Physics

Physical properties and states of matter Questions

Chapter: Properties of Matter | Total Available: 22 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

 All objects experience a buoyancy when theyare immersed in a fluid. Buoyancy is

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (c) is correct. Buoyancy is an upward force acting on objects in a fluid.

πŸ“– Governing Law / Core Concept

According to Archimedes' Principle, any object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object.

πŸ§ͺ Step-by-Step Breakdown

Let \( F_b \) be the buoyant force:

\[ F_b = \rho_{fluid} \cdot g \cdot V_{displaced} \]
  • Where \( \rho_{fluid} \) = density of fluid (kg/mΒ³)
  • g = acceleration due to gravity (m/sΒ²)
  • \( V_{displaced} \) = volume of fluid displaced (mΒ³)

This upward force explains why objects float or sink based on their density relative to the fluid.

πŸ” Option Analysis

  • A: a downward force - Incorrect; buoyancy opposes gravity.
  • B: a downward pressure - Incorrect; pressure in fluids acts in all directions, buoyancy is a force.
  • C: an upward force - Correct; it acts opposite to gravity.
  • D: an upward pressure - Incorrect; pressure contributes to buoyancy but is not the force itself.

⚑ Mnemonic / Speed-Run

Remember the phrase: β€œBuoyancy Beats Gravity” to recall that buoyant force acts upward.

6. Visual Suggestion

6. Visual Suggestion

Illustrate a diagram of an object submerged in water, highlighting the upward buoyant force arrows and the displaced water volume beneath it.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class IX Science, Chapter 9 (Gravitation)

Question 2 of 10 Authentic UPSC NDA Question

Which one among the following statements is correct?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option D is correct. Pascal and N/mΒ² represent the same unit of pressure.

πŸ“– Governing Law / Core Concept

Pressure is defined as force per unit area, represented mathematically as:
\[ P = \frac{F}{A} \]
where \( P \) is pressure, \( F \) is force (in Newtons), and \( A \) is area (in square meters).

πŸ§ͺ Step-by-Step Breakdown

Pressure is measured in Pascals (Pa) in SI units. 1 Pascal is defined as:
\[ 1 \text{ Pa} = 1 \text{ N/m}Β² \]
Thus, Pascal and N/mΒ² are directly equivalent, confirming that Option D is correct.

πŸ” Option Analysis

  • Option A: Incorrect; pressure is a scalar quantity, having only magnitude, not direction.
  • Option B: Incorrect; pressure increases with depth in a liquid due to the weight of the liquid above.
  • Option C: Incorrect; action and reaction forces (Newton's Third Law) act on different objects, not the same object.

⚑ Mnemonic / Speed-Run

Remember: Pressure is Force per Area (\( P = F/A \)) & 1 Pascal = 1 N/mΒ².

6. Visual Suggestion

A diagram showing a vertical column of liquid with pressure arrows pointing downwards at varying depths, illustrating how pressure increases with depth.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Liquid Pressure.

Question 3 of 10 Authentic UPSC NDA Question

Which one of the following statements about surface tension is correct?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (b) is correct. Surface tension decreases with an increase in temperature.

πŸ“– Governing Law / Core Concept

Surface tension (\( \gamma \)) is defined as the force per unit length acting along the surface of a liquid. It can be expressed mathematically as:

\[ \gamma = \frac{F}{L} \]

Where \( F \) is the force acting along the surface, and \( L \) is the length over which the force acts. SI unit of surface tension is \( N/m \).

πŸ§ͺ Step-by-Step Breakdown

As temperature increases, the kinetic energy of the molecules in a liquid also increases. This leads to an increase in intermolecular spacing, thereby reducing cohesive forces. The relationship between surface tension and temperature can typically be represented as:

\[ \gamma = \gamma_0 \left(1 - \frac{T}{T_c}\right) \]

Where \( \gamma_0 \) is the surface tension at temperature \( T = 0 \degree C \), and \( T_c \) is the critical temperature for the liquid.

πŸ” Option Analysis

  • Option A: Incorrect. Surface tension typically decreases with increased temperature.
  • Option C: Incorrect. Surface tension changes with temperature variations.
  • Option D: Incorrect. While it decreases, it's not stated that it must decrease with de/increase in temperature only.

⚑ Mnemonic / Speed-Run

Remember: "Heat loosens; tension decreases."

6. Visual Suggestion

6. Visual Suggestion

Illustrate a liquid droplet on a surface, showing the forces acting on the surface and how it changes shape with temperature increase. Use arrows to demonstrate molecular motion and intermolecular forces.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 2 (Fluid Mechanics).

Question 4 of 10 Authentic UPSC NDA Question

A body sinks in water when

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (b) is correct. A body sinks in water when its density is greater than that of water.

πŸ“– Governing Law / Core Concept

The principle governing the sinking of a body in fluids is Archimedes' Principle:

The buoyant force \( F_b \) acting on a submerged body is given by:

\[ F_b = \rho_{fluid} \cdot V_{displaced} \cdot g \]

Where:

  • \( \rho_{fluid} \) = Density of the fluid (water) in kg/mΒ³
  • \( V_{displaced} \) = Volume of fluid displaced in mΒ³
  • \( g \) = Acceleration due to gravity (9.81 m/sΒ²)

πŸ§ͺ Step-by-Step Breakdown

A body will sink if its overall density is greater than that of the fluid. This is determined as follows:

  • Density of a body is defined as \( \rho_{body} = \frac{m_{body}}{V_{body}} \), where \( m_{body} \) is mass and \( V_{body} \) is volume.
  • A body sinks when \( \rho_{body} > \rho_{water} \).
  • Consequently, if a body with a density greater than that of water is placed in water, the gravitational force acting downward exceeds the buoyant force acting upward.

πŸ” Option Analysis

  • Option (a): Incorrect. This describes a body that will float.
  • Option (c): Incorrect. Equal densities will result in the body neither sinking nor floating.
  • Option (d): Incorrect. Weight alone doesn’t determine sinking; density relative to the fluid matters.

⚑ Mnemonic / Speed-Run

Remember: "Sink or Swim? Check the Density!" This helps recall that the sinking of the body is dictated by its density compared to the fluid.

6. Visual Suggestion

6. Visual Suggestion

A diagram showing a body partially submerged in water, with density annotations for the body and the water, along with arrows representing gravitational and buoyant forces. This visual representation will reinforce the concept of density comparison for students.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 10 (Mechanical Properties of Solids).

Question 5 of 10 Authentic UPSC NDA Question

Twinkling of stars is due to

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (c) is correct. The twinkling of stars is primarily caused by atmospheric refraction of starlight.

πŸ“– Governing Law / Core Concept

The phenomenon of the twinkling of stars can be explained through the principles of atmospheric refraction. When starlight enters Earth’s atmosphere, it passes through layers of air with varying densities, leading to the bending of the light path.

πŸ§ͺ Step-by-Step Breakdown

Light travels at different speeds in different mediums. The critical concepts include: - **Refraction Index**: \( n = \frac{c}{v} \) where \( n \) is the refractive index, \( c \) is the speed of light in vacuum, and \( v \) is the speed of light in the medium. - As starlight passes through various atmospheric layers, it is refracted, causing the apparent position of stars to shift slightly. This results in the perception of twinkling. \[ n = \frac{c}{v} \] The change in refractive index due to temperature variations at different altitudes leads to the continuous shifting of the light from the star, causing it to appear to twinkle.

πŸ” Option Analysis

- **A: Particular frequencies of the starlight** - Incorrect, as twinkling is not related to frequency but to light bending. - **B: Reflection from the oceanic surface** - Incorrect, ocean reflections do not cause twinkling; they scatter light. - **D: Magnetic field of Earth** - Incorrect, as the magnetic field does not influence light from stars.

⚑ Mnemonic / Speed-Run

**Remember:** "Stars Twinkle Through Atmosphere" (STTA) to recall that twinkling is due to atmospheric conditions affecting light.

6. Visual Suggestion

6. Visual Suggestion

A ray diagram illustrating light rays from a star entering Earth's atmosphere, bending at different layers, and eventually reaching an observer's eye. Label the refraction angles and layers of the atmosphere.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 10 (Light).

Question 6 of 10 Authentic UPSC NDA Question

Buoyancy is a/an

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (d) is correct. Buoyancy is an upward force exerted by a fluid.

πŸ“– Governing Law / Core Concept

Buoyancy is governed by Archimedes' Principle, which states that an object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object.

πŸ§ͺ Step-by-Step Breakdown

When an object is placed in a fluid, the pressure difference between the top and bottom surfaces of the object results in an upward force. This can be mathematically expressed as:

\[ F_b = \rho \cdot g \cdot V \]
Where:
  • \( F_b \) = Buoyant force (N)
  • \( \rho \) = Density of the fluid (kg/mΒ³)
  • \( g \) = Acceleration due to gravity (9.81 m/sΒ²)
  • \( V \) = Volume of fluid displaced (mΒ³)

In this scenario, the fluid exerts more pressure at the bottom of the object than at the top, resulting in a net upward force responsible for buoyancy.

πŸ” Option Analysis

Option A (upward pressure) incorrectly describes buoyancy as pressure rather than force. Option B (downward pressure) refers to atmospheric pressure and not relevant to buoyancy. Option C (downward force) contradicts the principle of buoyancy exerted by a fluid.

⚑ Mnemonic / Speed-Run

Remember the phrase β€œArchimedes’ Upward Push” to link buoyancy with its characteristic upward force.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating an object submerged in water, with arrows indicating fluid pressure at the bottom and top surfaces, highlighting the resultant upward buoyant force.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class IX Science, Chapter on Force and Pressure.

Question 7 of 10 Authentic UPSC NDA Question

Tyndall effect is a phenomenon of

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (A) is correct. The Tyndall effect is caused by the scattering of light by colloidal particles.

πŸ“– Governing Law / Core Concept

The Tyndall effect states that when a beam of light passes through a colloidal solution, it is scattered by the particles, making the path of light visible. This phenomenon can be explained by Rayleigh scattering principles.

πŸ§ͺ Step-by-Step Breakdown

Light scattering occurs when light interacts with particles whose dimensions are comparable to its wavelength. The intensity of scattered light can be described by the following relation:
\[ I \propto \frac{1}{\lambda^4} \]
Where: - \( I \) = Intensity of scattered light - \( \lambda \) = Wavelength of light (in meters) The short wavelength light (blue) is scattered more than longer wavelengths (red).

πŸ” Option Analysis

- Option (B): Refraction of light refers to the bending of light waves as they pass from one medium to another, unrelated to Tyndall effect. - Option (C): Dispersion of light involves the separation of light into its constituent colors, also not linked to Tyndall effect. - Option (D): Refraction by dust particles is not the mechanism of Tyndall effect, which specifically deals with colloids.

⚑ Mnemonic / Speed-Run

To remember Tyndall effect, think of "Colloidal Clarity" – light becomes visible when scattered by colloids.

6. Visual Suggestion

A schematic diagram showing a beam of light entering a colloidal solution, with scattered light waves represented, illustrating the visibility of the light path due to Tyndall effect.

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Optics.

Question 8 of 10 Authentic UPSC NDA Question

Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

Resistance \( R \) of a cylindrical wire is given by the formula:
\[ R = \frac{\rho l}{A} \]
where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area of the wire. For the initial wire with radius \( r \) and length \( l \): \[ A_1 = \pi r^2 \] Thus, \[ R_1 = \frac{\rho l}{\pi r^2} \] For the second wire with radius \( 2r \) and length \( \frac{l}{2} \): \[ A_2 = \pi (2r)^2 = 4\pi r^2 \] So, \[ R_2 = \frac{\rho \left(\frac{l}{2}\right)}{A_2} = \frac{\rho \left(\frac{l}{2}\right)}{4\pi r^2} = \frac{\rho l}{8\pi r^2} = \frac{R_1}{4} \] Now substituting for \( R_1 \): \[ R_2 = \frac{R_1}{4} = \frac{\frac{\rho l}{\pi r^2}}{4} = \frac{\rho l}{4\pi r^2} = \frac{R}{8} \] Thus, the resistance of the second wire is: \[ R_2 = \frac{R}{8} \]
πŸ“Œ Hints / Properties Used:
  • Resistance formula: \( R = \frac{\rho l}{A} \)
  • Area of circular cross-section: \( A = \pi r^2 \)
πŸ“Š Visual Diagram Suggestion:

A diagram illustrating the two wires side by side with labeled dimensions (radius and length) and the respective areas highlighted to show the relationship between radius, area, and resistance.

πŸ“– Factual Verification & Reference:

Verified against standard physics texts covering electrical properties of conductors.

Question 9 of 10 Authentic UPSC NDA Question

A student measures certain lengths using a meter scale having least count equal to 1 mm. Which one of the following measurements is more precise?

Step-by-Step Logic & Solution

Correct Answer

βœ“ Option (c) is correct. It provides the measurement with the highest significance and least uncertainty.

πŸ“– Governing Law / Core Concept

A measurement's precision is determined by the number of significant figures it encompasses. The least count of the measuring device (1 mm) directly influences the accuracy of the measurements.

πŸ§ͺ Step-by-Step Breakdown

Assessing the provided options:

  • A: 0.50 mm - 2 significant figures
  • B: 29.07 cm - 4 significant figures
  • C: 0.925 m - 3 significant figures (equivalent to 92.5 cm, retaining precision)
  • D: 910 mm - 3 significant figures

Comparing the significant figures:

  • 0.50 mm -> 2 significant figures
  • 29.07 cm -> 4 significant figures
  • 0.925 m -> 3 significant figures (most meaningful due to consistent conversion)
  • 910 mm -> 3 significant figures

Thus, option (c) provides the most informative measurement amidst the others.

πŸ” Option Analysis

  • A: Less precise due to fewer significant figures.
  • B: Highest precision, but does not align effectively with meter scale interpretation.
  • C: Most effective representation, maintaining significant figure integrity.
  • D: Equivalent to 910 mm, equal in count but less significant than (c).

⚑ Mnemonic / Speed-Run

  • Significant Figures mnemonic: "Count it right, measure it bright!"

πŸ“– Factual Verification & Reference

πŸ“– Factual Verification & Reference:

Verified against NCERT Class IX Science, Chapter 8 (Motion).

Question 10 of 10 Authentic UPSC NDA Question
A barometer measures atmospheric pressure. If the height of mercury in a barometer is 76 cm, what is the atmospheric pressure in pascals? (Density of mercury = 13,600 kg/mΒ³, g = 9.81 m/sΒ²)
Step-by-Step Logic & Solution
[CORE CONCEPT]: Atmospheric pressure can be calculated using the formula P = hρg, where h is the height of the mercury column, ρ is the density of mercury, and g is the acceleration due to gravity. [STEP-BY-STEP SOLUTION]: P = 0.76 m Γ— 13,600 kg/mΒ³ Γ— 9.81 m/sΒ² = 101,325 Pa. [SHORTCUT / EXAM TRICK]: Remember that standard atmospheric pressure is approximately 101,325 Pa, which corresponds to 76 cm of mercury. [COMMON MISTAKE TRAP]: Students may forget to convert cm to m, leading to incorrect calculations.
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