JEE Main 2026 April 4 Shift 1 Physics Question Paper with Solutions PDF

Concept:

For a soap bubble, there are two surfaces (inner and outer). Therefore, the work done in expanding the bubble equals the increase in surface energy of both surfaces.

Surface energy: \[ U = 2 \gamma A \]

Work done in expansion: \[ W = \Delta U = 2\gamma (A_2 - A_1) \]

Area of a sphere: \[ A = 4\pi r^2 \]

Thus, \[ W = 2\gamma \left[4\pi r_2^2 - 4\pi r_1^2 \right] \]

Step 1: Write the expression for work done.
\[ W = 2\gamma \left[4\pi (r_2^2 - r_1^2)\right] \]
\[ W = 8\pi\gamma (r_2^2 - r_1^2) \]

Step 2: Substitute the given values.
\[ \gamma = 7.2 \times 10^{-2}\, N/m \]
\[ r_1 = 1\,cm = 1\times10^{-2}\,m \]
\[ r_2 = 2\,cm = 2\times10^{-2}\,m \]
\[ W = 8\pi \times 7.2 \times 10^{-2} \left[(2\times10^{-2})^2 - (1\times10^{-2})^2\right] \]

Step 3: Simplify the expression.
\[ (2\times10^{-2})^2 = 4\times10^{-4} \]
\[ (1\times10^{-2})^2 = 1\times10^{-4} \]
\[ r_2^2 - r_1^2 = 3\times10^{-4} \]
\[ W = 8\pi \times 7.2 \times 10^{-2} \times 3\times10^{-4} \]
\[ W = 542.6 \times 10^{-6}\,J \] Quick Tip: For a \textbf{soap bubble}, two surfaces exist (inside and outside), so the surface energy is \(U = 2\gamma A\). Always multiply by \(2\) when calculating work done due to surface tension in soap bubbles.

Concept:




For motion down an inclined plane:

Acceleration on a smooth incline \[ a = g\sin\theta \]

Acceleration on a rough incline \[ a = g(\sin\theta - \mu\cos\theta) \]

If a body starts from rest and moves a distance \(S\), \[ S = \frac{1}{2}at^2 \]

Thus, \[ t = \sqrt{\frac{2S}{a}} \]

Hence time is inversely proportional to the square root of acceleration.
\[ t \propto \frac{1}{\sqrt{a}} \]

Step 1: Acceleration of block \(A\) (smooth plane).
\[ a_A = g\sin45^\circ \]
\[ a_A = \frac{g}{\sqrt{2}} \]

Step 2: Acceleration of block \(B\) (rough plane).
\[ a_B = g(\sin45^\circ - \mu\cos45^\circ) \]
\[ a_B = g\left(\frac{1}{\sqrt{2}} - \mu\frac{1}{\sqrt{2}}\right) \]
\[ a_B = \frac{g}{\sqrt{2}}(1-\mu) \]

Step 3: Using the relation between time and acceleration.
\[ t \propto \frac{1}{\sqrt{a}} \]
\[ \frac{t_B}{t_A} = \sqrt{\frac{a_A}{a_B}} \]

Given \(B\) takes \(50%\) more time:
\[ t_B = 1.5\, t_A \]
\[ \frac{t_B}{t_A} = \frac{3}{2} \]

Step 4: Substitute accelerations.
\[ \frac{3}{2} = \sqrt{\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}}(1-\mu)}} \]
\[ \frac{3}{2} = \sqrt{\frac{1}{1-\mu}} \]

Step 5: Solve for \(\mu\).
\[ \left(\frac{3}{2}\right)^2 = \frac{1}{1-\mu} \]
\[ \frac{9}{4} = \frac{1}{1-\mu} \]
\[ 1-\mu = \frac{4}{9} \]
\[ \mu = \frac{5}{9} \] Quick Tip: For bodies starting from rest on an inclined plane, \(S=\frac{1}{2}at^2\). If distances are same, \(t \propto \frac{1}{\sqrt{a}}\). This relation greatly simplifies comparison problems involving different accelerations.

Concept:

For mirrors, the mirror formula relates object distance \(u\), image distance \(v\), and focal length \(f\):
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]

Using the mirror sign convention:

Object distance \(u\) is negative.
Focal length \(f\) is negative for a concave mirror.


The image length of the rod equals the difference between the image positions of its two ends.

Step 1: Image position of point \(A\).

Distance of \(A\) from mirror: \[ u_A = -20\,cm, \quad f = -10\,cm \]

Using mirror formula:
\[ \frac{1}{v_A} + \frac{1}{u_A} = \frac{1}{f} \]
\[ \frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10} \]
\[ \frac{1}{v_A} = -\frac{1}{10} + \frac{1}{20} \]
\[ \frac{1}{v_A} = -\frac{1}{20} \]
\[ v_A = -20\,cm \]

Step 2: Image position of point \(B\).

Since \(AB = 10\,cm\),
\[ u_B = -30\,cm \]

Applying mirror formula:
\[ \frac{1}{v_B} + \frac{1}{u_B} = \frac{1}{f} \]
\[ \frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10} \]
\[ \frac{1}{v_B} = -\frac{1}{10} + \frac{1}{30} \]
\[ \frac{1}{v_B} = -\frac{2}{30} \]
\[ v_B = -15\,cm \]

Step 3: Length of the image of the rod.
\[ Image length = |v_A - v_B| \]
\[ = |-20 - (-15)| \]
\[ = 5\,cm \]

Thus, the length of the image of the rod is:
\[ \boxed{5\,cm} \] Quick Tip: When an extended object is placed along the principal axis of a mirror, calculate the image position of both ends separately using the mirror formula. The difference between their image positions gives the length of the image.

Concept:

Energy released in a nuclear reaction equals the difference between the total binding energy of the products and the total binding energy of the reactants.
\[ Q = BE_{products} - BE_{reactants} \]

Binding energy of a nucleus:
\[ BE = (Binding energy per nucleon) \times (Number of nucleons) \]

Step 1: Calculate total binding energy of reactants.

For nucleus \(A^3\):
\[ BE = 3 \times 3 = 9\,MeV \]

Since there are two \(A\) nuclei:
\[ BE_{A} = 2 \times 9 = 18\,MeV \]

For nucleus \(B^4\):
\[ BE_{B} = 4 \times 7 = 28\,MeV \]

Total binding energy of reactants:
\[ BE_{LHS} = 18 + 28 = 46\,MeV \]

Step 2: Calculate total binding energy of product.

For nucleus \(C^{10}\):
\[ BE_{RHS} = 10 \times 6 = 60\,MeV \]

Step 3: Calculate energy released.
\[ Q = BE_{RHS} - BE_{LHS} \]
\[ Q = 60 - 46 \]
\[ Q = 14\,MeV \] Quick Tip: Energy released in nuclear reactions is calculated using binding energies: \(Q = BE_{products} - BE_{reactants}\). If the result is positive, energy is released.

Concept:

Power delivered by a force is given by the dot product of force and velocity.
\[ P = \vec{F} \cdot \vec{v} \]

Acceleration is related to force by Newton's second law:
\[ \vec{a} = \frac{\vec{F}}{m} \]

Velocity can be obtained by integrating acceleration.

Step 1: Find acceleration.
\[ \vec{a} = \frac{\vec{F}}{m} \]
\[ \vec{a} = \frac{2t}{2}\hat{i} + \frac{3t^2}{2}\hat{j} \]
\[ \vec{a} = t\,\hat{i} + \frac{3t^2}{2}\hat{j} \]

Step 2: Find velocity by integrating acceleration.
\[ \vec{v} = \int \vec{a}\,dt \]
\[ \vec{v} = \int t\,dt\,\hat{i} + \int \frac{3t^2}{2}\,dt\,\hat{j} \]
\[ \vec{v} = \frac{t^2}{2}\hat{i} + \frac{t^3}{2}\hat{j} \]

Step 3: Compute power.
\[ P = \vec{F} \cdot \vec{v} \]
\[ P = (2t\,\hat{i} + 3t^2\,\hat{j}) \cdot \left(\frac{t^2}{2}\hat{i} + \frac{t^3}{2}\hat{j}\right) \]
\[ P = t^3 + \frac{3}{2}t^5 \]

Step 4: Substitute \(t = 2\,s\).
\[ P = 2^3 + \frac{3}{2}(2^5) \]
\[ P = 8 + \frac{3}{2}\times 32 \]
\[ P = 8 + 48 \]
\[ P = 56\,W \] Quick Tip: Instantaneous power delivered by a force is \(P = \vec{F}\cdot\vec{v}\). First find velocity by integrating acceleration obtained from \( \vec{F} = m\vec{a} \), then take the dot product.

Concept:

The horizontal range of a projectile launched with speed \(u\) at an angle \(\theta\) with the horizontal is given by
\[ R = \frac{u^2 \sin 2\theta}{g} \]

Thus, when the initial speeds are the same, the ratio of ranges depends only on \(\sin 2\theta\).

Step 1: Write the range expressions for both projectiles.
\[ R_A = \frac{u^2 \sin 2\theta_A}{g} \]
\[ R_B = \frac{u^2 \sin 2\theta_B}{g} \]

Step 2: Take the ratio of ranges.
\[ \frac{R_A}{R_B} = \frac{\frac{u^2 \sin 2\theta_A}{g}} {\frac{u^2 \sin 2\theta_B}{g}} \]
\[ \frac{R_A}{R_B} = \frac{\sin 2\theta_A}{\sin 2\theta_B} \]

Step 3: Substitute the given angles.
\[ \theta_A = 15^\circ, \qquad \theta_B = 30^\circ \]
\[ \frac{R_A}{R_B} = \frac{\sin 30^\circ}{\sin 60^\circ} \]
\[ = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \]
\[ \frac{R_A}{R_B} = \frac{1}{\sqrt{3}} \] Quick Tip: For projectiles with the same initial speed, the range depends on \( \sin 2\theta \). Angles \( \theta \) and \( (90^\circ - \theta) \) produce the same range.

Concept:

When several cells with internal resistances are connected in parallel, the equivalent emf and equivalent resistance are given by
\[ E_{eq} = \frac{\sum \frac{E_i}{r_i}}{\sum \frac{1}{r_i}} \]
\[ r_{eq} = \frac{1}{\sum \frac{1}{r_i}} \]

This allows the complex network between \(A\) and \(B\) to be reduced to a single equivalent source.

Step 1: Write the equivalent emf expression.
\[ E_{eq} = \frac{\frac{27}{3}+\frac{27}{3}+\frac{27}{3}+\frac{0}{6}+\frac{27}{3}} {\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{3}} \]
\[ E_{eq} = \frac{36}{9/6} \]
\[ E_{eq} = 24\,V \]

Thus,
\[ V_{AB} = 24\,V \]

Step 2: Find equivalent resistance.
\[ r_{eq} = \frac{1} {\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{3}} \]
\[ r_{eq} = \frac{2}{3}\,\Omega \]

Step 3: Find current \(i_{BA}\).

The external resistor between the equivalent source and point \(B\) is \(3\,\Omega\).

Total resistance:
\[ R_{total} = 3 + \frac{2}{3} \]

Using Ohm's law:
\[ i_{BA} = \frac{E_{eq}}{R_{total}} \]
\[ i_{BA} = \frac{24}{3+\frac{2}{3}} \]
\[ i_{BA} = \frac{24}{\frac{11}{3}} \]
\[ i_{BA} \approx 1\,A \]

Thus,
\[ V_{AB} = 24\,V, \qquad i_{BA} = 1\,A \] Quick Tip: For cells connected in parallel with internal resistances, use \(E_{eq} = \frac{\sum (E/r)}{\sum (1/r)}\) and \(r_{eq} = \frac{1}{\sum (1/r)}\). This converts a complicated network into a single equivalent source.

Concept:

For a series AC circuit, the power factor is
\[ \cos\phi = \frac{R}{Z} \]

where \(Z\) is the impedance. Also,
\[ \tan\phi = \frac{|X_L - X_C|}{R} \]

These relations allow determination of the net reactance.

Step 1: Use the given power factor.
\[ \cos\phi = \frac{1}{2} \]
\[ \phi = 60^\circ \]

Step 2: Use the tangent relation.
\[ \tan\phi = \frac{|X_L - X_C|}{R} \]
\[ \tan 60^\circ = \frac{|X_L - X_C|}{100} \]
\[ \sqrt{3} = \frac{|X_L - X_C|}{100} \]

Step 3: Calculate net reactance.
\[ |X_L - X_C| = 100\sqrt{3} \]

Given
\[ |X_L - X_C| = \alpha\sqrt{3} \]

Thus,
\[ \alpha = 100 \] Quick Tip: In a series AC circuit, power factor relations are very useful: \( \cos\phi = R/Z \) and \( \tan\phi = (X_L - X_C)/R \). These directly connect reactance with resistance.

Concept:

Moment of inertia formulas:
\[ I_{disc} = \frac{1}{2}MR^2 \]
\[ I_{sphere} = \frac{2}{5}MR^2 \]

Mass and volume relations are used to determine the radius of the smaller sphere.

Step 1: Find density of the original sphere.
\[ M = \frac{4}{3}\pi R^3 \rho \]
\[ \rho = \frac{3M}{4\pi R^3} \]

Step 2: Find radius of smaller sphere of mass \(M/8\).
\[ \frac{M}{8} = \frac{4}{3}\pi r^3 \rho \]

Substitute \(\rho\):
\[ \frac{M}{8} = \frac{4}{3}\pi r^3 \left(\frac{3M}{4\pi R^3}\right) \]
\[ \frac{M}{8} = M\frac{r^3}{R^3} \]
\[ r^3 = \frac{R^3}{8} \]
\[ r = \frac{R}{2} \]

Step 3: Moment of inertia of disc formed from mass \(7M/8\).
\[ I_1 = \frac{1}{2}\left(\frac{7M}{8}\right)(2R)^2 \]
\[ I_1 = \frac{7MR^2}{4} \]

Step 4: Moment of inertia of smaller sphere.
\[ I_2 = \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 \]
\[ I_2 = \frac{MR^2}{80} \]

Step 5: Find the ratio.
\[ \frac{I_1}{I_2} = \frac{\frac{7MR^2}{4}}{\frac{MR^2}{80}} \]
\[ \frac{I_1}{I_2} = 140 \] Quick Tip: When objects are reshaped, mass remains constant but geometry changes. Use density and volume relations to determine new dimensions before applying moment of inertia formulas.

Concept:

Extension of a wire under load is given by
\[ \Delta L = \frac{FL}{AY} \]

where \(F\) = applied force, \(L\) = length of wire, \(A\) = cross-sectional area, \(Y\) = Young's modulus.

For rods connected in series, total extension equals the sum of individual extensions.
\[ \Delta \ell = \Delta \ell_1 + \Delta \ell_2 \]

Step 1: Write extension expressions for both rods.
\[ \Delta \ell_1 = \frac{MgL_1}{A\gamma} \]
\[ \Delta \ell_2 = \frac{MgL_2}{A(2\gamma)} \]

Step 2: Add the extensions.
\[ \Delta \ell = \frac{MgL}{A\gamma} + \frac{Mg(2L)}{A(2\gamma)} \]
\[ \Delta \ell = \frac{Mg}{A}\left(\frac{L}{\gamma}+\frac{2L}{2\gamma}\right) \]
\[ \Delta \ell = \frac{2MgL}{A\gamma} \]

Step 3: Substitute numerical values.

Mass \(m = 0.8\,kg\)
\[ A = \pi R^2 \]
\[ R = 0.2\,mm = 2\times10^{-4}\,m \]
\[ A = \pi (2\times10^{-4})^2 \]

Substitute:
\[ \Delta \ell = \frac{0.8\times10}{\pi\times4\times10^{-6}} \left(\frac{2\times0.314}{2\times10^9}\right) \]
\[ \Delta \ell = 2\times10^{-4}\,m \]
\[ \Delta \ell = 0.2\,mm \] Quick Tip: For rods connected in series under the same load, total extension is simply the sum of individual extensions: \(\Delta L_{total} = \Delta L_1 + \Delta L_2\).

Concept:

According to Einstein’s photoelectric equation,
\[ K_{max} = eV_s = hf - \phi \]

where \(K_{max}\) = maximum kinetic energy of emitted electrons, \(V_s\) = stopping potential, \(h\) = Planck’s constant, \(f\) = frequency of incident light, \(\phi\) = work function of the metal.

The threshold frequency \(f_0\) is related to work function as:
\[ \phi = hf_0 \]

Thus,
\[ K_{max} = h(f - f_0) \]

For a fixed incident frequency, the metal with the lowest threshold frequency (or lowest work function) will give the maximum kinetic energy.

Step 1: Interpret the graph.

In a stopping potential vs frequency graph:


All metals have the same slope \( \frac{h}{e} \).
The intercept on the frequency axis represents the threshold frequency.


Step 2: Identify the smallest threshold frequency.

From the graph, the line corresponding to \(X_1\) intersects the frequency axis first.

Thus,
\[ f_{0(X_1)} < f_{0(X_2)} < f_{0(X_3)} \]

Step 3: Determine which metal gives maximum kinetic energy.

Since
\[ K_{max} = h(f - f_0) \]

smaller \(f_0\) gives larger \(K_{max}\).

Therefore,
\[ X_1 produces the maximum kinetic energy. \] Quick Tip: In a stopping potential vs frequency graph, the slope is constant \(h/e\). The metal with the \textbf{smallest threshold frequency} (leftmost intercept) produces electrons with the \textbf{maximum kinetic energy}.

Concept:

Capacitance of a parallel plate capacitor is given by
\[ C = \frac{\varepsilon_0 A}{x} \]

where \(x\) is the separation between the plates.

Energy stored in a capacitor connected to a constant voltage source is
\[ U = \frac{1}{2}CV^2 \]

Since the capacitor remains connected to the battery, the voltage \(V\) remains constant.

Step 1: Substitute the capacitance expression into energy formula.
\[ U = \frac{1}{2}\left(\frac{\varepsilon_0 A}{x}\right)V^2 \]
\[ U = \frac{1}{2}\frac{\varepsilon_0 A V^2}{x} \]

Step 2: Differentiate with respect to time.
\[ \frac{dU}{dt} = \frac{1}{2}\varepsilon_0 A V^2 \frac{d}{dt}\left(\frac{1}{x}\right) \]
\[ \frac{dU}{dt} = \frac{1}{2}\varepsilon_0 A V^2 \left(-\frac{1}{x^2}\right)\frac{dx}{dt} \]

Step 3: Use constant velocity condition.

Since the plate moves with constant velocity,
\[ \frac{dx}{dt} = constant \]

Thus,
\[ \frac{dU}{dt} \propto \frac{1}{x^2} \] Quick Tip: For a capacitor connected to a battery, voltage remains constant. Since \(C = \frac{\varepsilon_0 A}{x}\), the stored energy \(U = \frac{1}{2}CV^2\) varies inversely with plate separation.

Concept:

From the circuit:


Input \(B\) first passes through a NOT gate giving \(\overline{B}\).
The output then enters a NAND gate with \(A\).


Thus,
\[ Y = \overline{A \cdot \overline{B}} \]

Using De Morgan’s theorem:
\[ Y = \overline{A} + B \]

Step 1: Write the given binary numbers.
\[ A = 1101 \]
\[ B = 1010 \]

Step 2: Find complement of \(B\).
\[ \overline{B} = 0101 \]

Step 3: Apply the Boolean expression.
\[ Y = A + \overline{B} \]

Performing OR operation:

Quick Tip: A NAND gate with one complemented input often simplifies using De Morgan’s theorem. \(\overline{A\cdot B} = \overline{A} + \overline{B}\).

Concept:

Resolving power of a telescope is given by Rayleigh’s criterion:
\[ \theta = \frac{1.22\lambda}{D} \]

where \(\theta\) = angular resolution, \(\lambda\) = wavelength of light, \(D\) = diameter of the objective lens.

Step 1: Substitute the given values.
\[ \theta = 5\times10^{-7}\,rad \]
\[ \lambda = 500\,nm = 500\times10^{-9}\,m \]
\[ 5\times10^{-7} = \frac{1.22\times500\times10^{-9}}{D} \]

Step 2: Solve for \(D\).
\[ D = \frac{1.22\times500\times10^{-9}}{5\times10^{-7}} \]
\[ D = 1.22\,m \] Quick Tip: Rayleigh’s criterion for telescope resolution is \(\theta = \frac{1.22\lambda}{D}\). Smaller \(\theta\) (better resolution) requires a larger aperture \(D\).

Concept:

For an ideal gas,
\[ PV=nRT \]

Here \(n=2\). Thus temperature can be found from
\[ T=\frac{PV}{2R} \]

Step 1: Find temperature at state \(A\).

At \(V=V_0\),
\[ P=\frac{P_0}{1+(V_0/V_0)^2} =\frac{P_0}{2} \]

Using ideal gas equation:
\[ \frac{P_0}{2}\times V_0 = 2R\,T_A \]
\[ T_A=\frac{P_0V_0}{4R} \]

Step 2: Find temperature at state \(B\).

At \(V=3V_0\),
\[ P=\frac{P_0}{1+\left(\frac{V_0}{3V_0}\right)^2} =\frac{P_0}{1+\frac{1}{9}} =\frac{9P_0}{10} \]

Using ideal gas equation:
\[ \frac{9P_0}{10}\times3V_0 =2R\,T_B \]
\[ T_B=\frac{27P_0V_0}{20R} \]

Step 3: Find temperature difference.
\[ T_B-T_A= \frac{27P_0V_0}{20R}-\frac{P_0V_0}{4R} \]
\[ =\frac{27P_0V_0}{20R}-\frac{5P_0V_0}{20R} \]
\[ =\frac{11P_0V_0}{10R} \] Quick Tip: Whenever pressure is given as a function of volume, substitute the specific volume into the relation first to obtain \(P\), then use the ideal gas law \(PV=nRT\) to find the temperature.

Concept:

For single slit diffraction, the position of minima on the screen is
\[ y_n=\frac{nD\lambda}{a}, \qquad n=1,2,3,... \]

Step 1: Position of the first minima.
\[ y_1=\frac{D\lambda}{a} \]

Step 2: Position of the third minima.
\[ y_3=\frac{3D\lambda}{a} \]

Step 3: Find the separation.
\[ Separation=y_3-y_1 \]
\[ =\frac{3D\lambda}{a}-\frac{D\lambda}{a} \]
\[ =\frac{2D\lambda}{a} \] Quick Tip: For single slit diffraction minima, use \(y_n=\frac{nD\lambda}{a}\). The separation between two minima is simply the difference of their positions.

Concept:

Least count of a screw gauge is given by
\[ Least Count = \frac{Pitch}{Number of divisions on circular scale} \]

Pitch is the linear distance moved by the screw in one complete rotation.

Step 1: Find the pitch.

Given 5 rotations move the screw by \(2.5\,mm\):
\[ Pitch=\frac{2.5}{5} \]
\[ Pitch=0.5\,mm \]
\[ =5\times10^{-4}\,m \]

Step 2: Calculate the least count.

Number of divisions on circular scale \(=100\)
\[ L.C.=\frac{Pitch}{Number of divisions} \]
\[ =\frac{5\times10^{-4}}{100}\,m \]
\[ =5\times10^{-6}\,m \]

Step 3: Convert into millimetres.
\[ 5\times10^{-6}\,m=5\times10^{-3}\,mm \] Quick Tip: For screw gauges, always remember: Pitch \(=\) distance moved in one rotation, and Least Count \(=\frac{Pitch}{Number of circular scale divisions}\).

Concept:

Magnetic moment of a current loop is
\[ M = I \times A \]

For a spiral coil with continuously varying radius, the total magnetic moment is obtained by integrating the contribution of small circular loops.



Step 1: Magnetic moment of a small ring.

For a small ring of radius \(r\),
\[ dM = I\,dA \]
\[ dM = I(\pi r^2) \]

If the number of turns changes continuously,
\[ dM = I(dN)\pi r^2 \]

Step 2: Relate \(dN\) with radius.

Since \(N\) turns are distributed between \(r_i\) and \(r_{ext}\),
\[ dN=\frac{N}{r_{ext}-r_i}\,dr \]

Substitute:
\[ dM = I\pi r^2 \frac{N}{r_{ext}-r_i}dr \]

Step 3: Integrate between the limits.
\[ M = \int_{r_i}^{r_{ext}} I\pi r^2 \frac{N}{r_{ext}-r_i}dr \]
\[ M = \frac{I\pi N}{r_{ext}-r_i}\int_{r_i}^{r_{ext}} r^2 dr \]
\[ M = \frac{I\pi N}{r_{ext}-r_i}\left[\frac{r^3}{3}\right]_{r_i}^{r_{ext}} \]
\[ M = \frac{I\pi N}{3(r_{ext}-r_i)}(r_{ext}^3-r_i^3) \]

Step 4: Substitute numerical values.
\[ I = 20\times10^{-3}\,A \]
\[ N = 200 \]
\[ r_{ext}=6\times10^{-2}\,m \]
\[ r_i=3\times10^{-2}\,m \]
\[ M= \frac{20\times10^{-3}\times3.14\times200}{3(6-3)\times10^{-2}} (6^3-3^3)\times10^{-6} \]
\[ M = 2.64\times10^{-2}\,A m^2 \] Quick Tip: Magnetic moment of a loop is \(M=IA\). For spiral coils with varying radius, integrate \(I\pi r^2\) over the turns.

Concept:

Work done by a force is given by the dot product of force and displacement.
\[ W = \vec{F}\cdot\vec{S} \]

If multiple forces act simultaneously, the resultant force is
\[ \vec{F}_{net}=\vec{F}_1+\vec{F}_2 \]

Step 1: Find the resultant force.
\[ \vec{F}_{net} = (3+8)\hat{i}+(-5+2)\hat{j}+(2-3)\hat{k} \]
\[ \vec{F}_{net}=11\hat{i}-3\hat{j}-\hat{k} \]

Step 2: Find the displacement vector.

Direction vector:
\[ 3\hat{i}-4\hat{j} \]

Magnitude of this direction:
\[ \sqrt{3^2+4^2}=5 \]

Unit vector:
\[ \frac{3\hat{i}-4\hat{j}}{5} \]

Since displacement magnitude is \(25\,m\),
\[ \vec{S}=25\left(\frac{3\hat{i}-4\hat{j}}{5}\right) \]
\[ \vec{S}=15\hat{i}-20\hat{j} \]

Step 3: Calculate the work done.
\[ W=\vec{F}_{net}\cdot\vec{S} \]
\[ =(11\hat{i}-3\hat{j}-\hat{k})\cdot(15\hat{i}-20\hat{j}) \]
\[ W=11\times15 + (-3)(-20) + (-1)(0) \]
\[ W=165+60 \]
\[ W=225\,J \] Quick Tip: When displacement is given along a direction, first convert the direction into a unit vector and multiply it by the magnitude to obtain the displacement vector.

Concept:

The time constant of an RC circuit is
\[ \tau = RC \]

where \(R\) is the effective resistance in the circuit and \(C\) is the capacitance.

Step 1: Determine effective resistance.

The circuit contains a resistor \(R=100\Omega\) and a diode.
Only one diode conducts in forward bias, contributing a resistance of \(10\Omega\).

Thus,
\[ R_{eq}=100+10=110\Omega \]

Step 2: Calculate the time constant.
\[ \tau = RC \]
\[ \tau = 110 \times 0.2\times10^{-6} \]
\[ \tau = 22\times10^{-6}\,s \]
\[ \tau = 22\,\mu s \] Quick Tip: Time constant of an RC circuit is \( \tau = RC \). When diodes are present, first determine which diode is forward biased and include its resistance in the effective resistance.

Concept:

For an ideal gas,
\[ PV = nRT \]

Thus,
\[ n = \frac{PV}{RT} \]

When two gases mix, the total number of moles is conserved.
\[ n = n_1 + n_2 \]

Step 1: Write number of moles for each gas.
\[ n_1 = \frac{P_1V_1}{RT_1} \]
\[ n_2 = \frac{P_2V_2}{RT_2} \]

Step 2: Write total number of moles after mixing.
\[ n = \frac{PV}{RT_f} \]

Step 3: Use mole conservation.
\[ \frac{PV}{RT_f} = \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} \]

Cancel \(R\):
\[ \frac{PV}{T_f} = \frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2} \]

Step 4: Solve for \(T_f\).
\[ T_f = \frac{PV}{\frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2}} \] Quick Tip: For mixing of ideal gases, total number of moles remains conserved. Always express moles using \(n=\frac{PV}{RT}\) and then apply \(n=n_1+n_2\).

Concept:

For simple harmonic motion (SHM), velocity as a function of displacement is
\[ v = \omega\sqrt{A^2 - x^2} \]

where \(A\) = amplitude, \(\omega\) = angular frequency.

Step 1: Rewrite the given equation.
\[ v^2 = 100 - x^2 \]
\[ v = \sqrt{100 - x^2} \]

Step 2: Compare with SHM velocity relation.
\[ v = \omega\sqrt{A^2 - x^2} \]

Comparing,
\[ A^2 = 100 \]
\[ A = 10 \]

and
\[ \omega = 1 \]

Step 3: Find the time period.
\[ T = \frac{2\pi}{\omega} \]
\[ T = \frac{2\pi}{1} \]
\[ T = 2\pi \] Quick Tip: In SHM, velocity-displacement relation is \(v=\omega\sqrt{A^2-x^2}\). Comparing the given equation with this form directly gives amplitude and angular frequency.