Physical Properties of Materials

21. Which one of the following regarding density of water at atmospheric p

Which one of the following regarding density of water at atmospheric pressure is correct ?

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Density of water at 4°C is 1000 kg/m³.
Density of water at 0°C is 1000 kg/m³.
Density of water at 0°C is 100 kg/m³.
Density of water at 4°C is 10 kg/m³.
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Density of water at atmospheric pressure is maximum at 4°C, and its value is approximately 1000 kg/m³.
Water exhibits anomalous expansion; its density increases from 0°C to 4°C, where it reaches its maximum density, and then decreases as the temperature rises above 4°C. At 0°C, water is less dense than at 4°C (whether in liquid or solid state).
The density of water at 0°C (liquid) is about 999.84 kg/m³, and the density of ice at 0°C is about 916.7 kg/m³. The value 1000 kg/m³ is a commonly used approximation for the density of liquid water at 4°C and is equivalent to 1 g/cm³.

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22. Which one of the following can charge an insulator ?

Which one of the following can charge an insulator ?

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Current electricity
Static electricity
Magnetic field
Gravitational field
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UPSC NDA-1 – 2019
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The correct answer is (B) Static electricity.
Insulators are materials that resist the flow of electric current. However, they can be charged through the accumulation of static electric charge on their surface or within their volume. This typically happens through processes like friction (triboelectric effect), contact with a charged object, or induction, which involve the transfer or redistribution of electrons, creating an imbalance of charge.
Current electricity involves the continuous flow of charge, which is impeded by insulators. Magnetic fields and gravitational fields are fundamental forces but do not directly cause the accumulation of charge on an insulator in the way static electricity does. Charging an insulator creates an electrostatic field around it.

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23. Which one of the following metals is used in the filaments of photo-el

Which one of the following metals is used in the filaments of photo-electric cells that convert light energy into electric energy ?

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Tungsten
Copper
Rubidium
Aluminium
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UPSC NDA-1 – 2018
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Photo-electric cells, which convert light energy into electrical energy via the photoelectric effect, utilize materials with low work functions. Alkali metals are known for their low work functions, meaning they easily emit electrons when illuminated by light of sufficient frequency. Among the given options, Rubidium is an alkali metal and is used in photoemissive surfaces in some types of photocells or photomultiplier tubes. Caesium is also very commonly used, often alloyed with other elements.
Alkali metals have low work functions, making them suitable for use as photoemissive materials in devices that rely on the photoelectric effect.
Tungsten is a refractory metal with a high melting point and is used as a filament in incandescent bulbs, but it has a high work function. Copper and Aluminium are conductors but are not typically used as primary photoemissive surfaces for converting light to electricity in this manner. Photocathodes in modern photomultiplier tubes often use complex alloys or semiconductor materials for optimized sensitivity to different wavelengths of light.

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24. Matter around us can exist in three different states, namely, solid, l

Matter around us can exist in three different states, namely, solid, liquid and gas. The correct order of their compressibility is

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[amp_mcq option1=”Liquid < Gas < Solid" option2="Solid < Liquid < Gas" option3="Gas < Liquid < Solid" option4="Solid < Gas < Liquid" correct="option2"]

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The correct answer is B) Solid < Liquid < Gas.
Compressibility is the measure of how much the volume of a substance decreases under pressure.
– Solids: Particles are tightly packed and held in fixed positions. They are very difficult to compress.
– Liquids: Particles are close together but can move past each other. They are only slightly compressible, much less so than gases.
– Gases: Particles are far apart and move randomly. There are large spaces between particles, making them highly compressible.
Therefore, the order of increasing compressibility is Solid < Liquid < Gas.
The difference in compressibility between the states of matter is due to the varying distances between the particles and the strength of intermolecular forces. Gases have the largest intermolecular distances, allowing them to be compressed significantly. Liquids have much smaller distances, and solids have the smallest, making them almost incompressible under normal pressures.

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25. Density of water is

Density of water is

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maximum at 0°C
minimum at 0°C
maximum at 4°C
minimum at – 4°C
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Water exhibits an anomalous expansion property. Unlike most substances which become denser as they cool (contracting), water’s density increases as it cools from boiling point down to 4°C. However, when cooled further from 4°C down to 0°C, it expands and its density decreases. At 0°C, when it freezes into ice, its density drops significantly. Therefore, the density of water is maximum at 4°C.
Water has maximum density at 4°C. This anomalous expansion is crucial for aquatic life in cold climates, as ice floats and water at 4°C remains at the bottom of lakes, preventing them from freezing solid.
The density of pure water at standard atmospheric pressure is approximately 999.97 kg/m³ at 0°C, 1000 kg/m³ at 4°C, and decreases gradually as temperature increases beyond 4°C. Ice (solid water) has a density of about 917 kg/m³.

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26. Thermal conductivity of aluminium, copper and stainless steel increase

Thermal conductivity of aluminium, copper and stainless steel increases in the order

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[amp_mcq option1=”Copper < Aluminium < Stainless Steel" option2="Stainless Steel < Aluminium < Copper" option3="Aluminium < Copper < Stainless Steel" option4="Copper < Stainless Steel < Aluminium" correct="option2"]

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Thermal conductivity is a measure of a material’s ability to conduct heat. Generally, pure metals are good thermal conductors, while alloys and non-metals are poorer conductors. Among the given options, copper is an excellent thermal conductor, aluminium is also good but less so than copper, and stainless steel (an alloy of iron, chromium, nickel, etc.) is a relatively poor conductor compared to pure metals. Typical thermal conductivity values (in W/m·K) are approximately: Copper ≈ 400, Aluminium ≈ 205, Stainless Steel ≈ 15-20. Therefore, the order of increasing thermal conductivity is Stainless Steel < Aluminium < Copper.
Different materials have different thermal conductivity values; pure metals like copper and aluminium are typically better conductors than alloys like stainless steel.
Good thermal conductors are used in applications like heat sinks and cooking utensils, while poor conductors (thermal insulators) are used for insulation purposes.

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27. Which one among the following does not wet the walls of the glass vess

Which one among the following does not wet the walls of the glass vessel in which it is kept ?

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Water
Alcohol
Mercury
Phenol
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UPSC NDA-1 – 2015
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Whether a liquid wets a solid surface depends on the balance between adhesive forces (attraction between liquid molecules and solid surface) and cohesive forces (attraction between liquid molecules). Water, alcohol, and phenol have adhesive forces with glass that are stronger than their internal cohesive forces, causing them to spread out and wet the glass surface. Mercury, on the other hand, has very strong cohesive forces (due to metallic bonding) that are much stronger than the adhesive forces between mercury and glass. Therefore, mercury minimizes its contact area with the glass, forming a convex meniscus and does not wet the glass.
Wetting is determined by the relative strengths of adhesive forces between the liquid and the solid and cohesive forces within the liquid.
A liquid wets a surface if the angle of contact between the liquid and the solid is less than 90 degrees. A liquid does not wet a surface if the angle of contact is greater than 90 degrees (as is the case with mercury on glass, where it is obtuse).

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28. Which one of the following is not a result of surface tension ?

Which one of the following is not a result of surface tension ?

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Nearly spherical drop of rain
Capillary rise
Removal of dirt by soap or detergent
Flow of a liquid
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The correct option is D. Flow of a liquid is not a result of surface tension.
Surface tension is a property of the surface of a liquid that allows it to resist an external force due to the cohesive nature of water molecules. It causes liquid surfaces to behave like stretched elastic membranes.
A) Nearly spherical drop of rain: Surface tension minimizes the surface area of the liquid, and a sphere has the minimum surface area for a given volume. This is a direct result of surface tension.
B) Capillary rise: This phenomenon, where a liquid rises in a narrow tube, is caused by the combined effects of surface tension and adhesion between the liquid and the tube walls. Surface tension contributes to the upward force.
C) Removal of dirt by soap or detergent: Soaps reduce the surface tension of water. This reduced surface tension allows the water to wet fabrics and dirt particles more effectively, helping in their removal. While it involves the *change* in surface tension, the ability to remove dirt in this way is linked to the surface properties facilitated by the detergent’s effect on surface tension.
D) Flow of a liquid: The bulk flow of a liquid (like water flowing in a pipe or a river) is driven by pressure differences or gravity, and it is resisted by viscosity. Surface tension primarily affects the liquid-air interface and phenomena occurring there (like drop formation, ripples), but it is not the fundamental cause or main characteristic of the bulk flow itself.
While surface tension might have secondary effects on flow dynamics (e.g., influencing the shape of a liquid jet), the primary forces driving and resisting bulk liquid flow are pressure gradients, gravity, and viscosity, not surface tension.

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29. Two dielectric media D1 and D2 have dielectric constants 3 and 2, resp

Two dielectric media D1 and D2 have dielectric constants 3 and 2, respectively. They are separated by x – z plane. A uniform electric field $\vec{E} = 3\hat{i} + 2\hat{j}$ exists in D1. Which one among the following is the correct electric field in D2 at the xz plane ?

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$ ec{E} = 2hat{j}$
$ ec{E} = 3hat{i} + 3hat{j}$
$ ec{E} = 3hat{i} - 2hat{j}$
$ ec{E} = 2hat{i} + 3hat{j}$
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Correct Answer: B
– The interface between the two dielectric media D1 and D2 is the x-z plane, which is defined by $y=0$. The normal vector to this interface is along the y-axis, i.e., $\hat{j}$.
– Medium D1 has dielectric constant $\epsilon_{r1} = 3$, and medium D2 has dielectric constant $\epsilon_{r2} = 2$. Let’s assume D1 is in the region $y<0$ and D2 is in the region $y>0$.
– The electric field in D1 is given as $\vec{E}_1 = 3\hat{i} + 2\hat{j}$.
– At the boundary between two dielectric media, two boundary conditions for the electric field and displacement field must be satisfied:
1. The tangential component of the electric field ($\vec{E}_{tan}$) is continuous across the boundary: $\vec{E}_{1,tan} = \vec{E}_{2,tan}$.
2. The normal component of the electric displacement field ($\vec{D}_{norm}$) is continuous across the boundary (assuming no free charges on the surface): $\vec{D}_{1,norm} = \vec{D}_{2,norm}$.
– The tangential components of the electric field are those parallel to the x-z plane (along $\hat{i}$ and $\hat{k}$). From $\vec{E}_1 = 3\hat{i} + 2\hat{j}$, the tangential component is $\vec{E}_{1,tan} = 3\hat{i}$.
– Therefore, the tangential component of the electric field in D2 is $\vec{E}_{2,tan} = 3\hat{i}$. This means the x-component of $\vec{E}_2$ is 3 ($E_{2x} = 3$) and the z-component is 0 ($E_{2z} = 0$).
– The normal component of the electric field is along the y-axis ($\hat{j}$). From $\vec{E}_1$, the normal component is $E_{1y} = 2$.
– The electric displacement field is given by $\vec{D} = \epsilon_0 \epsilon_r \vec{E}$.
– The normal component of the displacement field in D1 is $D_{1y} = \epsilon_0 \epsilon_{r1} E_{1y} = \epsilon_0 \times 3 \times 2 = 6\epsilon_0$.
– The normal component of the displacement field in D2 is $D_{2y} = \epsilon_0 \epsilon_{r2} E_{2y} = \epsilon_0 \times 2 \times E_{2y}$.
– Applying the boundary condition $\vec{D}_{1,norm} = \vec{D}_{2,norm}$, we have $D_{1y} = D_{2y}$.
– $6\epsilon_0 = 2\epsilon_0 E_{2y}$.
– $6 = 2 E_{2y} \implies E_{2y} = 3$.
– The electric field in D2 is $\vec{E}_2 = E_{2x} \hat{i} + E_{2y} \hat{j} + E_{2z} \hat{k}$.
– Substituting the values we found: $\vec{E}_2 = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3\hat{i} + 3\hat{j}$.
These boundary conditions are fundamental in electrostatics at interfaces between different dielectric materials. The tangential component of E is continuous because a discontinuity would imply an infinite tangential force on a charge, which is unphysical. The normal component of D is continuous because the electric flux must be conserved, and there are no free charges accumulating at the interface. The normal component of E changes by a factor of $\epsilon_{r1}/\epsilon_{r2}$.

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30. A steel rod having radius r and length L gets stretched along its leng

A steel rod having radius r and length L gets stretched along its length by ΔL, when a force F is applied to it. If another rod made of the same material having radius 2r and length L is subjected to the same force F, then the elongation observed for the second rod is

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4ΔL
2ΔL
ΔL /4
ΔL /2
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According to Hooke’s Law, for a material under tensile stress, Young’s Modulus (Y) is given by Y = Stress / Strain.
Stress = F/A, where F is the force applied and A is the cross-sectional area.
Strain = ΔL/L, where ΔL is the elongation and L is the original length.
So, Y = (F/A) / (ΔL/L) = (F * L) / (A * ΔL).
Rearranging for elongation, ΔL = (F * L) / (A * Y).
The cross-sectional area of a rod with radius r is A = πr².
For the first rod: ΔL₁ = (F * L) / (πr² * Y).
For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F.
The cross-sectional area of the second rod is A₂ = π(2r)² = 4πr².
The elongation for the second rod is ΔL₂ = (F * L) / (A₂ * Y) = (F * L) / (4πr² * Y).
Substituting the expression for ΔL₁ into the equation for ΔL₂:
ΔL₂ = (1/4) * [(F * L) / (πr² * Y)] = (1/4) * ΔL₁.
Thus, the elongation observed for the second rod is ΔL/4.
Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.
This problem assumes the material behaves elastically and obeys Hooke’s Law. Young’s modulus is a property of the material. The force applied must be within the elastic limit of the material.

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