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NEET Part Test - 3 - NEET MCQ


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30 Questions MCQ Test NEET Mock Test Series 2025 - NEET Part Test - 3

NEET Part Test - 3 for NEET 2024 is part of NEET Mock Test Series 2025 preparation. The NEET Part Test - 3 questions and answers have been prepared according to the NEET exam syllabus.The NEET Part Test - 3 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Part Test - 3 below.
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NEET Part Test - 3 - Question 1

A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

Detailed Solution for NEET Part Test - 3 - Question 1

In an ideas gas internal energy 

NEET Part Test - 3 - Question 2

Maxwell’s velocity distribution curve is given for two different temperatures. For the given curves.

Detailed Solution for NEET Part Test - 3 - Question 2

Higher is the temperature greater is the most probable velocity.

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NEET Part Test - 3 - Question 3

The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then the internal energy of one mole gas at that temperature is [R = 8.3 J/mol-K]

Detailed Solution for NEET Part Test - 3 - Question 3

According to law of equipartition of energy, energy equally distributes among its degree of freedom, Let translational and rotational degree of freedom be f1 and f2
∴  and 

Hence the ratio of translational to rotational degrees of freedom is 3:2. Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2. 
∴ Internal energy (U) = 
 = U 2075 J

NEET Part Test - 3 - Question 4

12 gm He and 4 gm H2 is filled in a container of volume 20 litre maintained at temperature 300  K. The pressure of the mixture is nearly : 

Detailed Solution for NEET Part Test - 3 - Question 4

PV = nRT

= 6.25 x 105 Pa

NEET Part Test - 3 - Question 5

Which of the following will have maximum total kinetic energy at temperature 300 K. 

Detailed Solution for NEET Part Test - 3 - Question 5

Total KE = U = f/2nRT

In case of H2 degree of freedom is greatest and  number of moles n is highest. So this is the case of maximum kinetic energy.

NEET Part Test - 3 - Question 6

A ring shaped tube contains two ideal gases with equal masses and atomic mass numbers M1 = 32 and M2 = 28. The gases are separated by one fixed partition P and another movable conducting partition S which can move freely without friction inside the ring. The angle a as shown in the figure in equilibrium is:

Detailed Solution for NEET Part Test - 3 - Question 6

 

P1 = P2      T1 = T2  

 or 

NEET Part Test - 3 - Question 7

In an experiment the speeds of any five molecules of an ideal gas are recorded. The experiment is repeated N times where N is very large. The average of recorded values, is :

Detailed Solution for NEET Part Test - 3 - Question 7

When speed of 5 molecules which are selected randomly are recorded, then the average is most likely to be equal to the most probable speed. 
∴ The average of these values is most likely equal to 

NEET Part Test - 3 - Question 8

Temperature at which Fahrenheit and Kelvin pair of scales give the same reading will be:

Detailed Solution for NEET Part Test - 3 - Question 8


 ⇒ 5θ - 32 x 5 = 9θ - 273 x 9 
= θ = 574.25

NEET Part Test - 3 - Question 9

20 gm ice at –10 ºC is mixed with m gm steam at 100 ºC. The minimum value of m so that finally all ice and steam converts into water is : (Use sice = 0.5 cal/gmºC,swater = 1 cal/gmºC,L (melting) = 80 cal/gm and L (vaporization) = 540 cal/gm)    

Detailed Solution for NEET Part Test - 3 - Question 9

For minimum value of m, the final temperature of the mixture must be 0°C. 
∴  = m 540 + m.1. 100
∴ 

NEET Part Test - 3 - Question 10

An ice block at 0°C is dropped from height ‘h’ above the ground. What should be the value of ‘h’ so that it melts completely by the time it reaches the bottom assuming  the loss of whole gravitational potential energy is used as heat by the ice ? [Given : Lf = 80 cal/gm]

Detailed Solution for NEET Part Test - 3 - Question 10

Applying energy conservation : mgh = mLf 
⇒   
= 33.6 km

NEET Part Test - 3 - Question 11

n moles of a gas filled in a container at temperature T is in thermodynamic equilibrium initially.  If the gas is compressed slowly and isothermally to half its initial volume the work done by the atmosphere on the piston is:

Detailed Solution for NEET Part Test - 3 - Question 11

 

Work done by atmosphere = Patm ΔV = 
...........(i)
Initially gas in container is in thermodynamic equilibrium with its surroundings.
∴ Pressure inside cylinder = Patm 
& PV = nRT ⇒ Patm V = nRT or 

NEET Part Test - 3 - Question 12

In a process the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas: 

Detailed Solution for NEET Part Test - 3 - Question 12

As p ∝ V2
T ∝ V3
i.e. if temperature increases, volume also increases hence work done will be positive.

NEET Part Test - 3 - Question 13

A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is:

Detailed Solution for NEET Part Test - 3 - Question 13

For process at constant pressure 

NEET Part Test - 3 - Question 14

5 moles of Nitrogen gas is enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal/sec to the nitrogen gas. A power of 30 cal /sec is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly:

Detailed Solution for NEET Part Test - 3 - Question 14

Net power given to N2 gas = 100 - 30 = 70 calls
The nitrogen gas expands isobarically.
∴ 

NEET Part Test - 3 - Question 15

The gas law PV/2 =constant for a given amount of a gas is true for :

Detailed Solution for NEET Part Test - 3 - Question 15

As PV = nRT
For n = constant :  = constant for all changes Hence (C)

NEET Part Test - 3 - Question 16

All the rods have same conductance ‘K’ and same area of cross section S. If  ends A and C are maintained at temperature 2T0 and T0 respectively then which of the following is/are correct:

Detailed Solution for NEET Part Test - 3 - Question 16

By symmetry 
IAB = IBC & IAD = IDC
∴ No current in BO and OD
∴ TB = TO = TD

NEET Part Test - 3 - Question 17

A hollow copper sphere and a hollow copper cube, of same surface area and negligible thickness, are filled with warm water of same temperature and placed in an enclosure of constant temperature, a few degrees below that of the bodies. Then in the beginning :

Detailed Solution for NEET Part Test - 3 - Question 17

[Newton's law of cooling]
∴ Rate of loss of heat is same

NEET Part Test - 3 - Question 18

The colour of a star indicates its :

Detailed Solution for NEET Part Test - 3 - Question 18

The colour of an object indicates the rate at which energy is emitted and hence indicates the temperature.

NEET Part Test - 3 - Question 19

Two identical solid spheres have the same temperature. One of the sphere is cut into two identical pieces. The intact sphere radiates an energy Q during a given small time interval. During the same interval, the two hemispheres radiate a total energy Q'. The ratio Q'/Q is equal to :

Detailed Solution for NEET Part Test - 3 - Question 19

Heat radiated (at temp same temp) ∝ A
⇒ Q ∝ 4πR2 and Q' ∝ (4πR2 + 2 x πR2)
⇒ 
Here πR2 is extra surface area of plane surface of the hemisphere

NEET Part Test - 3 - Question 20

A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then  the water equivalent of the calorimeter is :

Detailed Solution for NEET Part Test - 3 - Question 20

assuming rate of heat transfer to be constant.
s (m1 + w) ΔT = Q1 where s = specific heat of water = 1 g/cc s
s (m2 + w)ΔT = Q2 w = water equivalent of calorimeter 

NEET Part Test - 3 - Question 21

Heat is flowing through two cylindrical rods made of same materials whose ends are maintained at similar temperatures. If diameters of the rods are in ratio 1 : 2 and lengths in ratio 2 : 1, then the ratio of thermal current through them in steady state is :

Detailed Solution for NEET Part Test - 3 - Question 21

Thermal resistance 
for same temperature difference, thermal current 

NEET Part Test - 3 - Question 22

A balloon containing an ideal gas has a volume of 10 litre and temperature of 17ºC. If it is heated slowly to 75ºC, the work done by the gas inside the balloon is (neglect elasticity of the balloon and take atmospheric pressure as 105 Pa)        

Detailed Solution for NEET Part Test - 3 - Question 22

Since elasticity of balloon is negligible, pressure inside ballon outside baloon = Patm.
= Vin = 10 litre. 

NEET Part Test - 3 - Question 23

In the P-V diagram shown. The gas does 5 J of work in isothermal process a b and 4 J in adiabatic process b c. What will be the change in internal energy of the gas in straight path c to a?

Detailed Solution for NEET Part Test - 3 - Question 23

ΔU in a - b process = 0 (isothermal process)
in b - c process Q = 0 (adiabatic process)
∴ ΔU = - W = - 4 J
ΔU in cyclic process = 0
∴ ΔU in c - a process = 4J 

NEET Part Test - 3 - Question 24

A monoatomic ideal gas is filled in a nonconducting container. The gas can be compressed by a movable nonconducting piston. The gas is compressed slowly to 12.5% of its initial volume. The percentage increase in the temperature of the gas is 

Detailed Solution for NEET Part Test - 3 - Question 24

Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the final temperature and volume is  (γ = 5/3 for monoatomic gas)
∴ 
∴ percentage increase in temperature of gas is 4 x 100 - 100 = 300% 

NEET Part Test - 3 - Question 25

A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is: 

Detailed Solution for NEET Part Test - 3 - Question 25

Let initial pressure, volume, temperature be P0, V0, To indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0) isobarically. 

Total heat absorbed by 1 mole of gas


Total change in temperature from state A to C is ΔT = 3T0
∴ Molar heat capacity  

NEET Part Test - 3 - Question 26

The ratio of final root mean square velocity to initial root mean square velocity of nitrogen molecules if nitrogen gas is compressed adiabatically from a pressure of one atmosphere to a pressure of two atmosphere is : 

Detailed Solution for NEET Part Test - 3 - Question 26

 and  
= 21/7

NEET Part Test - 3 - Question 27

Four particles have velocities 1, 0, 2, 3 m/s. The root mean square velocity of the particles is: (in m/s) 

Detailed Solution for NEET Part Test - 3 - Question 27

NEET Part Test - 3 - Question 28

V–T diagram for a process of a given mass of ideal gas is as shown in the figure. During the process pressure of gas. 

Detailed Solution for NEET Part Test - 3 - Question 28

PV = nRT   P(aT - b) = nRT 
⇒ 

now as T increases then  increases hence P decreases

NEET Part Test - 3 - Question 29

A slab X of thickness ‘t’, thermal conductivity ‘K’ and area of cross-section ‘A’ is placed in thermal contact with another slab Y which is 2n2 times thicker, 4n times conductive and having n times larger cross section area. If the outside face of X is maintained at 100°C, the outside face of Y at 0°C, then the temperature of the junction θ is represented by the graph (n > 0) :

Detailed Solution for NEET Part Test - 3 - Question 29

At equilibrium : 
⇒ 
⇒ θ = 100/3.

NEET Part Test - 3 - Question 30

Two conducting movable smooth pistons are kept inside a non conducting, adiabatic container with initial positions as shown. Gas is present in the three parts A, B & C having initial pressures as shown. Now the pistons are released. Then the final equilibrium position length of part A will be     

Detailed Solution for NEET Part Test - 3 - Question 30

Moles in 'A' initially & finally will be same
 ......(1)


Moles in 'B' remain same
 ......(2)
Divided (1) by (2) 

⇒ x = L/4

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