NEET Physics - MCQ Practice Questions

According to the second law, entropy of an isolated system always increases for irreversible processes.

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

ΔG = ΔH - TΔS = 100000 - T(200). At low T, TΔS is small, so ΔG ≈ 100000 J > 0, non-spontaneous. Only at high T (T > 500 K) is it spontaneous.

For Carnot engine: η = 1 - Tc/Th = 1 - $\frac{300}{400}$ = 0.25. W = ηQh, so 100 = 0.25 × Qh, Qh = 400 J

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 → $\frac{600}{1200}$ = $\frac{300}{600}$ ✓

# Solution: Isothermal Expansion of Ideal Monatomic Gas

In an isothermal process, temperature remains constant, which has important implications for internal energy and entropy changes.

Step 1: Change in Internal Energy

For any ideal gas, internal energy depends only on temperature; since temperature is constant in an isothermal process, the change in internal energy must be zero.

Alternatively, using the first law of thermodynamics:

For an isothermal process: $Q=W=nRT\ln \left(\frac{V_f}{V_i}\right)$

Therefore:

Step 2: Change in Entropy

Entropy change during an isothermal expansion is calculated using the reversible heat transfer divided by temperature.

Substituting values: $n=5$ mol, $R=8.314$ J/mol·K, $V_f=50$ L, $V_i=10$ L

ΔU = 0 J and ΔS = 67.3 J/K

Answer: (A)

# Microscope Magnification Solution

The magnification of a microscope depends on the focal lengths of its lenses and the tube length, following the standard formula for compound microscopes.

Step 1: Identify the Magnification Formula

For a compound microscope, the total magnification is the product of objective magnification and eyepiece magnification.

where the tube length $L=v_o+u_e$ (approximately), $D$ = distance of distinct vision = 25 cm, and $f_e$ = focal length of eyepiece.

Step 2: Use the Standard Microscope Formula

For a microscope with tube length $L$, objective focal length $f_o$, and eyepiece focal length $f_e$, the magnification is:

Substituting the given values: $L=40$ cm, $f_o=2$ cm, $f_e=4$ cm, $D=25$ cm:

The magnification of the microscope is 125.

Answer: (B) 125

For a G.P. with first term $a_1$ and common ratio $r$, the n-th term is $a_n=a_1{r}^{n-1}$. We use the two given conditions to find $r$, then calculate $a_6$.

Step 1: Express terms using G.P. formula

**Step 2: Apply condition $a_1{a}_5=28$

Since all terms are positive: $a_1{r}^2=\sqrt{28}=2\sqrt{7}$

**Step 3: Apply condition $a_2+a_4=29$

**Step 4: Solve for $r$

Divide the equation from Step 3 by the result from Step 2:

Using the quadratic formula (or inspection): $r=\frac{29}{2\sqrt{7}}÷\frac{2\sqrt{7}}{1}$ gives $r=\frac{\sqrt{7}}{2}$ or $r=2\sqrt{7}$.

Since the sequence is increasing: $r=2\sqrt{7}$

**Step 5: Calculate $a_6$

From $a_1{r}^2=2\sqrt{7}$:

**Answer: $a_6=784$ (Option D)**