8 Apr 2024 Shift 2

We need to find the dimensions of $\epsilon_0 E^2$, where $\epsilon_0$ is the permittivity of free space and $E$ is the electric field. The electrostatic energy density stored \in an electric field is $u = \frac{1}{2}\epsilon_0 E^2$. Since $\frac{1}{2}$ is dimensionless, $\epsilon_0 E^2$ has the same dimensions as energy density. Energy density is defined as energy per unit volume, so $[u] = \frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$ Alternatively, we can verify this result by considering the dimensions of $\epsilon_0$ and $E$ separately. From Coulomb's law, $F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$, so $\epsilon_0 = \frac{q^2}{Fr^2} \implies [\epsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^4A^2$ The electric field has dimensions $[E] = \frac{[F]}{[q]} = \frac{MLT^{-2}}{AT} = MLT^{-3}A^{-1}$ Therefore, $[\epsilon_0 E^2] = M^{-1}L^{-3}T^4A^2 \times M^2L^2T^{-6}A^{-2} = ML^{-1}T^{-2}$ This confirms that $[\epsilon_0 E^2] = ML^{-1}T^{-2}$. The correct answer is Option (4): $[ML^{-1}T^{-2}]$.

R = u²sin2θ/g, H = u²sin²θ/(2g). R=H: sin2θ = sin²θ/2 → 2sinθcosθ = sin²θ/2 → 4cosθ = sinθ → tanθ = 4. Option (1): tan⁻¹(4).

We compare the descent of an object along a smooth incline with that along a rough incline, both inclined at 45° and covering the same distance s. In the absence of friction, the component of gravitational acceleration along the plane is $a_s = g\sin 45° = \frac{g}{\sqrt{2}}$, so that from $s = \tfrac{1}{2}a_s t_s^2$ it follows that $t_s = \sqrt{\frac{2s}{a_s}}\,.$ When kinetic friction is present, the net acceleration down the plane becomes $a_r = g\sin 45° - \mu_k g\cos 45° = \frac{g}{\sqrt{2}}\,(1 - \mu_k)\,,$ and the time to travel the same distance s satisfies $t_r = \sqrt{\frac{2s}{a_r}}\,.$ Because the object takes n \times as long on the rough incline, we have $\frac{t_r}{t_s} = n\,.$ Squaring both sides yields $\frac{t_r^2}{t_s^2} = n^2\quad\Longrightarrow\quad \frac{a_s}{a_r} = n^2\,.$ Substituting the expressions for $a_s$ and $a_r$ gives $\frac{g/\sqrt{2}}{(g/\sqrt{2})(1-\mu_k)} = \frac{1}{1-\mu_k} = n^2\,,$ from which $1 - \mu_k = \frac{1}{n^2}\quad\Longrightarrow\quad \mu_k = 1 - \frac{1}{n^2}\,.$ Therefore, the coefficient of kinetic friction is $\mu_k = 1 - \frac{1}{n^2}\,.$

A disc of mass $M$ and radius $R$ rotates with angular velocity $\omega$. When another disc of mass $M/2$ and the same radius is placed gently on it coaxially, no external torque acts on the system and therefore angular momentum is conserved: $L_{\text{before}} = L_{\text{after}}$. The moment of inertia of a disc about an axis through its centre perpendicular to its plane is given by $I = \frac{1}{2}MR^2$. Hence the initial moment of inertia is $I_1 = \frac{1}{2}MR^2$ while the moment of inertia of the second disc is $I_2 = \frac{1}{2} \cdot \frac{M}{2} \cdot R^2 = \frac{MR^2}{4}$. The combined moment of inertia therefore becomes $I_1 + I_2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3MR^2}{4}$. Since angular momentum is conserved, we have $I_1 \omega = (I_1 + I_2)\omega'$. Substituting the expressions for the moments of inertia gives $\frac{MR^2}{2} \cdot \omega = \frac{3MR^2}{4} \cdot \omega'$, which simplifies to $\omega' = \frac{MR^2/2}{3MR^2/4} \cdot \omega = \frac{2}{3}\omega$. Therefore, the new angular velocity is $\frac{2}{3}\omega$.

The kinetic energy at bottom is the potential energy when it was at the top of inclined plane $W_{KE}$= mgh =$5 \times 10 \times (10\sin30^\circ)$ =$250\text{ J}$ Friction acts opposite to the motion of block,so this is negative work $W_{fr}$=$\mu mgd$ =$0.5 \times 5 \times 10 \times 2$ =$50\text{ J}$ the leftover energy now, is stored completely \in spring as final kinetic energy of mass is 0. $W_{spring}$= $\frac{1}{2}(100)x^2$=$50x^2$ 250 - 50 = $50x^2$ 200=$50x^2$ 4=$x^2$ $x$=2

Two satellites A and B with orbital radii 4R and R. Speed of A = 3v. Find speed of B. For a circular orbit, the orbital speed formula is $v = \sqrt{\frac{GM}{r}}$, so $v \propto \frac{1}{\sqrt{r}}$. Using this relation, $\frac{v_B}{v_A}=\sqrt{\frac{r_A}{r_B}}=\sqrt{\frac{4R}{R}}=2$ and hence $v_B=2v_A=2\times3v=6v$. The correct answer is Option (2): 6v .

Let $V_w$ be the volume of ice \in water and $V_k$ be the volume \in kerosene. The total volume of the ice is $V = V_w + V_k$. $\text{Weight of Ice} = \text{Buoyant force (Water)} + \text{Buoyant force (Kerosene)}$ $\rho_i (V_w + V_k)g = \rho_w V_w g + \rho_k V_k g$ Using specific gravities ($\rho_w = 1$, $\rho_i = 0.9$, and $\rho_k = 0.8$): $0.9(V_w + V_k) = 1(V_w) + 0.8(V_k)$ $0.9V_w + 0.9V_k = 1V_w + 0.8V_k$ $0.1V_k = 0.1V_w$ $\frac{V_w}{V_k} = \frac{0.1}{0.1} = \frac{1}{1}$ The ratio of the volume immersed \in water to that \in kerosene is $1 : 1$ .

A diatomic gas ($\gamma=1.4$) does 100 J of work \in isobaric expansion. To find the heat given, recall that at constant pressure $Q = nC_p\Delta T$ and $W = nR\Delta T$. $\frac{Q}{W}=\frac{C_p}{R}=\frac{\gamma R/(\gamma-1)}{R}=\frac{\gamma}{\gamma-1}=\frac{1.4}{0.4}=\frac{7}{2}=3.5$ Therefore, $Q=3.5\times W=3.5\times100=350$ J. The correct answer is Option (3): 350 J .

Statement I: Mean free path is $\lambda =\frac{1}{\sqrt{\ 2}n\pi\ d^2}$ So $\lambda\propto\frac{1}{d^2}$ It is inversely proportional to square of molecular diameter. So Statement I is true. Statement II: Average kinetic energy per molecule is $\frac{3}{2}kT$ So $\text{Average KE}\propto T$ directly proportional to absolute temperature. So Statement II is also true.

Given $y=2\cos 2\pi(330t-x)$ m. The wave equation is $y = A\cos(2\pi ft - 2\pi x/\lambda)$. Comparing with $y=2\cos 2\pi(330t-x)=2\cos(2\pi\cdot330\cdot t-2\pi x)$, we get $2\pi f = 2\pi \times 330 \implies f = 330$ Hz. The correct answer is Option (1): 330 Hz.

Find the dielectric constant of a slab inserted in a parallel plate capacitor. Plate area $A = 12$ cm$^2$, separation $d = 0.6$ cm, and the original capacitance (air) is $C_0 = \frac{\varepsilon_0 A}{d}$. When a dielectric slab of thickness $t = 0.6$ cm fills the gap and one plate is moved by 0.2 cm to maintain the same capacitance, the new separation becomes 0.6 + 0.2 = 0.8 cm. The dielectric slab still has thickness 0.6 cm, and the remaining 0.2 cm is air. The capacitance with the dielectric and air gap is $C = \frac{\varepsilon_0 A}{\frac{t}{\kappa} + d_{\text{air}}} = \frac{\varepsilon_0 A}{\frac{0.6}{\kappa} + 0.2}$. Setting $C = C_0$ gives $\frac{\varepsilon_0 A}{\frac{0.6}{\kappa} + 0.2} = \frac{\varepsilon_0 A}{0.6}$, so $\frac{0.6}{\kappa} + 0.2 = 0.6 \implies \frac{0.6}{\kappa} = 0.4 \implies \kappa = \frac{0.6}{0.4} = 1.5$. The correct answer is Option (4): 1.50 .

Water boils in 20 min. How should the heating element length change to boil in 15 min? Energy to boil water is fixed: $Q = Pt$. For the same $Q$, $P_1 t_1 = P_2 t_2$, which yields $P_2 = P_1 \times \frac{t_1}{t_2} = P_1 \times \frac{20}{15} = \frac{4}{3}P_1$. Since $P = \frac{V^2}{R}$ for the same supply voltage, increasing power requires decreasing resistance, so $R_2 = R_1 \times \frac{P_1}{P_2} = R_1 \times \frac{3}{4}$. Because $R = \frac{\rho L}{A}$ implies $R \propto L$, it follows that $L_2 = \frac{3}{4}L_1$. The length must be decreased to $\frac{3}{4}$ of its original value. The correct answer is Option (1): decreased, 3/4 .

We need to find the ratio of the magnetic field at distances $\frac{a}{2}$ and $2a$ from the axis of a long straight wire of radius $a$ carrying a steady, uniformly distributed current $I$. Recall that Ampere's law states: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ where $I_{enc}$ is the current enclosed by the Amperian loop. Inside the wire at a distance $r$ from the axis, the current is uniformly distributed over the cross-section of radius $a$. The enclosed current is proportional to the enclosed area: $I_{enc} = I \cdot \frac{\pi r^2}{\pi a^2} = I \cdot \frac{r^2}{a^2}$. Applying Ampere's law with a circular Amperian loop of radius $r$ gives $B \cdot 2\pi r = \mu_0 \cdot I \cdot \frac{r^2}{a^2}$, so $B_{inside} = \frac{\mu_0 I r}{2\pi a^2}$. At $r = \frac{a}{2}$, this becomes $B_1 = \frac{\mu_0 I \cdot \frac{a}{2}}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$. Outside the wire at a distance $r$ from the axis, the entire current $I$ is enclosed, so Ampere's law gives $B \cdot 2\pi r = \mu_0 I$ and hence $B_{outside} = \frac{\mu_0 I}{2\pi r}$. At $r = 2a$, this becomes $B_2 = \frac{\mu_0 I}{2\pi \cdot 2a} = \frac{\mu_0 I}{4\pi a}$. The ratio of these fields is $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1}$, so $B_1 : B_2 = 1 : 1$. The correct answer is Option (2): 1 : 1 .

The series circuit consists of a pure resistor $R = 90\,\Omega$ and a coil whose resistance is negligible (so it behaves as a pure inductor with inductive reactance $X_L$). The applied rms supply is $V = 120\,\text{V}$ at frequency $f = 60\,\text{Hz}$. The voltmeter shows the rms voltage across the resistor to be $V_R = 36\,\text{V}$. For a pure resistor, voltage and current are \in phase, so the rms current \in the entire series circuit is $I = \frac{V_R}{R} = \frac{36}{90} = 0.40\,\text{A}$ In an $R$-$L$ series combination, the resistor voltage $V_R$ is \in phase with the current while the inductor voltage $V_L$ leads the current by $90^{\circ}$. Therefore, the supply voltage is the vector (phasor) \sum of $V_R$ and $V_L$, giving the rms relation $V = \sqrt{V_R^{2} + V_L^{2}} \quad -(1)$ Rearranging $(1)$ for $V_L$: $V_L = \sqrt{V^{2} - V_R^{2}} = \sqrt{(120)^{2} - (36)^{2}}$ $V_L = \sqrt{14400 - 1296} = \sqrt{13104} \approx 1.145 \times 10^{2}\,\text{V}$ Thus, $V_L \approx 114\,\text{V}$. The inductive reactance is then obtained from Ohm's law for the inductor: $X_L = \frac{V_L}{I} = \frac{114}{0.40} \approx 285\,\Omega$ For a pure inductor, $X_L = 2\pi f L$. Therefore, the inductance $L$ is $L = \frac{X_L}{2\pi f} = \frac{285}{2\pi \times 60}$ $L = \frac{285}{120\pi} \approx \frac{285}{376.99} \approx 0.756\,\text{H}$ Rounding to two significant figures, $L \approx 0.76\,\text{H}$. Answer: Option B (0.76 H)

The position of the final image is determined by calculating the image formed by each lens sequentially.

For the first convex lens ($f_1 = 10$ cm), the object is at $u_1 = -30$ cm. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, the image distance is $v_1 = 15$ cm. This image forms 15 cm to the right of the first lens.

This image acts as a virtual object for the second, concave lens ($f_2 = -10$ cm). The object distance is $u_2 = +(15 \text{ cm} - 5 \text{ cm}) = +10$ cm. Applying the lens formula, $\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-10} + \frac{1}{10} = 0$, so $v_2 = \infty$. This means the rays emerging from the second lens are parallel.

These parallel rays are incident on the third convex lens ($f_3 = 30$ cm). For a convex lens, parallel incident rays converge at its principal focus. Therefore, the final image is formed at a distance equal to the focal length of the third lens.

The final image is formed at $v_3 = f_3 = 30$ cm to the right of the third lens.

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is momentum.
Kinetic energy $K$ is related to momentum $p$ and mass $m$ by $K = \frac{p^2}{2m}$, which implies $p = \sqrt{2mK}$.

Given that the de Broglie wavelengths are equal:
$\lambda_p = \lambda_e$
$\frac{h}{p_p} = \frac{h}{p_e}$
This means $p_p = p_e$.

Substituting the momentum in terms of kinetic energy and mass:
$\sqrt{2m_p K_p} = \sqrt{2m_e K_e}$
Squaring both sides gives:
$2m_p K_p = 2m_e K_e$
$m_p K_p = m_e K_e$
Therefore, the ratio of kinetic energies is:
$\frac{K_p}{K_e} = \frac{m_e}{m_p}$
Since the mass of a proton ($m_p$) is much greater than the mass of an electron ($m_e$), i.e., $m_p > m_e$, it follows that $\frac{m_e}{m_p} < 1$.
Thus, $K_p < K_e$.

We need to find the nuclear binding energy of the isotope $^{12}_5 B$ (Boron-12), given that $M_o$ is its mass, $M_p$ is the mass of a proton, and $M_n$ is the mass of a neutron. The notation $^{12}_5 B$ tells us that the atomic number (Z) is 5, meaning there are 5 protons , and the mass number (A) is 12, indicating a total of 12 nucleons. Therefore, the number of neutrons is A - Z = 12 - 5 = 7 neutrons . The mass defect ($\Delta m$) is the difference between the total mass of the individual nucleons (protons and neutrons) when separated and the actual mass of the nucleus. This "missing mass" has been converted into binding energy that holds the nucleus together, according to Einstein's mass-energy equivalence. The total mass of individual nucleons is: $m_{nucleons} = 5M_p + 7M_n$ The mass defect is: $\Delta m = m_{nucleons} - M_o = 5M_p + 7M_n - M_o$ Note that $\Delta m$ is positive because the bound nucleus has less mass than the individual nucleons (energy has been released during binding). Using Einstein's relation $E = \Delta m \cdot C^2$, the binding energy is: $\text{Binding Energy} = (5M_p + 7M_n - M_o)C^2$ The correct answer is Option (2) : $(5M_p + 7M_n - M_o)C^2$.

We are given the hypothetical fission reaction: $_{92}X^{236} \rightarrow \, _{56}Y^{141} + \, _{36}Z^{92} + 3R$ We need to identify the emitted particle $R$. Applying conservation of mass number (A), \in any nuclear reaction the total mass number must be conserved (total on left = total on right). On the left side $A = 236$, and on the right side $141 + 92 + 3A_R = 233 + 3A_R$. Setting them equal gives $236 = 233 + 3A_R$, hence $3A_R = 3$ and $A_R = 1$, so each particle $R$ has mass number 1. Applying conservation of atomic number (Z), the total charge (atomic number) must also be conserved. On the left side $Z = 92$, and on the right side $56 + 36 + 3Z_R = 92 + 3Z_R$. Equating these yields $92 = 92 + 3Z_R$, so $3Z_R = 0$ and $Z_R = 0$. Each particle $R$ therefore has atomic number 0 (no charge). A particle with mass number $A = 1$ and atomic number $Z = 0$ is a neutron ($^1_0 n$). This makes physical sense, as neutron emission is very common in nuclear fission reactions. The emitted neutrons can go on to trigger further fission events, which is the basis of a nuclear chain reaction. The correct answer is Option (2): Neutron .

LC = 1/(20N) cm. MSD = 1mm = 1/10 cm. LC = MSD - VSD. VSD = MSD - LC = 1/10 - 1/(20N) = (2N-1)/(20N) cm. N×VSD = x×MSD. x = N×VSD/MSD = N×(2N-1)/(20N)÷(1/10) = (2N-1)/2. Option (3).

We need to find the diameter of a wire measured using a screw gauge that has a positive zero error. Since the least count of a screw gauge is given by $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$, we have determined the smallest measurable length increment. The problem states that with no measuring quantity between the jaws, the zero of the circular scale lies 5 divisions below the reference line. From this we infer a positive zero error, which is calculated as $\text{Zero error} = +5 \times \text{LC} = +5 \times 0.01 = +0.05 \text{ mm}$. Since a positive zero error indicates that the instrument reads more than the actual value, we must subtract it from the measured reading. Next, the observed reading consists of a main scale reading and a circular scale reading. For the main scale reading, 4 linear scale divisions are visible, and since the pitch is 1 mm, each linear division equals 1 mm. Hence $\text{MSR} = 4 \times 1 = 4 \text{ mm}$. The circular scale reading corresponds to the 60th division coinciding with the reference line, giving $\text{CSR} = 60 \times \text{LC} = 60 \times 0.01 = 0.60 \text{ mm}$. Altogether, the observed reading is $\text{Observed reading} = \text{MSR} + \text{CSR} = 4 + 0.60 = 4.60 \text{ mm}$. Finally, applying the zero error correction yields $\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 4.60 - 0.05 = 4.55 \text{ mm}$. The correct answer is Option (3): 4.55 mm .

We need to find the horizontal range of a body thrown horizontally from a tower, given modified initial conditions. Recall the formula for horizontal range in projectile motion (horizontal throw). When a body is thrown horizontally with velocity $v$ from a height $H$, the horizontal range $R$ is given by $R = v \times t$ where $t$ is the time to fall height $H$. Since vertical motion is free fall (initial vertical velocity = 0), we have $H = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2H}{g}}$. Substituting this into the expression for $R$ gives $R = v\sqrt{\frac{2H}{g}}$, showing that the range does not depend on the mass of the body. Applying the formula to the first case, where the mass is $M$, the horizontal velocity is $v$, and the height is $H$, and given that the range is 100 m, we get $100 = v\sqrt{\frac{2H}{g}} \quad \ldots (1)$. In the second case, the mass becomes $2M$ (which is irrelevant since the range is independent of mass), the horizontal velocity is $\frac{v}{2}$, and the height is $4H$. Hence, $R_2 = \frac{v}{2}\sqrt{\frac{2 \times 4H}{g}} = \frac{v}{2}\sqrt{\frac{8H}{g}} = \frac{v}{2} \times 2\sqrt{\frac{2H}{g}} = v\sqrt{\frac{2H}{g}}$. From equation (1), $v\sqrt{\frac{2H}{g}} = 100$ m, so $R_2 = 100 \text{ m}$. The halving of velocity is exactly compensated by the doubling of fall time (due to the four times greater height), resulting in the same horizontal range. The answer is 100 m.

The outward centrifugal force on the ball is $F = m\omega^2r$ $a_r = \omega^2 r$ $a = v \frac{dv}{dr}$ $v \frac{dv}{dr} = \omega^2 r$ Integrate both sides from the point where the ball is placed (r = 1 m) to the edge of the table (r = 3 m): $\int_{0}^{v} v \, dv = \omega^2 \int_{1}^{3} r \, dr$ $\frac{v^2}{2} = \omega^2 \left[ \frac{r^2}{2} \right]_1^3$ $v^2 = \omega^2 (3^2 - 1^2)$ $v^2 = 8\omega^2$ $v = \sqrt{8}\omega = 2\sqrt{2}\omega$ $x = 2$

We are given that small water droplets with radius $r = 0.01$ mm fall with terminal velocity $v_t = 10$ cm/s. When eight such droplets coalesce to form a larger drop, we need to find the new terminal velocity. Recall that for a small sphere falling through a viscous medium, Stokes' law gives the terminal velocity as $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, where $r$ is the radius, $\rho$ is the density of the sphere, $\sigma$ is the density of the medium, $g$ is the acceleration due to gravity, and $\eta$ is the viscosity of the medium. From this expression, it follows that $v_t \propto r^2$. When eight small droplets merge, volume conservation implies $\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$, which simplifies to $R^3 = 8r^3$ and hence $R = 2r$. This shows that the radius of the larger drop is twice that of each small droplet. Since the terminal velocity varies as the square of the radius, the ratio of the new terminal velocity to the original one is $\frac{v'}{v_t} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$. From this it follows that $v' = 4 \times v_t = 4 \times 10 = 40\text{ cm/s}$. The answer is 40 cm/s.

Total energy E = KE+PE = 0.5+0.4 = 0.9 J. E = ½mω²A². ω = 2πf = 2π×25/π = 50 rad/s. 0.9 = ½×0.2×2500×A². A² = 0.0036. A = 0.06 m = 6 cm. The answer is 6.

At point P, net electric field along y-axis is zero. So y-components of fields due to $q_2\ and\ q_3$​ must cancel. Distance of P from charges: For $q_2$​, $r_2=\sqrt{2^{2+}4^2}$ $=\sqrt{20}$ For $q_3$​, $r_3=\sqrt{3^{2+}4^2}$ =5 Electric field magnitude due to a point charge: $E=\frac{kq}{r^2}$ Y-component is $E_y=E\sin⁡\theta$ For $q_2$​, $\sin\theta_2=\frac{4}{\sqrt{20}}$ So $E_{2y}=\frac{kq_2}{20}\cdot\frac{4}{\sqrt{20}}$ For $q_3$​, $\sin\theta_3=\frac{4}{5}$ So $E_{3y}=\frac{kq_3}{25}\cdot\frac{4}{5}$ For cancellation: $|E_{2y}|=|E_{3y}|$ $\frac{|q_2|}{20}\cdot\frac{4}{\sqrt{20}}=\frac{|q_3|}{25}\cdot\frac{4}{5}$ Cancel 4: $\frac{|q_2|}{20\sqrt{20}}=\frac{|q_3|}{125}​$ Thus $\frac{|q_2|}{|q_3|}=\frac{20\sqrt{20}}{125}$ $x=\frac{8}{5\sqrt{\ 5}}$ so x=5

First, calculate the heater's resistance ($R_h$):
$R_h = \frac{V_{rated}^2}{P_{rated}} = \frac{(100 \text{ V})^2}{1000 \text{ W}} = 10 \Omega$
For the heater to operate at $62.5 \text{ W}$, the voltage across it ($V_h$) must be:
$V_h = \sqrt{P_{op} \cdot R_h} = \sqrt{62.5 \text{ W} \cdot 10 \Omega} = \sqrt{625} = 25 \text{ V}$
The voltage drop across the $10 \Omega$ resistor is $V_{10\Omega} = V_{mains} - V_h = 100 \text{ V} - 25 \text{ V} = 75 \text{ V}$.
The total current ($I_{total}$) flowing from the mains is $I_{total} = \frac{V_{10\Omega}}{10 \Omega} = \frac{75 \text{ V}}{10 \Omega} = 7.5 \text{ A}$.
The current through the heater is $I_h = \frac{V_h}{R_h} = \frac{25 \text{ V}}{10 \Omega} = 2.5 \text{ A}$.
The current through resistor $R$ is $I_R = I_{total} - I_h = 7.5 \text{ A} - 2.5 \text{ A} = 5 \text{ A}$.
Finally, the value of $R$ is:
$R = \frac{V_h}{I_R} = \frac{25 \text{ V}}{5 \text{ A}} = 5 \Omega$

We need to find the current required to demagnetise a magnet placed inside a solenoid, given the coercivity of the magnet. Coercivity is the intensity of the magnetising field ($H$) required to reduce the magnetisation of a material to zero after it has been magnetised. In other words, to demagnetise the magnet, we need to apply a magnetic field intensity equal to its coercivity, which is $H = 5 \times 10^3$ A/m. Since the magnetic field intensity inside a solenoid is given by $H = nI = \frac{N}{L} \times I$, where $N$ is the total number of turns, $L$ is the length of the solenoid, and $I$ is the current, we can use this relation to determine the required current. Next, substituting $N = 150$ turns, $L = 30$ cm $= 0.3$ m, and $H = 5 \times 10^3$ A/m into the expression gives $5 \times 10^3 = \frac{150}{0.3} \times I$, which simplifies to $5 \times 10^3 = 500 \times I$. From this, $I = \frac{5 \times 10^3}{500} = 10 \text{ A}$. The answer is 10 A.

We need to find the RMS value of current when an alternating EMF $E = 110\sqrt{2} \sin 100t$ volts is applied to a capacitor of $2 \, \mu F$. Since the general form of an AC voltage is $E = E_0 \sin \omega t$, comparing with the given expression yields the peak EMF $E_0 = 110\sqrt{2}$ V and the angular frequency $\omega = 100$ rad/s. Substituting the peak EMF into the RMS relation gives the RMS voltage. From the equation $E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110 \text{ V}$, we obtain an RMS voltage of 110 V. Next, the capacitive reactance is given by $X_C = \frac{1}{\omega C}$. Substituting $\omega = 100$ rad/s and $C = 2 \, \mu F = 2 \times 10^{-6}$ F into this expression, we find $X_C = \frac{1}{100 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega$. Then the RMS current \in a purely capacitive circuit follows from $I_{rms} = \frac{E_{rms}}{X_C} = \frac{110}{5000} = 0.022 \text{ A} = 22 \text{ mA}$. Therefore, the current is 22 mA. The answer is 22 mA.

Given: slit separation $d = 1 \text{ mm} = 10^{-3} \text{ m}$, screen distance $D = 1 \text{ m}$, wavelength $\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$. Width of central maximum of single-slit diffraction pattern: The angular half-width of the central maximum is $\theta_0 = \frac{\lambda}{w}$. The linear half-width on the screen is $y_0 = \frac{D\lambda}{w}$. The full width of the central maximum is $2y_0 = \frac{2D\lambda}{w}$. Fringe width of double-slit interference pattern: The fringe spacing (distance between consecutive maxima) is $\beta = \frac{D\lambda}{d}$. Number of interference maxima within the central diffraction maximum: The number of double-slit fringes that fit within the central diffraction maximum is: $N = \frac{\text{Width of central maximum}}{\text{Fringe width}} = \frac{2D\lambda / w}{D\lambda / d} = \frac{2d}{w}$ We are told $N = 10$: $\frac{2d}{w} = 10$ $w = \frac{2d}{10} = \frac{d}{5}$ $w = \frac{10^{-3}}{5} = 2 \times 10^{-4} \text{ m}$ The answer is $2$.

This is a potentiometer (potential divider) setup. Total length PR = 20 cm Total voltage = 20 V So potential gradient: $k=\frac{20}{20}=1\text{ V/cm}$ step 1: condition for LED to just glow LED forward voltage = 1.8 V Zener breakdown = 3.2 V Both are \in series (\in conduction path), so total required: $V=1.8+3.2=5\text{ V}$ step 2: required length $l=\frac{V}{k}=\frac{5}{1}=5\text{ cm}$

We need to identify the correct statements about p-block elements and their compounds from the given list. Analysis of each statement: (A) Non-metals have higher electronegativity than metals. This is TRUE . Electronegativity is the tendency of an atom to attract shared electrons towards itself. Non-metals, being on the right side of the periodic table, have higher electronegativity because they need fewer electrons to complete their octet and have smaller atomic sizes. (B) Non-metals have lower ionisation enthalpy than metals. This is FALSE . Non-metals generally have higher ionisation enthalpy than metals. Non-metals hold their electrons more tightly due to their smaller atomic size and higher effective nuclear charge, making it harder to remove an electron. (C) Compounds formed between highly reactive non-metals and highly reactive metals are generally ionic. This is TRUE . Highly reactive metals (like Na, K) have very low ionisation enthalpies and readily lose electrons, while highly reactive non-metals (like F, Cl) have high electron affinity and readily gain electrons. The large electronegativity difference between them leads to the formation of ionic bonds through complete transfer of electrons. (D) The non-metal oxides are generally basic in nature. This is FALSE . Non-metal oxides are generally acidic in nature. For example, $CO_2$ forms carbonic acid, $SO_3$ forms sulphuric acid, $P_4O_{10}$ forms phosphoric acid when dissolved \in water. (E) The metal oxides are generally acidic or neutral \in nature. This is FALSE . Metal oxides are generally basic \in nature. For example, $Na_2O$, $CaO$, $MgO$ are all basic oxides that form hydroxides when dissolved in water. The correct statements are (A) and (C) only. The correct answer is Option (2): (A) and (C) only .

We need to determine the shape of a carbocation. Understand the electronic structure of a carbocation A carbocation is a positively charged carbon species, such as $CH_3^+$. The carbon atom \in a carbocation has only 6 electrons (3 bonding pairs) instead of the usual 8. It has: - 3 bond pairs (from three bonds to hydrogen or other groups) - 0 lone pairs - 1 empty (vacant) p-orbital Determine the hybridisation With only 3 regions of electron density (3 bond pairs and no lone pairs), the carbon \in a carbocation undergoes sp$^2$ hybridisation . The three sp$^2$ hybrid orbitals form bonds, while the unhybridised p-orbital remains empty and is perpendicular to the plane of the three bonds. Determine the geometry According to VSEPR theory and the sp$^2$ hybridisation, three bond pairs with no lone pairs arrange themselves at $120°$ angles \in a plane to minimise repulsion. This gives a trigonal planar geometry. The bond angles are all $120°$, and all three bonds lie in the same plane with the carbon atom at the centre. The correct answer is Option (2): trigonal planar .

We need to determine the mathematical representation of an antibonding sigma molecular orbital ($\sigma^*$). Recall the LCAO (Linear Combination of Atomic Orbitals) method When two atomic orbitals $\psi_A$ and $\psi_B$ combine, they form two molecular orbitals: - Bonding MO ($\sigma$) : formed by constructive (\in-phase) combination - Antibonding MO ($\sigma^*$) : formed by destructive (out-of-phase) combination Write the mathematical expressions The bonding molecular orbital is formed by the addition of the two wave functions: $\sigma = \psi_A + \psi_B$ (with a normalisation constant, which is omitted here for simplicity) The antibonding molecular orbital is formed by the subtraction of the two wave functions: $\sigma^* = \psi_A - \psi_B$ Understand the physical meaning In the bonding MO, the wave functions reinforce each other \in the internuclear region, leading to increased electron density between the nuclei and a stable bond. In the antibonding MO ($\sigma^*$), the wave functions cancel \in the internuclear region, creating a node (region of zero electron density) between the nuclei. This results \in a destabilising effect that weakens bonding. The correct answer is Option (2): $\psi_A - \psi_B$ .

We need to determine in which medium the equilibrium $Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$ shifts to the right. Write the complete ionic equilibrium The interconversion between dichromate and chromate ions \in aqueous solution involves $H^+$ and $OH^-$ ions. The complete equation is: $Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$ Alternatively, \in terms of $H^+$: $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$ Apply Le Chatelier's Principle The forward reaction (dichromate to chromate) consumes $OH^-$ ions. According to Le Chatelier's Principle: - In a basic medium (excess $OH^-$): The high concentration of $OH^-$ drives the equilibrium to the right , favouring the formation of yellow $CrO_4^{2-}$ (chromate) ions. - In an acidic medium (excess $H^+$): The $OH^-$ concentration is low, so the equilibrium shifts to the left , favouring the orange $Cr_2O_7^{2-}$ (dichromate) ions. Conclusion This is why dichromate solutions appear orange in acidic conditions and turn yellow (chromate) in basic conditions. The equilibrium shifts to the right in a basic medium. The correct answer is Option (2): a basic medium .