JEE Main Part Test - 3 Free Online Test 2026

Explanation:

The theory for ideal gases makes the following assumptions

1. Gases consist of particles in constant, random motion. They continue in a straight line until they collide with something—usually each other or the walls of their container.

2. Particles are point masses with no volume. The particles are so small compared to the space between them, that we do not consider their size in ideal gases.

3. No molecular forces are at work. This means that there is no attraction or repulsion between the particles.

4. Gas pressure is due to the molecules colliding with the walls of the container. All of these collisions are perfectly elastic, meaning that there is no change in energy of either the particles or the wall upon collision. No energy is lost or gained from collisions.

5. The time it takes to collide is negligible compared with the time between collisions.

6. The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas molecules have different speeds, but the temperature and kinetic energy of the gas refer to the average of these speeds.

7. The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move.

8. All gases at a given temperature have the same average kinetic energy.

9. Lighter gas molecules move faster than heavier molecules.

Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm² = 0.065 × 10^-4 m²
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω²
= 14.5 × 9.8 + 14.5 × 1 × (12.56)²
= 2429.53 N
Young’s modulus = Stress / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 10¹¹ Pa
∆l = 2429.53 × 1 / (0.065 × 10^-4 × 2 × 10¹¹)   =   1.87 × 10^-3 m
Hence, the elongation of the wire is 1.87 × 10^–3 m
Hence 1.87 × 10^–3 m

In projectile motion, the time taken for an object to reach the ground is independent of its horizontal velocity. The only factor that determines the time it takes for an object to reach the ground is the vertical motion (due to gravity).

Since both bullets A and B are fired horizontally, their initial vertical velocities are zero. The time taken for both bullets to fall will only depend on the height from which they are fired and the acceleration due to gravity.

The time t taken for an object to reach the ground from a certain height h is given by:

Where:

• h is the height from which the bullets are fired,
• g is the acceleration due to gravity.

This equation shows that the time to reach the ground depends only on the height and gravitational acceleration, not the horizontal velocity.

Thus, both bullets A and B will reach the ground in the same time. Therefore, option A is correct.

Explanation:

for monoatomic gas 

from conservation of energy

Explanation:the mean free path is the average distance traveled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas the internal energy is zero because at 0K nearly all molecular motion stops.

The value of Cp is 30.3
and as Cp-Cv = R(8.3)
hence Cv = 30.3-8.3
Cv is 22
change in internal energy = no of moles × Cv × change in temperature
hence
change in internal energy = 22 × 4 × 10
= 880j
Hence Option D is correct.
 

The work done by the gas in taking it from state A to state B = PΔV where ΔW is the increase in volume at constant pressure P. 

We have PV = μRT where p is the number of moles in the sample of the gas and R is the universal gas constant. 

Therefore we have PΔV = μR ΔT = 3 xR(450 - 250) = 600R 

• Time Period, T = 2π √(l/g')where,
  l = Length of seconds pendulum 
  g’ = Apparent Gravity
• For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
• Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.

When the displacement of a SHM is:
y=a sin wt+ b cos wt

• Amplitude of the SHM will be:
  A=√a²+b²

Here, a = 3, b = 4
Amplitude, A= √(3²+4²) = 5 cm

Hence option B is correct.

   
∴ We get, ω = √3 s^-1
   T = 2π / √3

K. E=1/2 K(A²-x²)
Max of mean position,
K. E=1/2 KA²
=1/2 x4x10⁵x(3x10^-2)²
=180J
T.M. E=180+P.E
230=180+P.E
P.E=230-180
P.E=50J

Maximum velocity: v = ωA, where ω is uniform angular velocity and a is the radius of the circle in which a reference particle executing S.H.M.
Velocity is maximum at mean positions. The maximum value of velocity is called velocity amplitude in SHM.
Acceleration is maximum at extreme position given by A = - ω²A. The maximum value of acceleration is called acceleration amplitude in SHM.

Energy is supplied to the damped oscillatory system at the same rate at which it is dissipating energy, and then the amplitude of such oscillations would become constant. Such oscillations are called maintained oscillations. By the definition of maintained oscillations.

Natural frequency is the frequency at which a body tends to oscillate in the absence of any driving or damping force.
Free vibrations of any elastic body are called natural vibration and happen at a frequency called natural frequency. Natural vibrations are different from forced vibration which happen at frequency of applied force .

In an ideal environment where there is no resistance to oscillatory motion, that is, damping is zero, when we oscillate a system at its resonant frequency, since there is no opposition to oscillation, a very large value of amplitude will be recorded. Forced oscillation is when you apply an external oscillating force.

The necessary condition for phenomenon of interference to occur are:
1. There should be two coherent sources.
2. The frequency and amplitude of both the waves should be same.
3. The propagation of waves should be simultaneously and in same direction.
These are the conditions, no explanation.

Given the wave equation:

Y = 10^-2 sin 2π (160t – 0.5x + π/4)

Comparing this equation with the general form:

Y = Asin (2π(ft - kx + φ))

We can identify the wave number k = 0.5 m^-1 and the frequency f = 160 Hz. 

The wave speed v can be found using the relationship between wave number, wave speed, and frequency: 

Now, we can calculate the wave speed:

Now, we need to convert the wave speed from meters per second to kilometers per hour:

So, the speed of the wave is 1152 Km/h

The correct answer is Option B.

The C-C-C bond angles in cyclopropane (60^o) and cyclobutane (90^o) are much different than the ideal bond angle of 109.5^o.This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5^o.

The correct answer is option D
Cyclopropane is a cycloalkane molecule with the molecular formula C₃H₆, consisting of three carbon atoms linked to each other to form a ring, with each carbon atom bearing two hydrogen atoms resulting in D₃H molecular symmetry. The small size of the ring creates substantial ring strain in the structure.

Priority of functional groups is
- COOH > - CHO >>C = O.
Hence, the lUPAC name is 4-formyl-2-oxocyclohexane carboxylicacid.

CH₃F - sp³ hybridised carbon, tetrahedral shape
(ii) HC = N - sp hybridised carbon, linear shape.

• Both have all their C—C bonds of equal length due to conjugation.
• I does not decolorises brown colour of bromine water solution but II does as The π bonds in Cyclooctatetraene (Compound II) react as usual for olefins, rather than as aromatic ring systems.
• I is planar but II is not as it adopts a tub conformation.
• Cyclooctatetraene shows various other addition reactions including Sulfonation.

Hence, Option A, B and D are correct.

1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.
Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:
2 K + C8H8 → K2C8H8
The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.

Metals like Potassium tend to loose their own electrons and the excess electrons will complete the aromaticity of the system.Thus, Deprotonation of the above compound converts it into an aromatic anion witn 6 pi electrons.