CBSE Physics — Practice Paper 1

Electrostatics, current electricity, magnetism, electromagnetic induction, optics and modern physics — with derivations set out the way CBSE awards marks for them.

How to use this: CBSE Physics gives marks for the statement of the principle, the diagram, the derivation steps and the final result separately. A derivation that jumps to the answer typically scores 1 of 5. Draw the diagram even when it is not explicitly demanded.

Section A — 1 mark each

1. Define electric field intensity and state its SI unit (1 mark)

The force experienced by a unit positive test charge placed at a point in the field.

$$\vec{E} = \frac{\vec{F}}{q_0}$$

SI unit: N/C (or V/m)

Where the mark sits: the words unit positive charge. "Force on a charge" is insufficient — the magnitude depends on the charge unless it is normalised.

2. State Lenz's law (1 mark)

The direction of the induced emf (and current) is such that it opposes the change that produced it.

The follow-up CBSE asks: Lenz's law is a consequence of the conservation of energy — if the induced current aided the change, energy would be created from nothing.

3. Write the relation between the focal length and radius of curvature of a spherical mirror (1 mark)

$$f = \frac{R}{2}$$

Section B — 2 marks each

4. Two charges of $+4\ \mu\text{C}$ and $-4\ \mu\text{C}$ are 0.2 m apart. Find the electric dipole moment (2 marks)

p = q × 2a
  = (4 × 10⁻⁶) × 0.2
  = 8 × 10⁻⁷ C m

Direction: from the negative charge towards the positive charge.

Marks: 1 for the magnitude, 1 for the direction. The direction is half the answer — dipole moment is a vector, and CBSE awards it separately.

5. A wire of resistance 10 Ω is bent into a circle. Find the resistance between two diametrically opposite points (2 marks)

Bending into a circle gives two semicircular halves in parallel,
each of resistance 10/2 = 5 Ω.

 1     1     1     2
--- = --- + --- = ---
 R     5     5     5

R = 2.5 Ω

Answer: 2.5 Ω

The insight being tested: the two halves form a parallel combination between those points. Candidates who answer 10 Ω have not recognised the circuit.

6. State the two conditions for sustained interference of light (2 marks)

1. The two sources must be coherent — a constant phase difference between them
2. They must have equal or nearly equal amplitudes (for good contrast), and the same frequency

Note: two independent bulbs never produce sustained interference, because their phase relationship fluctuates randomly — which is why Young used a single source split into two.

Section C — 3 marks each

7. Derive an expression for the electric field on the axis of a dipole (3 marks)

Let the dipole have charges ±q separated by 2a, and let P lie on the
axis at distance r from the centre.

Field due to +q (towards P):    E₊ =  kq / (r - a)²
Field due to -q (away from P):  E₋ =  kq / (r + a)²

Net field, directed from -q to +q:

E = E₊ - E₋
                  1              1
  =  kq  [  ------------  -  ------------  ]
             (r - a)²        (r + a)²

           (r + a)² - (r - a)²          4ar
  =  kq · ---------------------- = kq --------------
              (r² - a²)²               (r² - a²)²

Since p = q(2a):

           2 k p r
E  =  -----------------
        (r² - a²)²

For a short dipole (r >> a):

        2 k p        2p
E  ≈  --------  =  ---------
          r³        4πε₀ r³

Marks: 1 for the individual fields with correct distances, 1 for the algebra, 1 for the final result including the short-dipole approximation.

Where marks are lost: writing both fields as $kq/r^2$ — ignoring the $\pm a$ offsets makes them cancel exactly and gives zero, which should itself alert you.

8. Using Ampère's circuital law, derive the magnetic field inside a long solenoid (3 marks)

Take a rectangular Amperian loop of length L, partly inside the solenoid
and partly outside.

∮ B⃗ · dl⃗ = μ₀ I_enclosed

Contributions:
  - Along the inside length L:        B·L
  - Along the two perpendicular sides: 0  (B ⊥ dl)
  - Along the outside length:          0  (B ≈ 0 outside)

So:      B·L = μ₀ (nL) I        where n = turns per unit length

∴        B = μ₀ n I

Marks: 1 for choosing a valid Amperian loop, 1 for evaluating the four contributions, 1 for the result.

Frequently omitted, frequently penalised: stating explicitly that $B \approx 0$ outside a long solenoid. Without it the derivation is incomplete.

9. A converging lens of focal length 20 cm forms an image of an object placed 30 cm away. Find the image distance and magnification (3 marks)

Using the lens formula with the Cartesian sign convention
(u negative for a real object):

 1     1     1
--- - --- = ---
 v     u     f

 1      1        1
--- - ----- = ------
 v    (-30)     20

 1      1      1        3 - 2        1
--- = ----- - ----- = ---------- = ------
 v      20     30         60         60

v = +60 cm     (positive ⟹ real image, on the far side)

           v        60
m  =  ----- =  --------  =  -2
           u      (-30)

Answer: $v = 60\ \text{cm}$, $m = -2$ — real, inverted, magnified twice.

Marks: 1 for the formula with correct signs, 1 for $v$, 1 for $m$ interpreted.

Sign convention is the whole question. Getting $u = -30$ right determines everything downstream; treating it as $+30$ gives $v = 12$ cm and loses all three marks.

Section D — 5 marks

10. Explain the working of an AC generator and derive the expression for the induced emf (5 marks)

Principle: electromagnetic induction — when the magnetic flux through a coil changes, an emf is induced (Faraday's law).

Construction: a rectangular coil of $N$ turns and area $A$ rotates with angular velocity $\omega$ in a uniform magnetic field $B$, with slip rings and brushes taking the current out.

Derivation:

At time t, the coil has rotated through θ = ωt from the plane
perpendicular to B.

Flux through the coil:       Φ = N B A cos ωt

By Faraday's law:

           dΦ         d
e  =  -  ------ = - ---- (N B A cos ωt)
            dt        dt

   =  N B A ω sin ωt

Peak emf:   e₀ = N B A ω

∴  e = e₀ sin ωt

Marks: 1 for the principle, 1 for the labelled diagram, 1 for the flux expression, 1 for the differentiation, 1 for the final result and identifying $e_0$.

The diagram earns its own mark even when the question does not say "draw". Label the coil, the field direction, the slip rings and the brushes.

What CBSE Physics rewards

1. State the principle first. Almost every derivation carries a mark for naming the law being applied before any algebra.
2. Draw the diagram. In the 5-mark questions it is an independent mark, and it takes thirty seconds.
3. Respect the sign convention. In optics it decides the entire answer, not just its sign.
4. Finish with interpretation. "$m = -2$" is worth more when you add "real, inverted, magnified" — several mark schemes require it explicitly.

Where to go next

• JEE Main Physics Practice Set 1 — the same syllabus at entrance-exam speed
• NEET Biology Practice Set 1 — for the medical entrance route
• CBSE Class 10 Mathematics Practice Paper 1 — the algebra these derivations assume

Ask StudyAI to mark your derivations step by step against the way CBSE allocates the marks.