Work done \(W\) by a variable force is given by the integral \(\int F \cdot dx\).


Given \(F = 2x + 3x^2\) acting along the x-axis.


The limits of integration are from \(x_1 = 2\) to \(x_2 = 4\).

\(W = \int_{2}^{4} (2x + 3x^2) \, dx\).


Integration yields: \([\frac{2x^2}{2} + \frac{3x^3}{3}]_2^4 = [x^2 + x^3]_2^4\).


Substitute upper limit (\(x=4\)): \((4^2 + 4^3) = 16 + 64 = 80\).


Substitute lower limit (\(x=2\)): \((2^2 + 2^3) = 4 + 8 = 12\).

\(W = 80 - 12 = 68\) J.
Quick Tip: For variable forces, always integrate \(F \cdot dx\). Remember that \(\int x^n dx = \frac{x^{n+1}}{n+1}\).

Let \(n_1 = 1\) mol of O\(_2\) (Diatomic) and \(n_2 = 2\) mol of He (Monoatomic).


For O\(_2\): \(C_{V1} = \frac{5}{2}R\) and \(C_{P1} = \frac{7}{2}R\).


For He: \(C_{V2} = \frac{3}{2}R\) and \(C_{P2} = \frac{5}{2}R\).


The molar heat capacity of the mixture at constant volume is \(C_{V(mix)} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}\).

\(C_{V(mix)} = \frac{1(\frac{5}{2}R) + 2(\frac{3}{2}R)}{1+2} = \frac{2.5R + 3R}{3} = \frac{5.5R}{3} = \frac{11}{6}R\).


The molar heat capacity of the mixture at constant pressure is \(C_{P(mix)} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2}\).

\(C_{P(mix)} = \frac{1(\frac{7}{2}R) + 2(\frac{5}{2}R)}{3} = \frac{3.5R + 5R}{3} = \frac{8.5R}{3} = \frac{17}{6}R\).


The ratio \(\gamma_{mix} = \frac{C_{P(mix)}}{C_{V(mix)}} = \frac{17/6 R}{11/6 R} = \frac{17}{11}\).
Quick Tip: Remember degrees of freedom (\(f\)): Monoatomic \(f=3\), Diatomic \(f=5\). \(C_V = \frac{f}{2}R\) and \(C_P = (1 + \frac{f}{2})R\).

The magnitude of the electric field intensity is directly proportional to the density of the electric field lines.


In the given figure, the field lines are crowded (closer together) near points A and C.


The field lines are spread out (farther apart) near point B.


Therefore, the electric field is stronger at A and C compared to B (\(E_A > E_B\) and \(E_C > E_B\)).


Assuming symmetry in the diagram, the density at A and C appears equal, so \(E_A = E_C\).


Combining these observations: \(E_A = E_C > E_B\).
Quick Tip: Crowded lines \(\Rightarrow\) Strong Field. Spaced out lines \(\Rightarrow\) Weak Field.

The color code sequence is: 1st digit, 2nd digit, Multiplier, Tolerance.


Red corresponds to the digit 2.


Black corresponds to the digit 0.


Blue corresponds to the multiplier \(10^6\).


Gold corresponds to a tolerance of \(\pm 5%\).


Combining these, the resistance is \(20 \times 10^6 \, \Omega\) with \(\pm 5%\) tolerance.


Since \(10^6 \, \Omega = 1 \, M\Omega\), the value is \(20 \, M\Omega \pm 5%\).
Quick Tip: Mnemonic: B B R O Y G B V G W (Black Brown Red Orange Yellow Green Blue Violet Grey White) \(\rightarrow\) 0 1 2 3 4 5 6 7 8 9. Gold=5%, Silver=10%.

Consider a small element \(dl\) of the circular loop placed in a uniform magnetic field \(\vec{B}\) perpendicular to the plane of the loop.
The magnetic force acting on this element is given by \[ dF = I\, dl \, B \]
This force acts radially outward and tends to stretch the loop, producing tension in the wire.

For a circular loop of radius \(r\), the total outward force is balanced by the tension \(T\) in the loop.
The standard result for the tension in a circular current-carrying loop placed in a uniform magnetic field is \[ T = B I r \]

Now, examine each option:


Option (A): If current \(I\) is doubled, \[ T' = B (2I) r = 2T \]
Hence, the tension is doubled. \checkmark

Option (B): If \(B\) is halved, \[ T' = \frac{B}{2} I r = \frac{T}{2} \]
The tension decreases, not doubles. \(\times\)

Option (C): If \(r\) is doubled, \[ T' = B I (2r) = 2T \]
Although this mathematically doubles the tension, doubling the radius changes the physical size (and length) of the loop. The question refers to the same loop, so this is not the intended operational change. \(\times\)

Option (D): If both \(B\) and \(I\) are doubled, \[ T' = (2B)(2I)r = 4T \]
The tension becomes four times, not twice. \(\times\)


Hence, the correct and physically appropriate option is (A) \(I\) is doubled. Quick Tip: The tension \(T\) balances the magnetic force. Derivation from small element \(dl\): \(2T\sin(d\theta/2) = B I (dl) \approx B I (r d\theta) \Rightarrow T = B I r\).

The magnitude of induced emf \(|\varepsilon|\) in a self-inductor is given by \(|\varepsilon| = L \frac{dI}{dt}\).


Given: \(|\varepsilon| = 40\) V.


Change in current \(dI = 2 - 0 = 2\) A.


Time interval \(dt = 0.05\) s.


Substituting the values: \(40 = L \times \frac{2}{0.05}\).

\(40 = L \times 40\).

\(L = 1\) H.
Quick Tip: Use the formula \(\varepsilon = -L \frac{dI}{dt}\). Ignore the negative sign if only magnitude is required.

Refractive index \(n = \frac{Speed in vacuum}{Speed in medium} = \frac{Wavelength in vacuum}{Wavelength in medium}\).

\(\frac{v_{air}}{v_{water}} = \frac{\lambda_{air}}{\lambda_{water}}\).


Given \(\lambda_{air} = 6000\) \AA\ and \(\lambda_{water} = 4500\) \AA.


We assume \(v_{air} \approx c = 3 \times 10^8\) m/s.

\(\frac{3 \times 10^8}{v_{water}} = \frac{6000}{4500} = \frac{4}{3}\).

\(v_{water} = 3 \times 10^8 \times \frac{3}{4}\).

\(v_{water} = 2.25 \times 10^8\) m/s.
Quick Tip: Frequency remains constant when light travels between media. \(v = f\lambda\), so \(v \propto \lambda\).

The energy of a photon emitted during a transition is given by \(E = \frac{hc}{\lambda}\), so \(\lambda \propto \frac{1}{E}\).


Minimum wavelength corresponds to maximum energy difference.


The energy difference is \(\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2})\) eV.


Let's check the energy gaps for the options (proportional to difference of inverse squares):


(A) \(5 \to 4\): \(\frac{1}{16} - \frac{1}{25} \approx 0.02\).


(B) \(4 \to 3\): \(\frac{1}{9} - \frac{1}{16} \approx 0.05\).


(C) \(3 \to 2\): \(\frac{1}{4} - \frac{1}{9} \approx 0.14\).


(D) \(2 \to 1\): \(1 - \frac{1}{4} = 0.75\).


The transition \(n=2\) to \(n=1\) involves the largest energy change, thus the shortest (minimum) wavelength.
Quick Tip: Energy gaps decrease as 'n' increases. The jump to n=1 (Lyman series) always has much higher energy than jumps to higher shells.

Activity \(A = \lambda N\), where \(\lambda\) is the decay constant and \(N\) is the number of nuclei.


Number of nuclei in 1g of Ra-226: \(N = \frac{mass}{molar mass} \times N_A = \frac{1}{226} \times 6.02 \times 10^{23}\).

\(N \approx 2.66 \times 10^{21}\) nuclei.


Given Activity \(A = 4 \times 10^{10}\) decays/s.

\(\lambda = \frac{A}{N} = \frac{4 \times 10^{10}}{2.66 \times 10^{21}} \approx 1.5 \times 10^{-11}\) s\(^{-1}\).


Half-life \(T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.5 \times 10^{-11}}\).

\(T_{1/2} \approx 4.6 \times 10^{10}\) s.
Quick Tip: Remember: Activity = \(\lambda N\). \(T_{1/2} = 0.693 / \lambda\). Don't forget to calculate N using Avogadro's number.

First, simplify the Boolean expression \(Y = A + \bar{A}B + \bar{A}\bar{B}C\).


Using the absorption law (\(X + \bar{X}Y = X + Y\)), we simplify stepwise:


Step 1: \(A + \bar{A}B = A + B\).


Step 2: Substitute back: \(Y = (A + B) + \bar{A}\bar{B}C\).


Since \(\bar{A}\bar{B} = \overline{A+B}\), let \(X = A+B\). Then \(Y = X + \bar{X}C\).


Again using absorption law: \(X + \bar{X}C = X + C = A + B + C\).


So, the expression simplifies to a 3-input OR gate: \(A + B + C\).


To implement an OR operation (\(A+B\)) using universal NAND gates, 3 gates are required.


To implement a 3-input OR (\( (A+B)+C \)) using 2-input NAND gates, we treat (\(A+B\)) as one input and \(C\) as the other, cascading the structure.


Total NAND gates = 3 (for \(A+B\)) + 3 (to OR the result with \(C\)) = 6 gates.
Quick Tip: Simplify the Boolean expression first. Implementing \(A+B\) (OR) takes 3 NANDs. Implementing \(AB\) (AND) takes 2 NANDs.