JEE Main 2026 April 6 Shift 1 Physics Question Paper with Solutions PDF

Concept:

Dimensional analysis is used to determine the relationship between physical quantities.

Key ideas used here:


Dimension of potential energy: \( [U] = ML^2T^{-2} \)
Quantities added together must have the same dimensions
If \(x\) represents displacement, then \( [x] = L \)


Step 1: Determine the dimension of \(B\).

Given expression:
\[ u=\frac{A\sqrt{x}}{B+x} \]

Since \(B\) is added to \(x\),
\[ [B] = [x] \]

Because \(x\) is displacement,
\[ [B] = L \]

Step 2: Use dimensional consistency of the equation.

Dimension of potential energy:
\[ [U] = ML^2T^{-2} \]

Now taking dimensions of the given equation:
\[ [U] = \left[\frac{A\sqrt{x}}{B+x}\right] \]

Since \(B+x\) has dimension \(L\),
\[ [U] = \frac{[A][x]^{1/2}}{L} \]

Substitute \( [x]=L \):
\[ ML^2T^{-2} = \frac{[A]L^{1/2}}{L} \]
\[ ML^2T^{-2} = [A]L^{-1/2} \]

Step 3: Solve for the dimension of \(A\).
\[ [A] = ML^2T^{-2}\times L^{1/2} \]
\[ [A] = ML^{5/2}T^{-2} \]

Final Result
\[ [A] = ML^{5/2}T^{-2}, \qquad [B] = L \] Quick Tip: Whenever two quantities are added or subtracted (like \(B+x\)), they must have the same dimensions. This rule is extremely useful in dimensional analysis problems.

Concept:

In an AC circuit containing resistance and inductance (R–L circuit), the power factor is given by
\[ \cos\phi = \frac{R}{Z} \]

where


\(R\) = Resistance
\(Z\) = Impedance of the circuit
\(Z = \sqrt{R^2 + X_L^2}\)
\(X_L\) = Inductive reactance


Step 1: Find the impedance of the circuit.
\[ Z = \sqrt{R^2 + X_L^2} \]

Substituting \(R = 80\,\Omega\) and \(X_L = 80\,\Omega\):
\[ Z = \sqrt{80^2 + 80^2} \]
\[ Z = \sqrt{6400 + 6400} \]
\[ Z = \sqrt{12800} \]
\[ Z = 80\sqrt{2}\,\Omega \]

Step 2: Calculate the power factor.
\[ \cos\phi = \frac{R}{Z} \]

Substituting the values:
\[ \cos\phi = \frac{80}{80\sqrt{2}} \]
\[ \cos\phi = \frac{1}{\sqrt{2}} \]

Final Result
\[ Power factor = \frac{1}{\sqrt{2}} \] Quick Tip: For a series \(R-L\) circuit, impedance is always \(Z=\sqrt{R^2+X_L^2}\). The power factor \( \cos\phi \) becomes smaller when the inductive reactance increases.

Concept:

For a uniformly charged semicircular ring, the electric field at the center is
\[ E=\frac{2k\lambda}{r} \]

where


\(k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\,N m^2/C^2\)
\(\lambda\) = linear charge density
\(r\) = radius of the semicircle


Also,
\[ \lambda=\frac{q}{\pi r} \]

since the length of a semicircle is \( \pi r \).

Step 1: Substitute \( \lambda = \dfrac{q}{\pi r} \) into the electric field formula.
\[ E=\frac{2k}{r}\left(\frac{q}{\pi r}\right) \]
\[ E=\frac{2kq}{\pi r^2} \]

Step 2: Substitute the given values.
\[ E=100\,N/C, \quad k=9\times10^9, \quad r=10\,cm=10\times10^{-2}\,m \]
\[ 100=\frac{2\times 9\times10^9 \times q}{\pi (10\times10^{-2})^2} \]
\[ 100=\frac{18\times10^9 q}{\pi \times 10^{-2}} \]

Step 3: Solve for the charge \(q\).
\[ q=\frac{100 \times \pi \times 10^{-2}}{18\times10^9} \]
\[ q=\frac{\pi}{18}\times10^{-9}\,C \]

Final Result
\[ q=\frac{\pi}{18}\times10^{-9}\,C \] Quick Tip: For a uniformly charged semicircular ring, the electric field at the center is \(E=\dfrac{2k\lambda}{r}\). The horizontal components cancel due to symmetry, leaving only the vertical component.

Concept:

When a particle moves in a vertical circular path, the centripetal force at the top of the circle is provided by the sum of weight and the normal reaction.
\[ \frac{mv^2}{R} = mg + N \]

Also, since the track is smooth, mechanical energy is conserved.
\[ Loss of potential energy = Gain in kinetic energy \]

Step 1: Apply the centripetal force condition at the highest point.

Given that the normal reaction is three times the weight:
\[ N = 3mg \]

Substitute into the centripetal force equation:
\[ mg + N = \frac{mv^2}{R} \]
\[ mg + 3mg = \frac{mv^2}{R} \]
\[ 4mg = \frac{mv^2}{R} \]
\[ v^2 = 4gR \]

Step 2: Apply conservation of energy.

The particle starts from height \(h\) and reaches the highest point of the circular track which is at height \(2R\).
\[ mg(h-2R) = \frac{1}{2}mv^2 \]

Substitute \(v^2 = 4gR\):
\[ mg(h-2R) = \frac{1}{2}m(4gR) \]
\[ mg(h-2R) = 2mgR \]

Cancel \(mg\):
\[ h-2R = 2R \]
\[ h = 4R \]

Final Result
\[ h = 4R \] Quick Tip: For motion in a vertical circle, always analyze the \textbf{top point carefully}. The centripetal force there is provided by \(mg + N\). Combine this with energy conservation to solve height or velocity problems quickly.

Concept:

For refraction at a spherical surface,
\[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \]

Transverse magnification is
\[ m=\frac{n_1 v}{n_2 u} \]

Step 1: Apply refraction formula at spherical surface.

Given:
\[ n_1=1, \quad n_2=1.4, \quad u=-4R \]
\[ \frac{1.4}{v}-\frac{1}{-4R}=\frac{1.4-1}{R} \]
\[ \frac{1.4}{v}+\frac{1}{4R}=\frac{0.4}{R} \]

Step 2: Solve for \(v\).
\[ \frac{1.4}{v}=\frac{0.4}{R}-\frac{1}{4R} \]
\[ \frac{1.4}{v}=\frac{4}{10R}-\frac{1}{4R} \]
\[ \frac{1.4}{v}=\frac{6}{40R} \]
\[ v=\frac{56R}{6}=\frac{28}{3}R \]

Step 3: Find transverse magnification.
\[ m=\frac{n_1 v}{n_2 u} \]
\[ m=\frac{1 \times \frac{28R}{3}}{1.4 \times (-4R)} \]
\[ m=-1.67 \]

Final Result
\[ m=-1.67 \] Quick Tip: For refraction at a spherical surface always use \( \dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R} \) and magnification \( m=\dfrac{n_1 v}{n_2 u} \).

Concept:

Heat produced in a resistor is given by Joule's law:
\[ H=\frac{V^2}{R}t \]

To find voltage across \(6\Omega\), node voltage method is used.

Step 1: Apply node voltage equation.

Let node potential be \(x\).
\[ \frac{x-2}{2}+\frac{x-0}{4}+\frac{x-3}{6}=0 \]

Multiply by \(12\):
\[ 6x-12+3x+2x-6=0 \]
\[ 11x-18=0 \]
\[ x=\frac{18}{11} volt \]

Step 2: Find voltage across \(6\Omega\).
\[ V_{6\Omega}=3-\frac{18}{11} \]
\[ V_{6\Omega}=\frac{33-18}{11}=\frac{15}{11} \]

Step 3: Calculate heat dissipated.
\[ H=\frac{V^2}{R}t \]
\[ H=\frac{\left(\frac{15}{11}\right)^2}{6}\times100 \]
\[ H=30.99 \approx 31\,J \]

Final Result
\[ H=31\,J \] Quick Tip: For complicated resistor circuits, the node voltage method quickly gives the potential difference across any resistor, after which Joule's law \(H=\frac{V^2}{R}t\) can be applied.

Concept:

When a thin glass slab of thickness \(t\) and refractive index \(\mu\) is inserted in front of one slit in YDSE, the central fringe shifts by
\[ y_0 = (\mu - 1)t\frac{D}{d} \]

Position of minima in YDSE is
\[ y_n = \left(n+\frac{1}{2}\right)\frac{D\lambda}{d} \]

Step 1: Central maxima shifts to the position of 4th minima.

For minima,
\[ y_n=\left(n+\frac{1}{2}\right)\frac{D\lambda}{d} \]

4th minima corresponds to \(n=3\)
\[ y_1=\left(3+\frac{1}{2}\right)\frac{D\lambda}{d} \]
\[ y_1=\frac{7}{2}\frac{D\lambda}{d} \]

Since the central maxima shifts to this position,
\[ y_0 = y_1 \]

Step 2: Equate shift expression.
\[ (\mu-1)t\frac{D}{d}=\frac{7}{2}\frac{D\lambda}{d} \]

Cancel \( \frac{D}{d} \):
\[ (\mu-1)t=\frac{7}{2}\lambda \]

Step 3: Substitute given values.
\[ t=8\times10^{-6}\,m \]
\[ \lambda=500\times10^{-9}\,m \]
\[ (\mu-1)=\frac{7}{2}\times\frac{500\times10^{-9}}{8\times10^{-6}} \]
\[ (\mu-1)=3.5\times62.5\times10^{-3} \]
\[ (\mu-1)=0.218 \]
\[ \mu=1.218 \approx 1.22 \]

Final Result
\[ \mu \approx 1.22 \] Quick Tip: Insertion of a glass slab in YDSE produces a fringe shift \(y_0=(\mu-1)t\frac{D}{d}\). Equating this shift with the position of a given bright or dark fringe quickly gives the refractive index.

Concept:

In Simple Harmonic Motion (SHM),
\[ x = A\sin\theta \]

where \( \theta \) is the phase angle.

Time corresponding to phase change is
\[ t = \frac{\theta}{\omega} \]

and
\[ \omega = \frac{2\pi}{T} \]

Step 1: Find phase corresponding to given positions.

At mean position,
\[ x = 0 \Rightarrow \theta = 0 \]

At \( x = \frac{A}{\sqrt{2}} \),
\[ \frac{A}{\sqrt{2}} = A\sin\theta \]
\[ \sin\theta = \frac{1}{\sqrt{2}} \]
\[ \theta = \frac{\pi}{4} \]

Thus phase covered
\[ \Delta\theta = \frac{\pi}{4} \]

Step 2: Find time taken.
\[ t=\frac{\Delta\theta}{\omega} \]
\[ t=\frac{\pi/4}{2\pi/T} \]
\[ t=\frac{\pi}{4}\times\frac{T}{2\pi} \]
\[ t=\frac{T}{8} \]

Step 3: Substitute \(T=5\) s.
\[ t=\frac{5}{8} sec \]

Final Result
\[ t=\frac{5}{8} sec \] Quick Tip: In SHM, motion from mean position to extreme takes \(T/4\). Use \(x=A\sin\theta\) or \(x=A\cos\theta\) to convert displacement into phase difference and then calculate time.

Concept:

For a system to remain at rest in an accelerating frame, pseudo force must be considered.

Pseudo force:
\[ F_p = ma \]

Maximum friction:
\[ f = \mu N \]

Step 1: Consider forces on the larger block.

Given acceleration
\[ a=\frac{g}{2} \]

Pseudo force on \(2\,kg\) block
\[ F_p = 2 \times \frac{g}{2} = g \]

Pseudo force on \(1.5\,kg\) block
\[ F_p = 1.5 \times \frac{g}{2} \]

Normal reaction on smaller block
\[ N = \frac{1.5g}{2} \]

Step 2: Friction force.
\[ f = \mu N \]
\[ f = \mu \times \frac{1.5g}{2} \]

Step 3: Apply equilibrium condition.
\[ 1.5g = g + \mu \frac{1.5g}{2} \]

Divide by \(g\):
\[ 1.5 = 1 + \mu \frac{1.5}{2} \]
\[ 0.5 = \mu \frac{1.5}{2} \]
\[ \mu = \frac{2}{3} \]

Final Result
\[ \mu = \frac{2}{3} \] Quick Tip: In accelerating systems, always analyze motion in the non-inertial frame by adding pseudo force \(ma\) opposite to acceleration. Then apply friction and equilibrium conditions.

Concept:

In an \(RL\) circuit, the applied voltage is shared between the resistor and the inductor.
\[ V = V_R + V_L \]

where
\[ V_R = iR \]

Thus,
\[ V_L = V - iR \]

Step 1: When current \(i = 2\,mA\).
\[ V_R = iR \]
\[ V_R = (2\times10^{-3})(6) \]
\[ V_R = 0.012\,V \]

Voltage across inductor:
\[ V_L = 3 - 0.012 \]
\[ V_L = 2.988\,V \]

Step 2: When current \(i = 4\,mA\).
\[ V_R = (4\times10^{-3})(6) \]
\[ V_R = 0.024\,V \]

Voltage across inductor:
\[ V_L = 3 - 0.024 \]
\[ V_L = 2.976\,V \]

Step 3: Find the ratio.
\[ Ratio = \frac{2.988}{2.976} \]
\[ Ratio \approx 1.004 \approx 1 \]

Final Result
\[ Ratio = 1 \] Quick Tip: In an \(RL\) circuit, the voltage across the inductor is obtained using \(V_L = V - iR\). Small current changes produce very small change in \(V_L\) if the resistance is small.

Concept:

Moment of inertia of a solid sphere about diameter:
\[ I = \frac{2}{5}MR^2 \]

Torque relation:
\[ \tau = I\alpha \]

Rotational kinematics:
\[ \omega_f = \omega_i + \alpha t \]

Step 1: Convert angular speed to rad/s.
\[ \omega_i = 1200\,rpm \]
\[ \omega_i = 1200 \times \frac{2\pi}{60} \]
\[ \omega_i = 40\pi \,rad/s \]

Step 2: Find angular acceleration.

Final angular velocity
\[ \omega_f = 0 \]
\[ 0 = 40\pi + \alpha(10) \]
\[ \alpha = -4\pi \,rad/s^2 \]

Step 3: Find moment of inertia.
\[ R = 4\,cm = 0.04\,m \]
\[ I = \frac{2}{5}(5)(0.04)^2 \]
\[ I = 2 \times 0.0016 \]
\[ I = 0.0032\,kg\,m^2 \]

Step 4: Calculate torque.
\[ \tau = I\alpha \]
\[ \tau = 0.0032 \times 4\pi \]
\[ \tau \approx 0.04\,N-m \]

Step 5: Find angular displacement.
\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
\[ 0 = (40\pi)^2 + 2(-4\pi)\theta \]
\[ \theta = 200\pi \]

Number of revolutions:
\[ n = \frac{\theta}{2\pi} \]
\[ n = \frac{200\pi}{2\pi} \]
\[ n = 100 \]

Final Result
\[ \tau = 0.04\,N-m, \quad n = 100 \] Quick Tip: For rotational stopping problems, first convert rpm to rad/s, then find angular acceleration using \( \omega_f = \omega_i + \alpha t \). Finally apply \( \tau = I\alpha \) and rotational kinematics.

Concept:

Resultant intensity due to interference of two waves is
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi \]

where phase difference
\[ \phi = \frac{2\pi}{\lambda}\Delta x \]

Since both sources have equal intensity
\[ I_1 = I_2 = I \]

Step 1: Intensity at point \(A\).

Path difference
\[ \Delta x_A = \frac{\lambda}{6} \]

Phase difference
\[ \phi_A = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} \]
\[ \phi_A = \frac{\pi}{3} \]

Now resultant intensity
\[ I_A = I + I + 2\sqrt{I\cdot I}\cos\left(\frac{\pi}{3}\right) \]
\[ I_A = 2I + 2I\left(\frac{1}{2}\right) \]
\[ I_A = 3I \]

Step 2: Intensity at point \(B\).

Path difference
\[ \Delta x_B = \frac{\lambda}{3} \]

Phase difference
\[ \phi_B = \frac{2\pi}{\lambda}\times\frac{\lambda}{3} \]
\[ \phi_B = \frac{2\pi}{3} \]
\[ I_B = I + I + 2\sqrt{I\cdot I}\cos\left(\frac{2\pi}{3}\right) \]
\[ I_B = 2I + 2I\left(-\frac{1}{2}\right) \]
\[ I_B = I \]

Step 3: Find ratio.
\[ \frac{I_A}{I_B}=\frac{3I}{I} \]
\[ \frac{I_A}{I_B}=3 \]

Final Result
\[ \frac{I_A}{I_B}=3 \] Quick Tip: For two coherent sources with equal intensity, the interference intensity simplifies to \(I = 2I(1+\cos\phi)\). Convert path difference to phase difference using \( \phi=\frac{2\pi}{\lambda}\Delta x \).

Concept:

Power relation:
\[ P = VI \]

Series resistor limits current through the LED. The remaining voltage appears across the resistor.

Step 1: Find voltage across LED.

Given LED power rating
\[ P = 2\times10^{-3} \,W \]

Current through LED
\[ i = 0.5\,mA = 0.5\times10^{-3} A \]
\[ V_{LED} = \frac{P}{i} \]
\[ V_{LED} = \frac{2\times10^{-3}}{0.5\times10^{-3}} \]
\[ V_{LED} = 4V \]

Step 2: Find voltage across series resistor.

Supply voltage
\[ V = 5V \]
\[ V = V_{Rs} + V_{LED} \]
\[ 5 = V_{Rs} + 4 \]
\[ V_{Rs} = 1V \]

Step 3: Find required resistance.
\[ V_{Rs} = iR_s \]
\[ 1 = 0.5\times10^{-3} R_s \]
\[ R_s = 2\Omega \]

Final Result
\[ R_s = 2\Omega \] Quick Tip: To protect LEDs, always use a series resistor. First determine the LED voltage from \(P=VI\), then use the remaining supply voltage to calculate the required resistor using Ohm's law.

Concept:

Density of a cylinder
\[ \rho = \frac{Mass}{Volume} \]

Volume of cylinder
\[ V = \pi r^2 l = \frac{\pi d^2 l}{4} \]

Thus
\[ \rho = \frac{4M}{\pi d^2 l} \]

For error propagation
\[ \frac{d\rho}{\rho} = \frac{dM}{M} + \frac{dl}{l} + \frac{2\,dd}{d} \]

Step 1: Substitute given errors.
\[ \frac{d\rho}{\rho} = \left( \frac{0.02}{19.42} + \frac{0.02}{10.10} + \frac{2\times0.02}{20.20} \right) \]

Step 2: Calculate percentage error.
\[ \frac{d\rho}{\rho} = (0.00103 + 0.00198 + 0.00198) \]
\[ \frac{d\rho}{\rho} \approx 0.005 \]
\[ % error = 0.005 \times 100 \]
\[ % error = 0.5% \]

Final Result
\[ % error in density = 0.5% \] Quick Tip: For quantities in product or division form, percentage errors add. If a quantity is squared (like \(d^2\)), its relative error is multiplied by the power.

Concept:

Change in potential energy is related to work done by electrostatic field.
\[ \Delta U = -W_{electric} \]

Work done by electric field
\[ W = q \int \vec{E}\cdot d\vec{r} \]

Step 1: Compute line integral.
\[ \vec{E}\cdot d\vec{r} = 2x\,dx + 3y^2\,dy + 4\,dz \]
\[ W = \int 2x\,dx + \int 3y^2\,dy + \int 4\,dz \]

Step 2: Integrate each term.
\[ \int 2x\,dx = x^2 \]
\[ \int 3y^2\,dy = y^3 \]
\[ \int 4\,dz = 4z \]

Thus
\[ W = (x^2 + y^3 + 4z) \]

Step 3: Evaluate between limits.
\[ W = (x^2 + y^3 + 4z)_{(0,-1,-3)}^{(5,1,2)} \]

At \( (5,1,2) \)
\[ 25 + 1 + 8 = 34 \]

At \( (0,-1,-3) \)
\[ 0 -1 -12 = -13 \]
\[ W = 34 - (-13) \]
\[ W = 47 \]

Since
\[ \Delta U = -W \]

Magnitude
\[ |\Delta U| = 47\,J \]

Final Result
\[ |\Delta U| = 47\,J \] Quick Tip: For position-dependent electric fields, compute work using the line integral \( \int \vec{E}\cdot d\vec{r} \). Then use \( \Delta U = -W \).

Concept:

For an isothermal process in an ideal gas,
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]

For an ideal gas undergoing isothermal change, bulk modulus
\[ B = P \]

Thus initial pressure \(P_0 = B\).

Step 1: Use work done expression for isothermal process.
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]

But
\[ nRT = P_0V_0 \]
\[ W = P_0V_0 \ln\left(\frac{V_2}{V_1}\right) \]

Step 2: Substitute volume change.
\[ V_1 = V_0, \qquad V_2 = \frac{V_0}{3} \]
\[ W = P_0V_0 \ln\left(\frac{V_0/3}{V_0}\right) \]
\[ W = P_0V_0 \ln\left(\frac{1}{3}\right) \]

Step 3: Replace \(P_0\) by \(B\).
\[ W = BV_0 \ln\left(\frac{1}{3}\right) \]

Final Result
\[ W = BV_0 \ln\left(\frac{1}{3}\right) \] Quick Tip: For isothermal process of ideal gas, bulk modulus \(B = P\). Work done is \(W = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)\).

Concept:

Magnetic field at centre of circular loop:
\[ B = \frac{\mu_0 I}{2R} \]

Magnetic energy density:
\[ u = \frac{B^2}{2\mu_0} \]

Total magnetic energy stored:
\[ U = uV \]

Step 1: Calculate magnetic field at centre.
\[ B = \frac{\mu_0 I}{2R} \]
\[ B = \frac{4\pi\times10^{-7}\times2}{2\times(10\times10^{-2})} \]
\[ B = \frac{8\pi\times10^{-7}}{0.2} \]
\[ B = 4\pi\times10^{-6}\,T \]

Step 2: Find magnetic energy density.
\[ u = \frac{B^2}{2\mu_0} \]
\[ u = \frac{(4\pi\times10^{-6})^2}{2\times4\pi\times10^{-7}} \]

Step 3: Find volume of cube.
\[ a = 1\,mm = 10^{-3} m \]
\[ V = a^3 = (10^{-3})^3 = 10^{-9} m^3 \]

Step 4: Calculate total energy stored.
\[ U = \frac{B^2}{2\mu_0}\times V \]
\[ U = \frac{(4\pi\times10^{-6})^2}{2\times4\pi\times10^{-7}}\times10^{-9} \]
\[ U = 2\pi\times10^{-14}\,J \]
\[ U \approx 6.28\times10^{-14}\,J \]

Final Result
\[ U = 6.28\times10^{-14}\,J \] Quick Tip: Magnetic energy stored in a region is calculated using energy density \(u = \frac{B^2}{2\mu_0}\) and multiplying it by the volume of that region.

Concept:

When unpolarised light passes through a polariser, its intensity becomes half.
\[ I' = \frac{I_0}{2} \]

When this polarised light passes through an analyser, intensity follows Malus' law:
\[ I = I' \cos^2 \theta \]

Step 1: Apply Malus' law.

Given final intensity
\[ I = \frac{3I_0}{8} \]

Substitute \(I' = \frac{I_0}{2}\)
\[ \frac{3I_0}{8} = \frac{I_0}{2}\cos^2\theta \]

Step 2: Solve for \(\theta\).

Cancel \(I_0\):
\[ \frac{3}{8} = \frac{1}{2}\cos^2\theta \]
\[ \cos^2\theta = \frac{3}{4} \]
\[ \cos\theta = \frac{\sqrt{3}}{2} \]
\[ \theta = 30^\circ \]

Final Result
\[ \theta = 30^\circ \] Quick Tip: Unpolarised light intensity becomes \(I_0/2\) after passing through a polariser. Then apply Malus' law \(I = I_0 \cos^2\theta\) for analyser.

Concept:

Rydberg formula for hydrogen spectrum:
\[ \frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

Shortest wavelength of Lyman series corresponds to transition
\[ n_2 = \infty \rightarrow n_1 = 1 \]

Thus
\[ x = \frac{1}{R} \]

Step 1: First Balmer line \((n_2=3 \rightarrow n_1=2)\).
\[ \frac{1}{\lambda_1} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \]
\[ \frac{1}{\lambda_1} = R\left(\frac{1}{4}-\frac{1}{9}\right) \]
\[ \frac{1}{\lambda_1} = R\left(\frac{5}{36}\right) \]
\[ \lambda_1 = \frac{36}{5R} \]

Since \(x = \frac{1}{R}\)
\[ \lambda_1 = \frac{36}{5}x \]

Step 2: Second Balmer line \((n_2=4 \rightarrow n_1=2)\).
\[ \frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{16}\right) \]
\[ \frac{1}{\lambda_2} = R\left(\frac{3}{16}\right) \]
\[ \lambda_2 = \frac{16}{3R} \]
\[ \lambda_2 = \frac{16}{3}x \]

Step 3: Find difference.
\[ \lambda_1 - \lambda_2 = \left(\frac{36}{5}-\frac{16}{3}\right)x \]
\[ = \left(\frac{108-80}{15}\right)x \]
\[ = \frac{28}{15}x \]

Final Result
\[ \lambda_1 - \lambda_2 = \frac{28}{15}x \] Quick Tip: Remember spectral transitions: Lyman \(n_1=1\), Balmer \(n_1=2\). Shortest wavelength occurs when \(n_2=\infty\).

Concept:

Using work–energy theorem:
\[ W_g + W_{res} = \Delta K \]

where
\[ W_g = mgh \]
\[ \Delta K = \frac{1}{2}mv^2 \]

Step 1: Write work–energy equation.
\[ mgh + W_{res} = \frac{1}{2}mv^2 \]

Step 2: Solve for work done by resistance.
\[ W_{res} = \frac{1}{2}mv^2 - mgh \]

Step 3: Substitute given values.
\[ m = 1\,g = 10^{-3}\,kg \]
\[ h = 1\,km = 1000\,m \]
\[ v = 5\,m/s \]
\[ W_{res} = 10^{-3}\left(\frac{25}{2} - 10^4\right) \]
\[ |W_{res}| = 987.5 \times 10^{-3} J \]

Thus
\[ x = 987.5 \]

Final Result
\[ x = 987.5 \] Quick Tip: Whenever resistive forces act, mechanical energy is not conserved. Use work–energy theorem and treat work done by resistive forces as negative.

Concept:

Centre of mass coordinates:
\[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]
\[ y_{cm} = \frac{\sum m_i y_i}{\sum m_i} \]

Total mass
\[ M = 2 + 15 + 3 = 20\,kg \]





Step 1: Assign coordinates.

Left mass \(2kg\)
\[ (0,0) \]

Right mass \(3kg\)
\[ (10\sqrt{3},0) \]

Top mass \(15kg\)
\[ (5\sqrt{3},5) \]

Step 2: Find \(x_{cm}\).
\[ x_{cm} = \frac{2(0) + 15(5\sqrt{3}) + 3(10\sqrt{3})}{20} \]
\[ x_{cm} = \frac{75\sqrt{3} + 30\sqrt{3}}{20} \]
\[ x_{cm} = \frac{21\sqrt{3}}{4} \]

Step 3: Find \(y_{cm}\).
\[ y_{cm} = \frac{2(0) + 15(5) + 3(0)}{20} \]
\[ y_{cm} = \frac{75}{20} \]
\[ y_{cm} = \frac{15}{4} \]

Step 4: Coordinates of midpoint of median \(P\).
\[ P = (5\sqrt{3}, 2.5) \]

Step 5: Find distance between COM and \(P\).
\[ d = \sqrt{(x_{cm}-x_P)^2 + (y_{cm}-y_P)^2} \]
\[ d = \sqrt{(5.25\sqrt{3} - 5\sqrt{3})^2 + (3.75 - 2.5)^2} \]
\[ d = \sqrt{(0.25\sqrt{3})^2 + (1.25)^2} \]
\[ d = \sqrt{1.75} \]
\[ d \approx 1.32 \]

Final Result
\[ d = 1.32 \] Quick Tip: For systems of point masses, first assign coordinates carefully. Then use COM formulas and finally apply distance formula to find separation between points.

Concept:

Elongation of a wire under tension is given by
\[ \Delta \ell = \frac{T\ell}{YA} \]

where


\(T\) = applied force
\(\ell\) = length of wire
\(Y\) = Young's modulus
\(A\) = cross-sectional area


Step 1: Condition for equal elongation.

Since both wires experience the same force and have same area,
\[ \Delta \ell_1 = \Delta \ell_2 \]
\[ \frac{T\ell_1}{Y_1A} = \frac{T\ell_2}{Y_2A} \]

Step 2: Simplify equation.

Cancel \(T\) and \(A\):
\[ \frac{\ell_1}{Y_1} = \frac{\ell_2}{Y_2} \]
\[ \frac{\ell_1}{\ell_2} = \frac{Y_1}{Y_2} \]

Step 3: Substitute given ratio.
\[ \frac{Y_1}{Y_2} = \frac{20}{11} \]

Thus
\[ \frac{\ell_1}{\ell_2} = \frac{20}{11} \]

Final Result
\[ \frac{\ell_1}{\ell_2} = \frac{20}{11} \] Quick Tip: For wires under the same force and area, elongation \( \Delta \ell \propto \frac{\ell}{Y} \). Equal elongation means \( \frac{\ell_1}{\ell_2} = \frac{Y_1}{Y_2} \).

Concept:

Momentum of a photon is given by
\[ p=\frac{h}{\lambda} \]

Hence
\[ p \propto \frac{1}{\lambda} \]

Using Rydberg formula for hydrogen spectrum
\[ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

For Balmer series \(n_1=2\).

Step 1: First Balmer line \((n_2=3 \rightarrow n_1=2)\).
\[ \frac{1}{\lambda_1} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \]
\[ = R\left(\frac{1}{4}-\frac{1}{9}\right) \]
\[ = R\left(\frac{5}{36}\right) \]

Step 2: Second Balmer line \((n_2=4 \rightarrow n_1=2)\).
\[ \frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{16}\right) \]
\[ = R\left(\frac{3}{16}\right) \]

Step 3: Find ratio of momentum.
\[ \frac{p_1}{p_2} = \frac{\lambda_2}{\lambda_1} \]
\[ = \frac{\frac{1}{(3R/16)}}{\frac{1}{(5R/36)}} = \frac{16}{3R}\times\frac{5R}{36} \]
\[ = \frac{20}{27} \]

Final Result
\[ \frac{p_1}{p_2}=\frac{20}{27} \] Quick Tip: Photon momentum \(p=h/\lambda\). Thus momentum ratio is inverse of wavelength ratio.

Concept:

Intensity of light from a point source is given by
\[ I=\frac{P}{4\pi r^2} \]

where \(r\) is the distance from the source.

Step 1: Analyze distances.

Points \(A\) and \(B\) lie on the same spherical surface.

Thus distance from the source to both points is the same:
\[ r_A=r_B \]

Step 2: Apply inverse square law.
\[ I=\frac{P}{4\pi r^2} \]

Since \(r\) is same for both points,
\[ I_A=I_B \]

Final Result
\[ I_B=I \] Quick Tip: For a point source, intensity depends only on distance \(r\). All points on the same spherical surface receive equal intensity.