Q: Convert $ 25^{0} C $ to Kelvin scale:

$ 298.15 K $. Kelvin and Celsius scales differ by a fixed offset of 273.15. So 25°C becomes 25 + 273.15 = 298.15 K, which matches option C.

Question 1

An instrument used to measure high temperature is

Accepted Answer

Pyrometer. A pyrometer is used to measure very high temperatures, often without direct contact. The other options measure different things: a pyrheliometer measures solar radiation, a pyknometer measures density, and a technometer is not for temperature.

Question 2

Ocean currents are caused by the setting up of in water

Accepted Answer

convection current. Ocean currents are driven by density differences caused by heating and cooling of water. This sets up circulating motion in the water, which is called a convection current.

Question 3

For the same mass, which one of the following has the maximum thermal capacity?

Accepted Answer

water. For the same mass, thermal capacity is proportional to specific heat capacity. Water has a much higher specific heat than wood, copper, or ice, so it can absorb the most heat for the same temperature rise.

Question 4

The excess temperature of a body falls from $ 12^{\circ} \mathrm{C} $ to $ 6^{\circ} \mathrm{C} $ in 5 minutes, then the time to fall the excess temperature from $ 6^{\circ} \mathrm{C} $ to $ 3^{\circ} \mathrm{C} $ is (assume the Newton's cooling is valid)

Accepted Answer

5 minutes. Newton’s law of cooling gives exponential decay of excess temperature, so the time to halve the excess temperature is constant. Since the excess temperature halves from 12 to 6 in 5 minutes, it takes the same time to halve again from 6 to 3.

Question 5

Bunty mixed 440 gm of ice at $ 0^{\circ} \mathrm{C} $ with $ 540 \mathrm{gm} $ of water at $ 80^{\circ} \mathrm{C} $ in a bowl. Then what would remain after sometime in the bowl?

Accepted Answer

only water. The 540 g of water cooling from 80°C to 0°C releases enough heat to melt all 440 g of ice at 0°C. Since the ice fully melts and the final mixture is above 0°C, only water remains.

Question 6

Mercury boils at $ 356^{\circ} \mathrm{C} $. However mercury thermometers are made such that they can measure temperatures up to $ 500^{\circ} \mathrm{C} $. This is done by :

Accepted Answer

filling nitrogen gas at high pressure above the mercury column. Mercury would normally boil at 356°C, so to measure higher temperatures the pressure above it must be increased. High-pressure nitrogen is inert and suppresses boiling, allowing the thermometer to work up to 500°C.

Question 7

Name the process associated with the following. Dry ice is kept at room temperature and at one atmospheric pressure.

Accepted Answer

Sublimation. Dry ice is solid carbon dioxide. At 1 atm and room temperature, solid CO2 changes directly from solid to gas, which is called sublimation. It does not pass through the liquid state under these conditions.

Question 8

Choose the wrong statements from the following. Two spheres made of the same material have the same diameter. One sphere is hollow and the other is solid. If they are heated through the same range of temperature, Then: This question has multiple correct options

Accepted Answer

Hollow sphere will expand more than solid one. The correct statement is that both spheres expand equally in terms of outer dimensions, because linear and volumetric expansion depend on the material and initial size, not on whether the interior is hollow. So the claim that the hollow sphere expands more is wrong, and the others are also not supported by thermal expansion laws.

Question 9

Convert $ 25^{0} C $ to Kelvin scale:

Accepted Answer

$ 298.15 K $. Kelvin and Celsius scales differ by a fixed offset of 273.15. So 25°C becomes 25 + 273.15 = 298.15 K, which matches option C.

Question 10

Calculate the heat energy required to convert completely $ 10 \mathrm{kg} $ of water at $ 50^{\circ} \mathrm{C} $ into steam at $ 100^{\circ} \mathrm{C} $, given that the specific heat capacity of water is $ 4200 J /\left(k g^{\circ} C\right) $ and the specific latent heat of vapor

Accepted Answer

24700 KJ. First, the 10 kg of water must be heated from 50°C to 100°C, which takes Q = mcΔT. Then it must be converted to steam at 100°C using the latent heat of vaporisation; adding both parts gives the total energy.

Question 11

For the construction of a thermometer one of the essential requirements is a thermometric substance which

Accepted Answer

has a property that varies linearly with temperature. A thermometric substance must have a property that changes uniformly with temperature so the thermometer scale can be calibrated accurately. Linear variation makes the relationship between the measured property and temperature simple and reliable.

Question 12

The pendulum of a clock is made of brass. If the clock keeps correct time at $ 20^{\circ} \mathrm{C} $. Calculate how many seconds per day will it loose at $ 35^{\circ} \mathrm{C} $ ( $ \alpha $ for brass = $ \left.2 \times 10^{-5 o} C\right) $

Accepted Answer

1.29 s. For a pendulum, the period varies as the square root of its length, so heating the brass rod makes the clock run slower. The small increase in period over 24 hours gives a loss of about 1.29 s per day.

Question 13

The unit of thermal conductivity is:

Accepted Answer

W m⁻¹ K⁻¹. Thermal conductivity is defined as the rate of heat transfer per unit area per unit temperature gradient. Its SI unit is W m⁻¹ K⁻¹.

Question 14

The unit of the Stefan-Boltzmann's constant is:

Accepted Answer

W/m²K⁴. The Stefan-Boltzmann constant relates the total energy radiated per unit surface area of a black body to the fourth power of its temperature, so its unit is derived as W/m²K⁴.

Question 15

A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is: (αCu = 1.7 × 10⁻⁵ K⁻¹ and αAl = 2.2 × 10⁻⁵ K⁻¹)

Accepted Answer

113.9 cm. For the increase in length to be independent of temperature, the product of the coefficient of linear expansion (α) and the length (L) must be the same for both rods. Using the relation αCu × LCu = αAl × LAl, we substitute αCu = 1.7 × 10⁻⁵, LCu = 88 cm, and αAl = 2.2 × 10⁻⁵. Solving for LAl gives LAl = (αCu × LCu) / αAl = (1.7 × 10⁻⁵ × 88) / (2.2 × 10⁻⁵) = 113.9 cm.

Question 16

Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ − l₁) is maintained same at all temperatures, which one of the following relations holds good?

Accepted Answer

α₁l₁ = α₂l₂. For the difference in lengths (l₂ − l₁) to remain constant at all temperatures, the thermal expansions of both rods must be equal. This leads to the condition α₁l₁ = α₂l₂.

Question 17

If the cold junction of a thermo-couple is kept at 0°C and the hot junction is kept at T°C then the relation between neutral temperature (Tn) and temperature of inversion (Ti) is:

Accepted Answer

Tn = Ti/2. The neutral temperature (Tn) is the average of the cold junction temperature (0°C) and the temperature of inversion (Ti). Hence, Tn = Ti/2.

Question 18

The density of water at 20°C is 998 kg/m³ and at 40°C 992 kg/m³. The coefficient of volume expansion of water is:

Accepted Answer

2 × 10⁻⁴/°C. The coefficient of volume expansion is calculated using the formula β = (Δρ / ρ₀) / ΔT. Substituting the given values: Δρ = 998 - 992 = 6 kg/m³, ρ₀ = 998 kg/m³, and ΔT = 40°C - 20°C = 20°C, we get β = (6 / 998) / 20 ≈ 2 × 10⁻⁴/°C.

Question 19

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale?

Accepted Answer

117° W. The W scale is linear, so we can use the relation: (T_W - 39)/(239 - 39) = (T_C - 0)/(100 - 0). Substituting T_C = 39°C, we get (T_W - 39)/200 = 39/100. Solving for T_W, T_W = 117° W.

Question 20

The value of coefficient of volume expansion of glycerine is 5 × 10⁻⁴ K⁻¹. The fractional change in the density of glycerine for a rise of 40°C in its temperature, is:

Accepted Answer

0.020. The fractional change in density is approximately equal to the negative of the coefficient of volume expansion multiplied by the temperature change. Using Δρ/ρ = -βΔT, we get -5 × 10⁻⁴ × 40 = -0.020.

NEET Physics: Thermal Properties of Matter questions with solutions

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