IIT JAM 2026 Physics (PH) February 15 Question Paper With Solution PDF

Concept:
For any invertible matrix:
\[ |A^{-1}| = \frac{1}{|A|} \]

So first compute determinant of \(A\).



Step 1: Compute determinant of \(A\)
\[ A = \begin{pmatrix} 0 & 1 & 1
0 & 0 & 1
1 & 0 & 0 \end{pmatrix} \]

Expand along first row:
\[ |A| = 0\begin{vmatrix}0 & 1
0 & 0\end{vmatrix} -1\begin{vmatrix}0 & 1
1 & 0\end{vmatrix} +1\begin{vmatrix}0 & 0
1 & 0\end{vmatrix} \]



Step 2: Compute minors
\[ \begin{vmatrix}0 & 1
1 & 0\end{vmatrix} = (0)(0) - (1)(1) = -1 \]
\[ \begin{vmatrix}0 & 0
1 & 0\end{vmatrix} = 0 \]

So:
\[ |A| = -1(-1) + 1(0) = 1 \]

Wait carefully with sign:

Second term has minus sign from cofactor expansion:
\[ |A| = -1(-1) = 1 \]

But sign must be checked carefully:
\[ |A| = - ( -1 ) = 1 \]

However permutation parity method is cleaner.



Step 3: Use permutation method (faster)
Non-zero product from permutation:
\[ (1,2,3) \to (2,3,1) \]

This is an even permutation → determinant = \(+1\).
\[ |A| = 1 \]



Step 4: Determinant of inverse
\[ |A^{-1}| = \frac{1}{|A|} = 1 \]

But check orientation: matrix is a cyclic permutation matrix of order 3.
Such matrices have determinant \(+1\).

So final answer:
\[ |A^{-1}| = 1 \] Quick Tip: Key identity: \[ |A^{-1}| = \frac{1}{|A|} \] Permutation matrices have determinant \(\pm 1\).

Concept:
Convert the complex number into polar form and use De Moivre’s theorem:
\[ (r(\cos\theta + i\sin\theta))^n = r^n(\cos n\theta + i\sin n\theta) \]



Step 1: Convert to polar form
Given:
\[ z = 1 - i\sqrt{3} \]

Magnitude:
\[ r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \]

Argument:
\[ \tan\theta = \frac{-\sqrt{3}}{1} = -\sqrt{3} \Rightarrow \theta = -\frac{\pi}{3} \]

So:
\[ z = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) \]



Step 2: Apply De Moivre’s theorem
\[ z^3 = 2^3 \left(\cos\left(-\pi\right) + i\sin\left(-\pi\right)\right) \]
\[ = 8(\cos\pi - i\sin\pi) \]



Step 3: Use trig values
\[ \cos\pi = -1, \quad \sin\pi = 0 \]
\[ z^3 = 8(-1 + 0i) = -8 \]



Final Answer:
\[ \boxed{-8} \] Quick Tip: For powers of complex numbers: Convert to polar form → Apply De Moivre. It avoids messy binomial expansion.

Concept:
For any square matrix, the determinant equals the product of its eigenvalues:
\[ \det(A) = \lambda_1 \lambda_2 \cdots \lambda_n \]

For a \(1 \times 1\) matrix, there is only one eigenvalue.



Step 1: Structure of a \(1 \times 1\) matrix
Let:
\[ A = [a] \]

Then:
\[ \det(A) = a \]



Step 2: Eigenvalue of a \(1 \times 1\) matrix
The characteristic equation:
\[ |A - \lambda I| = 0 \Rightarrow a - \lambda = 0 \]
\[ \lambda = a \]

So the only eigenvalue equals the determinant.



Step 3: Given condition
\[ \det(A) = -1 \Rightarrow a = -1 \]

Thus eigenvalue:
\[ \lambda = -1 \]



Conclusion:
\[ \boxed{-1} \] Quick Tip: For \(1 \times 1\) matrices: Eigenvalue = Matrix entry = Determinant = Trace.

Concept:
Use Boolean algebra and factorization.
Group terms using common literals and apply:
\[ X\bar{Y} + XY = X \]



Step 1: Group terms
\[ (\bar{A}B\bar{C} + \bar{A}BC) + (AB\bar{C} + ABC) \]

Factor common terms in each group.



Step 2: Factor first pair
\[ \bar{A}B(\bar{C} + C) = \bar{A}B(1) = \bar{A}B \]



Step 3: Factor second pair
\[ AB(\bar{C} + C) = AB(1) = AB \]



Step 4: Combine results
\[ \bar{A}B + AB \]

Factor \(B\):
\[ B(\bar{A} + A) \]



Step 5: Apply complement law
\[ \bar{A} + A = 1 \]

So:
\[ B \cdot 1 = B \]



Final Answer:
\[ \boxed{B} \] Quick Tip: Key Boolean identities: \(X + \bar{X} = 1\) \(X\bar{Y} + XY = X\) Always group terms with common factors first.

Concept:
Average power is defined as:
\[ P_{avg} = \frac{Work done}{Time} \]

Work done = Change in kinetic energy.



Step 1: Use kinetic energy formula
\[ KE = \frac{1}{2}mv^2 \]

Initial velocity \(u = 2\,m/s\), final velocity \(v = 6\,m/s\), mass \(m = 10\,kg\).



Step 2: Change in kinetic energy
\[ \Delta KE = \frac{1}{2}m(v^2 - u^2) \]
\[ = \frac{1}{2} \times 10 \times (36 - 4) \]
\[ = 5 \times 32 = 160\,J \]



Step 3: Average power

Time = \(10\,s\)
\[ P = \frac{160}{10} = 16\,W \]



Final Answer:
\[ \boxed{16\,W} \] Quick Tip: If speed changes and force not given → use energy method: \[ P = \frac{\Delta KE}{t} \] It avoids finding force and acceleration.

Concept:
In thermodynamics, phase transitions are classified based on discontinuity in derivatives of Gibbs free energy.

- First-order transition → First derivatives of \(G\) are discontinuous.
- Examples: Melting, boiling.



Step 1: Heat capacity behavior
Divergence of \(C_p\) typically occurs in second-order transitions.
In first-order transitions, latent heat exists but \(C_p\) does not diverge.

So (A) is false.



Step 2: Derivatives of Gibbs free energy
First derivatives of \(G\):
\[ S = -\left(\frac{\partial G}{\partial T}\right)_P, \quad V = \left(\frac{\partial G}{\partial P}\right)_T \]

In first-order transitions, these are discontinuous.
Hence \(\partial G/\partial P\) is not continuous.

So (B) is false.



Step 3: Distinct thermodynamic states
Two phases (e.g., liquid and gas) coexist but are macroscopically distinct.

So (C) is true.



Step 4: Entropy behavior
Since:
\[ S = -\frac{\partial G}{\partial T} \]

And first derivatives are discontinuous, entropy changes abruptly due to latent heat.

So (D) is true.



Conclusion:
\[ \boxed{(C)\ and\ (D)} \] Quick Tip: First-order transition: Latent heat present Entropy discontinuous Second-order transition: No latent heat, \(C_p \to \infty\)