Physical Properties of Materials

11. At which temperature does liquid water show maximum density ?

At which temperature does liquid water show maximum density ?

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299 K
277 K
285 K
373 K
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UPSC NDA-2 – 2024
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Liquid water exhibits its maximum density at 4°C. Converting this to Kelvin, we get 4 + 273.15 = 277.15 K. The option closest to this value is 277 K.
– Unlike most substances which become denser as they are cooled, water behaves unusually in the temperature range of 0°C to 4°C.
– As liquid water cools from higher temperatures, its density increases until it reaches a maximum density at approximately 4°C (3.98°C to be precise).
– Below 4°C, the density of liquid water starts to decrease as it approaches its freezing point (0°C). When it freezes to form ice at 0°C, its density decreases significantly, which is why ice floats on water.
– This anomalous expansion of water is crucial for aquatic life in cold climates, as the densest water (at 4°C) sinks to the bottom of lakes, allowing life to survive below the surface ice.
The Kelvin scale is an absolute thermodynamic temperature scale where 0 K is absolute zero. The relationship between Celsius (°C) and Kelvin (K) is K = °C + 273.15. The boiling point of water at standard atmospheric pressure is 100°C or 373.15 K.

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12. A pumpkin weighs 7.5 N. On submerging it completely in water, ¾ L of w

A pumpkin weighs 7.5 N. On submerging it completely in water, ¾ L of water gets displaced. The acceleration due to gravity at the place where the pumpkin was weighed is 10 m/s². Which one of the following is the correct value of the density of the pumpkin ?

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10 kg/m³
100 kg/m³
1000 kg/m³
10000 kg/m³
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UPSC NDA-2 – 2024
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The weight of the pumpkin is given as 7.5 N. Using the formula Weight = mass × gravity, the mass of the pumpkin can be calculated. When the pumpkin is completely submerged, the volume of water displaced is equal to the volume of the pumpkin. Given the volume of displaced water, we can calculate the volume of the pumpkin. Density is then calculated as mass divided by volume.
Mass of pumpkin (m) = Weight / gravity = 7.5 N / 10 m/s² = 0.75 kg.
Volume of water displaced (V_displaced) = 3/4 L = 0.75 L.
Converting litres to cubic meters: 1 L = 0.001 m³. So, V_displaced = 0.75 × 0.001 m³ = 0.00075 m³.
When completely submerged, Volume of pumpkin (V_pumpkin) = V_displaced = 0.00075 m³.
Density of pumpkin (ρ) = Mass / Volume = 0.75 kg / 0.00075 m³.
ρ = 0.75 / (7.5 × 10⁻⁴) = (7.5 × 10⁻¹) / (7.5 × 10⁻⁴) = 10³ kg/m³ = 1000 kg/m³.
The density of water is approximately 1000 kg/m³. A pumpkin with a density equal to or slightly less than water would float or remain suspended when fully submerged. The fact that it displaces ¾ L of water upon complete submersion means its volume is ¾ L. The calculation of density confirms it is very close to the density of water, which is reasonable for a pumpkin.

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13. How many of the following materials can be attracted by a magnet ? 1.

How many of the following materials can be attracted by a magnet ?
1. Plastic
2. Carbon
3. Aluminium
4. Stainless Steel
Select the correct answer using the code given below :

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1
2
3
None
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UPSC NDA-2 – 2023
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The correct answer is A. Only one of the listed materials, Stainless Steel (depending on its specific composition), can be attracted by a magnet in the conventional sense.
Magnetically permeable materials are attracted by magnets. Ferromagnetic materials like iron, nickel, cobalt, and some alloys (including certain types of stainless steel) are strongly attracted. Paramagnetic materials like aluminium are weakly attracted. Diamagnetic materials like carbon and plastic are weakly repelled. When a general question asks if a material is “attracted by a magnet,” it usually implies strong attraction (ferromagnetism).
Plastic is typically diamagnetic or paramagnetic, not attracted in practice. Carbon (like graphite) is diamagnetic, weakly repelled. Aluminium is paramagnetic, very weakly attracted, usually considered non-magnetic for everyday purposes. Stainless steel exists in different grades; austenitic stainless steels (e.g., 304, 316) are generally non-magnetic, while ferritic (e.g., 430) and martensitic (e.g., 410) stainless steels are magnetic due to their higher iron content and crystalline structure. Therefore, some forms of stainless steel *can* be attracted by a magnet.

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14. The heating element in an electric iron is usually made of

The heating element in an electric iron is usually made of

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Constantan
Tungsten
Nichrome
Copper
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The heating element in an electric iron is usually made of Nichrome.
Nichrome is an alloy primarily composed of nickel and chromium. It is chosen for heating elements because it has a high electrical resistance, allowing it to efficiently convert electrical energy into heat. It also has a high melting point and resistance to oxidation at high temperatures, ensuring durability and safety during operation.
Other materials like Constantan also have high resistance but are less commonly used for high-temperature heating elements compared to Nichrome. Tungsten has a very high melting point and is used for filaments in incandescent light bulbs, but not typically for resistive heating elements like those in irons or toasters. Copper has low resistance and is used for electrical wiring, not for generating heat.

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15. Which one of the following metals is most commonly used for making fil

Which one of the following metals is most commonly used for making filament of incandescent electric bulbs?

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Aluminium
Silver
Copper
Tungsten
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Tungsten is the metal most commonly used for making the filament of incandescent electric bulbs.
– The filament in an incandescent bulb needs to glow white-hot (reach a very high temperature) to produce light.
– This requires a material with a very high melting point that can withstand such extreme temperatures without melting or vaporizing quickly.
– Tungsten (chemical symbol W) has the highest melting point of all known metals (3422°C or 6192°F).
– It also has high tensile strength, even at high temperatures, and relatively low vapor pressure, reducing evaporation of the filament.
Early incandescent bulbs used carbon filaments, but tungsten proved to be significantly more efficient and durable. The bulb is often filled with inert gases like argon or nitrogen to reduce tungsten evaporation and prolong the filament’s life.

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16. An object is made of two equal parts by volume; one part has density $

An object is made of two equal parts by volume; one part has density $\rho_0$ and the other part has density $2\rho_0$. What is the average density of the object?

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$3 ho_0$
$ rac{3}{2} ho_0$
$ ho_0$
$ rac{1}{2} ho_0$
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Let V be the total volume of the object. The object is made of two equal parts by volume, so the volume of each part is $V_1 = V_2 = V/2$.
Let $\rho_1$ be the density of the first part and $\rho_2$ be the density of the second part.
Given: $\rho_1 = \rho_0$ and $\rho_2 = 2\rho_0$.
The mass of the first part is $m_1 = \rho_1 \times V_1 = \rho_0 \times (V/2)$.
The mass of the second part is $m_2 = \rho_2 \times V_2 = 2\rho_0 \times (V/2) = \rho_0 V$.
The total mass of the object is $M = m_1 + m_2 = \rho_0 (V/2) + \rho_0 V = \rho_0 V (\frac{1}{2} + 1) = \rho_0 V (\frac{3}{2})$.
The total volume of the object is $V_{\text{total}} = V_1 + V_2 = V/2 + V/2 = V$.
The average density of the object is $\rho_{\text{avg}} = \frac{M}{V_{\text{total}}} = \frac{\rho_0 V (3/2)}{V} = \frac{3}{2}\rho_0$.
When calculating average density for parts of equal volume, the average density is the simple arithmetic mean of the densities. However, in this case, the masses are different. The calculation involves finding the total mass and dividing by the total volume.
If the parts were of equal mass instead of equal volume, the calculation would be different, involving the reciprocal of the average of reciprocals (harmonic mean of densities).

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17. In an incandescent electric bulb, the filament of the bulb is made up

In an incandescent electric bulb, the filament of the bulb is made up of which metal ?

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Aluminium
Copper
Tungsten
Silver
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The filament in an incandescent electric bulb is made up of tungsten. Tungsten is used because it has the highest melting point (3422 °C) among all pure metals and remains solid at the very high temperatures required to produce light efficiently (around 2000-2500 °C). It also has high tensile strength and low vapor pressure at high temperatures.
The principle behind an incandescent bulb is resistive heating: an electric current passes through the filament, heating it to incandescence (glowing white hot) and emitting light. The bulb is usually filled with an inert gas (like argon or nitrogen) to reduce the evaporation of the tungsten filament, prolonging its life.
Aluminium, Copper, and Silver are good conductors but have significantly lower melting points than tungsten and would melt or vaporize rapidly at the temperatures required for incandescence.

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18. Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal v

Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal volume and their relative density is 4. When they are mixed in equal masses, relative density is 3. The values of $\rho_1$ and $\rho_2$ respectively are

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6, 2
3, 5
12, 4
9, 3
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UPSC NDA-2 – 2019
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Let the densities be $\rho_1$ and $\rho_2$. Relative density is numerically equal to density assuming water density is 1.
Case 1: Mixed in equal volume ($V$). Total mass $m = m_1 + m_2 = \rho_1 V + \rho_2 V = (\rho_1 + \rho_2)V$. Total volume $V_{total} = V + V = 2V$. Density $\rho_{mix, V} = \frac{(\rho_1 + \rho_2)V}{2V} = \frac{\rho_1 + \rho_2}{2}$. Given relative density is 4, so $\frac{\rho_1 + \rho_2}{2} = 4 \implies \rho_1 + \rho_2 = 8$.
Case 2: Mixed in equal masses ($m$). Volume $V_1 = \frac{m}{\rho_1}$, $V_2 = \frac{m}{\rho_2}$. Total mass $m_{total} = m + m = 2m$. Total volume $V_{total} = \frac{m}{\rho_1} + \frac{m}{\rho_2} = m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)$. Density $\rho_{mix, m} = \frac{2m}{m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)} = \frac{2\rho_1 \rho_2}{\rho_1 + \rho_2}$. Given relative density is 3, so $\frac{2\rho_1 \rho_2}{\rho_1 + \rho_2} = 3$.
Substitute $\rho_1 + \rho_2 = 8$ into the second equation: $\frac{2\rho_1 \rho_2}{8} = 3 \implies \frac{\rho_1 \rho_2}{4} = 3 \implies \rho_1 \rho_2 = 12$.
We need two numbers whose sum is 8 and product is 12. These are the roots of the quadratic equation $x^2 – 8x + 12 = 0$, which factors as $(x-2)(x-6) = 0$. The roots are 2 and 6. So, the densities are 2 and 6. Option A is (6, 2), which satisfies both conditions: (6+2)/2 = 4 and (2*6*2)/(6+2) = 24/8 = 3.
– Density of a mixture depends on the densities of the components and the proportions in which they are mixed (by mass or by volume).
– Average density when mixed by volume is the arithmetic mean of the densities.
– Average density when mixed by mass is the harmonic mean of the densities, weighted by mass proportion (or a form related to it). The formula derived $\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2}$ is 2 times the harmonic mean of $\rho_1$ and $\rho_2$.
– Relative density (or specific gravity) is the ratio of the density of a substance to the density of a reference substance (usually water).
– The problem effectively provides the arithmetic mean and a form of the harmonic mean of the two densities, allowing us to solve for the densities.

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19. Which one among the following statements with reference to the propert

Which one among the following statements with reference to the properties of water is not correct ?

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The specific heat of water is abnormally high.
Latent heat of fusion of water is very low.
Density of water is higher than ice.
Pure water is a non-conductor of electricity.
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The statement that the latent heat of fusion of water is very low is not correct.
Water has an unusually high latent heat of fusion (approximately 334 J/g or 80 cal/g) and a very high latent heat of vaporization (approximately 2260 J/g or 540 cal/g). These high values are due to the extensive hydrogen bonding in water.
Other statements are correct: Water’s specific heat is abnormally high (~4.18 J/g°C), which helps moderate temperatures. Liquid water is denser than ice due to the open crystalline structure of ice. Pure water has very low electrical conductivity as it contains very few free ions, making it a poor conductor.

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20. A spherical shell of outer radius R and inner radius R/2 contains a so

A spherical shell of outer radius R and inner radius R/2 contains a solid sphere of radius R/2 (see figure). The density of the material of the solid sphere is ρ and that of the shell is ρ/2. What is the average mass density of the larger sphere thus formed?

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3ρ/4
9ρ/16
7ρ/8
5ρ/8
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UPSC NDA-1 – 2024
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The average mass density of the larger sphere is the total mass divided by the total volume.
Let R be the outer radius. The solid sphere has radius R/2 and density ρ. The shell has outer radius R, inner radius R/2, and density ρ/2. The total volume of the larger sphere is (4/3)πR³.
Mass of solid sphere (M_sphere) = Density * Volume = ρ * (4/3)π(R/2)³ = ρ * (4/3)π(R³/8) = (1/6)πR³ρ.
Volume of the shell material (V_shell) = Volume of sphere with radius R – Volume of sphere with radius R/2 = (4/3)πR³ – (4/3)π(R/2)³ = (4/3)πR³ – (1/6)πR³ = (7/6)πR³.
Mass of the shell (M_shell) = Density * Volume = (ρ/2) * (7/6)πR³ = (7/12)πR³ρ.
Total mass (M_total) = M_sphere + M_shell = (1/6)πR³ρ + (7/12)πR³ρ = (2/12)πR³ρ + (7/12)πR³ρ = (9/12)πR³ρ = (3/4)πR³ρ.
Average density = M_total / V_total = ((3/4)πR³ρ) / ((4/3)πR³) = (3/4) * (3/4) * ρ = 9ρ/16. This calculation represents the overall density if the entire composite structure were considered a single sphere of radius R with uniform density.

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