Class 12 Exam Preparation | 2026 | Physics

Worked Problems

Class 12 Physics · NEB

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1.

The graph of a wave motion is given below. Observe the graph and answer the following questions.

(a) What properties of the medium are relevant for the propagation of this type of wave?

The given wave is a transverse mechanical wave. For its propagation, the medium should have:

• Elasticity
• Inertia

In addition, the frictional/resistive force of medium should be minimum to minimize energy loss.

(b) What do PQ and PR indicate with respect to the wave?

PQ represents the amplitude (i.e. maximum displacement of a particle from mean position) and PR represents the wavelength (i.e. distance between two consecutive troughs/crests).

(c) What do you mean by phase in a wave? Why are particles at 'S' and 'R' in anti-phase?

Phase in a wave refers to the state of motion of a vibrating particle at a given instant, which defines both the position and direction of motion.

Particles S and R are in anti-phase (i.e. phase difference of 180°) because their displacements are opposite in direction despite being equal in magnitude. Here, S moves upward and R moves downward.

(d) Two particles 'Q' and 'T' are positioned 5 cm and 18 cm away respectively from the origin particle 'O'. The angular wave number is 1/13 rad cm⁻¹. What is the phase difference between particles 'Q' and 'T' in their vibrations?

Here, x₁ = 5 cm
x₂ = 18 cm
K = 1/13 rad cm⁻¹

Then, phase diff. (Δφ) = K Δx
               = 1/13 × 13
∴ Δφ = 1 rad

2.

(a) Derive an expression for the torque experienced by a rectangular coil placed in a uniform magnetic field.

Figure: A current carrying rectangular coil placed in a uniform magnetic field.

Let us consider a rectangular coil PQRS carrying current 'I' placed in a uniform magnetic field B such that the plane of PQRS makes an angle 'θ' with the magnetic field.

Now, force experienced by arm PS is,

\(\vec{F₁} = I(\vec{l} × \vec{B})\)
or, F₁ = IlB sin90°   [∵ l ⊥ B]
∴ F₁ = IlB  …(i)

The direction of F₁ is shown in the figure, as per Fleming's Left Hand Rule.

Similarly, force experienced by arm RQ is,

∴ F₂ = IlB  …(ii)

Force experienced by arm SR is,

\(\vec{F₃} = I(\vec{b} × \vec{B})\)
∴ F₃ = IbB sinθ  …(iii)

The direction of F₃ is towards downwards (F.L.H.R.).

Similarly, the direction the force on arm QP is,

F₄ = IbB sinθ  …(iv), with F₄ directing upwards.

Here, F₁ and F₂ form a couple and hence produce a torque. However, F₃ and F₄ do not contribute to the rotational motion.

The top view of the given arrangement gives:

Torque produced = (magnitude of either force) × (⊥ distance between the forces)
           = F₁ b cosθ
           = IlB b cosθ
∴ τ = IAB cosθ   [∵ A = lb]

For a coil of N turns,
τ = NIAB cosθ
∴ τ = BINA cosθ

(b) A horizontal straight wire 5 cm long weighing 1.2 g/m is placed perpendicular to a uniform horizontal magnetic flux density 0.6 T. If the resistance per unit length of the wire is 3.8 Ω m⁻¹, calculate the p.d. that has to be applied between the ends of the wire to make it just self-supporting.

Given,
l = 5 cm = 0.05 m
m/l = 1.2 g/m = 1.2×10⁻³ kg/m
B = 0.6 T
R/l = 3.8 Ω m⁻¹
V = ?

For a self-supporting wire, force due to magnetic field = weight

i.e. IlB sinθ = mg   {θ = 90° as the field is ⊥ to the wire}
i.e. IlB = mg
i.e. IB = (m/l)g
i.e. I × 0.6 = 1.2×10⁻³ × 10
i.e. I = 2×10⁻² A

Then, V = IR
     = 2×10⁻² × 3.8 × l  {∵ R = 3.8 × l}
     = 2×10⁻² × 3.8 × 0.05
     = 0.0038
∴ V = 3.8×10⁻³ V

3.

(a) Derive an expression for the magnitude and direction of force experienced by a charged particle moving through a uniform magnetic field. Express with conditions of maximum and minimum force it experiences.

Figure: Force on a moving charge due to magnetic field

Let us consider a charge '+q' moving with velocity v in a magnetic field B as in figure.

Experimentally, it has been found that the force due to the magnetic field is:

1. Directly proportional to the magnitude of charge, F ∝ q  …(i)
2. Directly proportional to the speed of charge, F ∝ v  …(ii)
3. Directly proportional to the strength of magnetic field, F ∝ B  …(iii)
4. Directly proportional to the sine of the angle between v & B, F ∝ sinθ  …(iv)

Combining these equations,

F ∝ qvB sinθ
∴ F = KqvB sinθ

In SI units, K = 1. So,

F = qvB sinθ

In vector form, \(\vec{F} = q(\vec{v} × \vec{B})\). So, the direction of this force is ⊥ to the plane containing \(\vec{v}\) \(\vec{B}\). In our case, \(\vec{F}\) is directed in the 'y' direction.

Conditions:

1. Maximum force: for θ = 90°, F_max = qvB.

2. Minimum force: for θ = 0° or 180°, F_min = 0.

(b) An α-particle (doubly ionized He atom) of mass 6.65×10⁻²⁷ kg travels at right angle to a magnetic field of 0.2 T with a speed of 6×10⁵ m/s. Find the acceleration of the particle.

Given,
m = 6.65×10⁻²⁷ kg
q = 2e = 3.2×10⁻¹⁹ C
B = 0.2 T
V = 6×10⁵ m/s
θ = 90° (v ⊥ B)

We know, F = qvB
or, ma = qvB
or, a = qvB/m
a = (3.2×10⁻¹⁹ × 6×10⁵ × 0.2) / 6.65×10⁻²⁷
∴ a = 5.7 \(\times\) 10¹² m/s²

4.

(a) What happens when an electron beam enters a uniform electric field? Explain with mathematical details how to find the trajectory and also determine the angle at which the beam emerges out from the field.

Suppose a horizontal beam of electrons moving with a velocity v enters midway between the two parallel plates as shown in figure. Upper plate is at higher potential and lower is at lower potential. If V be the potential difference between the plates d distance apart, then the electric field intensity is,

\[E=\frac{V}{d}\]

Hence the force on an electron is,

\[F=eE=e\frac{V}{d}\]

which is directed towards the positive plate.

If m be the mass of electron, the force acting on an electron is,

\[F=ma\]

So, the vertical acceleration is,

\[\begin{align*} a&=\frac{F}{m}\\ &=\frac{eE}{m}\\ &=\frac{eV}{md}\text{ ... (i)}\\ \end{align*}\]

Let, the electron reaches at point P(x,y) at a time t , then,

For horizontal motion,

\[\begin{align*} x&=u_x t +\frac{1}{2} a_x t^2\\ &=vt \because a_x=0 \text{as there is no horizontal force}\\ t&=\frac{x}{v}\hspace{0.1cm} \text{ ... (ii)}\\ \end{align*}\]

For vertical motion,

\[\begin{align*} y&=u_yt+\frac{1}{2}a_y t^2\\ &=\frac{1}{2} at^2\\ &=\frac{1}{2} \times \frac{eV}{md} t^2 \hspace{0.1cm} \text{(from (i))}\\ &=\frac{1}{2} \frac{eV}{md}\left(\frac{x}{v}\right)^2 \hspace{0.1cm} \text{(from (ii))}\\ \therefore y&=\left(\frac{eV}{2mdv^2}\right)x^2 \hspace{0.1cm} \text{ ... (iii)}\\ \end{align*}\]

Comparing equation (iii), with the equation of parabola

\[y=Ax^2+bx+c\]

we find that equation (iii) is a parabolic equation with

\[A=\frac{eV}{2mdv^2}\] \[B=0\] \[C=0\]

Hence, the trajectory of electron beam in an electric field is parabolic in nature.

For the angle,

\[\begin{align} tan\theta&=\frac{v_y}{v_x}\\ &=\frac{a_yt}{v}\\ &=\frac{eVt}{mdv}\\ \therefore \theta&=tan^{-1}(\frac{eVD}{mdv^2})\because t=\frac{D}{v}\\ \end{align}\]

(b) A beam of protons with a velocity of \(4 \times 10^5\) m/s enters a uniform magnetic field of 0.3 T at an angle of 60⁰. Find the radius of the helical path taken by the proton beam. Also, find the pitch of the helix.

Given

• Velocity of proton (\(v\)) = \(4 \times 10^5 \text{ m/s}\)
• Magnetic field (\(B\)) = \(0.3 \text{ T}\)
• Angle (\(\theta\)) = \(60^\circ\)
• Mass of proton (\(m\)) = \(1.67 \times 10^{-27} \text{ kg}\)
• Charge of proton (\(q\)) = \(1.6 \times 10^{-19} \text{ C}\)

• Perpendicular velocity component: \(v_\perp = v \sin 60^\circ = 4 \times 10^5 \times \frac{\sqrt{3}}{2} \approx 3.464 \times 10^5 \text{ m/s}\)
• Parallel velocity component: \(v_\parallel = v \cos 60^\circ = 4 \times 10^5 \times 0.5 = 2 \times 10^5 \text{ m/s}\)

Radius of the Helical Path (\(r\)):

The centripetal force is provided by the magnetic force:

\[r = \frac{m v_\perp}{q B}\]

\[r = \frac{1.67 \times 10^{-27} \times 3.464 \times 10^5}{1.6 \times 10^{-19} \times 0.3} \approx 0.01205 \text{ m} = \mathbf{1.21 \text{ cm}}\]

Pitch of the Helix (\(p\)):

The time period for one complete revolution is:

\[T = \frac{2\pi m}{q B} = \frac{2 \pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3} \approx 2.186 \times 10^{-7} \text{ s}\]

The pitch is the linear distance covered along the field during time \(T\):

\[p = v_\parallel \times T = (2 \times 10^5) \times (2.186 \times 10^{-7}) \approx 0.0437 \text{ m} = \mathbf{4.37 \text{ cm}}\]

The radius of the helical path is 1.21 cm and the pitch of the helix is 4.37 cm.

5.

(a) What do you mean by cross-field?

A cross field (or crossed fields) refers to a region in space where an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) exist simultaneously and act perpendicular (at right angles) to each other. i.e., \(\vec{E} \perp \vec{B}\).

Under this condition,

• The net force on a charged particle is 0.
• The particle passes through the cross field without deflections regardless of its mass and charge.

(b)Describe and give theory of J.J. Thomson's method to determine the ratio of the charge to mass of an electron.

Construction:

Figure shows the basic design of Thomson's experiment to measure specific charge of an electron \(\frac{e}{m}\). E is the electric field in the region between the plates which is directed downward direction in figure and exerts a force in the upward direction on the electrons (towards S₂).

A magnetic field B can also be applied in the region between the plates by passing electric currents in circular coils (represented by dashed circle in figure). This field is perpendicular to the electric field as well as to the undeviated path of the cathode rays. If magnetic field alone is present, the electrons moves in a circular arc and deviates in a downward direction (towards S₁) in the figure).

Theory:

If both the electric fields and the magnetic fields are switched on and the values are chosen such that,

\[\begin{align*} F_m&=F_e\\ Bev&=eE\\ \therefore v&=\frac{E}{B}\hspace{0.1cm} \text{ ... (i)}\\ \end{align*}\]

In this case, the electron beam will pass undeflected and pass towards S in the figure.

If V be the potential difference between the anode A and the cathode C, the speed of the electrons coming out of A is,

\[\begin{align*} eV&=\frac{1}{2} mv^2\\ eV&=\frac{1}{2} m \left(\frac{E}{B}\right)^2 \hspace{0.1cm} \text{(from (i))}\\ \frac{e}{m}&=\frac{E^2}{2B^2 V}\\ \end{align*}\]

If V^' be the potential difference between the plates P₁ and P₂. Then,

\[E=\frac{V'}{d}\]

Thus, above equation becomes

\[\frac{e}{m}=\frac{V'^2}{2B^2d^2 V}\]

This relation gives the value of specific charge of an electron.

(c) Why specific charge for positive rays much smaller than that of cathode rays?

The specific charge of a charged particle is given by \(\frac{q}{m}\). For positive rays, the denominator (i.e., mass) is exceptionally higher than that of the cathode rays. Due to this reason, the specific charge for positive rats is much smaller than that of cathode rays.

6.

(a) All SHM is periodic motion, not all periodic motion is SHM. Express your understanding of this statement with an example.

A motion is periodic if it repeats itself at regular intervals of time. Simple H.M. is a special type of periodic motion where the restoring force (or acceleration) is directly proportional to displacement and directed towards the mean position, i.e. F ∝ −y or, a ∝ −y. Since every SHM repeats at regular time intervals, all SHM is periodic motion. However, periodic motions like circular motion or planetary motion lack a restoring force towards a central position.

(b) Explain the graphical & mathematical variation of velocity with displacement from mean position of a particle in SHM.

The displacement of a particle in SHM is,

y = r sinωt   where r is the amplitude.

Now,

v = dy/dt = rω cosωt
or, v = rω √(1 − sin²ωt)
or, v = rω √(1 − y²/r²)
or, v² = r²ω²(1 − y²/r²)
or, v² = ω²(r² − y²)
or, v² = ω²r² − ω²y²
or, v² + ω²y² = ω²r²
or, v²/(ω²r²) + ω²y²/(ω²r²) = 1
or, y²/r² + v²/(ω²r²) = 1

Comparing with x²/a² + y²/b² = 1 (i.e. eqn. of an ellipse), the graph becomes:

(c) What is a simple pendulum? Show that the motion of a simple pendulum is simple harmonic and find the expression for its time period.

Simple pendulum is an ideal pendulum that consists of a heavy point mass (bob) suspended from a rigid, frictionless support by a light, inextensible, and flexible string.

Figure: Simple Pendulum

Let us consider a simple pendulum as shown in the figure. Let 'y' be the displacement at time 't'. Resolving weight 'mg' into two components:

• mgcosθ balances the tension on the string
• mgsinθ provides the restoring force

So, Restoring force (f) = −mgsinθ   {∵ force is directed opposite to the displacement}
So, ma = −mgsinθ
or, a = −g sinθ

For small angle 'θ':
sinθ ≈ θ = y/l
So, a = −gy/l  …(i)
∴ a ∝ −y — this shows the motion of simple pendulum is S.H.M.

For time period, comparing eqn. (i) with a = −ω²y:

ω² = g/l
or, 2π/T = √(g/l)
∴ T = 2π√(l/g)

(d) A SHM has time period 5 seconds and amplitude (r) = 10 cm. Find the time it takes to travel 3/2 r starting from the mean position.

Given,

Amplitude (r)= 10 cm
Time Period (T)= 5 seconds
Total Distance to Travel (3/2)r = 1.5 × 10 cm = 15 cm

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The displacement equation for SHM is,

\(y = r sin(\omega t)\)

where \(\omega = \frac{2\pi}{T}\)

Initially, from y=0 cm to r

The particle travels from the mean position (x = 0) to the extreme position (x = r = 10 cm).

• Distance covered = 10 cm
• Time taken (t₁) = T / 4 = 5 / 4 = 1.25 seconds

For the return path,

Remaining distance to complete 15 cm = 15 cm - 10 cm = 5 cm (which is r / 2).

After turning back from x = 10 cm, moving 5 cm towards the mean position leaves the displacement at x = 10 cm - 5 cm = 5 cm (i.e., x = r / 2).

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Then,

Substitute x = r / 2 into the displacement equation:

r / 2 = r · sin((2π / T) · t)

sin((2π / T) · t) = 1 / 2

Since the particle is on its return journey in the second quarter cycle (where π/2 < ωt < π):

(2π / T) · t = π - (π / 6) = 5π / 6

t = (5 / 12) · T

Substituting T = 5 seconds:

t = (5 / 12) × 5 = 25 / 12 seconds

The time taken is ²⁵⁄₁₂ seconds (≈ 2.083 seconds).

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