JEE Main Physics Mock Test- 1 - JEE MCQ

For an AC supply, the peak voltage is V_₀ = V_rms × √2.
Given V_rms = 220 V,
V_₀ = 220 × √2 ≈ 220 × 1.414 ≈ 311 V.
The closest option is 310 V (option C).

For an ideal transformer, power is conserved:
V_p × i_p = V_s × i_s.
Given V_p = 220 V, i_p = 5 A, V_s = 22000 V,
the secondary current is i_s = (V_p × i_p) / V_s
= (220 × 5) / 22000
= 0.05 A

For the body to not slip, the centripetal force mω²r equals the frictional force μmg.
Thus, mω²r = μmg ⇒ ω² = μg/r.
Given μ = 0.5, g = 9.8 m/s², r = 10 m,
ω = √((0.5 × 9.8) / 10)
= √(4.9/10) ≈ √0.49
= 0.7 rad/s.
The correct answer is option B.

The kinetic energy of emitted electrons (cathode rays) in the photoelectric effect is KE = hν – φ, where hν is the photon energy (proportional to voltage in some setups) and φ is the work function.
Higher voltage increases hν, increasing KE; higher φ reduces KE.
Thus, KE depends on both voltage and work function

For a semicircular arc of radius a with charge per unit length λ, consider a small segment dθ.
The charge is dq = λa dθ, and the electric field at the center is dE = (K dq / a²) cos θ
= (Kλa dθ / a²) cos θ
= (Kλ / a) cos θ, where K = 1/(4πε₀).
Integrating from θ = –π/2 to π/2: E
= ∫(-π/2 to π/2) (Kλ / a) cos θ dθ
= (Kλ / a) [sin θ](-π/2 to π/2)
= (Kλ / a) (1 – (–1))
= 2Kλ / a
= 2 (λ / (4πε₀a))
= λ / (2πε₀a).

The answer is option B."

For a dipole along the X-axis, the electric field at point P (r = 0.2 m, θ = π/3) has components E_r = (2Kp cos θ) / r³ and E_θ = (Kp sin θ) / r³.
The angle of E with the X-axis is θ + α, where tan α = E_θ / E_r
= (sin θ) / (2 cos θ).
For θ = π/3, sin (π/3)
= √3/2, cos (π/3)
= 1/2, tan (π/3)
= √3, so tan α = (√3/2) / (2·1/2) = √3 / 2.
Thus, α = tan⁻¹(√3 / 2), and the total angle is θ = π/3 + tan⁻¹(√3 / 2)

Thrust is the force due to the rate of change of momentum.
Each bullet has mass m = 10 g = 0.01 kg and velocity v = 50 m/s.
With n = 1000 bullets per second,
the thrust is F = (m × v) × n
= (0.01 × 50) × 1000
= 0.5 × 1000
= 500 N

The metre stick (length L = 1 m) is initially horizontal, with its center of mass at L/2 = 0.5 m above the vertical position.
Potential energy is PE = mgL/2 = m × 10 × 0.5 = 5m J.
When vertical, PE = 0, and the energy is kinetic:
KE = (1/2)Iω², where I = (1/3)mL² = (1/3)m × 1² = m/3 kg·m² for a rod about one end.
By conservation of energy: mgL/2 = (1/2)Iω²
5m = (1/2)(m/3)ω²
 = ω² / 6
ω² = 30
ω = √30 ≈ 5.477 rad/s.

The speed of end B is v = ωL = √30 × 1 ≈ 5.4 m/s

The OR gate has a curved input and pointed output, the NOT gate is a triangle with a circle at the output, and the NAND gate is an AND gate with a circle at the output. Without specific symbols, we assume option B (iv, ii, i) correctly matches OR, NOT, and NAND based on standard representations. The exact symbols (i, ii, iii, iv) need to be provided for confirmation.

Answer: Correct (B: (iv), (ii), (i)), but requires symbol clarification.

Apply Ampère's law:

∮ B · dl = μ₀i_enclosed
For a hollow copper pipe carrying current, consider an Amperian loop:
Inside the pipe: No current is enclosed (i_enclose = 0), so B = 0.
Outside the pipe: The total current is enclosed (i_enclose ≠ 0), so B ≠ 0 (like a straight wire).
Thus, the magnetic field is produced only outside the pipe.

Answer: Correct (B: outside the pipe only).

The height of liquid in a capillary tube is given by:
h = (2T cos θ) / (Rρg)

Where T is surface tension, θ is the contact angle, R is the radius, ρ is the liquid density, and g is gravitational acceleration.

Since h ∝ 1/R, for two tubes with radii R₁ and R₂:
h₂ / h₁ = R₁ / R₂

Given h₁ = 3 mm, R₂ = R₁ / 3, then:
h₂ / 3 = R₁ / (R₁ / 3) → h₂ = 9 mm.

Answer: Correct (D: 9 mm).

A nuclear chain reaction in uranium fission is sustained because each fission event releases more than one neutron (typically 2–3), which can trigger further fissions. Option (A) is true but not the cause, (B) is irrelevant to chain reaction, and (D) is not necessary for sustaining the reaction.
Answer: Correct (C: More than one neutron is given out in each fission).

Centripetal acceleration is a = v² / r, where v = 2πr / T (same period T for both bodies).

Thus, 

For the 10 kg body (radius R): a₁₀ = 4π²R / T².
For the 5 kg body (radius r): a₅ = 4π²r / T².
Ratio: a₁₀ / a₅ = R / r.

When current flows through a conducting wire, an electric field E exists inside the wire, driving the charge flow.
The field is parallel to the wire (along its length), as the current (charge motion) is along the wire.
(A) Inside, parallel: Correct. 
(B) Inside, normal: Incorrect, as the field aligns with current. 
(C, D) Outside: Incorrect, as the field is primarily inside for a steady current.

Answer: Correct (A: inside the wire but parallel to it).

Assertion: Not all radioactive elements decay to lead (e.g., the 4n+1 Neptunium series ends in bismuth). False.
Reason: Elements above lead (atomic number >82) are generally unstable, which is true.

Thus, the assertion is false, but the reason is true.

Initial moment of inertia of the first disc: I₁ = 1 kg·m², angular velocity: ω₁ = 100 rad/s.

Second identical disc: I₂ = 1 kg·m², initially at rest (ω₂ = 0).

By conservation of angular momentum: I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω.

1 × 100 + 1 × 0 = (1 + 1)ω → 100 = 2ω → ω = 50 rad/s.

Initial kinetic energy: Kᵉᵢ = 1/2 I₁ω₁² = 1/2 × 1 × 100² = 5000 J.

Final kinetic energy: Kᵉf = 1/2 (I₁ + I₂)ω² = 1/2 × 2 × 50² = 2500 J.

Loss in kinetic energy: Kᵉᵢ - Kᵉf = 5000 - 2500 = 2500 J = 2.5 kJ.

Answer: Correct (A: 2.5 kJ).

For an ideal gas, internal energy U = nC_v T depends on temperature.
By the first law, ΔU = Q + W. In adiabatic processes (Q = 0), ΔU = W.
In adiabatic expansion, the gas does work (W < 0), so ΔU < 0, and temperature decreases.
In adiabatic compression, work is done on the gas (W > 0), so ΔU > 0, and temperature increases.
In isothermal processes (ΔT = 0), ΔU = 0.

Thus, internal energy increases in adiabatic compression (option B)

Magnetic flux Φ = B × A.
For magnetic field B, use F = qvB: [B]
= [F] / ([q] [v])
= [M L T⁻²] / ([I T] [L T⁻¹])
= M T⁻² I⁻¹.
Flux: [Φ] = [B] [L²]
= (M T⁻² I⁻¹) (L²)
= M L² T⁻² I⁻¹.
The answer is option A.

Reverberation time T is given by Sabine’s formula: T = (0.161 V) / A,
where V is the auditorium’s volume and A is the total absorption.
For a given material, A ∝ S (surface area), and S ∝ V^(2/3) for a room.
Thus, T ∝ V / V^(2/3) = V^(1/3).
In simplified models, T is often proportional to V.

The answer is T ∝ V (option A).

Isotopes are atoms of the same element with different neutron numbers. Analyzing the statements:
(i) False: Isotopes have the same proton number but different neutron numbers.
(ii) False: Many elements have multiple stable isotopes (e.g., C-12, C-13).
(iii) True: All elements have isotopes (natural or synthetic).
(iv) True: All carbon isotopes (C-12, C-13, C-14) form CO and CO₂ with O-16, as chemical properties depend on protons.

Only (iii) and (iv) are correct (option C).

Orbital velocity for a satellite in a circular orbit is given by:

v = √(GM/r)

Where:

• G is the gravitational constant
• M is Earth’s mass
• r is the orbital radius

For a satellite just above Earth’s surface, r ≈ Rₑ, so:

v₀ = √(GM/Rₑ)

At an altitude of 2Rₑ, the orbital radius becomes:

r = Rₑ + 2Rₑ = 3Rₑ

So, the velocity at this altitude is:

v = √(GM/(3Rₑ)) = √(GM/Rₑ) / √3 = v₀ / √3

Thus, the ratio n is:

n = 3 

Given:

P_c = 300 W (carrier power)

μ = 0.75 (modulation depth)

Formula for total power in amplitude modulation: P_t = P_c (1 + μ² / 2)

Substituting the values: P_t = 300 (1 + (0.75)² / 2)

First, calculate (0.75)²: (0.75)² = 0.5625

Now, substitute this value into the formula: P_t = 300 (1 + 0.5625 / 2)
= 300 (1 + 0.28125) P_t
= 300 × 1.28125 P_t
= 384.375 W

So, the total power is approximately 384 W.

The rod’s weight is mg = 1 kg × 10 m/s² = 10 N.
The magnetic force F = I L B, where
I is the current, L = 1 m (rod length),
and B = 2 T (magnetic field normal to the rod).
For zero tension in the wires, the magnetic force balances the weight:
I × 1 × 2 = 10
 = 5 A.

The required current is 5 A.

Given the formula for the distance traveled in the nth second:

sₙ = u - g(n - 1/2)

Where:
u is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s²)
n is the nth second

The condition s₅ = s₆ indicates that the ball is at the same height at 5 seconds and 6 seconds. This symmetry suggests that the ball is at its peak at t = 5 s.

At the peak of its trajectory, the velocity is zero (v = 0), so: v = u - g × t = 0

Solving for u: u = g × t

Substitute t = 5 s and g = 9.8 m/s²:

u = 9.8 × 5 = 49 m/s

The time of flight T can be calculated using:

T = 2u / g = 2 × 49 / 9.8 = 10 s

Thus, the initial velocity u is 49 m/s, and the time of flight is 10 seconds.

So, the correct value for u is 49 m/s.

Answer: 49 m/s.

By the first law of thermodynamics, we have:
ΔU = Q + W

Where:
ΔU is the change in internal energy.
Q is the heat absorbed or released by the gas.
W is the work done by or on the gas.

Given:
Q = –30 J (heat released by the gas, negative since heat is released).
W = +18 J (work done on the gas, positive as work is done on the system).

Now, calculate the change in internal energy:
ΔU = Q + W = –30 J + 18 J = –12 J
Next, we use the initial internal energy to find the final internal energy. The initial internal energy is given as U_i = 60 J.
U_f = U_i + ΔU = 60 J – 12 J = 48 J

Thus, the final internal energy U_f is 48 J.