NEET Previous Year Questions (2014-24): Thermal Properties of Matter | Physics Class 11 PDF Download

2024

Q1: The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P₁,P₂ and P₃ compared with those of Charles's law represented as dotted lines. [2024]
Then the correct relation is :
(a) P₃ > P₂> P₁
(b) P₁ > P₃ > P₂
(c) P₂ > P₁ > P₃
(d) P₁ > P₂ > P₃
Ans:

Explanation:
At same temperature, curve with higher volume corresponds to lower pressure.

(We draw a straight line parallel to volume axis to get this)

2022

Q1: The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is  
(a) 36 × 10⁴ J
(b) 36 × 10⁵ J
(c) 1 × 10⁵ J
(d) 36 × 10⁷ J                   [2022]
Ans: (d) 36 × 10⁷ J  

Explanation:

Given:

• Power of the transmitter, P = 100 kW = 100 × 10³ W
• Time, t = 1 hour = 3600 seconds

The energy radiated is calculated using the formula:
Energy (E) = Power (P) × Time (t)

Substitute the given values:
E = (100 × 10³) × 3600

Simplify:
E = 36 × 10⁷ J

Final Answer: (d) 36 × 10⁷ J 

2021

Q1: A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C.     [2021]
(a) 10/13t
(b) 5/13t
(c) 13/10t
(d) 13/5t
Ans: (d) 13/5 t

Explanation:
According to Newton's law of cooling:

T₁ - T₂t = K (T₁ + T₂)2 - T₀
For the 1^st cup of coffee:

90 - 80t = K (90 + 80)2 - 20 ....(1)
For the 2^nd cup of coffee:

80 - 60t' = K (80 + 60)2 - 20 ....(2)
Divide equation (1) by equation (2):

t't = 6550

⇒ t' = 135 t

2020

Q1: The quantities of heat required to raise the temperature of two solid copper spheres of radii r₁ and r₂ (r₁ = 1.5 r₂) through 1 K are in the ratio: [2020]
(a) 3 : 2
(b) 5 : 3
(c) 27 : 8
(d) 9 : 4
Ans: (c) 27/8

Explanation:

We know that heat required,

Q = msΔT = 43 πr³ × ρ × sΔT

[∵ mass = density × volume]

For same material and for same temperature change:

Q ∝ r³ .......(1)

So, Q₁Q₂ = (r₁)³(r₂)³

Substituting the given data, we get:

Q₁Q₂ = (1.5 r₂)³r₂³ = 278
⇒ Q₁ : Q₂ = 27 : 8

2019

Q1: 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:       [2019]
(^αC_u = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)
(a) 6.8 cm
(b) 113.9 cm
(c) 88 cm
(d) 68 cm
Ans: (d) 68 cm

Explanation:

CONCEPT:

The change in the length of any rod is due to the thermal expansion, and it is written as:

α = 1T Δl / ΔT ....(1)

Here, Δl is the change in the length, ΔT the change in temperature, α is the thermal expansion, and L is the length of the rod.

CALCULATION:

Given:

• The thermal expansion of copper rod, α_Cu = 1.7 × 10^-5 K^-1
• The thermal expansion of aluminum rod, α_Al = 2.2 × 10^-5 K^-1
• Length of copper rod, L_Al = 88 cm

Here it is given that an aluminum rod of unknown length has an increase in length independent of an increase in temperature. Therefore, we can write from equation (1):

α_Cu × L_Cu = α_Al × L_Al

1.7 × 10^-5 × 88 cm = 2.2 × 10^-5 × L_Al

⇒ 1.7 × 882.2 = L_Al

⇒ L_Al = 68 cm

2018

Q1: The power radiated by a black body is P and it radiates maximum energy at wavelength  If the temperature of the black body is now changed so that it radiates maximum energy at wavelength  , the power radiated by it becomes nP. The value of n is:- [2018]
(a) 3/4
(b) 4/3
(c) 256/81
(d) 81/256
Ans: (c)

Explanation:
P = σAT⁴ ⇒ P ∝ T⁴
According to Wein's law, T ∝ 1λ_m
⇒ P ∝ ( 1λ_m)⁴
⇒ P₂P₁ = ( λ_m1λ_m2)⁴
⇒ P₂P₁ = ( λ₀34λ₀)⁴
⇒ nPP = 25681 ⇒ n = 25681

2017

Q1: A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:-    [2017]
(a) 450
(b) 1000
(c) 1800
(d) 225
Ans: (d) 1800

Explanation:

The energy radiated per second by a black body is given by Stefan's Law as:

Et = σT⁴ × A

where A is the surface area of the black body.

For a sphere, A = 4πr²:

Et = σT⁴ × 4πr²

Case (i):

Given:

• E/t = 450
• T = 500K
• r = 0.12m

Substituting into the equation:
450 = 4πσ(500)⁴(0.12)² ....(i)

Case (ii):

Given:

• E/t = ?
• T = 1000K
• r = 0.06m

Dividing (ii) by (i), we get:

E/t450 = (1000)⁴ × (0.06)²(500)⁴ × (0.12)²

Simplifying:

E/t450 = 2⁴1 = 4

⇒ E/t = 450 × 4 = 1800W

(d) is the correct option.

Q2: Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K₁ and K₂. The thermal conductivity of the composite rod will be:-    [2017]

(a)
(b) K₁ + K₂
(c) 2(K₁ + K₂)
(d)
Ans: (d)

Explanation:
Concept:
When two rods with different thermal conductivities are placed in series (end-to-end), the equivalent thermal conductivity K_eq of the composite rod can be calculated using the formula for thermal conductivity in series:
1K_eq = d₁K₁ + d₂K₂
Where d₁ and d₂ are the lengths of the respective rods.
However, when the rods are placed side by side, as shown in the figure, the thermal conductivities act in parallel. The equivalent thermal conductivity for parallel conduction is:
K_eq = K₁ + K₂2
This formula applies since the rods are assumed to have the same cross-sectional area.
Solution
For the rods placed side by side as in the figure, the effective thermal conductivity is the average of the two conductivities:
K_eq = K₁ + K₂2
Thus, the correct answer is: (d) K₁ + K₂2

2016

Q1: 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 ? [2016]
(a) α₁ℓ₁ = α₂ℓ₂
(b) α₁ℓ₂ = α₂ℓ₁
(c) α₁ℓ₂² = α₂ℓ₁²
(d) α₁²ℓ₂ = α₂²ℓ₁
Ans: (a) α₁ℓ₁ = α₂ℓ₂

Explanation:
Coefficient of linear expansion of brass = α₁
Coefficient of linear expansion = α₂
Length of brass and steel rods are l₁ and l₂ respectively.
Given,
Increase in length (l₂'-l₁' ) is same for all temperature.
So,

'l₂ - l_1' = l₂ - l₁

⇒ l₂(1 + α₂Δt) - l₁(1 + α₁Δt) = l₂ - l₁

⇒ l₂α₂ = l₁α₁

Q2: A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted in to heat during its fall. The value of h is : [Latent heat of ice is 3.4 x 10⁵ J/Kg and g = 10 N/kg] [2016]
(a) 68 km
(b) 34 km
(c) 544 km
(d) 136 km
Ans: (d) 136 Km

Explanation:

Given:

• Latent heat of ice (L) = 3.4 × 10⁵ J/kg
• Gravitational acceleration (g) = 10 N/kg
• Only one-quarter of the heat produced is absorbed by the ice.

Let h be the height from which the ice falls.

The energy produced due to the fall is given by:
Potential energy = m × g × h

Only one-quarter of this energy is absorbed by the ice:
Energy absorbed by ice = (1/4) × m × g × h

The energy required to melt the ice completely is:
Q = m × L

Equating the energy absorbed by the ice to the energy required to melt it:
(1/4) × m × g × h = m × L

Canceling m from both sides:
(1/4) × g × h = L

Substitute the given values:
(1/4) × 10 × h = 3.4 × 10⁵

Simplify:
2.5 × h = 3.4 × 10⁵

Solve for h:
h = (3.4 × 10⁵) / 2.5
h = 1.36 × 10⁵ meters

Convert meters to kilometers:
h = 136 kilometers

Q3: A block body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁ at wavelength 500 nm is U₂ and that at 1000 nm is U₃. Wien's constant, b = 2.88 x 10⁶ nmK. Which of the following is correct? [2016]
(a) U₁ > U₂
(b) U₁ = 0
(c) U₃ = 0
(d) U₂ > U₁
Ans: (d) U₂ > U₁

Explanation:

Given, temperature, T₁ = 5760 K

Given that energy of radiation emitted by the body at wavelength 250 nm in U₁, at wavelength 500 nm is U₂ and that at 1000 nm is U₃.

Now, according to Wein's law, we get
λ_m T = b
where, b = Wien's constant = 2.88 × 10⁶ nmK
⇒ λ_m = bT
⇒ λ_m = 2.88 × 10⁶ nmK5760 K
⇒ λ_m = 500 nm

λ_m is the wavelength corresponding to maximum energy, so U₂ > U₁.

Q4: Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is    [2016]
(a) 50°C
(b) more than 50°C
(c) less than 50°C but greater than 0°C
(d) 0°C
Ans: (b) more than 50°C

Explanation:
Since, heat capacity of material increases with increase in temperature so, body at 100°C has more heat capacity than body at 0°C. Hence, final common temperature of the system will be closer to 100°C.
∴ Tc > 50°C

Q5: A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be     [2016]
(a) 7/4T
(b) 3/2T
(c) 4/3T
(d) T
Ans: (b) 3/2T

Explanation:
According to Newton’s law of cooling,

dTdt = K(T - T_s)

For two cases:

dT₁dt = K(T₁ - T_s) and dT₂dt = K(T₂ - T_s)

Here, T_s = T,

T₁ = 3T + 2T2 = 2.5T

dT₁dt = 3T - 2T10 = T10

T₂ = 2T + T'2 and dT₂dt = 2T - T'10

So:

T10 = K(2.5T - T) ....(i)

2T - T'10 = K (2T + T')2 - T ....(ii)

Dividing eqn. (i) by eqn. (ii), we get:

T2T - T' = 2.5T - T2T + T'2 - T

or, 2T + T'2 = (2T - T') × 32

T' = 3(2T - T') or, 4T' = 6T; ∴ T' = 32T

2015

Q1: The two ends of a metal rod are maintained at temperatures 100^oC and 110^oC. the rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200^oC and 210^oC, the rate of heat flow will be: [2015]
(a) 4.0 J/s
(b) 44.0 J/s
(c) 16.8 J/s
(d) 8.0 J/s
Ans: (a) 4.0 J/s

Explanation:

Concept:
When there is a temperature difference between two points, heat flow occurs from higher temperatures to lower temperatures. This phenomenon is called heat transfer.

dQdt = k A dTdx

Where:

• dQ/dt = rate of heat transfer
• k = constant
• A = area
• dT/dx = rate of temperature gradient

Explanation:

Given:

• Case 1: T₁ = 100°C, T₂ = 110°C, ΔT = 10°C, rate of heat flow = 4 J/s.
• Case 2: T₁ = 200°C, T₂ = 210°C, ΔT = 10°C.

As the temperature difference and length of the rod are constant, the heat flow will remain constant.

So, the value of heat flow in case 2 will also be equal to 4 J/s.

Q2: The value of coefficient of volume expansion of glycerin is 5 ×10^–4 K^–1. The fractional change in the density of glycerin for a rise of 40°C in its temperature, is [2015]
(a) 0.025
(b) 0.010
(c) 0.015
(d) 0.020
Ans:  (d) 0.020

Explanation:
Let ρ₀ and ρ_T be densities of glycerin at 0°C and T °C respectively. Then,
ρ_T = ρ₀(1 - γΔT)

where γ is the coefficient of volume expansion of glycerine and ΔT is the rise in temperature.

ρ_Tρ₀ = 1 - γΔT or γΔT = 1 - ρ_Tρ₀

Thus, ρ₀ - ρ_Tρ₀ = γΔT

Here:

• γ = 5 × 10^-4 K^-1
• ΔT = 40°C = 40 K

∴ The fractional change in the density of glycerine:
ρ₀ - ρ_Tρ₀ = γΔT = (5 × 10^-4 K^-1)(40 K) = 0.020

Q3: On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If T_P, T_Q and T_R are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that: [2015]
(a) T_P < T_Q < T_R
(b) T_P > T_Q > T_R
(c) T_P > T_R > T_Q
(d) T_P < T_R < T_Q
Ans: (b)  T_P > T_Q > T_R

Explanation:

According to Wien's law:

λ ∝ 1T and λ_V < λ_G < λ_R

∴ T_P > T_Q > T_R

2014

Q1: Light with an energy flux of 25 x 10⁴Wm⁻² falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm², the average force exerted on the surface is [2014]
(a) 1.20 x 10⁻⁶ N
(b) 3.0 x 10⁻⁶ N
(c) 1.25 x 10⁻⁶ N
(d) 2.50 x 10⁻⁶ N
Ans: (d) 2.50 x 10⁻⁶ N

Explanation:
Step 1: Use the Formula for Radiation Pressure on a Perfectly Reflecting Surface

For a perfectly reflecting surface, the radiation pressure P is given by:

P = 2Ic

where:

• I is the energy flux
• c is the speed of light in vacuum, c = 3 × 10⁸ m/s

Step 2: Find the Radiation Pressure

Substitute the given energy flux into the formula:

P = 2 × 25 × 10⁴3 × 10⁸

P = 50 × 10⁴3 × 10⁸

P = 503 × 10^-4 N/m²

P = 16.67 × 10^-4 N/m²
Step 3: Find the Force Exerted
The force F exerted on the surface is given by:
F = P × A
Substitute the values for P and A:
F = (16.67 × 10^-4) × (15 × 10^-4) N
F = 250.05 × 10^-8 N
F = 2.50 × 10^-6 N

Q2: Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in the next 5 minutes. The temperature of the surroundings is    [2014]
(a) 42°C
(b) 10°C
(c) 45°C
(d) 20°C
Ans: (c) 45°C

Explanation:
Newton's Law of Cooling:

dTdt = -k(T - T_surroundings)

Where:

• T is the temperature of the object
• T_surroundings is the temperature of the surroundings
• k is a constant
• dTdt is the rate of change of temperature

The cooling of the water from 70°C to 60°C takes the same time as from 60°C to 54°C. We can use this information to estimate the temperature of the surroundings.
Step 1: Average temperature during each time interval

1. From 70°C to 60°C:
   • Average temperature T₁ = 70 + 602 = 65°C
2. From 60°C to 54°C:
   • Average temperature T₂ = 60 + 542 = 57°C

Step 2: Ratio of temperature drops

The temperature difference between the water and the surroundings governs the rate of cooling.

Let T_s be the temperature of the surroundings. According to Newton's Law of Cooling, the temperature drop is proportional to the difference in temperature between the water and the surroundings:

(70 - 60)(60 - 54) = (65 - T_s)(57 - T_s)

106 = 65 - T_s57 - T_s

Step 3: Solving the equation

Cross-multiply:

10(57 - T_s) = 6(65 - T_s)

Expand:

570 - 10T_s = 390 - 6T_s

Simplify:

570 - 390 = 10T_s - 6T_s

180 = 4T_s

Solve for T_s:

T_s = 1804 = 45°C

Explanation:
Step 1: Heat required to raise the temperature of water from 10°C to 80°C

Given:

• Mass of water = 20 g
• Initial temperature of water = 10°C
• Final temperature of water = 80°C
• Specific heat of water = 1 cal/g°C

The heat required to raise the temperature of water is:
Q_water = m_water × specific heat of water × ΔT
Q_water = 20 × 1 × (80 - 10) = 20 × 70 = 1400 cal
Step 2: Heat lost by the steam during condensation
The latent heat of steam is given as 540 cal/g. Let the mass of steam condensed be m_steam.
The heat released by steam upon condensation is:
Q_steam = m_steam × latent heat of steam
Q_steam = m_steam × 540
Step 3: Heat released by condensed steam to cool down from 100°C to 80°C
The condensed steam (now water) will also release some heat as it cools down from 100°C to 80°C. This heat is given by:
Q_cooling = m_steam × specific heat of water × ΔT
Q_cooling = m_steam × 1 × (100 - 80) = m_steam × 20
Step 4: Total heat lost by steam
The total heat lost by steam is:
Q_total = Q_steam + Q_cooling = m_steam × 540 + m_steam × 20 = m_steam × 560
Step 5: Using the principle of conservation of energy
The heat gained by water is equal to the heat lost by the steam:
1400 = m_steam × 560
Solve for m_steam:
m_steam = 1400560 = 2.5 g
Step 6: Total mass of water present
The mass of the water initially present was 20 g, and the steam condensed to form 2.5 g of water. Therefore, the total mass of water is:
Total mass of water = 20 g + 2.5 g = 22.5 g