Physics M.C.Q (1 Marks)

2. Equal to AB.
3. Less than AB.
4. Equal to zero.

Explanation:

The magnitude of the vector product of two vectors $\big|\overrightarrow{\text{A}}\big|$ and $\big|\overrightarrow{\text{B}}\big|$ may be less than or equal to AB or equal to zero, but cannot be greater than AB.

3. $\alpha<\beta \text{ if}\text{A}>\text{B}$

Explanation:

The resultant of two vectors is closer to the vector with the greater magnitude. Thus, $\alpha<\beta \text{ if }\text{A}>\text{B}$

2. It is possible to have $|\overrightarrow{\text{C}}|<|\overrightarrow{\text{A}}|$ and $|\overrightarrow{\text{C}}|<|\overrightarrow{\text{B}}|$

Explanation:

Statements (a), (c) and (d) are incorrect.

Given $\overrightarrow{\text{C}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$

Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

2. C must be less than $|\text{A}-\text{B}|$

Explanation:

Here, we have three vector A, B and C.

$\big|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\big|^2=\big|\overrightarrow{\text{A}}\big|^2+\big|\overrightarrow{\text{B}}\big|^2+2\overrightarrow{\text{A}}.\overrightarrow{\text{B}} \ ...{\text{(i)}}$

$\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=\big|\overrightarrow{\text{A}}\big|^2+\big|\overrightarrow{\text{B}}\big|^2-2\overrightarrow{\text{A}}.\overrightarrow{\text{B}} \ ...{\text{(ii)}}$

Subtracting (i) from (ii), we get:

$\big|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\big|^2-\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$

Using the resultant property $\overrightarrow{\text{C}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}},$ we get:

$\big|\overrightarrow{\text{C}}\big|^2-\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$

$\Rightarrow\big|\overrightarrow{\text{C}}\big|^2=\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2+4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$

$\Rightarrow\big|\overrightarrow{\text{C}}\big|^2=\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2+4\big|\overrightarrow{\text{A}}\big|.\big|\overrightarrow{\text{B}}\big|\cos120^{\circ}$

Since cosine is negative in the second quadrant, C must be less than $|\text{A}-\text{B}|.$

1. Along the west.

Explanation:

The vector product $\overrightarrow{\text{A}}\times\overrightarrow{\text{B}}$ will point towards the west. We can determine this direction using the right hand thumb rule.

4. It is slid parallel to itself.

Explanation:

A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.

Let the magnitude of a displacement vector $\big(\overrightarrow{\text{A}}\big)$ directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.

2. 14.1cm².

Explanation:

Area of a circle, $\text{A}=\pi\text{r}^2$

On putting the values, we get:

$\text{A}=\frac{22}{7}\times2.12\times2.12$

$\Rightarrow\text{A}=14.1\text{cm}^2$

The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12cm has a minimum of three significant digits. So, the answer must be written in three significant digits.