101. When two vectors are perpendicular, the resultant formula reduces to \(R=\sqrt{A^2+B^2}\) because:
ⓐ. \(\sin90^\circ=0\)
ⓑ. \(\cos90^\circ=1\)
ⓒ. \(\cos90^\circ=0\)
ⓓ. \(\tan90^\circ=0\)
Correct Answer: \(\cos90^\circ=0\)
Explanation: The general formula for the resultant is \(R=\sqrt{A^2+B^2+2AB\cos\theta}\). For perpendicular vectors, \(\theta=90^\circ\). Since \(\cos90^\circ=0\), the cross term \(2AB\cos\theta\) becomes zero. The expression then becomes \(R=\sqrt{A^2+B^2}\). This is the same form as the Pythagoras theorem because perpendicular vectors form a right triangle in the vector diagram.
102. A resultant of magnitude \(R=A+B\) is observed for two non-zero vectors of magnitudes \(A\) and \(B\). The angle between the two vectors must be:
ⓐ. \(60^\circ\)
ⓑ. \(90^\circ\)
ⓒ. \(180^\circ\)
ⓓ. \(0^\circ\)
Correct Answer: \(0^\circ\)
Explanation: The maximum possible resultant of two vectors is \(A+B\). This occurs only when both vectors point in the same direction. In the formula \(R=\sqrt{A^2+B^2+2AB\cos\theta}\), this maximum value requires \(\cos\theta=1\). The angle for which \(\cos\theta=1\) is \(0^\circ\). Any non-zero angle makes \(\cos\theta\) smaller than \(1\), so the resultant becomes less than \(A+B\).
103. For two vectors of magnitudes \(A\) and \(B\), the statement \(R=|A-B|\) is valid when the vectors:
ⓐ. are perpendicular to each other
ⓑ. act in opposite directions
ⓒ. act in exactly the same direction
ⓓ. have no definite direction
Correct Answer: act in opposite directions
Explanation: The minimum resultant of two vectors occurs when they point in opposite directions. In that case, the angle between them is \(180^\circ\), and \(\cos180^\circ=-1\). Substitution in the general formula gives \(R=\sqrt{A^2+B^2-2AB}\). This simplifies to \(R=\sqrt{(A-B)^2}=|A-B|\). The modulus is needed because resultant magnitude cannot be negative.
104. A record says that two vectors of magnitudes \(4\,\text{m}\) and \(9\,\text{m}\) have a resultant of \(3\,\text{m}\). This record is:
ⓐ. possible because \(3\,\text{m}\) lies below \(4\,\text{m}+9\,\text{m}\)
ⓑ. possible only when the vectors are perpendicular
ⓒ. impossible because the minimum possible resultant is \(5\,\text{m}\)
ⓓ. impossible because resultants of displacement vectors have no unit
Correct Answer: impossible because the minimum possible resultant is \(5\,\text{m}\)
Explanation: \( \textbf{Given magnitudes:} \) \(A=4\,\text{m}\) and \(B=9\,\text{m}\).
\( \textbf{Minimum possible resultant:} \)
\[
R_{\min}=|A-B|
\]
\( \textbf{Substitution:} \)
\[
R_{\min}=|9\,\text{m}-4\,\text{m}|=5\,\text{m}
\]
\( \textbf{Maximum possible resultant:} \)
\[
R_{\max}=9\,\text{m}+4\,\text{m}=13\,\text{m}
\]
\( \textbf{Allowed range:} \)
\[
5\,\text{m}\leq R\leq13\,\text{m}
\]
\( \textbf{Check the record:} \) The stated value \(3\,\text{m}\) is less than the minimum possible resultant.
\( \textbf{Final answer:} \) The record is impossible because the resultant cannot be smaller than \(5\,\text{m}\).
105. Equal vectors of magnitude \(8\,\text{N}\) make an angle of \(60^\circ\) with each other. Their resultant magnitude is:
ⓐ. \(8\,\text{N}\)
ⓑ. \(8\sqrt{2}\,\text{N}\)
ⓒ. \(8\sqrt{3}\,\text{N}\)
ⓓ. \(16\sqrt{3}\,\text{N}\)
Correct Answer: \(8\sqrt{3}\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(A=8\,\text{N}\), \(B=8\,\text{N}\), and \(\theta=60^\circ\).
\( \textbf{Required:} \) Resultant magnitude \(R\).
\( \textbf{Formula:} \)
\[
R=\sqrt{A^2+B^2+2AB\cos\theta}
\]
\( \textbf{Substitution:} \)
\[
R=\sqrt{8^2+8^2+2(8)(8)\cos60^\circ}
\]
\( \textbf{Angle value:} \)
\[
\cos60^\circ=\frac{1}{2}
\]
\( \textbf{Simplification inside the root:} \)
\[
R=\sqrt{64+64+128\left(\frac{1}{2}\right)}
\]
\( \textbf{Further simplification:} \)
\[
R=\sqrt{64+64+64}=\sqrt{192}
\]
\( \textbf{Root form:} \)
\[
\sqrt{192}=\sqrt{64\times3}=8\sqrt{3}
\]
\( \textbf{Final answer:} \) The resultant magnitude is \(8\sqrt{3}\,\text{N}\).
106. Two equal non-zero vectors have a resultant whose magnitude is equal to the magnitude of either vector. The angle between the two vectors is:
ⓐ. \(30^\circ\)
ⓑ. \(60^\circ\)
ⓒ. \(150^\circ\)
ⓓ. \(120^\circ\)
Correct Answer: \(120^\circ\)
Explanation: \( \textbf{Let:} \) Each vector have magnitude \(A\), and let the resultant magnitude also be \(A\).
\( \textbf{Resultant formula:} \)
\[
R^2=A^2+A^2+2A^2\cos\theta
\]
\( \textbf{Use \(R=A\):} \)
\[
A^2=2A^2+2A^2\cos\theta
\]
\( \textbf{Divide by \(A^2\):} \)
\[
1=2+2\cos\theta
\]
\( \textbf{Solve for \(\cos\theta\):} \)
\[
2\cos\theta=-1
\]
\[
\cos\theta=-\frac{1}{2}
\]
\( \textbf{Angle value:} \)
\[
\theta=120^\circ
\]
\( \textbf{Physical meaning:} \) The two equal vectors partly cancel each other strongly enough that the resultant returns to the size of one vector.
\( \textbf{Final answer:} \) The angle between the vectors is \(120^\circ\).
107. A zero resultant can be obtained by adding two non-zero vectors only when the two vectors:
ⓐ. have equal magnitudes and opposite directions
ⓑ. have unequal magnitudes and the same direction
ⓒ. are perpendicular with any magnitudes
ⓓ. have equal magnitudes and the same direction
Correct Answer: have equal magnitudes and opposite directions
Explanation: The resultant of two vectors becomes zero only when one vector exactly cancels the other. For exact cancellation, the magnitudes must be equal. The directions must also be opposite, so that the vectors have no remaining combined effect. If the vectors are in the same direction, their magnitudes add instead of canceling. If they are perpendicular, they form a non-zero resultant unless both vectors themselves are zero.
108. Study the special-case table for adding two vectors of magnitudes \(A\) and \(B\).
| Row | Angle between vectors | Resultant magnitude |
|---|
| P | \(0^\circ\) | \(A+B\) |
| Q | \(90^\circ\) | \(\sqrt{A^2+B^2}\) |
| R | \(180^\circ\) | \(|A-B|\) |
| S | \(0^\circ\) | \(|A-B|\) |
The row that needs correction is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: When the angle is \(0^\circ\), the two vectors point in the same direction, so the resultant magnitude is \(A+B\). When the angle is \(90^\circ\), the vectors are perpendicular and the Pythagorean form \(R=\sqrt{A^2+B^2}\) applies. When the angle is \(180^\circ\), the vectors oppose each other and the resultant becomes \(|A-B|\). Row S wrongly assigns the opposite-direction result to the same-direction case. The formula chosen must match the angle condition, not only the two magnitudes.
109. Assertion: If two vectors are perpendicular, their resultant magnitude is less than their arithmetic sum.
Reason: For perpendicular vectors, the cross term \(2AB\cos\theta\) becomes zero.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: For perpendicular vectors, \(\theta=90^\circ\), so \(\cos90^\circ=0\). The resultant becomes \(R=\sqrt{A^2+B^2}\). For two non-zero positive magnitudes, \(\sqrt{A^2+B^2}\) is less than \(A+B\). The reason correctly explains why the direct supporting term disappears in the perpendicular case. The arithmetic sum \(A+B\) is reserved for vectors acting in the same direction.
110. A graph description is given below.
For two fixed non-zero vector magnitudes \(A\) and \(B\), the resultant magnitude \(R\) is plotted against the angle \(\theta\) between the vectors from \(0^\circ\) to \(180^\circ\).
The endpoint values of the graph are:
ⓐ. \(R=|A-B|\) at \(0^\circ\) and \(R=A+B\) at \(180^\circ\)
ⓑ. \(R=\sqrt{A^2+B^2}\) at both \(0^\circ\) and \(180^\circ\)
ⓒ. \(R=0\) at both \(0^\circ\) and \(180^\circ\)
ⓓ. \(R=A+B\) at \(0^\circ\) and \(R=|A-B|\) at \(180^\circ\)
Correct Answer: \(R=A+B\) at \(0^\circ\) and \(R=|A-B|\) at \(180^\circ\)
Explanation: At \(\theta=0^\circ\), the two vectors point in the same direction. Their magnitudes add directly, giving \(R=A+B\). At \(\theta=180^\circ\), the vectors point in opposite directions. The smaller vector then cancels part of the larger one, giving \(R=|A-B|\). The graph’s endpoint values therefore represent the maximum and minimum possible resultants for fixed \(A\) and \(B\).
111. A report claims that two vectors of magnitudes \(6\,\text{m}\) and \(6\,\text{m}\) have a resultant of \(15\,\text{m}\). The report is:
ⓐ. possible when the vectors are in the same direction
ⓑ. possible when the vectors are perpendicular
ⓒ. impossible because the maximum resultant is \(12\,\text{m}\)
ⓓ. impossible because displacement cannot be measured in \(\text{m}\)
Correct Answer: impossible because the maximum resultant is \(12\,\text{m}\)
Explanation: \( \textbf{Given magnitudes:} \) \(A=6\,\text{m}\) and \(B=6\,\text{m}\).
\( \textbf{Maximum resultant condition:} \) The largest resultant occurs when the vectors point in the same direction.
\( \textbf{Maximum value:} \)
\[
R_{\max}=A+B
\]
\( \textbf{Substitution:} \)
\[
R_{\max}=6\,\text{m}+6\,\text{m}=12\,\text{m}
\]
\( \textbf{Reported value:} \) The report gives \(R=15\,\text{m}\).
\( \textbf{Range check:} \) A resultant cannot exceed the arithmetic sum of the two magnitudes.
\( \textbf{Final answer:} \) The report is impossible because \(15\,\text{m}\) is greater than the maximum possible resultant \(12\,\text{m}\).
112. A unit vector is used in vector notation mainly to represent:
ⓐ. magnitude only, with no direction
ⓑ. direction with unit magnitude
ⓒ. time interval only
ⓓ. a vector whose magnitude is always zero
Correct Answer: direction with unit magnitude
Explanation: A unit vector is a vector of magnitude \(1\). Its main purpose is to show direction. For example, a vector may be written as a magnitude multiplied by a unit vector in the required direction. The number \(1\) in a unit vector is not a physical length by itself; it is a direction marker with unit magnitude. A zero vector cannot be a unit vector because its magnitude is \(0\), not \(1\).
113. The symbols \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are commonly used as:
ⓐ. unit vectors along the \(x\)-, \(y\)-, and \(z\)-axes respectively
ⓑ. scalar distances from the origin
ⓒ. magnitudes of acceleration along any path
ⓓ. units of time, velocity, and acceleration respectively
Correct Answer: unit vectors along the \(x\)-, \(y\)-, and \(z\)-axes respectively
Explanation: The symbol \(\hat{i}\) denotes a unit vector along the positive \(x\)-axis. The symbol \(\hat{j}\) denotes a unit vector along the positive \(y\)-axis. The symbol \(\hat{k}\) denotes a unit vector along the positive \(z\)-axis. In motion in a plane, \(\hat{i}\) and \(\hat{j}\) are used most often because the motion is commonly described in the \(x\)-\(y\) plane. These symbols describe directions of axes, not separate units such as \(\text{s}\) or \(\text{m s}^{-1}\).
114. The expression \(7\,\text{m}\,\hat{i}\) represents:
ⓐ. a displacement of magnitude \(7\,\text{m}\) along the positive \(x\)-direction
ⓑ. a displacement of magnitude \(1\,\text{m}\) along the positive \(x\)-direction
ⓒ. a displacement of magnitude \(7\,\text{m}\) along the positive \(y\)-direction
ⓓ. a scalar distance of \(7\,\text{m}\) with no direction
Correct Answer: a displacement of magnitude \(7\,\text{m}\) along the positive \(x\)-direction
Explanation: The unit vector \(\hat{i}\) points along the positive \(x\)-axis. The coefficient \(7\,\text{m}\) gives the magnitude of the displacement along that direction. Therefore, \(7\,\text{m}\,\hat{i}\) is a vector, not a scalar distance without direction. Its magnitude is \(7\,\text{m}\), while its direction is fixed by \(\hat{i}\). The unit vector contributes direction, and the coefficient contributes size and unit.
115. A vector is written as \(-4\,\text{m}\,\hat{j}\). Its direction is:
ⓐ. positive \(x\)-direction
ⓑ. negative \(x\)-direction
ⓒ. positive \(y\)-direction
ⓓ. negative \(y\)-direction
Correct Answer: negative \(y\)-direction
Explanation: The unit vector \(\hat{j}\) points along the positive \(y\)-axis. The negative sign before \(4\,\text{m}\) reverses this direction. Therefore, \(-4\,\text{m}\,\hat{j}\) points along the negative \(y\)-direction. Its magnitude is \(4\,\text{m}\), not \(-4\,\text{m}\). The sign belongs to the chosen direction along the axis, while the magnitude remains non-negative.
116. Match the unit-vector symbols with their usual coordinate-axis directions.
| Symbol | Direction |
|---|
| P. \(\hat{i}\) | 1. positive \(z\)-axis |
| Q. \(\hat{j}\) | 2. positive \(x\)-axis |
| R. \(\hat{k}\) | 3. positive \(y\)-axis |
The suitable matching is:
ⓐ. P-3, Q-2, R-1
ⓑ. P-1, Q-2, R-3
ⓒ. P-2, Q-1, R-3
ⓓ. P-2, Q-3, R-1
Correct Answer: P-2, Q-3, R-1
Explanation: The unit vector \(\hat{i}\) is along the positive \(x\)-axis. The unit vector \(\hat{j}\) is along the positive \(y\)-axis. The unit vector \(\hat{k}\) is along the positive \(z\)-axis. In a two-dimensional plane problem, usually only \(\hat{i}\) and \(\hat{j}\) are needed. The hat notation signals a unit vector, so the symbol gives direction with magnitude \(1\).
117. Three statements about unit vectors are listed.
I. A unit vector has magnitude \(1\).
II. A unit vector is used to specify direction.
III. Every vector with magnitude \(1\) must point along the positive \(x\)-axis.
The supported statements are:
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is true because the defining feature of a unit vector is magnitude \(1\). Statement II is also true because unit vectors are used to indicate direction in vector notation. Statement III is false because a unit vector can point in any direction, not only along the positive \(x\)-axis. The special symbol \(\hat{i}\) is along positive \(x\), but other unit vectors may point along \(\hat{j}\), \(\hat{k}\), or any specified direction. Unit magnitude and direction choice are separate ideas.
118. A displacement in a plane is written as \(3\,\text{m}\,\hat{i}+0\,\text{m}\,\hat{j}\). This vector lies:
ⓐ. along the positive \(x\)-axis
ⓑ. along the positive \(y\)-axis
ⓒ. equally along positive \(x\) and positive \(y\)
ⓓ. opposite to the positive \(x\)-axis
Correct Answer: along the positive \(x\)-axis
Explanation: The coefficient of \(\hat{i}\) gives the part of the vector along the \(x\)-axis. Here, the \(\hat{i}\) part is \(3\,\text{m}\), so there is a positive \(x\)-direction displacement. The coefficient of \(\hat{j}\) gives the part along the \(y\)-axis. Since the \(\hat{j}\) coefficient is \(0\,\text{m}\), there is no displacement along \(y\). The vector therefore lies along the positive \(x\)-axis rather than in a slant direction.
119. A notation record contains the entries below.
| Row | Expression | Interpretation |
|---|
| P | \(\hat{i}\) | unit vector along positive \(x\) |
| Q | \(\hat{j}\) | unit vector along positive \(y\) |
| R | \(5\hat{i}\) | vector of magnitude \(5\) along positive \(x\) |
| S | \(\hat{i}\) | vector of magnitude \(5\) along positive \(x\) |
The row that misreads the notation is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P correctly identifies \(\hat{i}\) as a unit vector along the positive \(x\)-axis. Row Q correctly identifies \(\hat{j}\) as a unit vector along the positive \(y\)-axis. Row R correctly interprets \(5\hat{i}\) as a vector with magnitude \(5\) along positive \(x\). Row S is wrong because \(\hat{i}\) alone has magnitude \(1\), not \(5\). The coefficient written before a unit vector changes the magnitude, while the unit vector itself supplies direction.
120. A vector \(\vec{A}\) in a plane is written as \(\vec{A}=A_x\hat{i}+A_y\hat{j}\). The quantities \(A_x\) and \(A_y\) represent:
ⓐ. two separate vectors that have no relation to \(\vec{A}\)
ⓑ. signed components along \(x\)- and \(y\)-axes
ⓒ. the magnitudes of \(\hat{i}\) and \(\hat{j}\)
ⓓ. the time and speed of the moving body
Correct Answer: signed components along \(x\)- and \(y\)-axes
Explanation: The form \(\vec{A}=A_x\hat{i}+A_y\hat{j}\) represents a vector by its rectangular components. The coefficient \(A_x\) gives the component along the \(x\)-axis, and \(A_y\) gives the component along the \(y\)-axis. These components are signed scalars, so they may be positive, negative, or zero depending on direction along the chosen axes. The unit vectors \(\hat{i}\) and \(\hat{j}\) supply the axis directions. The components are not independent physical vectors by themselves unless they are combined with the corresponding unit vectors.
