Mixed Type — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Mixed Type MCQs with step-by-step solutions (36 questions). Part of Vectors. Practise online on Prepizo — no login needed.
▶ Practise Mixed Type online (free)
Questions with solutions
Q1 — Mixed Type · medium · theory
$(\vec A + \vec B)\times(\vec A - \vec B)$ equals
A. $2(\vec A\times\vec B)$
B. $2(\vec B\times\vec A)$ ✓ Correct
C. $\vec 0$
D. $A^2 - B^2$
Solution: Expanding: $\vec A\times\vec A - \vec A\times\vec B + \vec B\times\vec A - \vec B\times\vec B = 2(\vec B\times\vec A)$.
Q2 — Mixed Type · easy · numerical
The value of $\lambda$ for which $\vec A = \lambda\hat{i} + \hat{j}$ is perpendicular to $\vec B = \hat{i} - \hat{j}$ is
A. $2$
B. $0$
C. $-1$
D. $1$ ✓ Correct
Solution: $\vec A\cdot\vec B = \lambda - 1 = 0 \Rightarrow \lambda = 1$.
Q3 — Mixed Type · medium · theory
The unit vector parallel to $\vec A = 3\hat{i} + 4\hat{j}$ is
A. $\frac{3\hat{i}+4\hat{j}}{7}$
B. $\frac{4\hat{i}+3\hat{j}}{5}$
C. $3\hat{i}+4\hat{j}$
D. $\frac{3\hat{i}+4\hat{j}}{5}$ ✓ Correct
Solution: Divide by the magnitude $\sqrt{9+16} = 5$: $\hat{A} = \frac{3\hat{i}+4\hat{j}}{5}$.
Q4 — Mixed Type · medium · theory
If $\vec A\cdot\vec B = 0$ and $\vec A\cdot\vec C = 0$, then $\vec A$ is parallel to
A. $\vec B\times\vec C$ ✓ Correct
B. $\vec B$
C. $\vec B + \vec C$
D. $\vec B - \vec C$
Solution: $\vec A$ is perpendicular to both $\vec B$ and $\vec C$, and $\vec B\times\vec C$ is the direction perpendicular to both — so $\vec A \parallel \vec B\times\vec C$.
Q5 — Mixed Type · medium · theory
If $\vec A = \vec B + \vec C$ and $|\vec A|^2 = |\vec B|^2 + |\vec C|^2$, then
A. $|\vec B| = |\vec C|$
B. $\vec C = \vec 0$
C. $\vec B \parallel \vec C$
D. $\vec B \perp \vec C$ ✓ Correct
Solution: $A^2 = B^2 + C^2 + 2\vec B\cdot\vec C$; the given condition forces $\vec B\cdot\vec C = 0$, i.e. perpendicular.
Q6 — Mixed Type · medium · theory
The position of a particle is $\vec r = 2t\,\hat{i} + t^2\,\hat{j}$ (metres). Its speed at $t = 1$ s is
A. $2$ m/s
B. $\sqrt{5}$ m/s
C. $2\sqrt{2}$ m/s ✓ Correct
D. $4$ m/s
Solution: $\vec v = \frac{d\vec r}{dt} = 2\hat{i} + 2t\hat{j}$; at $t=1$: $\vec v = 2\hat{i}+2\hat{j}$, speed $= 2\sqrt{2}$ m/s.
Q7 — Mixed Type · medium · theory
For two vectors, $\vec A\cdot\vec B = 6$ and $|\vec A\times\vec B| = 6\sqrt{3}$. The angle between them is
A. $90^\circ$
B. $30^\circ$
C. $45^\circ$
D. $60^\circ$ ✓ Correct
Solution: $\tan\theta = \frac{|\vec A\times\vec B|}{\vec A\cdot\vec B} = \frac{6\sqrt{3}}{6} = \sqrt{3} \Rightarrow \theta = 60^\circ$.
Q8 — Mixed Type · medium · theory
If $\vec A\times\vec B = \vec C\times\vec B$ for a non-zero $\vec B$, then
A. $\vec A - \vec C$ is parallel to $\vec B$ (not necessarily $\vec A = \vec C$) ✓ Correct
B. $\vec B = \vec 0$
C. $\vec A \perp \vec C$
D. $\vec A = \vec C$ always
Solution: $(\vec A-\vec C)\times\vec B = \vec 0$ means $\vec A-\vec C$ is parallel to $\vec B$ — equality is only one special case.
Q9 — Mixed Type · easy · numerical
A body of mass $0.5$ kg has momentum $\vec p = (3\hat{i} + 4\hat{j})$ kg·m/s. Its kinetic energy is
A. $25$ J ✓ Correct
B. $12.5$ J
C. $50$ J
D. $5$ J
Solution: $p^2 = \vec p\cdot\vec p = 9+16 = 25$; $KE = \frac{p^2}{2m} = \frac{25}{1} = 25$ J.
Q10 — Mixed Type · easy · numerical
Forces $\vec F_1 = \hat{i}+\hat{j}$ N and $\vec F_2 = 2\hat{i}-\hat{j}$ N act on a body displaced by $\vec d = 3\hat{i}$ m. The total work done is
A. $12$ J
B. $3$ J
C. $6$ J
D. $9$ J ✓ Correct
Solution: Net force $= 3\hat{i}$ N; $W = (3\hat{i})\cdot(3\hat{i}) = 9$ J.
Q11 — Mixed Type · easy · theory
$\vec B$ is parallel to $\vec A = 4\hat{i} + 3\hat{j}$ and $|\vec B| = 10$. Then $\vec B$ is
A. $4\hat{i} + 6\hat{j}$
B. $8\hat{i} + 6\hat{j}$ ✓ Correct
C. $5\hat{i} + 5\hat{j}$
D. $40\hat{i} + 30\hat{j}$
Solution: $\hat{A} = \frac{4\hat{i}+3\hat{j}}{5}$; $\vec B = 10\hat{A} = 8\hat{i}+6\hat{j}$.
Q12 — Mixed Type · medium · theory
The unit vector along the bisector of the angle between $\hat{i}$ and $\hat{j}$ is
A. $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$
B. $\hat{i}+\hat{j}$
C. $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ ✓ Correct
D. $\frac{\hat{i}+\hat{j}}{2}$
Solution: The bisector of two unit vectors is along their sum: $\hat{i}+\hat{j}$, normalised to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
Q13 — Mixed Type · medium · numerical
If $|\vec A| = 3$, $|\vec B| = 4$ and $|\vec A + \vec B| = 5$, then $|\vec A \times \vec B|$ is
A. $6$
B. $7$
C. $12$ ✓ Correct
D. $0$
Solution: $5^2 = 3^2+4^2$ means $\vec A \perp \vec B$. So $|\vec A\times\vec B| = AB\sin 90^\circ = 12$.
Q14 — Mixed Type · medium · numerical
The value of $(2\hat{i}+3\hat{j}+\hat{k})\cdot[(\hat{i}+\hat{j})\times(\hat{j}+\hat{k})]$ is
A. $-2$
B. $6$
C. $1$
D. $0$ ✓ Correct
Solution: $(\hat{i}+\hat{j})\times(\hat{j}+\hat{k}) = \hat{i}-\hat{j}+\hat{k}$; dotting: $2-3+1 = 0$. The three vectors are coplanar.
Q15 — Mixed Type · easy · numerical
A force of $10$ N acts at $60^\circ$ to the direction of a $4$ m displacement. The work done is
A. $20\sqrt{3}$ J
B. $20$ J ✓ Correct
C. $40$ J
D. $10$ J
Solution: $W = Fd\cos\theta = 10\times4\times\cos 60^\circ = 20$ J.
Q16 — Mixed Type · medium · theory
$\vec A\times(\vec B + \vec C)$ equals
A. $\vec B\times\vec A + \vec C\times\vec A$
B. $(\vec A\cdot\vec B)\vec C$
C. $\vec A\times\vec B + \vec A\times\vec C$ ✓ Correct
D. $\vec A\times\vec B - \vec A\times\vec C$
Solution: The cross product distributes over addition (order preserved): $\vec A\times(\vec B+\vec C) = \vec A\times\vec B + \vec A\times\vec C$.
Q17 — Mixed Type · easy · numerical
A particle rotates with angular velocity $\omega = 2$ rad/s at position $r = 3$ m perpendicular to the rotation axis. Its speed $|\vec v| = |\vec\omega\times\vec r|$ is
A. $6$ m/s ✓ Correct
B. $12$ m/s
C. $5$ m/s
D. $1.5$ m/s
Solution: $v = \omega r\sin 90^\circ = 2\times3 = 6$ m/s.
Q18 — Mixed Type · easy · theory
What vector must be added to $\hat{i} + 2\hat{j}$ to get a resultant of $3\hat{i}$?
A. $-2\hat{i} + 2\hat{j}$
B. $2\hat{i} - 2\hat{j}$ ✓ Correct
C. $2\hat{i} + 2\hat{j}$
D. $4\hat{i} + 2\hat{j}$
Solution: Required vector $= 3\hat{i} - (\hat{i}+2\hat{j}) = 2\hat{i} - 2\hat{j}$.
Q19 — Mixed Type · easy · theory
The vectors $\vec A = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec B = 2\hat{i} - 2\hat{j}$ are
A. Parallel
B. At $45^\circ$
C. Antiparallel
D. Perpendicular ✓ Correct
Solution: $\vec A\cdot\vec B = (1)(2) + (1)(-2) + (-2)(0) = 0$ — perpendicular.
Q20 — Mixed Type · medium · numerical
If $|\vec A| = 2$, $|\vec B| = 3$ and the angle between them is $60^\circ$, then $|\vec A\times\vec B|$ is
A. $3$
B. $6$
C. $3\sqrt{3}$ ✓ Correct
D. $\sqrt{19}$
Solution: $|\vec A\times\vec B| = 2\times3\times\sin 60^\circ = 6\times\tfrac{\sqrt{3}}{2} = 3\sqrt{3}$.
Q21 — Mixed Type · medium · theory
A unit vector in the xy-plane perpendicular to $\vec A = 3\hat{i} + 4\hat{j}$ is
A. $\frac{4\hat{i}+3\hat{j}}{5}$
B. $\hat{k}$
C. $\frac{3\hat{i}-4\hat{j}}{5}$
D. $\frac{4\hat{i}-3\hat{j}}{5}$ ✓ Correct
Solution: Swap components and negate one: $(4, -3)$ gives $\vec A\cdot\vec B = 12-12 = 0$; normalise by $5$. ($\hat{k}$ is perpendicular too but not in the xy-plane.)
Q22 — Mixed Type · medium · numerical
Points A and B have coordinates $(1, 2, 3)$ and $(4, 6, 3)$. The magnitude of $\vec{AB}$ is
A. $3$
B. $7$
C. $5$ ✓ Correct
D. $\sqrt{34}$
Solution: $\vec{AB} = 3\hat{i} + 4\hat{j} + 0\hat{k}$, magnitude $\sqrt{9+16} = 5$.
Q23 — Mixed Type · easy · theory
For unit vectors $\hat{p}$ and $\hat{q}$, $\hat{p}\cdot\hat{q}$ equals
A. $\sin\theta$
B. $1$ always
C. $\cos\theta$, where $\theta$ is the angle between them ✓ Correct
D. $\tan\theta$
Solution: $\hat{p}\cdot\hat{q} = (1)(1)\cos\theta = \cos\theta$.
Q24 — Mixed Type · medium · theory
A ball of mass $0.1$ kg moving with velocity $20\hat{i}$ m/s rebounds with velocity $-20\hat{i}$ m/s. The magnitude of the impulse on the ball is
A. $2$ N·s
B. $4$ N·s ✓ Correct
C. $40$ N·s
D. $0$
Solution: Impulse $= \Delta\vec p = m(\vec v_2 - \vec v_1) = 0.1\times(-40\hat{i})$, magnitude $4$ N·s.
Q25 — Mixed Type · medium · theory
If $\vec A = \hat{i} - \hat{j}$ and $\vec B = \hat{j} + \hat{k}$, then $|\vec A + 2\vec B|$ is
A. $\sqrt{5}$
B. $\sqrt{3}$
C. $3$
D. $\sqrt{6}$ ✓ Correct
Solution: $\vec A + 2\vec B = \hat{i} + \hat{j} + 2\hat{k}$, magnitude $\sqrt{1+1+4} = \sqrt{6}$.
Q26 — Mixed Type · easy · theory
If $(\vec a + \vec b)\cdot(\vec a + \vec b) = a^2 + b^2$, then $\vec a$ and $\vec b$ are
A. Perpendicular ✓ Correct
B. Parallel
C. Antiparallel
D. Equal
Solution: Expanding gives $a^2 + b^2 + 2\vec a\cdot\vec b$; equality forces $\vec a\cdot\vec b = 0$.
Q27 — Mixed Type · medium · numerical
A particle of mass $2$ kg at position $4\hat{i}$ m moves with velocity $3\hat{j}$ m/s. The magnitude of its angular momentum about the origin is
A. $24$ kg·m²/s ✓ Correct
B. $6$ kg·m²/s
C. $12$ kg·m²/s
D. $8$ kg·m²/s
Solution: $\vec L = \vec r\times m\vec v = 4\hat{i}\times(2)(3\hat{j}) = 24\hat{k}$; magnitude $24$ kg·m²/s.
Q28 — Mixed Type · medium · theory
If $\hat{a} + \hat{b}$ is itself a unit vector, the angle between the unit vectors $\hat{a}$ and $\hat{b}$ is
A. $90^\circ$
B. $0^\circ$
C. $120^\circ$ ✓ Correct
D. $60^\circ$
Solution: $|\hat{a}+\hat{b}|^2 = 1+1+2\cos\theta = 1 \Rightarrow \cos\theta = -\tfrac{1}{2} \Rightarrow \theta = 120^\circ$. (A famous exam pattern.)
Q29 — Mixed Type · medium · theory
If $|\hat{a} - \hat{b}| = \sqrt{2}$ for unit vectors $\hat{a}$ and $\hat{b}$, the angle between them is
A. $60^\circ$
B. $90^\circ$ ✓ Correct
C. $45^\circ$
D. $120^\circ$
Solution: $|\hat{a}-\hat{b}|^2 = 2 - 2\cos\theta = 2 \Rightarrow \cos\theta = 0 \Rightarrow \theta = 90^\circ$.
Q30 — Mixed Type · medium · theory
For any two vectors, $(\vec A\cdot\vec B)^2 \leq A^2B^2$. This statement is
A. Never true
B. True only for parallel vectors
C. Always true ✓ Correct
D. True only for perpendicular vectors
Solution: $(\vec A\cdot\vec B)^2 = A^2B^2\cos^2\theta \leq A^2B^2$ since $\cos^2\theta \leq 1$ (equality for parallel/antiparallel).