SRMJEE Practice Test (Engg) - 1 - JEE MCQ

In DC circuits, a capacitor initially allows current during charging but becomes fully charged quickly. After charging, it behaves like an open circuit, resulting in infinite resistance as no steady current flows through it.

To solve the equation for θ, we use the formula:

θ = (1/2)αt²

Given the values:

• θ = π
• α = 1 rad/s²

We can rearrange the equation to find t:

Substituting the values into the equation:

π = (1/2)(1 rad/s²)t²

This simplifies to:

t² = (2π) / (1 rad/s²)

Calculating t:

t = √(2π) = √6.28 ≈ 2.51 s

Current in the circuit

I = 2E​ / (R+r₁​+r₂​)

∴ Potential difference across first cell

V= E−Ir

= E − 2Er1 / (R+r1​+r2​)

On solving we get, 

R = (r₁​−r_2​)

Cathode rays and visible light rays are similar in that they both can affect photographic plates. This means both can expose photographic film, causing it to react when exposed.

• Option A is incorrect because while cathode rays (being charged particles) can be deflected by electric and magnetic fields, visible light cannot.
• Option B is incorrect because while visible light has a definite wavelength, cathode rays do not; their 'wavelength' depends on the velocity of the electrons.
• Option C is incorrect because only cathode rays, not visible light, can ionise gases.
• Option D is correct as both can affect photographic plates.

The breaking point of a wire depends on the cross-sectional area and the material's tensile strength. Since both wires are made from the same material, their tensile strengths are equal.

Tensile strength (σ) is given by the formula:

σ = F / A, where:

• F is the force
• A is the cross-sectional area

For Wire 1:

• Diameter d₁ = 1 mm → Radius r₁ = 0.5 mm
• Cross-sectional Area A₁ = π(r₁)²

For Wire 2:

• Diameter d₂ = 2 mm → Radius r₂ = 1 mm
• Cross-sectional Area A₂ = π(r₂)² = π(1)² = π mm²

Since A₂ is four times larger than A₁ (because area scales with radius squared), the force F required for Wire 2 to break is four times that of Wire 1.

Thus, we have:

F₂ = 4 × F₁ = 4 × 1000 N = 4000 N.

Therefore, the correct answer is D) 4000 N.

The decibel (dB) gain is calculated using the formula:

Gain (dB) = 10 * log(P_out / P_in)

Given a 20 dB gain, we can solve for the ratio:

• 20 = 10 * log(ratio)
• Leading to: ratio = 10² = 100

Thus, the power ratio is 100.

By Gauss's Law, the total electric flux through a closed surface is given by:

Φ_total = q_enclosed / ε₀.

For a cube with charge q at its centre, the symmetry implies that the flux is equally distributed among all six faces. Thus, the flux through one face is:

• Φ_face = Φ_total / 6 = q / (6ε₀).

This matches option A.

To determine the minimum force F required to move the cube up the rough inclined plane, we analyse the forces acting on it:

• Weight of the Cube: The weight is given as W = 10 N.
• Components of Weight:
  • Component along the incline: W sin(θ) = 10 × (3/5) = 6 N.
  • Component perpendicular to the incline: W cos(θ) = 10 × (4/5) = 8 N.
• Frictional Force: The maximum static friction is given by: f_max = μ × Normal Force = 0.6 × 8 N = 4.8 N.
• Total Resistance: To move the cube up, the applied force must overcome both the component of weight along the incline and friction: F_min = W sin(θ) + f_max = 6 N + 4.8 N = 10.8 N.

Thus, the minimum force required is 10.8 N, corresponding to option B.

The acceleration due to gravity, g, near the Earth's surface is not constant but varies with location. This variation is primarily due to two factors:

• Earth's Oblateness: The Earth is flatter at the poles and bulges at the equator. As a result, the gravitational acceleration is greater at the poles than at the equator because the distance from the centre of the Earth is smaller at the poles.
• Centrifugal Effect Due to Rotation: The Earth's rotation causes a centrifugal force that opposes gravity. This effect is most pronounced at the equator (latitude 0°), where the radius of rotation is largest, thus reducing the effective gravitational acceleration. At higher latitudes (approaching the poles), the radius of rotation decreases, and so does the centrifugal effect, leading to an increase in the effective gravitational acceleration.

Therefore, as latitude increases (moving from the equator toward the poles), the acceleration due to gravity increases. This means that g varies directly with latitude. Hence, the correct answer is B.

Boyle's Law states that the pressure of a gas is inversely proportional to its volume when temperature and mass are held constant. For this law to hold, the gas must:

• Behave ideally (perfect gas)
• Maintain a constant temperature
• Have a fixed mass (constant amount of gas)

Therefore, option A is correct as it satisfies all necessary conditions.

In a weightless environment, convection cannot occur because it relies on gravitational effects to drive fluid movement. However, heat can still be transmitted through the liquid via conduction, where energy is transferred directly between molecules in contact with each other.

Thus, the correct answer is:

• Conduction (Option A)

To determine the time period for which the force acts on the body, we use the impulse-momentum theorem, which states that impulse (force multiplied by time) equals the change in momentum.

When a current-carrying circular loop is freely suspended, it behaves like a magnetic dipole. The Earth's magnetic field exerts a torque on the loop, causing it to align such that its plane becomes perpendicular to the Earth's magnetic field lines.

Since the Earth's magnetic field near its surface runs approximately horizontal and north-south, the loop orients its plane in the east-west direction to minimise potential energy. Therefore, the correct answer is option C: East-West.

When a ferromagnetic material is heated above its Curie temperature, the following occurs:

• The thermal energy disrupts the alignment of magnetic domains.
• The magnetic domains begin to arrange randomly instead of in an ordered manner.

Using Torricelli's Law, the formula for velocity is given by:

v = √(2gh)

When the depth increases to 4h, the new velocity can be calculated as follows:

• Substituting into the formula, we have:
• v = √(2g(4h))
• This simplifies to:
• v = √(8gh)
• Which can be rewritten as:
• v = 2√(2gh)
• This shows that:
• v = 2v₁
• Given that v₁ = 5 m/s, we find:
• v = 10 m/s

Given that a body is moved with constant power, we need to determine the proportionality of distance with time.

• Power and Force Relation: Power P = F · v, where F is force and v is velocity.
• Newton's Second Law: F = m · a, so P = m · a · v.
• Acceleration as Derivative: a = dv/dt, thus P = m · v · dv/dt.
• Differential Equation: m · v dv = P dt. Integrating both sides:
• ∫₀^v m · v' dv' = ∫₀^t P dt'.
• Solving gives v = √(2Pt/m).
• Distance Integration: Integrate velocity to find distance:
• s(t) = ∫₀^t √(2P/m) t' dt'.
• This results in s(t) ∝ t^3/2.

Thus, the distance moved is proportional to t^3/2, making option C correct.

The system's acceleration is calculated using the formula for an Atwood machine:

a = [(m₂ - m₁)/(m₁ + m₂)] * g.

Substituting the values:

• m₁ = 3 kg
• m₂ = 4 kg
• g = 9.8 m/s²

This results in:

a ≈ 1.4 m/s².

The period T of a simple pendulum is given by the formula:

T = 2π√(L/g), where:

• L is the length of the pendulum
• g is the acceleration due to gravity

When the length L is made four times, the new period T' becomes:

T' = 2π√(4L/g)

This can be simplified as follows:

• T' = 2 × 2π√(L/g)
• T' = 2T

Thus, the period is doubled when the length is made four times.

Given that the horizontal range R is three times the maximum height H, we use the projectile motion formulas:

• R = (v² sin(2θ)) / g
• H = (v² sin²(θ)) / (2g)

Setting R = 3H:

(v² sin(2θ)) / g = 3 × (v² sin²(θ)) / (2g).

Simplifying, we get:

• sin(2θ) = (3/2) sin²(θ).

We have:

• 2 sin(θ) cos(θ) = (3/2) sin²(θ).

Dividing both sides by sin(θ) (assuming sin(θ) ≠ 0):

• 2 cos(θ) = (3/2) sin(θ).

Thus, we find:

• tan(θ) = 4/3.

Therefore, θ = arctan(4/3) ≈ 53.13°.

The closest option is D, 53°8'.

The surface tension (T) is calculated using the formula:

T = F / (2L),

where:

• F is the force,
• L is the length of the wire.

Given:

• F = 200 N,
• L = 0.1 m.

We find:

T = 200 / (2 * 0.1) = 1000 N/m.

Since none of the options match this result, the correct answer isn't listed among the choices provided.

Using Snell's Law and the given geometric condition, we can express the angle of refraction (φ) as:

φ = 90° - θ

Substituting this into Snell's Law leads to the conclusion that the refractive index is:

μ = tan(θ), confirming that option C is correct.

To find the moment of inertia (I) of the wheel, we use the relationship between torque (τ), angular acceleration (α), and moment of inertia:

• τ = I · α

Rearranging for I gives us:

• I = τ / α

Given:

• τ = 31.4 N-m
• α = 4π rad/s²

Substituting the values:

• I = 31.4 / (4π)

Calculate 4π:

• 4π ≈ 12.566

Now calculate I:

• I = 31.4 / 12.566 ≈ 2.5 kg-m²

Thus, the moment of inertia is approximately 2.5 kg-m², which corresponds to option A.

In 15s the hand moves through 90º.
If the speed of the tip is v, then

The soap bubble has two surfaces (inner and outer).

Initial surface area:

• Initial Surface Area = 8πr²

After doubling the radius, the surface area becomes:

• New Surface Area = 32πr²

The change in area (ΔA) is:

• Change in Area (ΔA) = 32πr² - 8πr² = 24πr²

The work done (energy) is calculated as:

• Work Done = T × ΔA
• Substituting ΔA, we find:
• Work Done = 24πr²T

When a copper sphere is heated, it undergoes thermal expansion. This phenomenon can be understood through the following key points:

• The percentage change in volume is proportional to three times the linear expansion coefficient.
• The percentage changes in radius are proportional to the linear expansion coefficient.
• The percentage changes in area are proportional to twice the linear expansion coefficient.

As a result, the maximum percentage change is observed in volume.