Mathematical Physics Questions and Answers

Answer: D

The vector that completes a basis is \(\vec{V}_{3} = 2\hat{\imath} + \hat{\jmath} + 4\hat{k}\).

Three vectors span three-dimensional space only if they are linearly independent, that is, if \(\vec{V}_{3}\) does not lie in the plane of \(\vec{V}_{1}\) and \(\vec{V}_{2}\). The normal to that plane is

\[ \begin{aligned} \vec{n} &= \vec{V}_{1} \times \vec{V}_{2} \\ &= -2\hat{\imath} - 2\hat{\jmath} + \hat{k} \end{aligned} \]

and \(\vec{V}_{3}\) works only if \(\vec{n} \cdot \vec{V}_{3} \ne 0\) (the scalar triple product):

• \(\hat{\imath} + \hat{\jmath} + 4\hat{k}\): \(-2 - 2 + 4 = 0\)
• \(2\hat{\imath} - \hat{\jmath} + 2\hat{k}\): \(-4 + 2 + 2 = 0\)
• \(\hat{\imath} + 2\hat{\jmath} + 6\hat{k}\): \(-2 - 4 + 6 = 0\)
• \(2\hat{\imath} + \hat{\jmath} + 4\hat{k}\): \(-4 - 2 + 4 = -2\)

The first three lie in the plane of \(\vec{V}_{1}\) and \(\vec{V}_{2}\) (for example \(\hat{\imath} + \hat{\jmath} + 4\hat{k} = 5\vec{V}_{1} + 2\vec{V}_{2}\)), so only the last makes a complete set.

Answer: B

The quantities that change sharply are \(\rho, C, \chi\).

• Resistivity drops abruptly to zero at \(T_{c}\), the defining property of a superconductor.
• Specific heat shows a discontinuous jump at \(T_{c}\), the signature of a second-order phase transition.
• Magnetic susceptibility changes abruptly to \(\chi = -1\), perfect diamagnetism, because of the Meissner effect.

Thermal conductivity does not jump at \(T_{c}\); it changes gradually below it as electrons condense into pairs that carry no entropy. Young’s modulus, a lattice property, is essentially unaffected. So options including \(\kappa\) or \(Y\) are wrong, and only \(\rho, C, \chi\) matches.

Answer: B

The tensor becomes matrix (B):

\[ \begin{pmatrix} \sin 2\theta & \cos 2\theta \\ \cos 2\theta & -\sin 2\theta \end{pmatrix} \]

A rank 2 contravariant tensor transforms as \(T' = R\,T\,R^{T}\). For axes rotated anticlockwise by \(\theta\),

\[ R = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \]

First

\[ R\,T = \begin{pmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{pmatrix} \]

then multiplying by \(R^{T}\) gives the entries \(2\sin\theta\cos\theta = \sin 2\theta\), \(\cos^{2}\theta - \sin^{2}\theta = \cos 2\theta\) (twice) and \(-\sin 2\theta\).

A quick check: the original matrix is symmetric with zero trace, and a rotation preserves both properties. Only matrix (B) is symmetric and traceless; (A) and (C) are not symmetric, and the last has trace \(\sin 2\theta - \cos 2\theta\).

Answer: A

The expression is \(\psi(\vec{\nabla} \times \vec{A}) - \vec{A} \times (\vec{\nabla}\psi)\).

The product rule for the curl of a scalar times a vector is

\[ \begin{aligned} \vec{\nabla} \times (\psi\vec{A}) &= \psi(\vec{\nabla} \times \vec{A}) + (\vec{\nabla}\psi) \times \vec{A} \\ &= \psi(\vec{\nabla} \times \vec{A}) - \vec{A} \times (\vec{\nabla}\psi) \end{aligned} \]

The second line uses \(\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}\). The option with a plus sign keeps \(\vec{A} \times \vec{\nabla}\psi\) without reversing the sign, and the last option is the negative of the correct result. The curl is not zero in general, so 'Null vector' is wrong.

Answer: C

The curl is non-zero and points into the plane.

The sketch shows a shear field: \(\vec{V} = V_{x}(y)\hat{x}\), with \(V_{x}\) positive above the x axis, negative below it, and growing with \(|y|\). Take it as \(V_{x} = ky\) with k > 0. Then

\[ \begin{aligned} \vec{\nabla} \times \vec{V} &= \left(\frac{\partial V_{y}}{\partial x} - \frac{\partial V_{x}}{\partial y}\right)\hat{z} \\ &= (0 - k)\hat{z} = -k\hat{z} \end{aligned} \]

With x to the right and y up, \(-\hat{z}\) points into the page. A paddle wheel placed anywhere would turn clockwise, because the flow above it is faster to the right than the flow below. The curl is the same everywhere, so it is not zero and does not reverse across the x axis, and its direction is into, not out of, the plane.

Answer: B

The curve is \(x = 2y\).

The Laplacian in two dimensions is the sum of the second partial derivatives:

\[ \begin{aligned} \frac{\partial^{2} f}{\partial x^{2}} &= 6x, \quad \frac{\partial^{2} f}{\partial y^{2}} = -12y \\ \nabla^{2} f &= 6x - 12y = 0 \\ x &= 2y \end{aligned} \]

The square-root options come from mistakes such as differentiating only once, and \(x = -y/2\) has the wrong sign and factor. Only \(x = 2y\) makes \(6x\) and \(12y\) cancel.

Answer: D

The unit tangent is \(\frac{\hat{\imath} + 2\hat{\jmath} + 3\hat{k}}{\sqrt{14}}\).

The tangent is the derivative of the position vector, divided by its length:

\[ \begin{aligned} \frac{d\vec{r}}{dt} &= \hat{\imath} + 2t\hat{\jmath} + 3t^{2}\hat{k} \\ \text{at } t = 1: &\ \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \\ \left|\,\cdot\,\right| &= \sqrt{1 + 4 + 9} = \sqrt{14} \end{aligned} \]

The first option is the position vector itself normalised, not its derivative. The others use wrong coefficients for the \(t^{2}\) and \(t^{3}\) terms; for example \(\hat{\imath} + 2\hat{\jmath} + 2\hat{k}\) forgets that \(d(t^{3})/dt = 3t^{2}\).

Answer: A

The line integral is 0.

An electrostatic field is the negative gradient of a potential, \(\vec{E} = -\nabla V\), so its line integral between two points depends only on the end points:

\[ \int_a^b \vec{E} \cdot d\vec{l} = V(a) - V(b) \]

The path in the figure is closed: it starts and ends at (0,0). So the integral is \(V(0,0) - V(0,0) = 0\), whatever the form of \(V\). Equivalently, \(\nabla \times \vec{E} = 0\), so by Stokes' theorem the circulation round any loop vanishes.

The specific powers \(x^8y^9\) are a distraction. The values 1, 2 and 1/2 would need a non-conservative field, or come from integrating only part of the loop: along the side from (1,0) to (1,1) alone, the integral is \(V(1,0) - V(1,1) = -1\).

Answer: A

The resultant vector lies in the \(xy\)-plane.

Use the identity below with \(\vec{V} = xyz\,\hat{k}\):

\[ \begin{aligned} & \nabla \times (\nabla \times \vec{V}) \\ &= \nabla(\nabla \cdot \vec{V}) - \nabla^{2}\vec{V} \end{aligned} \]

\[ \begin{aligned} \nabla \cdot \vec{V} &= \frac{\partial (xyz)}{\partial z} = xy \\ \nabla(xy) &= y\,\hat{\imath} + x\,\hat{\jmath} \\ \nabla^{2}(xyz) &= 0 \end{aligned} \]

So \(\nabla \times \nabla \times \vec{V} = y\,\hat{\imath} + x\,\hat{\jmath}\), which has no \(\hat{k}\) component and lies in the \(xy\)-plane. Working it directly gives the same: \(\nabla \times \vec{V} = xz\,\hat{\imath} - yz\,\hat{\jmath}\), whose curl is \(y\,\hat{\imath} + x\,\hat{\jmath}\).

It has both \(\hat{\imath}\) and \(\hat{\jmath}\) parts, so it is not along \(x\) alone, and it has no \(z\) component, so it cannot lie in the \(yz\)- or \(xz\)-plane in general.