Which of the following is considered a state variable in thermodynamics?
The THERMODYNAMICS_Physics chapter states, 'Examples of state variables are pressure (P), volume (V), temperature (T), and mass (m). Heat and work are not state variables.'
Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Which of the following is considered a state variable in thermodynamics?
The THERMODYNAMICS_Physics chapter states, 'Examples of state variables are pressure (P), volume (V), temperature (T), and mass (m). Heat and work are not state variables.'
If $P_1$, $V_1$, and $T_1$ represent the initial pressure, volume, and absolute temperature of a given sample of gas, and $P_2$, $V_2$, and $T_2$ represent its final state, which of the following relationships holds true for an ideal gas?
The THERMAL_PROPERTIES_OF_MATTER chapter mentions, 'since PV = constant and V/T = constant for a given quantity of gas, then PV/T should also be a constant. This relationship is known as ideal gas law.' So, $\frac{PV}{T}$ is constant.
The Boltzmann constant ($k_B$) serves as a link between which two physical domains?
The KINETIC_THEORY chapter states, 'This is a fundamental result relating temperature, a macroscopic measurable parameter of a gas (a thermodynamic variable as it is called) to a molecular quantity, namely the average kinetic energy of a molecule. The two domains are connected by the Boltzmann constant.'
For a constant volume gas thermometer, how is temperature read?
The THERMAL_PROPERTIES_OF_MATTER chapter states, 'Holding the volume of a gas constant, it gives P $\propto$ T. Thus, with a constant-volume gas thermometer, temperature is read in terms of pressure.'
What is the universal gas constant R equal to, using the Boltzmann constant ($k_B$) and Avogadro's number ($N_A$)?
The KINETIC_THEORY summary states, '$PV = \mu RT = k_B NT$ where N is the number of molecules. Since $N = \mu N_A$, then $R = k_B N_A$.
Which of the following statements correctly describes Gauss's law for magnetism?
According to the NCERT text 'Thus, Gauss’s law for magnetism is: The net magnetic flux through any closed surface is zero.' This law reflects the non-existence of isolated magnetic poles (monopoles).
The fundamental difference between Gauss's law for electrostatics and Gauss's law for magnetism arises from the fact that:
The NCERT text states: 'The difference between the Gauss’s law of magnetism and that for electrostatics is a reflection of the fact that isolated magnetic poles (also called monopoles) are not known to exist.' Since magnetic monopoles don't exist, magnetic field lines always form closed loops, implying no net flux through a closed surface. Electrostatic field lines originate from positive charges and end on negative charges.
If an isolated magnetic monopole were discovered, how would Gauss's law for magnetism need to be modified?
Gauss's law for electrostatics states that $\oint \vec{E} \cdot d\vec{A} = q/\epsilon_0$, where $q$ is the enclosed electric charge. If magnetic monopoles existed, they would act as sources or sinks for magnetic field lines, similar to electric charges. Therefore, Gauss's law for magnetism would become $\oint \vec{B} \cdot d\vec{A} = \mu_0 q_m$, where $q_m$ is the enclosed magnetic monopole charge.
Which of the following is NOT a characteristic of magnetic field lines?
Magnetic field lines form closed loops, originate from the North pole and end at the South pole (outside the magnet, and continue inside from South to North), and do not intersect. Magnetic fields can pass through conductors; for example, a current-carrying wire produces a magnetic field in the surrounding space, including within the wire itself if considered.
Gauss's law for magnetism mathematically can be expressed as:
The NCERT text explicitly states: 'Thus, Gauss’s law for magnetism is: The net magnetic flux through any closed surface is zero.' Mathematically, magnetic flux is given by the surface integral of the magnetic field, so $\oint \vec{B} \cdot d\vec{A} = 0$.
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.