M.C.Q (1 Marks)

MCQ 11 Mark

A vector is not changed if:

A
It is rotated through an arbitrary angle.
B
It is multiplied by an arbitrary scalar.
C
It is cross multiplied by a unit vector.
✓
It is slid parallel to itself.

Answer

Correct option: D.

It is slid parallel to itself.

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.

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MCQ 21 Mark

The $x-$component of the resultant of several vectors:

A
Is equal to the sum of the $x-$components of the vectors.
B
May be smaller than the sum of the magnitudes of the vectors.
C
May be equal to the sum of the magnitudes of the vectors.
✓
All of the above

Answer

Correct option: D.

All of the above

The $x-$component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

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MCQ 31 Mark

Which of the sets given below may represent the magnitudes of three vectors adding to zero?

A
2, 4, 8
B
4, 8, 16
✓
1, 2, 1
D
0.5, 1, 2

Answer

Correct option: C.

1, 2, 1

1, 2 and 1 may represent the magnitudes of three vectors adding to zero. For example one of the vector of length 1 should make an angle of 135° with x-axis and the other vector of length 1 makes an angle of 225° with x-axis. The third vector of length 2 should lie along x-axis.

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MCQ 41 Mark

A vector $\overrightarrow{\text{A}}$ points vertically upward and $\overrightarrow{\text{B}}$ points towards north. The vector product $\overrightarrow{\text{A}}\times\overrightarrow{\text{B}}$ is:

✓
Along west.
B
Along east.
C
Zero.
D
Vertically downward.

Answer

Correct option: A.

Along west.

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.

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MCQ 51 Mark

The magnitude of the vector product of two vectors $\big|\overrightarrow{\text{A}}\big|$ and $\big|\overrightarrow{\text{B}}\big|$ may be:

A
Equal to zero.
B
Equal to $AB.$
C
Less than $AB.$
✓
All of the above

Answer

Correct option: D.

All of the above

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 $AB$.

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MCQ 61 Mark

The resultant of $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ makes an angle a with $\overrightarrow{\text{A}}$ and $\beta$ with $\overrightarrow{\text{B}},$

A
$\alpha<\beta$
B
$\alpha<\beta \text{ if}\text{ A}<\text{B}$
✓
$\alpha<\beta \text{ if}\text{ A}>\text{B}$
D
$\alpha<\beta \text{ if}\text{ A}=\text{B}$

Answer

Correct option: C.

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

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

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MCQ 71 Mark

A situation may be described by using different sets of coordinate axes having different orientations. Which of the following do not depend on the orientation of the axes?

A
The value of a scalar.
B
The magnitude of a vector.
C
A vector.
✓
All of the above

Answer

Correct option: D.

All of the above

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

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MCQ 81 Mark

Let $\overrightarrow{\text{C}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}.$

A
$|\overrightarrow{\text{C}}|$ is always greater than $|\overrightarrow{\text{A}}|$
✓
It is possible to have $|\overrightarrow{\text{C}}|<|\overrightarrow{\text{A}}|$ and $|\overrightarrow{\text{C}}|<|\overrightarrow{\text{B}}|$
C
C is always equal to A + B
D
C is never equal to A + B.

Answer

Correct option: B.

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

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.

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MCQ 91 Mark

The component of a vector is:

A
Always less than its magnitude.
B
Always greater than its magnitude.
C
Always equal to its magnitude.
✓
None of these.

Answer

Correct option: D.

None of these.

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components.

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MCQ 101 Mark

The radius of a circle is stated as $2.12\ cm$. Its area should be written as:

A
$14 \mathrm{~cm}^2$.
✓
$14.1 \mathrm{~cm}^2$.
C
$14.11 \mathrm{~cm}^2$.
D
$14.1124 \mathrm{~cm}^2$.

Answer

Correct option: B.

$14.1 \mathrm{~cm}^2$.

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, $2.12\ cm$ has a minimum of three significant digits.
So, the answer must be written in three significant digits.

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MCQ 111 Mark

Let the angle between two nonzero vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ be 120° and its resultant be $\vec{\text{c}}:$

A
C must be equal to $|\text{A}-\text{B}|$
✓
C must be less than $|\text{A}-\text{B}|$
C
C must be greater than $|\text{A}-\text{B}|$
D
C may be equal to $|\text{A}-\text{B}|$

Answer

Correct option: B.

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

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 $|\text{A}-\text{B}|.$

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Physics M.C.Q (1 Marks)

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