JEE Main 2019 (Online) 12th January Morning Slot | Heat and Thermodynamics Question 348 | Physics

For the given cyclic process CAB as shown for a gas, the work done is :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Heat and Thermodynamics Question 346 English

JEE Main 2019 (Online) 12th January Morning Slot Physics - Heat and Thermodynamics Question 346 English

1 J

10 J

5 J

30 J

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $$\Delta $$T. The amount of heat absorbed by gas is (R is gas constant) :

$${1 \over 2}$$ KR$$\Delta $$T

$${1 \over 2}$$ R$$\Delta $$T

$${3 \over 2}$$ R$$\Delta $$T

$${2K \over 3}$$ $$\Delta $$T

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

Two rods A and B of identical dimensions are at temperature 30^°C. If A is heated upto 180^oC and B upto T^oC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is

200^oC

270^oC

230^oC

250^oC

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

When 100 g of a liquid A at 100^oC is added to 50 g of a liquid B at temperature 75^oC, the temperature of the mixture becomes 90^oC. The temperature of the mixture, if 100 g of liquid A at 100^oC is added to 50 g of liquid B at 50^oC, will be :

60^oC

70^oC

85^oC

80^oC

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