Explanation:
At same temperature, curve with higher volume corresponds to lower pressure.
(We draw a straight line parallel to volume axis to get this)
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:
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,
For 1st cup of coffee,
For 2nd cup of coffee,
Divide (1) by (2),
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:
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]
(αCu = 1.7 × 10-5 K-1 and αAl = 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:
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:
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:
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:
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,
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:
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
where, b = Wien's constant = 2.88 x 106 nmK
λm is the wavelength corresponding to maximum energy, so U2 > U1.
Q3: 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
Q4: 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,So,
Dividing eqn. (i) by eqn. (ii), we get
Q1: The two ends of a metal rod are maintained at temperatures 100oC and 110oC. the rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200oC and 210oC, 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:
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 ρT be densities of glycerin at 0°C and T °C respectively. Then,where γ is the coefficient of volume expansion of glycerine and ΔT is rise in temperature.
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 TP, TQ and TR are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that: [2015]
(a) TP < TQ < TR
(b) TP > TQ > TR
(c) TP > TR > TQ
(d) TP < TR < TQ
Ans: (b) TP > TQ > TR
Explanation:
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:
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:
Q3: Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be:
[Take specific heat of water = 1 cal g−1 °C−1 and latent heat of steam = 540 cal g−1] [2014]
(a) 42.5 g
(b) 22.5 g
(c) 24 g
(d) 31.5 g
Ans: (b) 22.5 g
Explanation:
97 videos|378 docs|103 tests
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1. What are the main thermal properties of matter that NEET students should focus on? |
2. How does temperature affect the state of matter in thermal properties? |
3. What is the significance of specific heat capacity in thermal properties? |
4. Can you explain thermal conductivity and its importance in real-world applications? |
5. What are the common numerical problems related to thermal properties of matter in NEET exams? |
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