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Explanation:

Uniform motion refers to the motion of an object moving at a constant speed along a straight line. It is represented graphically by a straight line on a distance-time graph, where the slope of the line indicates the speed of the object.​

• Option (a): This represents a straight line, which is characteristic of uniform motion.
• Option (b): This represent uniform motion as the graph is gtraight line, constant speed.
• Option (c): This does not represent uniform motion since the graph is non-linear, suggesting acceleration.
• Option (d): This graph is unrelated to uniform motion as it does not indicate a proportional relationship between distance and time.

Therefore, the graph in Option (a and b) correctly represents uniform motion.

Therefore, the correct answer is Option 1.

Explanation:

Energy is essential for all life processes and activities on Earth. The ultimate source of this energy can be traced back to a single primary source that drives all other energy forms.

• Radioactive elements: These provide energy through radioactive decay, but they are not the ultimate source.
• Fossil fuels: These are derived from ancient organic matter, but they originate from a more primary source.
• Water: Water is a medium for energy transfer (e.g., hydropower) but not the source.
• Sun: The Sun is the ultimate source of energy. It drives the water cycle, supports plant growth (photosynthesis), and influences Earth's climate.

Therefore, the correct answer is Option 4.

Explanation:

The motion of a sound wave in air is represented by a sinusoidal wave. 

• Wavelength (λ) is the distance between two successive compressions or rarefactions in a sound wave.
• Time period (T) is the time taken for one complete oscillation.
• Wavelength/2 is half the distance of a complete wave cycle.
• Time period/2 is half the time taken for one complete oscillation.

The interval BD in the diagram represents the time duration for half an oscillation of the wave. This can be identified as:

Time period T for one complete oscillation ⇒ Time period/2 for half an oscillation.

The interval BD does not represent the full wavelength (λ), full time period (T), or half of the wavelength (wavelength/2). Instead, it corresponds to time period/2.

Therefore, the correct answer is Option 3.

Calculation:

Given:

Distance traveled along the circumference = 25 m

Time taken = 5.0 s

First, calculate the speed of the girl:

v = distance/time

v = 25/5 = 5.0 m/s

Next, calculate the radius of the circular path:

One-fourth of the circumference corresponds to the distance traveled:

One-fourth of circumference = 25 m ⇒ Circumference = 4 × 25 = 100 m

Circumference of a circle is given by C = 2πr:

100 = 2πr

r = 100/(2π) = 100/6.28 ≈ 15.92 m

Now, calculate the centripetal acceleration:

a = v²/r

a = (5.0)²/15.92

a = 25/15.92 ≈ 1.6 m/s²

Therefore, the correct answer is Option 3.

Explanation:

Universal constants are physical quantities that remain the same across the universe and do not depend on external factors like location, time, or experimental conditions. They are fundamental to the laws of physics.

Option 1: Acceleration due to gravity on a planet depends on the planet's mass and radius. It is not a universal constant.

Option 2: Gravitational constant (G) is a universal constant. It appears in Newton's law of gravitation and has a fixed value of approximately 6.674 × 10^-11 N·m²/kg².

Option 3: Gas constant (R) is specific to the behavior of gases and depends on the chosen unit system. It is not universal.

Option 4: Speed of light (c) in a vacuum is a universal constant, approximately 3 × 10⁸ m/s. However, it is not listed as the correct answer here.

Therefore, the correct answer is Option 2.

Calculation:

The minimum speed corresponds to the point where all the kinetic energy at P is converted into potential energy at the height h:

(1/2) m v² = mgh

⇒ (1/2) v² = gh

⇒ v² = 2gh

⇒ v = √(2gh)

Substitute g = 10 m/s² and h = 5 m:

v = √(2 × 10 × 5)

v = √100

v = 10 m/s

Therefore, the correct answer is Option 3.

Calculation:

Given:

Mass of the object, m = 2 kg

Force towards north, F_north = 6.0 N

Force towards east, F_east = 8.0 N

The resultant force:

F_resultant = √(F_north² + F_east²)

⇒ F_resultant = √(6.0² + 8.0²)

⇒ F_resultant = √100

⇒ F_resultant = 10 N

Using F = ma, we have:

a = F_resultant/m

⇒ a = 10/2

⇒ a = 5 m/s²

the direction of acceleration:

θ = tan^-1(F_north/F_east)

⇒ θ = tan^-1(6.0/8.0)

⇒ θ = tan^-1(3/4)

Thus, the acceleration is 5 m/s² at an angle of tan^-1(3/4) from east.

Therefore, the correct answer is Option 4.

Calculation:

The mass of the hypothetical planet is 1/100 that of Earth, and its radius is 1/4 that of Earth.

Acceleration due to gravity on Earth is g_Earth = 10 m/s².

Acceleration due to gravity on the hypothetical planet:

g_planet = g_Earth × (M_planet/M_Earth) × (R_Earth/R_planet)²

⇒ g_planet = 10 × (1/100) × (4)²

⇒ g_planet = 160/100

⇒ g_planet = 1.6 m/s².

Weight of the person on the hypothetical planet:

W = m × g_planet

⇒ W = 50 × 1.6

⇒ W = 80 N.

Therefore, the correct answer is Option 2.

Concept:

Orbitals:

• The area where there is a maximum probability of finding electrons is called an orbital.
• As stated by de- Broglie, we know that matter has dual nature, a wave, and a particle.
• The electrons in an atom also possess this dual nature. The wave nature is significant in them due to their small size and negligible mass.
• An orbital describes the wave-like nature of the electron.
• The orbitals are mathematical functions that are used to calculate the probability of finding an electron.
• The probability of finding an electron is maximum in these orbitals.
• The square of the wave functions gives us the probability density of the electrons.
• The wave functions have two parts :
• Radial - giving us the spread or size of the orbitals
• Angular- giving us the orientation of the orbitals in space.
• Plotting the wave functions in a three-dimensional coordinate system, we get the shape, size, and orientation of the orbitals.

Explanation:

• Half-filled and fully filled orbitals are most stable.

Reasons for the stability of half-filled and full filled orbitals:

• Exchange energy:
• all the orbitals in a given subshell have equal energies, the electrons present in different orbitals of the same subshell can exchange their positions.
• Such an exchange of electrons results in the release of energy known as exchange energy.
• The greater the exchange energy, the more is stability associated with the orbital.
• Exchange energies are highest for full-filled and half-filled orbitals.
• Symmetry:
• Symmetry leads to stability.
• A subshell is most stable if it is fully filled or half-filled because, in this state, electrons are symmetrically distributed in their orbitals.
• If shifting of electrons results in the symmetrical distribution of electrons in a subshell, the configuration becomes more stable.
• ​p3 and d5 are both half-filled electronic configurations and halve the highest symmetry.

Hence, the most stable orbitals are​ p3 and d5.

Explanation:

Henry’s Law states that “the partial pressure applied by any gas on a liquid surface is directly proportional to its mole fraction present in a liquid solvent.”

Henry’s law is given by:

P ∝ C (or) P = k_H C

Where, ‘P’ is the partial pressure of the gas, ‘C’ is the concentration of the dissolved gas and ‘k_H’ is the Henry’s law constant of the gas.

Limitations of Henry’s law 

• It is applicable when the system is in equilibrium.
• It is applicable for those gases only which do not react with water (solvent).
• Gas shouldn’t cause any chemical change in the solution.
• Henry’s law is applicable only when
• Pressure is not high
• Temperature is not too low
• Gas is extremely soluble.

So, when pressure of gas is high, Henry law is not applicable.