NEET Physics - MCQ Practice Questions

At same T and P, the chemical potential per molecule is identical. Internal energy and entropy depend on amount (volume), heat capacity depends on mass.

In adiabatic process, Q = 0. From first law: ΔU = Q - W = 0 - (-500) = 500 J (internal energy increases)

When n = γ (ratio of specific heats), the polytropic process is adiabatic. For isothermal n=1, isobaric n=0, isochoric n=∞.

At equilibrium: ΔG = 0, so ΔH = TΔS. T = ΔH/ΔS = (-150000)/(-100) = 1500 K

For diatomic gas at constant pressure: Cp = ($\frac{7}{2}$)R. Q = nCpΔT = 3 × ($\frac{7}{2}$) × 8.314 × 300 = 37,413 J

Efficiency η = W/Qh = 0.30, so W = 0.30 × 5000 = 1500 J. Heat rejected: Qc = Qh - W = 5000 - 1500 = 3500 J

Isobaric process: V₁/T₁ = V₂/T₂. If V₂ = 2V₁, then T₂ = 2T₁ = 2 × 300 = 600 K

For adiabatic process: TV^(γ-1) = constant. T₁V₁^(γ-1) = T₂V₂^(γ-1). 400 × V^0.4 = T₂ × (2V)^0.4. T₂ = 400/(2^0.4) = $\frac{400}{1}$.3195 ≈ 252.1 K

Efficiency η = 1 - T_c/T_h = 1 - $\frac{300}{600}$ = 0.5. Work done W = η × Q_h = 0.5 × 1200 = 600 J. Heat rejected Q_c = Q_h - W = 1200 - 600 = 600 J. Verification: Q_c/Q_h = T_c/T_h → $\frac{600}{1200}$ = $\frac{300}{600}$ ✓

Understanding:

We need the point on the line joining the charges where the net electric potential is zero. Let the charges be $q_1=+4\mu \text{C}$ at origin and $q_2=-1\mu \text{C}$ at $3\text{m}$.

Formula:

Electric potential due to a point charge:

For the net potential to be zero:

Step 1: Set up the equation.

Let the point be at distance $x$ from $q_1$. For a point between the charges ($0<x<3$), the distance from $q_2$ is $(3-x)$:

Step 2: Solve for $x$ (between charges).

Step 3: Check the external point.

For a point beyond $q_2$ (at distance $x$ from $q_1$, $x>3$):

But the question asks for the point other than at infinity — both $2.4\text{m}$ and $4\text{m}$ are valid zeros; $2.4\text{m}$ is between the charges.

Answer:

The electric potential is zero at $2.4\text{m}$ from the $+4\mu \text{C}$ charge (between the two charges).

Quick Tip:

Potential is a scalar, so unlike field, you simply add values algebraically. For two charges of opposite sign, there is always an internal zero and an external zero (at finite distance), unlike the electric field.

Understanding:

We need the energy stored in a dielectric-filled parallel plate capacitor.

Formula:

Capacitance with dielectric:

Energy stored:

Step 1: Calculate capacitance.

Step 2: Calculate energy stored.

Answer:

The energy stored in the capacitor is $2.2125\mu \text{J}$.

Quick Tip:

A common mistake is forgetting to square the voltage. Also note that introducing a dielectric increases the capacitance by factor $K$, and hence increases the stored energy by the same factor when connected to a fixed voltage source.

Understanding:

A proton moves perpendicular to a magnetic field and follows a circular path. We need the radius.

Formula:

The magnetic force provides centripetal force:

Step 1: Substitute values.

Answer:

The radius of the circular path is approximately $0.0696\text{m}$.

Quick Tip:

The radius of the circular path increases with velocity and mass but decreases with larger charge or stronger magnetic field. For an electron, the radius would be about 1836 times smaller for the same velocity.

Understanding:

We need the force per unit length between two current-carrying parallel wires.

Formula:

Currents in the same direction → attractive force.

Step 1: Substitute values.

Step 2: Simplify.

Step 3: Determine nature of force.

Since both currents flow in the same direction, by Ampere's rule the wires attract each other.

Answer:

The force per unit length is $4.8\times 10^{-5}\text{N/m}$ and is attractive.

Quick Tip:

Remember: same-direction currents attract, opposite-direction currents repel. This is the principle behind the definition of the Ampere in SI units.

Understanding:

We need the electric field outside a uniformly charged spherical shell at a given point.

Formula:

By Gauss's law, for $r>R$, the shell behaves as a point charge:

Step 1: Confirm the point is outside the shell.

$r=0.5\text{m}>R=0.2\text{m}$, so the formula applies.

Step 2: Substitute values.

Answer:

The electric field at $r=0.5\text{m}$ from the centre is $3.6\times 10^5\text{N/C}$.

Quick Tip:

The electric field inside a uniformly charged spherical shell is exactly zero. The field outside is identical to that of an equivalent point charge at the centre — a direct consequence of Gauss's law.

Understanding:

We need the magnetic flux through one turn of a solenoid carrying current.

Formula:

Magnetic field inside a solenoid:

Flux through one turn:

Step 1: Calculate the number of turns per unit length.

Step 2: Calculate the magnetic field.

Step 3: Calculate the flux through one turn.

Answer:

The magnetic flux through one turn is $2.01\times 10^{-6}\text{Wb}$.

Quick Tip:

The total flux linkage of the solenoid is $N\Phi$, which is used to calculate the self-inductance $L=N\Phi /I$. Here $L=1000\times 2.01\times 10^{-6}/2\approx 1\text{mH}$.

Understanding:

Three capacitors in series are connected across a battery. We need the charge on each capacitor.

Formula:

Equivalent capacitance for series combination:

Charge on each capacitor (same in series):

Step 1: Find the equivalent capacitance.

Step 2: Calculate the charge.

Step 3: Key property of series combination.

In a series combination, the same charge $Q$ appears on every capacitor, regardless of individual capacitances.

Answer:

The charge on each capacitor is $12\mu \text{C}$.

Quick Tip:

In a series circuit, capacitors share the same charge but different voltages. The smallest capacitance has the largest voltage drop across it.

Understanding:

We need the magnetic field at the centre of a circular current-carrying loop.

Formula:

Step 1: Substitute values.

Answer:

The magnetic field at the centre of the circular loop is $\pi \times 10^{-5}\text{T}$.

Quick Tip:

For $N$ turns instead of one, the field at the centre is simply multiplied by $N$: $B=\frac{{\mu }_{0}NI}{2r}$. Don't forget the factor of 2 in the denominator — a very common error.

Understanding:

We need the torque on an electric dipole in a uniform electric field.

Formula:

Step 1: Substitute values.

Answer:

The torque acting on the dipole is $8.66\times 10^{-6}\text{Ncdotm}$.

Quick Tip:

The torque is maximum when $\theta =90°$ (dipole perpendicular to field) and zero when $\theta =0°$ or $180°$ (dipole along or opposite to field). The equilibrium at $\theta =0°$ is stable, while at $\theta =180°$ it is unstable.

Understanding:

We need the magnetic field inside a toroid.

Formula:

For a toroid, the magnetic field inside is derived from Ampere's law along a circular path of radius $R$:

Step 1: Substitute values.

Answer:

The magnetic field inside the toroid is $2\times 10^{-3}\text{T}$.

Quick Tip:

Unlike a solenoid, the field inside a toroid is not uniform — it varies as $B\propto 1/r$. However, at the mean radius, the formula $B={\mu }_{0}NI/(2\pi R)$ gives the average field. Also note that the field outside a toroid is exactly zero.

Understanding:

We need the orbital speed of a satellite at height $h$ above Earth's surface.

Formula:

For circular orbit, gravitational force provides centripetal force:

Step 1: Solve for orbital speed.

Step 2: Express $GM$ in terms of $g$ and $R$.

At Earth's surface: $g=\frac{GM}{R^2}$, so $GM=g{R}^2$.

Step 3: Substitute into the expression for $v$.

Answer:

The orbital speed of the satellite is:

Quick Tip:

For a satellite at the surface ($h=0$), this reduces to $v=\sqrt{gR}$, which is the first cosmic velocity — a value worth remembering.