x_1 = 2$ cm, so we have $3 \times 2 = 2 \times d_2$, giving $d_2 = \frac{6}{2} = 3 \text{ cm}$. Both particles move towards the CM, so both displacements reduce the distance to the CM. The correct answer is 3 cm.

Mercury fills a tube of radius 2 cm to height 30 cm. Find the force on the bottom. The total pressure at the bottom is atmospheric pressure plus the pressure due to the mercury column: $P = P_0 + \rho g h$. Substituting gives $P = 10^5 + (1.36 \times 10^4)(10)(0.3) = 10^5 + 40800 = 140800 \text{ N/m}^2$. Using $r = 2$ cm = 0.02 m, the area of the bottom is $A = \pi r^2 = \frac{22}{7} \times (0.02)^2 = \frac{22}{7} \times 4 \times 10^{-4} = \frac{88}{7} \times 10^{-4} \text{ m}^2$. The force on the bottom is $F = P \times A = 140800 \times \frac{88}{7} \times 10^{-4}$, which simplifies as $F = 140800 \times \frac{88}{70000} = 140800 \times 1.2571 \times 10^{-3}$, and further $F = \frac{140800 \times 88}{70000} = \frac{12390400}{70000} = 177.006 \approx 177 \text{ N}$. The correct answer is 177 N.

Given $x = 10\sin\left(\omega t + \frac{\pi}{3}\right)$ m with $T = 3.14$ s, we want to find the velocity at $t = 0$. First, we find the angular frequency $\omega$ by using $\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} = \frac{2 \times 3.14}{3.14} = 2 \text{ rad/s}$. The velocity is the time derivative of the displacement, so we have $v = \frac{dx}{dt} = 10\omega\cos\left(\omega t + \frac{\pi}{3}\right)$. Evaluating this expression at $t = 0$ gives $v(0) = 10 \times 2 \times \cos\frac{\pi}{3} = 20 \times \frac{1}{2} = 10 \text{ m/s}$. The correct answer is 10 m/s.

Initial capacitance of the parallel-plate capacitor $C = 12.5\ \text{pF} = 12.5 \times 10^{-12}\ \text{F}$ The plates are first charged by a battery to a potential difference $V = 12.0\ \text{V}$ Because the battery is disconnected before inserting the dielectric, the charge $Q$ on the plates remains constant. Step 1 : Calculate the initial charge Using $Q = C V$, $Q = \left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right) = 150 \times 10^{-12}\ \text{C}$ Step 2 : Initial electrostatic energy For a charged capacitor, the energy stored is $U = \frac{1}{2} C V^{2}$ $U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right)^{2}$ $U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\left(144\right)$ $U_i = 9.0 \times 10^{-10}\ \text{J}$ Step 3 : Capacitance after inserting the dielectric Relative permittivity of the slab ${\epsilon_r} = 6$. New capacitance $C' = \epsilon_r\,C = 6 \times 12.5\ \text{pF} = 75\ \text{pF}$ $C' = 75 \times 10^{-12}\ \text{F}$ Step 4 : Final electrostatic energy Charge stays the same, so use $U = \dfrac{Q^{2}}{2C}$: $U_f = \frac{\left(150 \times 10^{-12}\ \text{C}\right)^{2}}{2 \left(75 \times 10^{-12}\ \text{F}\right)}$ $U_f = \frac{2.25 \times 10^{-20}}{150 \times 10^{-12}}$ $U_f = 1.5 \times 10^{-10}\ \text{J}$ Step 5 : Change \in potential energy $\Delta U = U_f - U_i = 1.5 \times 10^{-10}\ \text{J} \;-\; 9.0 \times 10^{-10}\ \text{J}$ $\Delta U = -7.5 \times 10^{-10}\ \text{J}$ The negative sign shows that the energy decreases. Expressing the magnitude \in units of $10^{-12}\ \text{J}$: $|\Delta U| = 7.5 \times 10^{-10}\ \text{J} = 750 \times 10^{-12}\ \text{J}$ Hence, the change \in potential energy is $\mathbf{750} \times 10^{-12}\ \text{J}$ (decrease).

Given same material: Same resistivity ρ\rhoρ Same mass means same volume (since density same) So $A_Al_A=A_Bl_B$ where area $A=\pi r^2$ Since $r_A=2mm,r_B​=4mm$ $\frac{A_A}{A_B}=\frac{2^2}{4^2}=\frac{1}{4}$ Thus $A_Al_A=A_Bl_B$ gives $\frac{1}{4}l_A=l_B$ $l_A=4l_B$ Resistance: $R=\frac{\rho l}{A}$ So $\frac{R_A}{R_B}=\frac{l_A}{l_B}\cdot\frac{A_B}{A_A}$ $4\cdot4=16$ Given $R_B=2\Omega$ therefore $R_A=16(2)=32\Omega$

The magnetic field at a distance $d$ from a long straight wire carrying current $i$ is given by $B = \frac{\mu_0 i}{2\pi d}$. Let the currents be $I_1 = I$ and $I_2 = 2I$.
At point A, the distance from the top wire is $r$ and from the bottom wire is $3r$. The fields add up:
$B_A = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 (2I)}{2\pi (3r)} = \frac{\mu_0 I}{2\pi r} \left(1 + \frac{2}{3}\right) = \frac{5}{3} \frac{\mu_0 I}{2\pi r}$
At point C, the distance from the top wire is $3r$ and from the bottom wire is $r$. The fields also add up:
$B_C = \frac{\mu_0 I}{2\pi (3r)} + \frac{\mu_0 (2I)}{2\pi r} = \frac{\mu_0 I}{2\pi r} \left(\frac{1}{3} + 2\right) = \frac{7}{3} \frac{\mu_0 I}{2\pi r}$
The ratio is $\frac{B_A}{B_C} = \frac{5/3}{7/3} = \frac{5}{7}$. Comparing this with $\frac{x}{7}$, we get $x=5$.

We need to find the potential difference between the two ends of a rod rotating about its perpendicular bisector in a magnetic field parallel to the axis of rotation. Rod length = 60 cm = 0.6 m, angular velocity $\omega = 20$ rad/s, magnetic field $B = 0.5$ T. The rod rotates about its perpendicular bisector, and the magnetic field is parallel to the axis of rotation. When a rod rotates about its center (perpendicular bisector), the EMF induced between the center and one end is: $\varepsilon = \frac{1}{2}B\omega\left(\frac{L}{2}\right)^2 = \frac{B\omega L^2}{8}$ Both halves of the rod generate EMF from center to their respective ends. The EMF from center to each end is the same \in magnitude. Since both halves sweep \in opposite directions relative to the center, the induced EMF from one end to center equals that from center to the other end. End 1 is at higher potential than center by $\frac{B\omega(L/2)^2}{2}$, and end 2 is also at higher potential than center by the same amount. So the potential difference between the two ends = 0. Alternatively: By symmetry, both ends are at the same potential (both are at the same distance from the axis). The rod is symmetric about its center, so the motional EMF from center to end 1 equals the motional EMF from center to end 2. Both ends are at the same potential. The potential difference between the two ends is $0$ V. Therefore, the answer is 0.

The thickness of the glass slab is $t = 4\sqrt{3}\,\text{cm}$ and its refractive index with respect to air is $\mu = \sqrt{2}$. Step 1: Find the critical angle For a glass-air boundary, the critical angle $c$ satisfies $\sin c = \frac{\mu_{\text{air}}}{\mu_{\text{glass}}} = \frac{1}{\sqrt{2}} \quad -(1)$ Hence $c = 45^{\circ}$ because $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$. The question states that the light ray is incident on the slab at an angle equal to this critical angle, so $i = c = 45^{\circ}$. Step 2: Use Snell's law to obtain the refraction angle inside the slab Snell's law for the air-glass interface is $\mu_{\text{air}}\sin i = \mu_{\text{glass}}\sin r \quad -(2)$ With $\mu_{\text{air}} = 1$ and $\mu_{\text{glass}} = \sqrt{2}$, put $i = 45^{\circ}$ into $(2)$: $\sin 45^{\circ} = \sqrt{2}\,\sin r$ $\frac{\sqrt{2}}{2} = \sqrt{2}\,\sin r$ $\sin r = \frac{1}{2}$ Therefore $r = 30^{\circ}$. Step 3: Formula for lateral displacement For a parallel-sided slab, the lateral displacement $d$ is $d = t\,\frac{\sin (i - r)}{\cos r} \quad -(3)$ Step 4: Substitute the known values $i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$ Given $\sin 15^{\circ} = 0.25$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, insert all values into $(3)$: $d = 4\sqrt{3}\;\text{cm}\times \frac{\sin 15^{\circ}}{\cos 30^{\circ}}$ $d = 4\sqrt{3}\;\text{cm}\times \frac{0.25}{\frac{\sqrt{3}}{2}}$ $d = 4\sqrt{3}\;\text{cm}\times \frac{0.25 \times 2}{\sqrt{3}}$ $d = 4\sqrt{3}\;\text{cm}\times \frac{0.5}{\sqrt{3}}$ $d = 4 \times 0.5 \;\text{cm} = 2\;\text{cm}$. Final Answer: The lateral displacement of the ray after emerging from the glass slab is $2\;\text{cm}$.

Find the disintegration energy $Q$ for $^{235}U \rightarrow ^{140}Ce + ^{94}Zr + n$. $Q = (\text{mass of reactants} - \text{mass of products}) \times c^2$ In atomic mass units with $c^2 = 931$ MeV/u: $Q = (M_U - M_{Ce} - M_{Zr} - M_n) \times 931 \text{ MeV}$ $Q = (235.0439 - 139.9054 - 93.9063 - 1.0086) \times 931$ $\Delta m = 235.0439 - 139.9054 - 93.9063 - 1.0086$ $= 235.0439 - 234.8203 = 0.2236 \text{ u}$ $Q = 0.2236 \times 931 = 208.17 \approx 208 \text{ MeV}$ The correct answer is 208 MeV.

We need to identify the incorrect statement about Dalton's Atomic Theory. Dalton's Atomic Theory states: (i) Matter consists of indivisible atoms. (ii) All atoms of a given element have identical properties including identical mass. (iii) Chemical reactions involve reorganization of atoms (atoms are neither created nor destroyed). (iv) Compounds are formed when atoms of different elements combine in fixed, simple whole number ratios . Option 1: "Chemical reactions involve reorganization of atoms" - This is correct per Dalton's theory. Option 2: "Matter consists of indivisible atoms" - This is correct per Dalton's theory. Option 3: "Compounds are formed when atoms of different elements combine in any ratio " - This is INCORRECT. Dalton stated atoms combine in fixed simple ratios, not "any ratio." Option 4: Combines "any ratio" with "identical mass" - while the "any ratio" part is wrong, the "identical mass" part is actually a correct postulate of Dalton. Option 3 directly contradicts Dalton's law of definite proportions by stating "any ratio" instead of "fixed ratio." The correct answer is Option 3 .

We need to find the correct order of first ionization enthalpy for Group 13 elements. The first ionization enthalpies (in kJ/mol, approximate) for Group 13 elements are: B: 801, Al: 577, Ga: 579, In: 558, Tl: 589 The general trend down a group is decreasing IE due to increasing atomic size and shielding. However, anomalies exist: Ga vs Al: Ga (579) has slightly higher IE than Al (577) due to poor shielding by the 3d electrons in Ga, leading to higher effective nuclear charge. Tl vs Ga: Tl (589) has higher IE than Ga (579) due to the poor shielding effect of 4f electrons (lanthanide contraction) and relativistic effects on the 6s electrons. Among the given options, the order $Tl > Ga > Al$ (589 > 579 > 577) is correct. The correct answer is Option C: $Tl > Ga > Al$.

We need to verify two statements about ionization enthalpy and electron gain enthalpy. Na: 496 kJ/mol Li: 520 kJ/mol Cl: 1251 kJ/mol F: 1681 kJ/mol Order: Na (496) < Li (520) < Cl (1251) < F (1681) ✓ Statement I is TRUE. Negative electron gain enthalpy means how exothermic the process of gaining an electron is. A more negative ΔH_eg means a higher tendency to gain an electron. Na: −53 kJ/mol Li: −60 kJ/mol F: −328 kJ/mol Cl: −349 kJ/mol Order of negative electron gain enthalpy (magnitude): Na (53) < Li (60) < F (328) < Cl (349) ✓ Statement II is TRUE. Note: Chlorine has a more negative electron gain enthalpy than fluorine because fluorine's very small size causes high electron-electron repulsion in its compact 2p orbitals. The answer is Option D: Both Statement I and Statement II are true.

We evaluate each statement about hydrogen bonding. A: "H bonding exists when H is covalently bonded to a highly electronegative atom." TRUE. Hydrogen bonds form when H is bonded to F, O, or N (highly electronegative atoms). B: "Intermolecular H bonding is present in o-nitrophenol." FALSE. In o-nitrophenol, intramolecular H bonding occurs between the -OH and the -NO₂ group in the ortho position, forming a six-membered chelate ring. It is the para-isomer that shows intermolecular H bonding. C: "Intramolecular H bonding is present in HF." FALSE. HF shows strong intermolecular hydrogen bonding (not intramolecular). Intramolecular H bonding requires two functional groups within the same molecule. D: "The magnitude of H bonding depends on the physical state of the compound." TRUE. In the solid state, molecules are closer and more ordered, allowing stronger H bonding. In the gas phase, molecules are far apart and H bonding is minimal. E: "H bonding has powerful effect on the structure and properties of compounds." TRUE. H bonding significantly affects boiling points, solubility, viscosity, and molecular structure (e.g., DNA double helix, protein folding). Correct statements: A, D, E. The correct answer is Option (2): A, D, E only .

Count species with pyramidal geometry around the central atom from: $S_2O_3^{2-}$, $SO_4^{2-}$, $SO_3^{2-}$, $S_2O_7^{2-}$. $S_2O_3^{2-}$ (thiosulphate): The central sulfur atom is bonded to 3 oxygen atoms and 1 sulfur atom \in a tetrahedral arrangement (similar to $SO_4^{2-}$). Not pyramidal. $SO_4^{2-}$ (sulphate): S is surrounded by 4 oxygen atoms. Geometry: tetrahedral . Not pyramidal. $SO_3^{2-}$ (sulphite): S has 3 bond pairs (to O atoms) + 1 lone pair. Electron geometry is tetrahedral, but molecular geometry is pyramidal (trigonal pyramidal) . This is pyramidal. $S_2O_7^{2-}$ (disulphate / pyrosulphate): Two $SO_4$ tetrahedra share one oxygen. Each S is \in a tetrahedral environment. Not pyramidal. Only 1 species ($SO_3^{2-}$) has pyramidal geometry. The correct answer is Option (4): 1 .

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The stability of carbanions is determined by several factors. Electron-donating groups (like alkyl groups, +I effect) destabilize the carbanion, while electron-withdrawing groups and resonance stabilize it. The order of stability for simple alkyl carbanions is methyl > primary > secondary > tertiary. Resonance provides significant stabilization. The order $d > c > b > a$ suggests that 'd' is the most stable and 'a' is the least. This can be explained if 'd' and 'c' are resonance-stabilized (e.g., benzyl and allyl carbanions), while 'b' and 'a' are simple alkyl carbanions (e.g., primary and tertiary, respectively).

We need the common name of benzene-1,2-diol. Benzene-1,2-diol has two hydroxyl (-OH) groups on adjacent carbon atoms of the benzene ring (ortho positions). - Benzene-1,2-diol (ortho) = Catechol (also called pyrocatechol) - Benzene-1,3-diol (meta) = Resorcinol - Benzene-1,4-diol (para) = Hydroquinone (quinol) Note: o-Cresol is 2-methylphenol, not a diol. The correct answer is Option (1): Catechol .

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The reagents in option B perform specific chemical transformations.
Reagent "A" is $O_3, Zn/H_2O$, used for reductive ozonolysis. This reaction cleaves a carbon-carbon double bond (C=C) and replaces it with two carbon-oxygen double bonds (C=O), forming aldehydes and/or ketones without further oxidation.
Reagent "B" is $NaOH_{(alc)}/I_2$, which is used for the iodoform test. This test is positive for compounds containing a methyl ketone group ($CH_3-C(=O)-$) or alcohols that can be oxidized to a methyl ketone. A positive test results \in a yellow precipitate of iodoform ($CHI_3$).