NEET Thermal Properties of Matter Previous Year Questions with Solutions

For an isotropic metallic sphere of initial radius R:

• • gamma is the coefficient of cubical (volume) expansion, which equals 3 times the linear coefficient alpha.

The increase in volume for a small temperature rise Delta T is given by thermal expansion theory:

• • Valid for small temperature increments (Delta T << 1/alpha).

Substitute V = (4/3) * pi * R^3 into the expansion formula:

• • The factor of 3 in the numerator cancels the factor of 3 in the denominator.
• • Leaves the exact result: 4 * pi * R^3 * alpha * Delta T.
• • Matches Option (3) exactly.

Using the thermal resistance formula R_th = L / (k A), let R₀ = L / (K A) denote the thermal resistance of the middle rod:

• • Total thermal resistance in series: R_total = R₁ + R₂ + R₃ = R₀/2 + R₀ + R₀/2 = 2R₀.

In steady state, the same rate of heat flow H passes through all three rods in series:

• • Heat flow rate H is constant throughout the composite rod.

Apply heat conduction across each individual section:

• • Ratio T₁ / T₂ = (5/2 T) / (3/2 T) = 5/3, corresponding to Option (3).

When a metallic bar of length L is heated through temperature difference ΔT, its natural unrestricted thermal expansion would be ΔL = L α ΔT. Because the bar is rigidly held between immovable walls with no expansion allowed, a compressive mechanical strain is developed:

From Hooke’s law, compressive thermal stress σ is the product of Young’s modulus Y and thermal strain:

The compressive force F developed in the cross-section is the stress multiplied by the cross-sectional area A:

Substitute Y = 0.5 × 10¹¹ N m⁻², A = 10⁻³ m², α = 10⁻⁵ °C⁻¹, and ΔT = 100°C:

For a body cooling in surroundings at constant ambient temperature $T_0$, the average rate of fall of temperature over time interval $\Delta t$ is proportional to the excess of average temperature over surroundings:

For cooling from $90^\circ\text{C}$ to $80^\circ\text{C}$ in time $t$ with $T_0 = 20^\circ\text{C}$:

For cooling from $80^\circ\text{C}$ to $60^\circ\text{C}$ in time $t'$ in the same surroundings:

Divide equation (1) by equation (2) to eliminate the cooling constant $K$:

The heat required to raise the temperature of a body by $$\Delta T$$ is given by: $$Q = m s \Delta T$$, where $$m$$ is mass and $$s$$ is specific heat capacity.

For a solid sphere of uniform density $$\rho$$ and radius $$r$$: $$V = \frac{4}{3}\pi r^3$$ $$m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right)$$ Because both spheres are made of copper, density $$\rho$$ and specific heat $$s$$ are identical.

$$Q = \left[\rho \left(\frac{4}{3}\pi r^3\right)\right] s \Delta T$$ For identical $$\Delta T = 1\text{ K}$$: $$Q \propto r^3$$.

$$\frac{Q_1}{Q_2} = \left(\frac{r_1}{r_2}\right)^3 = (1.5)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$. Matches Option 1.