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Vector Analysis Previous Year Question with Solution

Pravegaa PYQ Series · Volume 1: Mathematical Physics

MP01: Vector Analysis

63 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 35
  • GATE: 7
  • CSIR NET: 4
  • JEST: 12
  • TIFR GS: 5
MP01.1IIT JAM 2026, Q27MCQ | 2 marks

Consider the following vector \(\vec{V}=xyz\,\hat{k}\). Which of the following statements is correct for the resultant vector \(\vec{\nabla}\times\vec{\nabla}\times\vec{V}\)?

  1. It lies in \(xy\)-plane

  2. It lies in \(yz\)-plane

  3. It lies in \(xz\)-plane

  4. It is along the \(x\) direction

Show solution

Use \(\vec\nabla\times(\vec\nabla\times\vec V)=\vec\nabla(\vec\nabla\cdot\vec V)-\nabla^2\vec V\): \[ \vec\nabla\cdot\vec V=\frac{\partial(xyz)}{\partial z}=xy,\qquad \vec\nabla(xy)=y\,\hat{\imath}+x\,\hat{\jmath},\qquad \nabla^2(xyz)=0 . \] \[ \vec\nabla\times\vec\nabla\times\vec V=y\,\hat{\imath}+x\,\hat{\jmath}, \] which has no \(z\)-component. \[ \boxed{y\,\hat{\imath}+x\,\hat{\jmath}\ \text{lies in the } xy\text{-plane}} \]

Answer: (A) It lies in \(xy\)-plane

MP01.2IIT JAM 2025, Q1MCQ | 1 mark

Consider a volume \(V\) enclosed by a closed surface \(S\) having unit surface normal \(\hat{n}\). For \(\mathbf{r}=x\hat{\imath}+y\hat{\jmath}+z\hat{k}\), the value of the surface integral \[\frac{1}{9}\oiint \mathbf{r}\cdot\hat{n}\,dS\] is

  1. \(V\)

  2. \(3V\)

  3. \(\dfrac{V}{3}\)

  4. \(\dfrac{V}{9}\)

Show solution

By the divergence theorem, with \(\vec\nabla\cdot\vec r = 3\), \[\frac{1}{9}\oint_S \vec r\cdot\hat n\,dS=\frac{1}{9}\int_V(\vec\nabla\cdot\vec r)\,d\tau=\frac{1}{9}\int_V 3\,d\tau=\boxed{\dfrac{V}{3}}\]

Answer: (C) \(V/3\)

MP01.3IIT JAM 2025, Q3MCQ | 1 mark

Which one of the following figures represents the vector field \(\vec A=y\hat{\imath}\)?

(\(\hat{\imath}\) is the unit vector along the \(x\)-direction)Figure

Show solution

The field \(\vec A=y\hat\imath\) always points along \(\pm x\). For \(y>0\) it points along \(+\hat\imath\), for \(y<0\) along \(-\hat\imath\), and its magnitude \(|y|\) grows with distance from the \(x\)-axis (zero on the axis). Hence arrows point to the right above the \(x\)-axis, to the left below it, and get longer away from the axis. This is figure \(\boxed{\text{(A)}}\).

Answer: (A)

MP01.4IIT JAM 2025, Q15MCQ | 2 marks

In Cartesian coordinates, consider the functions \(u(x,y)=\dfrac12\left(x^2-y^2\right)\) and \(v(x,y)=xy\). If \((r,\theta)\) are the polar coordinates, the Jacobian determinant \(\left|\dfrac{\partial(u,v)}{\partial(r,\theta)}\right|\) is

  1. \(r\)

  2. \(\dfrac1r\)

  3. \(r^2\)

  4. \(r^3\)

Show solution

With \(x=r\cos\theta\), \(y=r\sin\theta\): \[u=\frac{r^2}{2}\cos2\theta,\qquad v=\frac{r^2}{2}\sin2\theta .\] \[\frac{\partial(u,v)}{\partial(r,\theta)}=\begin{vmatrix}r\cos2\theta & -r^2\sin2\theta\\ r\sin2\theta & r^2\cos2\theta\end{vmatrix}=r^3\left(\cos^22\theta+\sin^22\theta\right)=\boxed{r^3}\]

Answer: (D) \(r^3\)

MP01.5IIT JAM 2025, Q54NAT | 2 marks

Consider a vector \(\vec F=\dfrac{1}{\pi}\left[-\sin y\,\hat\imath+x(1-\cos y)\hat\jmath\right]\). The value of the integral \(\oint\vec F\cdot d\vec r\) over a circle \(x^2+y^2=1\) evaluated in the anti-clockwise direction is . (in integer)

Show solution

By Stokes' theorem, \(\oint_C\vec F\cdot d\vec r=\iint_S(\vec\nabla\times\vec F)\cdot\hat k\,dS\), with \(S\) the unit disc and \(\hat n=\hat k\) for anticlockwise \(C\): \[(\vec\nabla\times\vec F)_z=\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=\frac1\pi\left[(1-\cos y)-(-\cos y)\right]=\frac1\pi .\] Hence \[\oint\vec F\cdot d\vec r=\frac1\pi\times(\text{area of unit disc})=\frac1\pi\times\pi=\boxed{1}\]

Answer: 1

MP01.6IIT JAM 2024, Q4MCQ | 1 mark

The divergence of a 3-dimensional vector \(\dfrac{\hat r}{r^3}\) (\(\hat r\) is the unit radial vector) is:

  1. \(-\dfrac{1}{r^4}\)

  2. zero

  3. \(\dfrac{1}{r^3}\)

  4. \(-\dfrac{3}{r^4}\)

Show solution

For a purely radial field \(\vec A=A_r\hat r\), \[ \nabla\cdot\vec A=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2A_r\right)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(\frac{1}{r}\right)=\frac{1}{r^2}\left(-\frac{1}{r^2}\right)=\boxed{-\frac{1}{r^4}} \]

Answer: (A) \(-1/r^4\)

MP01.7IIT JAM 2024, Q9MCQ | 1 mark

A vector field is expressed in the cylindrical coordinate system \((s,\phi,z)\) as, \[ \vec F=\frac{A}{s}\hat s+\frac{B}{s}\hat z . \] If this field represents an electrostatic field, then the possible values of \(A\) and \(B\), respectively, are:

  1. 1 and 0

  2. 0 and 1

  3. \(-1\) and 1

  4. 1 and \(-1\)

Show solution

An electrostatic field must be irrotational, \(\nabla\times\vec F=0\). With \(F_s=A/s\), \(F_\phi=0\), \(F_z=B/s\): \[ (\nabla\times\vec F)_s=\frac1s\frac{\partial F_z}{\partial\phi}-\frac{\partial F_\phi}{\partial z}=0,\qquad (\nabla\times\vec F)_\phi=\frac{\partial F_s}{\partial z}-\frac{\partial F_z}{\partial s}=\frac{B}{s^2},\qquad (\nabla\times\vec F)_z=0 . \] So we need \(B=0\), while \(A\) is unrestricted. Among the options only \(\boxed{A=1,\ B=0}\) works.

Answer: (A) 1 and 0

MP01.8IIT JAM 2024, Q14MCQ | 2 marks

The value of the line integral for the vector, \[ \vec v=2\hat x+yz^2\hat y+(3y+z^2)\hat z \] along the closed path OABO (as shown in the figure) is:Figure(Path \(AB\) is the arc of a circle of unit radius)

  1. \(\dfrac14(3\pi-1)\)

  2. \(3\pi-\dfrac14\)

  3. \(\dfrac{3\pi}{4}-1\)

  4. \(3\pi-1\)

Show solution

Use Stokes' theorem, \(\oint\vec v\cdot d\vec l=\int_S(\nabla\times\vec v)\cdot d\vec a\). The path lies in the \(yz\)-plane and is traversed \(O\to A\) (along \(y\)) \(\to B\) (along \(z\)), i.e. anticlockwise seen from \(+x\), so \(d\vec a=\hat x\,dy\,dz\). \[ (\nabla\times\vec v)_x=\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}=3-2yz . \] Over the quarter disc of unit radius (polar coordinates \(y=r\cos\theta\), \(z=r\sin\theta\)): \[ \int_S 3\,da=3\cdot\frac{\pi}{4},\qquad \int_S 2yz\,da=2\int_0^1 r^3dr\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta=2\cdot\frac14\cdot\frac12=\frac14 . \] \[ \oint\vec v\cdot d\vec l=\frac{3\pi}{4}-\frac14=\boxed{\frac14(3\pi-1)} \] (Direct check: \(OA\) gives 0, arc \(AB\) gives \(\tfrac{3\pi}{4}-\tfrac14+\tfrac13\), \(BO\) gives \(-\tfrac13\).)

Answer: (A) \(\frac14(3\pi-1)\)

MP01.9IIT JAM 2024, Q15MCQ | 2 marks

In the \(x-y\) plane, a vector is given by \[ \vec F(x,y)=\frac{-y\hat x+x\hat y}{x^2+y^2}. \] The magnitude of the flux of \(\vec\nabla\times\vec F\), through a circular loop of radius 2, centered at the origin, is:

  1. \(\pi\)

  2. \(2\pi\)

  3. \(4\pi\)

  4. 0

Show solution

In plane polar coordinates \(-y\hat x+x\hat y=s\,\hat\phi\), so \(\vec F=\hat\phi/s\). Its curl vanishes for \(s\neq0\), but the field is singular at the origin, which lies inside the loop, so the flux must be found with Stokes' theorem: \[ \int_S(\nabla\times\vec F)\cdot d\vec a=\oint_C\vec F\cdot d\vec l=\int_0^{2\pi}\frac{1}{s}\,(s\,d\phi)=\boxed{2\pi} \] (independent of the radius; equivalently \(\nabla\times\vec F=2\pi\,\delta^2(\vec r)\,\hat z\)).

Answer: (B) \(2\pi\)

MP01.10IIT JAM 2024, Q42NAT | 1 mark

The co-ordinate system \((x,y,z)\) is transformed to the system \((u,v,w)\), as given by: \[ u=2x+3y-z,\qquad v=x-4y+z,\qquad w=x+y \] The Jacobian of the above transformation is

Show solution

\[ J=\frac{\partial(u,v,w)}{\partial(x,y,z)}=\begin{vmatrix}2&3&-1\\1&-4&1\\1&1&0\end{vmatrix} =2(0-1)-3(0-1)+(-1)(1+4)=-2+3-5=\boxed{-4} \] The inverse transformation has Jacobian \(\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=\dfrac1J=-\dfrac14\), and the volume element scales as \(du\,dv\,dw=|J|\,dx\,dy\,dz=4\,dx\,dy\,dz\). The question does not specify which Jacobian is meant (\(-4\), \(-\tfrac14\), or the magnitude \(4\)), so \(\boxed{\text{Marks to All}}\).

Answer: Marks to All (ambiguous: \(J=-4\), inverse \(-1/4\))

MP01.11IIT JAM 2024, Q43NAT | 1 mark

Two sides of a triangle \(OAB\) are given by: \[ \overrightarrow{OA}=\hat x+2\hat y+\hat z,\qquad \overrightarrow{OB}=2\hat x-\hat y+3\hat z \] The area of the triangle is . (Rounded off to one decimal place)

Show solution

\[ \overrightarrow{OA}\times\overrightarrow{OB}=\begin{vmatrix}\hat x&\hat y&\hat z\\1&2&1\\2&-1&3\end{vmatrix} =\hat x(6+1)-\hat y(3-2)+\hat z(-1-4)=7\hat x-\hat y-5\hat z \] \[ \text{Area}=\frac12\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|=\frac12\sqrt{49+1+25}=\frac{\sqrt{75}}{2}=\boxed{4.33} \]

Answer: 4.3 (range 4.2 to 4.4)

MP01.12IIT JAM 2023, Q2MCQ | 1 mark

Which of the following fields has non-zero curl?

  1. \(x\hat i+y\hat j+z\hat k\)

  2. \((y+z)\hat i+(x+z)\hat j+(x+y)\hat k\)

  3. \(y^2\hat i+(2xy+z^2)\hat j+2yz\hat k\)

  4. \(xy\hat i+2yz\hat j+3xz\hat k\)

Show solution

Using \(\nabla\times\vec E=\left(\partial_yE_z-\partial_zE_y\right)\hat i+\left(\partial_zE_x-\partial_xE_z\right)\hat j+\left(\partial_xE_y-\partial_yE_x\right)\hat k\): \begin{align*} \text{(A)}\quad &\nabla\times\vec E=\hat i(0-0)+\hat j(0-0)+\hat k(0-0)=0,\\ \text{(B)}\quad &\nabla\times\vec E=\hat i(1-1)+\hat j(1-1)+\hat k(1-1)=0,\\ \text{(C)}\quad &\nabla\times\vec E=\hat i(2z-2z)+\hat j(0-0)+\hat k(2y-2y)=0,\\ \text{(D)}\quad &\nabla\times\vec E=\hat i(0-2y)+\hat j(0-3z)+\hat k(0-x)=-2y\,\hat i-3z\,\hat j-x\,\hat k\neq0 . \end{align*} \[\boxed{\text{(D)}}\]

Answer: (D)

MP01.13IIT JAM 2023, Q13MCQ | 2 marks

The Jacobian matrix for transforming from \((x,y)\) to another orthogonal coordinates system \((u,v)\) as shown in the figure isFigure

  1. \(\dfrac1{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}\)

  2. \(\dfrac1{\sqrt2}\begin{bmatrix}1&1\\-1&1\end{bmatrix}\)

  3. \(\dfrac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\)

  4. \(\dfrac1{\sqrt2}\begin{bmatrix}1&-1\\1&-1\end{bmatrix}\)

Show solution

From the figure, the \(v\)-axis is at \(+45^\circ\) and the \(u\)-axis at \(-45^\circ\) to the \(x\)-axis, so \[ \hat u=\frac{\hat x-\hat y}{\sqrt2},\qquad \hat v=\frac{\hat x+\hat y}{\sqrt2}. \] The new coordinates of a point \((x,y)\) are its projections on these axes: \[ u=\vec r\cdot\hat u=\frac{x-y}{\sqrt2},\qquad v=\vec r\cdot\hat v=\frac{x+y}{\sqrt2}. \] Hence \[ J=\begin{bmatrix}\partial u/\partial x&\partial u/\partial y\\[2pt]\partial v/\partial x&\partial v/\partial y\end{bmatrix} =\frac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}. \] \[\boxed{J=\tfrac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}}\]

Answer: (C)

MP01.14IIT JAM 2023, Q21MCQ | 2 marks

For a given vector \(\vec F=-y\hat i+z\hat j+x^2\hat k\), the surface integral \(\int_S(\vec\nabla\times\vec F)\cdot\hat r\,dS\) over the surface \(S\) of a hemisphere of radius \(R\) with the centre of the base at the origin isFigure

  1. \(\pi R^2\)

  2. \(\dfrac{2\pi R^2}{3}\)

  3. \(-\pi R^2\)

  4. \(-\dfrac{2\pi R^2}{3}\)

Show solution

By Stokes' theorem, the flux of \(\nabla\times\vec F\) through the curved hemispherical surface (outward normal \(\hat r\)) equals the circulation of \(\vec F\) around its boundary, the circle \(x^2+y^2=R^2\), \(z=0\), traversed anticlockwise seen from above: \[ \int_S(\vec\nabla\times\vec F)\cdot\hat r\,dS=\oint\vec F\cdot d\vec l=\oint\left(-y\,dx+z\,dy\right)_{z=0}=-\oint y\,dx . \] With \(x=R\cos\phi\), \(y=R\sin\phi\), \(dx=-R\sin\phi\,d\phi\): \[ -\oint y\,dx=\int_0^{2\pi}R^2\sin^2\phi\,d\phi=\pi R^2 . \] (Check: \(\nabla\times\vec F=-\hat i-2x\hat j+\hat k\), whose flux through the base disc with normal \(+\hat z\) is also \(\pi R^2\).) \[\boxed{\pi R^2}\]

Answer: (A) \(\pi R^2\)

MP01.15IIT JAM 2023, Q48NAT | 1 mark

Unit vector normal to the equipotential surface of \(V(x,y,z)=4x^2+y^2+z\) at \((1,2,1)\) is given by \((a\hat i+b\hat j+c\hat k)\). The value of \(|b|\) is (rounded off to two decimal places).

Show solution

The normal to the surface \(V=\) constant is along \(\nabla V\): \[ \nabla V=8x\,\hat i+2y\,\hat j+\hat k\ \xrightarrow{(1,2,1)}\ 8\hat i+4\hat j+\hat k,\qquad |\nabla V|=\sqrt{64+16+1}=9 . \] \[ \hat n=\frac{8\hat i+4\hat j+\hat k}{9}\quad\Rightarrow\quad|b|=\frac49=0.444\approx0.44 . \] \[\boxed{|b|\approx0.44}\]

Answer: 0.44 (range 0.43 to 0.45)

MP01.16IIT JAM 2022, Q8MCQ | 1 mark

Consider a two-dimensional force field \[\vec F(x,y)=(5x^2+ay^2+bxy)\,\hat x+(4x^2+4xy+y^2)\,\hat y .\] If the force field is conservative, then the values of \(a\) and \(b\) are

  1. \(a=2\) and \(b=4\)

  2. \(a=2\) and \(b=8\)

  3. \(a=4\) and \(b=2\)

  4. \(a=8\) and \(b=2\)

Show solution

For a conservative field \(\vec\nabla\times\vec F=0\). Only the \(z\)-component is non-trivial: \[ (\vec\nabla\times\vec F)_z=\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=(8x+4y)-(2ay+bx)=(8-b)x+(4-2a)y=0 \] for all \(x,y\). Hence \(b=8\) and \(a=2\). \[\boxed{a=2,\;b=8}\]

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