Question 11 Mark
Assertion $\quad$ : Vector addition is commutative.
Reason $\quad: \quad(\vec{A}+\vec{B}) \neq(\vec{B}+\vec{A})$.
Answer(c) Since vector addition is commutative, therefore $\vec{A}+\vec{B}=\vec{B}+\vec{A}$
View full question & answer→Question 21 Mark
Assertion : If $\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{C}$, then $\vec{A}$ may not always be equal to $\vec{C}$
Reason : The dot product of two vectors involves cosine of the angle between the two vectors.
Answer$ \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{C} \Rightarrow A B \cos \theta_1=B C \cos \theta_2 $
$ \therefore A=C, \text { only when } \theta_1=\theta_2$
So when angle between $\vec{A}$ and $\vec{B}$ is equal to angle between $\vec{B}$ and $\vec{C}$ only then $\vec{A}$ equal to $\vec{C}$
View full question & answer→Question 31 Mark
Assertion : The minimum number of non coplanar vectors whose sum can be zero, is four.
Reason : The resultant of two vectors of unequal magnitude can be zero.
Answer(c) The resultant of two vectors of unequal magnitude given by $R=\sqrt{A^2+B^2+2 A B \cos \theta}$ cannot be zero for any value of $\theta$.
View full question & answer→Question 41 Mark
Assertion : The cross product of a vector with itself is a null vector.
Reason : The cross-product of two vectors results in a vector quantity.
Question 51 Mark
Assertion $\quad$ : If dot product and cross product of $\vec{A}$ and $\vec{B}$ are zero, it implies that one of the vector $\vec{A}$ and $\vec{B}$ must be a null vector.
Reason $\quad$ : Null vector is a vector with zero magnitude.
Answer(b)$\begin{aligned}& \vec{A} \cdot \vec{B} \neq \vec{A} \| \quad \vec{B} \mid \cos \theta=0 \\& \vec{A} \times \vec{B} \neq \vec{A} \| \vec{B} \mid \sin \theta=0\end{aligned}$If $\vec{A}$ and $\vec{B}$ are not null vectors then it follows that $\sin \theta$ and (\cos \theta$ both should be zero simultaneously. But it cannot be possible so it is essential that one of the vector must be null vector.
View full question & answer→Question 61 Mark
Assertion : A null vector is a vector whose magnitude is zero and direction is arbitrary.
Reason $\quad$ : A null vector does not exist.
Answer(c) If two vectors equal in magnitude are in opposite direction, then their sum will be a null vector.A null vector has direction which is intermediate (or depends on direction of initial vectors) even its magnitude is zero.
View full question & answer→Question 71 Mark
Assertion : Multiplying any vector by an scalar is a meaningful operations.
Reason : In uniform motion speed remains constant.
Answer(b) We can multiply any vector by any scalar.For example, in equation $\vec{F}=m \vec{a}$ mass is a scalar quantity, but acceleration is a vector quantity.
View full question & answer→Question 81 Mark
Assertion : The scalar product of two vectors can be zero.
Reason : If two vectors are perpendicular to each other, their scalar product will be zero.
Answer(a) If $\theta$ be the angle between two vectors $\vec{A}$ and $\vec{B}$, then their scalar product, $\vec{A} \cdot \vec{B}=A B\cos \theta$If $\theta=90^{\circ}$ then $\vec{A} \cdot \vec{B}=0$i.e. if $\vec{A}$ and $\vec{B}$ are perpendicular to each other then their scalar product will be zero.
View full question & answer→Question 91 Mark
Assertion : Two vectors are said to be like vectors if they have same direction but different magnitude.
Reason $\quad:$ Vector quantities do not have specific direction.
Answer(c) If two vectors are in opposite direction, then they cannot be like vectors.
View full question & answer→Question 101 Mark
Assertion : The sum of two vectors can be zero.
Reason : The vector cancel each other, when they are equal and opposite.
Answer(a) Let $\vec{P}$ and $\vec{Q}$ are two vectors in opposite direction, then their sum $\vec{P}+(-\vec{Q})=\vec{P}-\vec{Q}$If $\vec{P}=\vec{Q}$ then sum equal to zero.
View full question & answer→Question 111 Mark
Assertion : A physical quantity cannot be called as a vector if its magnitude is zero.
Reason $\quad:$ A vector has both, magnitude and direction.
Answer(e) If a vector quantity has zero magnitude then it is called a null vector. That quantity may have some direction even if its magnitude is zero.
View full question & answer→Question 121 Mark
Assertion : A negative acceleration of a body is associated with a slowing down of a body.
Reason $\quad:$ Acceleration is vector quantity.
View full question & answer→Question 131 Mark
Assertion $\quad: \quad \vec{\tau}=\vec{r} \times F$ and $\vec{\tau} \neq F \times \vec{r}$
Reason $\quad:$ Cross product of vectors is commutative.
Answer(c) Cross-product of two vectors is anticommutative. i.e. $\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}$
View full question & answer→Question 141 Mark
Assertion $\quad: \quad \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}$
Reason : Dot product of two vectors is commutative.
View full question & answer→Question 151 Mark
Assertion : Vector addition of two vectors $\vec{A}$ and $\vec{B}$ is commutative.
Reason $\quad: \quad \vec{A}+\vec{B}=\vec{B}+\vec{A}$
Answer(b) Vector addition of two vectors is commutative i.e. $\vec{A}+\vec{B}=\vec{B}+\vec{A}$.
View full question & answer→Question 161 Mark
Assertion : Relative velocity of $A$ w.r.t. $B$ is greater than the velocity of either, when they are moving in opposite directions.
Reason : Relative velocity of $A$ w.r.t. $B=\vec{v}_A-\vec{v}_B$
Answer(a) Since velocities are in opposite direction, therefore$v_{A B} \neq \vec{v}_A-\vec{v}_B \mid=v_A+v_B \text {. }$Which is greater than $v_A$ or $v_B$
View full question & answer→Question 171 Mark
Assertion : Minimum number of non-equal vectors in a plane required to give zero resultant is three.
Reason $\quad$ : If $\vec{A}+\vec{B}+\vec{C}=\overrightarrow{0}$, then they must lie in one plane
Answer(b) For giving a zero resultant, it should be possible to represent the given vectors along the sides of a closed polygon and minimum number of sides of a polygon is three.
View full question & answer→Question 181 Mark
Assertion : Vector product of two vectors is an axial vector
Reason : If $\vec{v}=$ instantaneous velocity, $\vec{r}=$ radius vector and $\vec{\omega}=$ angular velocity, then $\vec{\omega}=\vec{v} \times \vec{r}$.
Answer(c) $\vec{v}=\vec{\omega} \times \vec{r}$The expression $\vec{\omega}=\vec{v} \times \vec{r}$ is wrong.
View full question & answer→Question 191 Mark
Assertion : If $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$, then angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$
Reason $\quad: \quad \vec{A}+\vec{B}=\vec{B}+\vec{A}$
Answer(b)$\begin{aligned}& |\vec{A}+\vec{B}|=|\vec{A}-\vec{B}| \\& \Rightarrow A^2+B^2+2 A B \cos \theta=A^2+B^2+2 A B \cos \theta\end{aligned$Hence $\cos \theta=0$ which gives $\theta=90^{\circ}$Also vector addition is commutative.Hence $\vec{A}+\vec{B}=\vec{B}+\vec{A}$.
View full question & answer→Question 201 Mark
Assertion : If $\theta$ be the angle between $\vec{A}$ and $\vec{B}$, then
$
\tan \theta=\frac{\vec{A} \times \vec{B}}{\vec{A} \cdot \vec{B}}
$
Reason $\quad: \quad \vec{A} \times \vec{B}$ is perpendicular to $\vec{A} \cdot \vec{B}$
Answer(d) $\frac{\vec{A} \times \vec{B}}{\vec{A} \cdot \vec{B}}=\frac{A B \sin \theta \hat{n}}{A B \cos \theta}=\tan \theta \hat{n}$where $\hat{n}$ is unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.However $\frac{|\vec{A} \times \vec{B}|}{\vec{A} \cdot \vec{B}}=\tan \theta$
View full question & answer→Question 211 Mark
Assertion $\quad:$ Angle between $\hat{i}+\hat{j}$ and $\hat{i}$ is $45^{\circ}$
Reason $: \hat{i}+\hat{j}$ is equally inclined to both $\hat{i}$ and $\hat{j}$ and the angle between $\hat{i}$ and $\hat{j}$ is $90^{\circ}$
Answer(a) $\quad \cos \theta=\frac{(\hat{i}+\hat{j}) \cdot(\hat{i})}{|\hat{i}+\hat{j}||\hat{i}|}=\frac{1}{\sqrt{2}}$. Hence (\theta=45^{\circ}$.
View full question & answer→Question 221 Mark
Assertion : $\vec{A} \times \vec{B}$ is perpendicular to both $\vec{A}+\vec{B}$ as well as $\vec{A}-\vec{B}$
Reason $\quad: \vec{A}+\vec{B}$ as well as $\vec{A}-\vec{B}$ lie in the plane containing $\vec{A}$ and $\vec{B}$, but $\vec{A} \times \vec{B}$ lies perpendicular to the plane containing $\vec{A}$ and $\vec{B}$.
Answer(a) Cross product of two vectors is perpendicular to the plane containing both the vectors.
View full question & answer→