JEE Main Physics Test- 2 - JEE MCQ

• Liquid drop assumes a spherical shape because drop always tends to acquire minimum surface area for surface tension.
• And forgiven volume, the sphere has a minimum surface area.

Concept:

• The property of a liquid due to which its free surface tries to have the minimum surface area and behaves as if it were under tension somewhat like a stretched elastic membrane is called surface tension.
• The surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of the liquid.

Explanation:

• In the case of small drops of a liquid, the gravitational potential energy is negligible in comparison to the potential energy due to surface tension.
• Consequently, to keep the drop-in equilibrium, the liquid drop’s surface tends to contract so that its surface area will be the least for a sphere and the drops will be spherical.
• So, small drops of liquid are spherical because surface tension predominates over that of gravity and bigger drops are oval.

To solve this problem, we need to use the formula for the height to which a liquid will rise in a capillary tube due to capillary action. The formula is derived from the balance of gravitational and surface tension forces and is given by:

Where:

• h is the height the liquid rises,
• T is the surface tension of the liquid (in dyn/cm, or erg/cm²),
• θ is the contact angle between the liquid and the tube (for water in a clean glass tube, we often approximate θ as 0°, implying cos(θ) = 1),
• ρ is the density of the liquid (for water, ρ is approximately 1 g/cm³),
• g is the acceleration due to gravity (approximately 980 cm/s²),
• r is the radius of the tube (in cm).

Given:

• Surface Tension, T = 98 dyn/cm,
• Radius, r = 1 mm = 0.1 cm,
• g = 980 cm/s²,
• ρ = 1 g/cm³ (for water).

Plugging these values into the formula:

Thus, the maximum height of water that the capillary tube can hold when it is taken out of the water after being immersed up to 5 cm is 2 cm.
The correct answer is B: 2 cm.

• Density Considerations: Ice is less dense than water, which is why it floats. When ice melts, it converts into water, having a higher density than its solid form. This means that the volume of water originally displaced by the ice is greater than the volume of water produced from the melted ice.
• Volume and Water Level: Since the volume of water from the melted ice is less than the volume of ice originally displacing the water, the overall water level in the vessel will decrease when the ice melts. This is under the assumption that the vessel does not lose or gain water other than by the ice melting.
• Temperature Effects: The temperature of the water decreases from 10°C to 3°C. Water generally expands slightly as it cools down until about 4°C, below which it starts to contract. However, this expansion and contraction due to temperature changes in this temperature range (from 10°C to 3°C) is relatively minor compared to the volume change from the melting ice.

Given these points, the primary factor affecting the water level in the vessel is the conversion of ice (less dense) to liquid water (more dense), which results in a decrease in the displaced volume. Therefore, the water level will fall as the ice melts, and not rise, because the melted ice occupies less volume than the ice itself did.

Answer: A: fall

The water level falls simply because the ice, when melted, becomes water that occupies less volume than it did as ice. Temperature effects on water expansion or contraction in this case are secondary and do not reverse the trend set by the change from solid to liquid state.

To find the ratio of the amplitudes of the two simple harmonic motions given by the equations y_1​ and y₂​, we first need to analyze each equation.
For y₁​: y₁ = 10sin⁡(3πt + π/8) Here, the amplitude is clearly 10, because the amplitude of a sine or cosine function in the form Asin(ωt + ϕ) or Acos⁡(ωt + ϕ) is |A|.
For This equation uses trigonometric identities. Recognize that the expression within the parentheses is a standard trigonometric identity: cos⁡²A − sin⁡²A = cos⁡2A. Applying this identity withThus, the equation becomes: y₂ = 5cos⁡(3πt) Here, the amplitude is 5.
Ratio of the amplitudes: The ratio of the amplitudes A₁​ and A₂​ from the given equations is:

Therefore, the amplitudes of the two simple harmonic motions are in the ratio of 2 : 1. This indicates that the first motion has an amplitude twice that of the second motion.

r = 12 mm

When the effective length of a pendulum is comparable to the radius of the earth, then its tangential acceleration cannot be considered the same throughout its motion, since the value of acceleration due to gravity varies with height above the surface of the earth.

Hence, a different general formula exists for the time period in this case.

For a pendulum with effective length l, comparable to the radius of the Earth R, the time period T is given by: 

where g is the acceleration due to gravity.
The radius of the earth is R.
Let the required time period of the pendulum be T.
We are given the effective length of the pendulum (l) is equal to the radius of the earth. Hence, l = R.
Therefore, putting this value and using equation (1), we get:

Hence, the time period of the pendulum is 

The wave represented is y = 3cos ⁡π(100t − x). The wave velocity can be deduced by analyzing the phase term π(100t − x). The phase speed V of the wave, where the argument of the cosine function is βt − kx (with β being angular frequency in time and k being wave number), is given by: V = β​/k
For this equation, β = 100π and k = π so: V = 100π/π = 100 cm/s
To find the maximum particle velocity, differentiate y with respect to t: dy/dt = −3π.
100 sin ⁡π(100t − x) The maximum value of sin⁡ is 1, so the maximum particle velocity is:
Max(dy/dt) = ∣−3π ⋅ 100∣ = 300π 
Hence, the wave velocity is 100 cm/s and the maximum particle velocity is 300π cm/s, corresponding to Answer D.

Let pressure at A = P.
Net pressure = P_₀ - P.
Net resultant pressure on the hemispherical depression = (P_₀ - P)*πr² along the axis of the tube.
The surface tension along the circular edge of the depression = 2πrS, along the axis of the tube. Equating we get, 

(P_₀ - P)*πr² = 2πrS 

⇒ (P_₀ - P) = 2S/r 

⇒ P = P_₀ - 2S/r 

To find the wavelength of ultrasonic waves emitted by dolphins in water, we use the basic wave equation:
λ = v​/f
Where:

• λ is the wavelength,
• v is the speed of sound in water,
• f is the frequency of the sound wave.

Given:

• Frequency (f) = 250 kHz = 250,000 Hz

The speed of sound in water is generally taken as approximately 1500 m/s (this value can vary slightly depending on temperature and salinity).
Using these values:

The closest answer in the options provided is A: 5.76 mm.

In an organ pipe closed at one end there will be an antinode at the open end and a node at closed end. The distance between an antinode and nearest node is 4λ​. If i₁​ be the length of pipe, and λ the wavelength then,

If n be the frequency of note emitted and v the velocity of sound in air, then

Also, closed pipe produces only odd harmonics, hence first overtone of closed pipe is

For open pipe antinodes are formed at both ends, frequency is given by

Also open pipe produces both even and odd harmonics,

hence we have freqency of third overtone of open pipe is

At resonance both the frequencies are equal

• A-Q: Longitudinal progressive waves typically involve oscillations along the direction of wave propagation, which is characteristic of sound waves traveling through mediums like air or fluids. In this case, vibrations of the air column of resonance accurately represent this type of wave as seen in wind instruments where longitudinal sound waves resonate within the instrument's air column.
• B-P: Transverse progressive waves involve oscillations perpendicular to the direction of wave propagation. Ripples formed on water surface are a classic example of transverse waves where the water surface displaces perpendicular to the direction of wave propagation, showcasing the characteristics of transverse progressive waves.
• C-S: While tuning forks are generally not the first example for stationary waves, they can still be associated with longitudinal stationary waves if we consider the concept of sound waves reflecting off surfaces and creating standing waves through interference. Thus, tuning fork vibrating in waves air is matched here, though it's traditionally more associated with progressive waves. In the context of this question set, it illustrates the longitudinal component of the sound waves produced by the tuning fork.
• D-R: Transverse stationary waves occur when the oscillations are perpendicular to the wave's direction, and there are points (nodes) that do not move, and points (antinodes) with maximum oscillation. A vibrations of stretched wire in a sonometer setup exhibit this behavior, where a wire under tension vibrates in segments, creating nodes and antinodes that demonstrate transverse stationary waves.

1. All the particles perform simple harmonic motion with a frequency which is four times that of the two component waves: This statement is incorrect. In stationary waves, the frequency of the particle motion matches the frequency of the component waves that form the stationary wave. The frequency does not multiply.
2. Particles on the opposite sides of a node vibrate with a phase difference of π: This statement is correct. Particles on opposite sides of a node in a stationary wave are out of phase by π (180 degrees). This means when one side is at its maximum positive displacement, the other side is at its maximum negative displacement.
3. The amplitude of vibration of a particle at an antinode is equal to that of either component wave: This statement is incorrect. The amplitude at an antinode is the sum of the amplitudes of the two interfering waves, typically resulting in a maximum amplitude that is twice that of each individual component wave, assuming they are of equal amplitude.
4. All the particles between two adjacent nodes vibrate in phase: This statement is correct. Within the span between two nodes (in one loop of the wave), all the particles move in phase—they reach their maximum and minimum displacements simultaneously.

The correct and logically consistent statements, upon re-evaluation, are 2 and 4, indicating that particles on opposite sides of a node are out of phase (statement 2), and particles between nodes are in phase with each other (statement 4).

Answer Option: A: 2 and 4 are correct.

For no slipping v₀ = ωR
Now v_A = v_B = 

Centripetal force = mω²r

Escape velocity is the minimum velocity with which a body is projected from the surface of the planet so as to reach infinity, by overcoming the pull by gravity.
Escape velocity at the surface of a planet is given by:

Where G = gravitational constant (6.67 × 10^-11 Nm²/kg²), M = mass of the planet and R = radius of the planet.
Explanation:

• An object cannot escape out of the earth's gravity unless it has a speed equal to or greater than the escape speed.
• The atmosphere is retained by the earth because the escape velocity is greater than the mean speed of the molecules of the atmospheric gases.

In case of pure rolling, ratio of rotational to translational kinetic energy is 2/5. Therefore, total kinetic energy is 7/5 times the translational kinetic energy. At maximum compression, whole of energy is elastic potential. Hence,

∴ The compression of the spring, 

Given Data:

• Moment of Inertia (I): 5 x 10^-3 kg m²
• Initial angular velocity (ω_i​): 10 revolutions per second
• Time to stop (t): 20 seconds

Conversion and Kinematic Equations:

First, we convert revolutions per second to radians per second for angular velocity:

• 1 revolution = 2π radians
• ω_i = 10 rev/s × 2π rad/rev = 20π rad/s

Since the wheel is stopped, the final angular velocity (ω_f) is:
ω_f​ = 0 rad/s

Using the first equation of motion for rotational motion, where angular acceleration (α\alphaα) is constant: ω_f = ω_i + αt

We rearrange to solve for: 
Calculation:

The angular retardation (α\alphaα) of the wheel is . This negative sign indicates that it is a retardation (or deceleration) since the wheel is slowing down.

Amount of work done in an isothermal expansion is given by

= 1728.98J ≈ 1728 J