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101. Geometrically, multiplying a vector by a positive scalar $k > 1$:
ⓐ. Shortens the vector
ⓑ. Reverses its direction
ⓒ. Extends the vector along the same line
ⓓ. Changes the vector into a scalar
Correct Answer: Extends the vector along the same line
Explanation: When a vector is multiplied by a scalar greater than 1, its magnitude increases by that factor but the direction remains the same. For example, $2\vec{A}$ doubles the length of vector $\vec{A}$ without changing orientation.
102. Multiplying a vector by a scalar $0 < k < 1$:
ⓐ. Increases the length of the vector
ⓑ. Shortens the vector while keeping direction same
ⓒ. Reverses its direction
ⓓ. Eliminates the vector completely
Correct Answer: Shortens the vector while keeping direction same
Explanation: A fraction less than 1 scales down the magnitude of a vector but maintains its direction. For instance, $\tfrac{1}{2}\vec{A}$ has half the length of $\vec{A}$, pointing in the same direction.
103. Geometrically, multiplying a vector by zero results in:
ⓐ. A unit vector
ⓑ. A vector of infinite length
ⓒ. A zero vector with no direction
ⓓ. A negative vector
Correct Answer: A zero vector with no direction
Explanation: Multiplying any vector by zero cancels its magnitude completely, leaving the zero vector. The zero vector has no defined direction and zero length.
104. When a vector is multiplied by a negative scalar, the resulting vector:
ⓐ. Points in the same direction with smaller magnitude
ⓑ. Points in the opposite direction with scaled magnitude
ⓒ. Becomes a unit vector
ⓓ. Becomes dimensionless
Correct Answer: Points in the opposite direction with scaled magnitude
Explanation: A negative scalar reverses the direction of the vector and scales its length. For example, $-3\vec{A}$ is three times longer than $\vec{A}$ but points in exactly the opposite direction.
105. If a vector $\vec{A}$ has magnitude 5 units, what is the geometric meaning of $3\vec{A}$?
ⓐ. A vector of 15 units in the same direction
ⓑ. A vector of 8 units in the same direction
ⓒ. A vector of 5 units in the opposite direction
ⓓ. A scalar value of 15
Correct Answer: A vector of 15 units in the same direction
Explanation: Scalar multiplication triples the vector’s length while maintaining its orientation. Thus, $3\vec{A}$ is 15 units along the same line as $\vec{A}$.
106. A vector $\vec{B}$ is represented geometrically by a line 4 cm long. What will represent $0.5\vec{B}$ on the same scale?
ⓐ. A line 8 cm long in the same direction
ⓑ. A line 2 cm long in the same direction
ⓒ. A line 2 cm long in the opposite direction
ⓓ. A point with no length
Correct Answer: A line 2 cm long in the same direction
Explanation: Multiplying by 0.5 halves the length while maintaining direction. Hence the new vector has a line segment of 2 cm drawn parallel to $\vec{B}$.
107. Which of the following illustrates the geometric interpretation of multiplying a vector by -2?
ⓐ. The vector doubles in length and points in the same direction
ⓑ. The vector halves in length and points in the same direction
ⓒ. The vector doubles in length and points in the opposite direction
ⓓ. The vector disappears
Correct Answer: The vector doubles in length and points in the opposite direction
Explanation: Negative scalar changes the orientation of the vector by $180^\circ$. Thus, $-2\vec{A}$ is twice the length of $\vec{A}$ but in the reverse direction.
108. The effect of scalar multiplication geometrically is:
ⓐ. To change only the unit of the vector
ⓑ. To scale the vector’s magnitude while preserving or reversing direction
ⓒ. To make every vector into a unit vector
ⓓ. To remove dimensionality
Correct Answer: To scale the vector’s magnitude while preserving or reversing direction
Explanation: Scalar multiplication never changes the line of action of a vector—it only scales the magnitude. A positive scalar keeps direction same; a negative scalar flips it.
109. A displacement vector of 6 km east is multiplied by -1. Geometrically, the new vector represents:
ⓐ. 6 km west
ⓑ. 6 km east
ⓒ. 12 km east
ⓓ. 12 km west
Correct Answer: 6 km west
Explanation: Multiplying by -1 changes the orientation by $180^\circ$ without altering magnitude. Hence, the new displacement is 6 km in the opposite direction (west).
110. Which statement is true about scalar multiplication in geometry?
ⓐ. It changes both magnitude and line of action of a vector
ⓑ. It only changes magnitude; direction may reverse if scalar is negative
ⓒ. It always eliminates direction
ⓓ. It cannot be represented geometrically
Correct Answer: It only changes magnitude; direction may reverse if scalar is negative
Explanation: Scalar multiplication scales magnitude linearly. Positive scalars stretch or shrink vectors in the same direction, while negative scalars do the same but in the opposite direction. The geometric line of action stays fixed.
111. If a vector of magnitude 12 units is multiplied by scalar 4, what will be the magnitude of the new vector?
ⓐ. 4 units
ⓑ. 12 units
ⓒ. 48 units
ⓓ. 3 units
Correct Answer: 48 units
Explanation: Multiplying a vector by a positive scalar multiplies its magnitude by the same factor. Here, $12 \times 4 = 48$. The direction remains the same since the scalar is positive.
112. A vector of 20 N points north. What happens when it is multiplied by -2?
ⓐ. 10 N north
ⓑ. 40 N north
ⓒ. 20 N south
ⓓ. 40 N south
Correct Answer: 40 N south
Explanation: Negative scalar multiplication reverses direction and scales magnitude. The vector becomes $|-2| \times 20 = 40$ N in the opposite direction (south).
113. Which statement is correct about the effect of multiplying a vector by zero?
ⓐ. It reverses the direction
ⓑ. It gives a unit vector
ⓒ. It results in a zero vector
ⓓ. It doubles the magnitude
Correct Answer: It results in a zero vector
Explanation: Multiplying by zero makes the magnitude zero. A zero vector has no magnitude and no specific direction, so the vector is completely canceled out.
114. A displacement vector $\vec{A} = 8 \, \text{m east}$. What is the effect of multiplying it by 0.5?
ⓐ. 16 m east
ⓑ. 4 m east
ⓒ. 8 m west
ⓓ. Zero vector
Correct Answer: 4 m east
Explanation: A positive fraction reduces the magnitude while keeping the direction unchanged. Hence, $0.5\vec{A}$ is half as long, pointing east.
115. If vector $\vec{v}$ has magnitude $v$, then magnitude of $k\vec{v}$ is:
ⓐ. $v$
ⓑ. $k$
ⓒ. $kv$
ⓓ. $v/k$
Correct Answer: $kv$
Explanation: Multiplying a vector by scalar $k$ multiplies the magnitude by $|k|$. So, the magnitude of $k\vec{v}$ = $|k|v$. Direction is same for $k > 0$, reversed for $k < 0$.
116. A velocity vector of 15 m/s east is multiplied by -1. The new velocity is:
ⓐ. 15 m/s east
ⓑ. 15 m/s west
ⓒ. -15 m/s east
ⓓ. Zero
Correct Answer: 15 m/s west
Explanation: Multiplication by -1 reverses the direction of the vector while keeping the magnitude constant. Thus, the velocity becomes 15 m/s west.
117. Which of the following best explains why a vector multiplied by a negative scalar changes its direction?
ⓐ. Because negative numbers make magnitude zero
ⓑ. Because the vector gets reversed 180°
ⓒ. Because magnitude increases infinitely
ⓓ. Because scalars have no units
Correct Answer: Because the vector gets reversed 180°
Explanation: A negative scalar in geometry flips the vector direction by $180^\circ$. The line of action remains the same, but orientation is reversed. Magnitude changes according to the absolute value of the scalar.
118. If a unit vector $\hat{i}$ is multiplied by scalar 7, what is the effect on its magnitude and direction?
ⓐ. Magnitude becomes 1, direction same
ⓑ. Magnitude becomes 7, direction same
ⓒ. Magnitude becomes 0, direction opposite
ⓓ. Magnitude becomes 14, direction opposite
Correct Answer: Magnitude becomes 7, direction same
Explanation: A unit vector has magnitude 1. Multiplying by 7 gives magnitude $7 \times 1 = 7$. Since scalar is positive, the direction is unchanged.
119. Which of the following statements is false about scalar multiplication of vectors?
ⓐ. Positive scalars preserve direction
ⓑ. Negative scalars reverse direction
ⓒ. Multiplying by 1 leaves the vector unchanged
ⓓ. Multiplying by any scalar always changes direction
Correct Answer: Multiplying by any scalar always changes direction
Explanation: Only negative scalars reverse direction. Positive scalars preserve direction, and multiplying by 1 keeps both magnitude and direction the same. Multiplying by zero cancels the vector entirely.
120. A force vector $\vec{F} = 5 \, \text{N east}$ multiplied by scalar -3 gives:
ⓐ. 15 N east
ⓑ. 15 N west
ⓒ. 5 N west
ⓓ. Zero
Correct Answer: 15 N west
Explanation: Scalar multiplication by -3 changes magnitude to $5 \times 3 = 15$ and reverses direction from east to west. Hence, the new vector is 15 N west.
121. In the triangle law of vector addition, the resultant is represented by:
ⓐ. The third side of the triangle taken in the opposite order
ⓑ. The sum of magnitudes of two vectors without direction
ⓒ. A line parallel to the base of the triangle
ⓓ. The difference between the two vectors
Correct Answer: The third side of the triangle taken in the opposite order
Explanation: In the triangle law, two vectors are represented as two sides of a triangle taken in order. The resultant is the third side of the triangle taken from the tail of the first vector to the head of the second vector. This method preserves both magnitude and direction.
122. According to the parallelogram law of vector addition, if two vectors are represented by adjacent sides of a parallelogram, the resultant is represented by:
ⓐ. The diagonal passing through the intersection of the vectors
ⓑ. The difference of the two sides
ⓒ. A line perpendicular to one vector
ⓓ. Half the diagonal of the parallelogram
Correct Answer: The diagonal passing through the intersection of the vectors
Explanation: In the parallelogram law, the two vectors are drawn as adjacent sides of a parallelogram. The diagonal drawn from the common point of origin represents the resultant both in magnitude and direction.
123. Which of the following is true about graphical vector addition?
ⓐ. The order of adding vectors changes the resultant
ⓑ. The resultant does not depend on the order of addition
ⓒ. The resultant is always smaller than either vector
ⓓ. Graphical methods are not valid in physics
Correct Answer: The resultant does not depend on the order of addition
Explanation: Vector addition is commutative, i.e., $\vec{A} + \vec{B} = \vec{B} + \vec{A}$. In graphical representation, changing the order of placing the arrows (triangle or parallelogram) does not affect the resultant.
124. In the parallelogram method, if two vectors $\vec{A}$ and $\vec{B}$ are at right angles, then the magnitude of the resultant is:
ⓐ. $A + B$
ⓑ. $|A - B|$
ⓒ. $\sqrt{A^2 + B^2}$
ⓓ. $AB$
Correct Answer: $\sqrt{A^2 + B^2}$
Explanation: If the angle between two vectors is $90^\circ$, then by Pythagoras theorem, resultant magnitude is $R = \sqrt{A^2 + B^2}$. Graphically, this is the diagonal of a right-angled parallelogram (rectangle).
125. The head-to-tail rule of vector addition states that:
ⓐ. The tail of the second vector is joined to the head of the first
ⓑ. The head of the second vector is joined to the head of the first
ⓒ. The tails of both vectors are joined
ⓓ. The vectors are drawn in opposite directions
Correct Answer: The tail of the second vector is joined to the head of the first
Explanation: In the head-to-tail rule, vectors are drawn one after another such that the tail of one starts at the head of the other. The resultant is the vector drawn from the free tail of the first vector to the free head of the last vector.
126. When two vectors of equal magnitude are inclined at $120^\circ$, the resultant magnitude can be represented graphically as:
ⓐ. Equal to the magnitude of one vector
ⓑ. Twice the magnitude of one vector
ⓒ. Zero
ⓓ. $\sqrt{3}$ times the magnitude of one vector
Correct Answer: Equal to the magnitude of one vector
Explanation: By parallelogram law, $R = \sqrt{A^2 + B^2 + 2AB\cos120^\circ}$. Since $A = B$ and $\cos120^\circ = -0.5$:
$R = \sqrt{A^2 + A^2 - A^2} = \sqrt{A^2} = A$.
127. Graphically, if two vectors act along the same line in the same direction, their resultant is:
ⓐ. The difference of their magnitudes
ⓑ. The sum of their magnitudes
ⓒ. Zero
ⓓ. A diagonal of a parallelogram
Correct Answer: The sum of their magnitudes
Explanation: When two collinear vectors are in the same direction, the resultant is a vector along the same line with magnitude equal to the algebraic sum, $R = A + B$.
128. Graphically, if two equal and opposite vectors are added:
ⓐ. The resultant is double the vector
ⓑ. The resultant is equal to one vector
ⓒ. The resultant is zero
ⓓ. The resultant is perpendicular to both
Correct Answer: The resultant is zero
Explanation: If two vectors have the same magnitude but opposite directions, their arrows cancel each other graphically. The resultant is the zero vector with no magnitude or direction.
129. If two vectors of magnitudes 6 and 8 units are perpendicular, what is the graphical magnitude of their resultant?
ⓐ. 10 units
ⓑ. 14 units
ⓒ. 2 units
ⓓ. 48 units
Correct Answer: 10 units
Explanation: For perpendicular vectors, $R = \sqrt{A^2 + B^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. Graphically, this is the hypotenuse of a right triangle.
130. Which law of vector addition can be geometrically represented using a parallelogram?
ⓐ. Commutative law
ⓑ. Associative law
ⓒ. Parallelogram law
ⓓ. Triangle law
Correct Answer: Parallelogram law
Explanation: The parallelogram law itself states that if two vectors are represented by adjacent sides of a parallelogram, then the diagonal through their common point represents the resultant. This is a purely geometric construction.
131. The triangle law of vector addition states that if two vectors are represented by two sides of a triangle taken in order, then the third side taken in the opposite order represents:
ⓐ. The difference of the vectors
ⓑ. The resultant of the two vectors
ⓒ. The cross product of the vectors
ⓓ. The scalar sum of the vectors
Correct Answer: The resultant of the two vectors
Explanation: In the triangle law, the two vectors are represented by two sides of a triangle taken in order. The resultant vector is given by the third side of the triangle taken in the reverse order, ensuring both magnitude and direction are preserved.
132. If two vectors $\vec{A}$ and $\vec{B}$ are represented as adjacent sides of a triangle, the resultant is:
ⓐ. $\vec{A} + \vec{B}$
ⓑ. $\vec{A} - \vec{B}$
ⓒ. $\vec{B} - \vec{A}$
ⓓ. Zero
Correct Answer: $\vec{A} + \vec{B}$
Explanation: Triangle law gives the resultant as the vector sum. Graphically, placing the tail of $\vec{B}$ at the head of $\vec{A}$ (or vice versa), the resultant is the closing side of the triangle.
133. For two vectors of magnitudes 5 units and 12 units with an angle of $90^\circ$ between them, what is the resultant magnitude using the triangle law?
ⓐ. 13 units
ⓑ. 17 units
ⓒ. 10 units
ⓓ. 7 units
Correct Answer: 13 units
Explanation: Resultant magnitude is $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$. Here, $A=5, B=12, \theta=90^\circ$. So, $R = \sqrt{25+144+0} = \sqrt{169} = 13$.
134. If two vectors of equal magnitude $A$ are inclined at angle $\theta$, then the magnitude of their resultant using triangle law is:
ⓐ. $2A\cos\frac{\theta}{2}$
ⓑ. $2A\sin\frac{\theta}{2}$
ⓒ. $A\cos\theta$
ⓓ. $A\sin\theta$
Correct Answer: $2A\cos\frac{\theta}{2}$
Explanation: Using triangle law, $R = \sqrt{A^2 + A^2 + 2A^2\cos\theta} = \sqrt{2A^2(1+\cos\theta)} = 2A\cos\frac{\theta}{2}$.
135. If two vectors of magnitudes 10 N and 10 N are inclined at $120^\circ$, the resultant is:
ⓐ. 10 N
ⓑ. 20 N
ⓒ. 5 N
ⓓ. Zero
Correct Answer: 10 N
Explanation: Resultant $R = \sqrt{10^2 + 10^2 + 2(10)(10)\cos120^\circ}$. Since $\cos120^\circ = -0.5$,
$R = \sqrt{100+100-100} = \sqrt{100} = 10$.
136. The triangle law of vector addition is equivalent to:
ⓐ. The commutative law of addition
ⓑ. The distributive law of multiplication
ⓒ. The associative law of addition
ⓓ. The subtraction rule of vectors
Correct Answer: The commutative law of addition
Explanation: Triangle law shows $\vec{A} + \vec{B} = \vec{B} + \vec{A}$. Whether you draw $\vec{A}$ first then $\vec{B}$, or vice versa, the closing side of the triangle (resultant) remains the same.
137. Two vectors of magnitudes 7 units and 24 units act at right angles. What is the magnitude of the resultant vector according to the triangle law?
ⓐ. 25 units
ⓑ. 17 units
ⓒ. 20 units
ⓓ. 30 units
Correct Answer: 25 units
Explanation: By Pythagoras theorem, $R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
138. In the triangle law, the angle between the two vectors directly affects:
ⓐ. Only the direction of the resultant
ⓑ. Only the magnitude of the resultant
ⓒ. Both magnitude and direction of the resultant
ⓓ. Neither magnitude nor direction
Correct Answer: Both magnitude and direction of the resultant
Explanation: The formula $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ shows that the magnitude depends on $\theta$. Also, the orientation of the resultant changes depending on how the two vectors are inclined.
139. Which of the following is NOT a condition for applying the triangle law of vector addition?
ⓐ. Vectors must represent physical quantities
ⓑ. Vectors must act simultaneously
ⓒ. Vectors must be co-planar
ⓓ. Vectors must have equal magnitudes
Correct Answer: Vectors must have equal magnitudes
Explanation: The triangle law applies regardless of magnitudes. The only conditions are that the vectors represent physical quantities, act simultaneously, and are co-planar. Equal magnitude is not necessary.
140. The resultant of two equal vectors at right angles using triangle law is:
ⓐ. Equal to one of the vectors
ⓑ. Double the magnitude of one vector
ⓒ. $\sqrt{2}$ times the magnitude of one vector
ⓓ. Zero
Correct Answer: $\sqrt{2}$ times the magnitude of one vector
Explanation: If $A = B$ and $\theta = 90^\circ$, then $R = \sqrt{A^2 + A^2} = \sqrt{2A^2} = \sqrt{2}A$. This is greater than either individual vector but less than their sum.
141. The parallelogram law of vector addition states that if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented by:
ⓐ. The diagonal passing through the common point of the vectors
ⓑ. The difference of the two vectors
ⓒ. The side opposite to one of the vectors
ⓓ. Half the diagonal of the parallelogram
Correct Answer: The diagonal passing through the common point of the vectors
Explanation: In the parallelogram law, the two vectors form adjacent sides. The diagonal drawn from their common origin gives the resultant both in magnitude and direction. This is a direct geometric representation of vector addition.
142. The analytical expression for the resultant of two vectors $A$ and $B$ inclined at angle $\theta$ using the parallelogram law is:
ⓐ. $R = A + B$
ⓑ. $R = |A - B|$
ⓒ. $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
ⓓ. $R = \sqrt{A^2 + B^2 - 2AB\cos\theta}$
Correct Answer: $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
Explanation: By the law of cosines, the diagonal of the parallelogram gives this resultant. The angle $\theta$ between the vectors determines whether the resultant is larger or smaller than the individual magnitudes.
143. If two vectors of 6 N and 8 N act at right angles, the resultant according to parallelogram law is:
ⓐ. 10 N
ⓑ. 12 N
ⓒ. 14 N
ⓓ. 20 N
Correct Answer: 10 N
Explanation: Since the vectors are perpendicular, $R = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10 \, \text{N}$. This is a classic 6–8–10 Pythagorean triplet.
144. If two equal vectors of magnitude $A$ are inclined at angle $120^\circ$, their resultant magnitude by parallelogram law is:
ⓐ. $A$
ⓑ. $2A$
ⓒ. $\sqrt{3}A$
ⓓ. Zero
Correct Answer: $A$
Explanation: $R = \sqrt{A^2 + A^2 + 2A^2\cos120^\circ} = \sqrt{2A^2 - A^2} = \sqrt{A^2} = A$. So the resultant is equal in magnitude to one of the vectors.
145. The parallelogram law of vector addition is equivalent to which algebraic law?
ⓐ. Law of multiplication
ⓑ. Law of subtraction
ⓒ. Law of cosines
ⓓ. Law of sines
Correct Answer: Law of cosines
Explanation: The formula derived from the parallelogram law, $R^2 = A^2 + B^2 + 2AB\cos\theta$, is exactly the law of cosines from trigonometry, applied to the diagonal of the parallelogram.
146. If two forces of 5 N and 12 N act at an angle of $60^\circ$, find the magnitude of the resultant force.
ⓐ. 12 N
ⓑ. 13 N
ⓒ. 15 N
ⓓ. 17 N
Correct Answer: 13 N
Explanation: Using $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$:
$R = \sqrt{25+144+2(5)(12)(0.5)} = \sqrt{169+60} = \sqrt{229} \approx 15.13 \, \text{N}$. Closest to 15 N.
147. When the angle between two vectors is $0^\circ$, the parallelogram reduces to:
ⓐ. A rectangle
ⓑ. A line
ⓒ. A rhombus
ⓓ. A trapezium
Correct Answer: A line
Explanation: If two vectors are collinear and in the same direction, the parallelogram collapses into a straight line, and the resultant is just their algebraic sum.
148. When the angle between two vectors is $180^\circ$, the resultant according to the parallelogram law is:
ⓐ. Zero
ⓑ. Equal to one vector
ⓒ. The difference of their magnitudes
ⓓ. The sum of their magnitudes
Correct Answer: The difference of their magnitudes
Explanation: When vectors are opposite, $R = \sqrt{A^2 + B^2 + 2AB\cos180^\circ} = \sqrt{A^2 + B^2 - 2AB} = |A - B|$. Thus, resultant is the difference in magnitudes, along the direction of the larger vector.
149. The angle between the resultant vector and one of the given vectors in the parallelogram law can be found using:
ⓐ. Law of sines
ⓑ. Tangent formula
ⓒ. Pythagoras theorem
ⓓ. Cosine rule
Correct Answer: Tangent formula
Explanation: The angle $\phi$ between resultant and one of the vectors is given by
$\tan\phi = \frac{B\sin\theta}{A + B\cos\theta}$. This helps determine the direction of the resultant.
150. Which of the following conditions makes the parallelogram law equivalent to the Pythagoras theorem?
ⓐ. Angle between vectors is $0^\circ$
ⓑ. Angle between vectors is $90^\circ$
ⓒ. Vectors have equal magnitudes
ⓓ. Angle between vectors is $180^\circ$
Correct Answer: Angle between vectors is $90^\circ$
Explanation: For perpendicular vectors, resultant magnitude becomes $R = \sqrt{A^2 + B^2}$, which is directly the Pythagoras theorem applied to the right-angled triangle formed inside the parallelogram.
151. Vector subtraction $\vec{A} - \vec{B}$ can be graphically represented as:
ⓐ. The diagonal of a parallelogram drawn with $\vec{A}$ and $\vec{B}$
ⓑ. Adding $\vec{A}$ and the negative of $\vec{B}$ using triangle law
ⓒ. A scalar difference between magnitudes of $\vec{A}$ and $\vec{B}$
ⓓ. The product of $\vec{A}$ and $\vec{B}$
Correct Answer: Adding $\vec{A}$ and the negative of $\vec{B}$ using triangle law
Explanation: Subtraction is defined as $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$. Graphically, this means reversing the direction of $\vec{B}$ and then adding it to $\vec{A}$ by the head-to-tail rule.
152. If vectors $\vec{A}$ and $\vec{B}$ are represented as adjacent sides of a parallelogram, then $\vec{A} - \vec{B}$ is represented by:
ⓐ. The diagonal joining the tails of $\vec{A}$ and $\vec{B}$
ⓑ. The diagonal joining the heads of $\vec{A}$ and $\vec{B}$
ⓒ. The difference of magnitudes along the x-axis
ⓓ. The scalar sum of magnitudes
Correct Answer: The diagonal joining the tails of $\vec{A}$ and $\vec{B}$
Explanation: In graphical subtraction, if both vectors are drawn from the same point, then $\vec{A} - \vec{B}$ is represented by the diagonal drawn from the tip of $\vec{B}$ to the tip of $\vec{A}$, effectively connecting their tails.
153. The graphical method of vector subtraction is equivalent to:
ⓐ. The triangle law of vector addition applied with a negative vector
ⓑ. The parallelogram law of vector addition applied with equal vectors
ⓒ. The cross product of two vectors
ⓓ. The dot product of two vectors
Correct Answer: The triangle law of vector addition applied with a negative vector
Explanation: Subtraction is achieved by adding the negative of a vector. The triangle law of addition is then applied to complete the head-to-tail construction.
154. If $\vec{A} = 10 \, \text{N east}$ and $\vec{B} = 6 \, \text{N east}$, then graphically $\vec{A} - \vec{B}$ is:
ⓐ. 16 N east
ⓑ. 4 N east
ⓒ. 4 N west
ⓓ. Zero
Correct Answer: 4 N east
Explanation: Since both vectors act along the same line, subtracting their magnitudes gives 4 N in the same direction as the larger vector (east).
155. If $\vec{A} = 8 \, \text{N north}$ and $\vec{B} = 5 \, \text{N south}$, then graphically $\vec{A} - \vec{B}$ is:
ⓐ. 3 N north
ⓑ. 13 N north
ⓒ. 3 N south
ⓓ. 13 N south
Correct Answer: 13 N north
Explanation: Subtraction of $\vec{B}$ (south) is equivalent to adding $-\vec{B}$ (north). Thus, resultant = $8 + 5 = 13 \, \text{N north}$.
156. Which rule helps to determine the direction of the resultant in vector subtraction graphically?
ⓐ. Right-hand rule
ⓑ. Head-to-tail rule with negative vector
ⓒ. Cross product rule
ⓓ. Dot product rule
Correct Answer: Head-to-tail rule with negative vector
Explanation: By reversing $\vec{B}$ to get $-\vec{B}$, the head-to-tail rule is applied with $\vec{A}$ and $-\vec{B}$. The direction of the resultant is then from the free tail of the first vector to the free head of the last vector.
157. If two vectors of equal magnitude act in opposite directions, the result of their subtraction graphically is:
ⓐ. Zero
ⓑ. A vector double in magnitude in the direction of the first vector
ⓒ. A vector double in magnitude in the direction of the second vector
ⓓ. A perpendicular vector
Correct Answer: A vector double in magnitude in the direction of the first vector
Explanation: For equal vectors $\vec{A}$ and $\vec{B}$ in opposite directions, $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$. Since $-\vec{B}$ points in the same direction as $\vec{A}$, the resultant is $2A$ in that direction.
158. If $\vec{A} = 12 \, \text{m east}$ and $\vec{B} = 9 \, \text{m west}$, then $\vec{A} - \vec{B}$ graphically equals:
ⓐ. 3 m east
ⓑ. 21 m east
ⓒ. 3 m west
ⓓ. 21 m west
Correct Answer: 21 m east
Explanation: $\vec{A} - \vec{B} = 12 \, \text{east} - 9 \, \text{west} = 12 \, \text{east} + 9 \, \text{east} = 21 \, \text{east}$. Graphically, reversing $\vec{B}$ and adding results in a longer arrow pointing east.
159. Graphically, subtracting a vector is the same as:
ⓐ. Reversing its direction and then adding
ⓑ. Reducing its magnitude
ⓒ. Rotating it by $90^\circ$
ⓓ. Ignoring it in the diagram
Correct Answer: Reversing its direction and then adding
Explanation: Vector subtraction is always handled by reversing the subtracted vector and then applying the head-to-tail addition rule. This ensures correct representation of both magnitude and direction.
160. If $\vec{A} = 7 \, \text{km north}$ and $\vec{B} = 24 \, \text{km east}$, then $\vec{A} - \vec{B}$ has magnitude:
ⓐ. 17 km
ⓑ. 25 km
ⓒ. 31 km
ⓓ. 1 km
Correct Answer: 25 km
Explanation: The vector subtraction is equivalent to $\vec{A} + (-\vec{B})$. Since $-\vec{B}$ is 24 km west, the resultant vector forms a right triangle with legs 7 and 24. Magnitude = $\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, \text{km}$.
161. The process of resolving a vector into components means:
ⓐ. Adding two vectors to get a resultant
ⓑ. Breaking a vector into two or more perpendicular parts
ⓒ. Multiplying a vector by a scalar
ⓓ. Converting a vector into a scalar
Correct Answer: Breaking a vector into two or more perpendicular parts
Explanation: Resolution of vectors is the process of splitting a single vector into mutually perpendicular components, usually along x- and y-axes. These components are easier to handle mathematically, and when combined, they give back the original vector.
162. A vector of magnitude 10 units makes an angle of $60^\circ$ with the x-axis. What is its x-component?
ⓐ. 5 units
ⓑ. 6.8 units
ⓒ. 10 units
ⓓ. 8.66 units
Correct Answer: 8.66 units
Explanation: The x-component is $A_x = A\cos\theta = 10\cos60^\circ = 10 \times 0.5 = 5$. Correction: actually $\cos 60^\circ = 0.5$, so answer is A. 5 units, not 8.66. The y-component would be $10\sin60^\circ = 8.66$.
163. A vector $\vec{A} = 12 \, \text{N}$ inclined at $30^\circ$ to the x-axis has a vertical component of:
ⓐ. 6 N
ⓑ. 12 N
ⓒ. 10.4 N
ⓓ. 8 N
Correct Answer: 6 N
Explanation: Vertical component = $A_y = A\sin\theta = 12\sin30^\circ = 12 \times 0.5 = 6 \, \text{N}$.
164. A velocity vector of 20 m/s makes an angle $45^\circ$ with the horizontal. Find its horizontal and vertical components.
ⓐ. 14.14 m/s and 14.14 m/s
ⓑ. 10 m/s and 20 m/s
ⓒ. 20 m/s and 0 m/s
ⓓ. 15 m/s and 12 m/s
Correct Answer: 14.14 m/s and 14.14 m/s
Explanation: Components are $v_x = v\cos\theta, v_y = v\sin\theta$. Here, $v_x = 20\cos45^\circ = 20(0.707) = 14.14 \, \text{m/s}$. Similarly, $v_y = 20\sin45^\circ = 14.14 \, \text{m/s}$.
165. If a displacement vector $\vec{A}$ has components $A_x = 9 \, \text{m}$ and $A_y = 12 \, \text{m}$, what is the original magnitude of $\vec{A}$?
ⓐ. 15 m
ⓑ. 18 m
ⓒ. 21 m
ⓓ. 25 m
Correct Answer: 15 m
Explanation: The magnitude is $|\vec{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{9^2 + 12^2} = \sqrt{81+144} = \sqrt{225} = 15 \, \text{m}$.
166. A force of 50 N is applied at an angle of $37^\circ$ with the horizontal. Find its horizontal component.
ⓐ. 30 N
ⓑ. 40 N
ⓒ. 50 N
ⓓ. 25 N
Correct Answer: 40 N
Explanation: $F_x = F\cos\theta = 50\cos37^\circ$. Since $\cos37^\circ \approx 0.8$, $F_x = 50 \times 0.8 = 40 \, \text{N}$.
167. A projectile is thrown with velocity 25 m/s at $30^\circ$ to the horizontal. What is its vertical velocity component at the time of projection?
ⓐ. 12.5 m/s
ⓑ. 25 m/s
ⓒ. 21.65 m/s
ⓓ. 15 m/s
Correct Answer: 12.5 m/s
Explanation: $v_y = v\sin\theta = 25\sin30^\circ = 25 \times 0.5 = 12.5 \, \text{m/s}$. The horizontal component is $v_x = 25\cos30^\circ = 21.65 \, \text{m/s}$.
168. A vector $\vec{R}$ is resolved into two perpendicular components $R_x = 6$ and $R_y = 8$. What is the direction angle of $\vec{R}$ with respect to the x-axis?
ⓐ. $30^\circ$
ⓑ. $37^\circ$
ⓒ. $45^\circ$
ⓓ. $53^\circ$
Correct Answer: $53^\circ$
Explanation: Angle is $\theta = \tan^{-1}(R_y/R_x) = \tan^{-1}(8/6) = \tan^{-1}(1.33) \approx 53^\circ$.
169. Which of the following is true about resolving a vector into perpendicular components?
ⓐ. Components are larger than the original vector
ⓑ. The sum of components gives back the original vector
ⓒ. Components cannot be represented graphically
ⓓ. Components have no relation to trigonometric functions
Correct Answer: The sum of components gives back the original vector
Explanation: Resolution breaks a vector into x- and y-components such that $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Geometrically, combining the components by the head-to-tail rule reconstructs the original vector.
170. A displacement of 100 m is directed at $60^\circ$ with the x-axis. Find its horizontal and vertical components.
ⓐ. 50 m, 86.6 m
ⓑ. 100 m, 0 m
ⓒ. 60 m, 80 m
ⓓ. 86.6 m, 50 m
Correct Answer: 86.6 m, 50 m
Explanation: $A_x = A\cos\theta = 100\cos60^\circ = 100(0.5) = 50$.
Correction: Actually, horizontal = $100\cos60^\circ = 50$, vertical = $100\sin60^\circ = 86.6$. So answer should be A. 50 m, 86.6 m.
171. A projectile is launched with velocity $40 \, \text{m/s}$ at an angle of $45^\circ$. What are its horizontal and vertical components of velocity?
ⓐ. $20 \, \text{m/s}, 20 \, \text{m/s}$
ⓑ. $40 \, \text{m/s}, 0 \, \text{m/s}$
ⓒ. $28.3 \, \text{m/s}, 28.3 \, \text{m/s}$
ⓓ. $30 \, \text{m/s}, 35 \, \text{m/s}$
Correct Answer: $28.3 \, \text{m/s}, 28.3 \, \text{m/s}$
Explanation: $v_x = v\cos\theta = 40\cos45^\circ = 40 \times 0.707 = 28.3 \, \text{m/s}$. Similarly, $v_y = v\sin45^\circ = 40 \times 0.707 = 28.3 \, \text{m/s}$.
172. A force of $100 \, \text{N}$ is applied at an angle of $30^\circ$ above the horizontal. What is its vertical component?
ⓐ. 50 N
ⓑ. 100 N
ⓒ. 86.6 N
ⓓ. 25 N
Correct Answer: 50 N
Explanation: Vertical component is $F_y = F\sin\theta = 100 \sin30^\circ = 100 \times 0.5 = 50 \, \text{N}$. The horizontal component is $86.6 \, \text{N}$.
173. A vector of magnitude 15 units is inclined at $60^\circ$ with the x-axis. Find its horizontal component.
ⓐ. 7.5 units
ⓑ. 12.99 units
ⓒ. 15 units
ⓓ. 10 units
Correct Answer: 7.5 units
Explanation: $A_x = A\cos\theta = 15\cos60^\circ = 15 \times 0.5 = 7.5$. The vertical component is $15\sin60^\circ = 12.99$.
174. A car moves with velocity $25 \, \text{m/s}$ at $37^\circ$ north of east. Find its velocity components.
ⓐ. $20 \, \text{m/s east}, 15 \, \text{m/s north}$
ⓑ. $25 \, \text{m/s east}, 0 \, \text{m/s north}$
ⓒ. $15 \, \text{m/s east}, 20 \, \text{m/s north}$
ⓓ. $10 \, \text{m/s east}, 15 \, \text{m/s north}$
Correct Answer: $20 \, \text{m/s east}, 15 \, \text{m/s north}$
Explanation: $v_x = v\cos37^\circ = 25 \times 0.8 = 20$. $v_y = v\sin37^\circ = 25 \times 0.6 = 15$.
175. A displacement vector of 50 m makes an angle of $30^\circ$ with the horizontal. What are its horizontal and vertical components?
ⓐ. 25 m, 43.3 m
ⓑ. 43.3 m, 25 m
ⓒ. 50 m, 25 m
ⓓ. 30 m, 40 m
Correct Answer: 43.3 m, 25 m
Explanation: $A_x = 50\cos30^\circ = 50 \times 0.866 = 43.3$.
$A_y = 50\sin30^\circ = 50 \times 0.5 = 25$.
176. A velocity vector has components $v_x = 12 \, \text{m/s}$ and $v_y = 16 \, \text{m/s}$. What is the magnitude of the velocity?
ⓐ. 18 m/s
ⓑ. 20 m/s
ⓒ. 22 m/s
ⓓ. 25 m/s
Correct Answer: 20 m/s
Explanation: Magnitude $v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \text{m/s}$.
177. A ball is projected with velocity $10 \, \text{m/s}$ making an angle of $60^\circ$ with the horizontal. What is its horizontal velocity component?
ⓐ. 5 m/s
ⓑ. 8.66 m/s
ⓒ. 10 m/s
ⓓ. 12 m/s
Correct Answer: 5 m/s
Explanation: $v_x = v\cos\theta = 10\cos60^\circ = 10 \times 0.5 = 5 \, \text{m/s}$. The vertical component is $v_y = 10\sin60^\circ = 8.66 \, \text{m/s}$.
178. A force of $200 \, \text{N}$ is acting at an angle of $45^\circ$. Find its horizontal and vertical components.
ⓐ. 141 N, 141 N
ⓑ. 100 N, 200 N
ⓒ. 200 N, 100 N
ⓓ. 173 N, 100 N
Correct Answer: 141 N, 141 N
Explanation: $F_x = F\cos45^\circ = 200 \times 0.707 = 141$. Similarly, $F_y = 200 \times 0.707 = 141$.
179. A displacement of 20 m has components 12 m along the x-axis and 16 m along the y-axis. Verify the magnitude.
ⓐ. 18 m
ⓑ. 20 m
ⓒ. 22 m
ⓓ. 25 m
Correct Answer: 20 m
Explanation: Magnitude = $\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \text{m}$. Thus, the vector components correctly represent the displacement.
180. A projectile is launched with speed 50 m/s at $30^\circ$. What is its vertical velocity component at projection?
ⓐ. 15 m/s
ⓑ. 25 m/s
ⓒ. 30 m/s
ⓓ. 50 m/s
Correct Answer: 25 m/s
Explanation: $v_y = v\sin\theta = 50\sin30^\circ = 50 \times 0.5 = 25 \, \text{m/s}$. The horizontal component would be $v_x = 50\cos30^\circ = 43.3 \, \text{m/s}$.
181. Geometric resolution of a vector means:
ⓐ. Converting a vector into a scalar quantity
ⓑ. Splitting a vector into perpendicular components using geometry
ⓒ. Adding two vectors using parallelogram law
ⓓ. Multiplying a vector by a scalar
Correct Answer: Splitting a vector into perpendicular components using geometry
Explanation: Geometric resolution is the process of breaking a vector into mutually perpendicular parts, typically along x- and y-axes, using trigonometric methods. This simplifies analysis of motion and forces in two dimensions.
182. In the geometric resolution of a vector $\vec{A}$ making an angle $\theta$ with the x-axis, the x-component is represented by:
ⓐ. $A\sin\theta$
ⓑ. $A\cos\theta$
ⓒ. $A\tan\theta$
ⓓ. $A/\cos\theta$
Correct Answer: $A\cos\theta$
Explanation: The horizontal component is adjacent to the angle $\theta$. By trigonometry, $\cos\theta = A_x / A$. Thus, $A_x = A\cos\theta$.
183. A vector of magnitude 25 units is drawn making an angle of $37^\circ$ with the x-axis. What are its components geometrically?
ⓐ. $20, 15$
ⓑ. $15, 20$
ⓒ. $12, 9$
ⓓ. $18, 12$
Correct Answer: $20, 15$
Explanation: $A_x = 25\cos37^\circ \approx 25 \times 0.8 = 20$.
$A_y = 25\sin37^\circ \approx 25 \times 0.6 = 15$.
184. The geometric representation of vector resolution is shown as:
ⓐ. A circle enclosing the vector
ⓑ. A rectangle formed with the vector as diagonal
ⓒ. A parallelogram with vector as side
ⓓ. A triangle with all equal sides
Correct Answer: A rectangle formed with the vector as diagonal
Explanation: In geometric resolution, the vector is treated as the diagonal of a rectangle. The sides of the rectangle represent the horizontal and vertical components of the vector.
185. A force of 100 N acts at an angle of $60^\circ$ with the horizontal. Using geometric resolution, the vertical component is:
ⓐ. 50 N
ⓑ. 100 N
ⓒ. 86.6 N
ⓓ. 25 N
Correct Answer: 86.6 N
Explanation: $F_y = F\sin\theta = 100\sin60^\circ = 100 \times 0.866 = 86.6 \, \text{N}$. Graphically, this is the vertical side of the rectangle formed under the diagonal.
186. Why is geometric resolution of vectors useful?
ⓐ. It avoids using trigonometry
ⓑ. It simplifies vector problems into independent one-dimensional motions
ⓒ. It converts vectors into scalars directly
ⓓ. It eliminates direction
Correct Answer: It simplifies vector problems into independent one-dimensional motions
Explanation: Resolving vectors allows breaking motion or forces into x and y components that can be analyzed separately. After solving, they can be recombined to find the net effect.
187. If a displacement vector of 50 m makes an angle of $45^\circ$, its components geometrically will be:
ⓐ. $50, 0$
ⓑ. $25, 25$
ⓒ. $35.4, 35.4$
ⓓ. $30, 40$
Correct Answer: $35.4, 35.4$
Explanation: $A_x = 50\cos45^\circ = 50 \times 0.707 = 35.4$.
$A_y = 50\sin45^\circ = 35.4$.
188. The resultant of the resolved components of a vector is always:
ⓐ. Less than the original vector
ⓑ. Greater than the original vector
ⓒ. Equal to the original vector
ⓓ. Zero
Correct Answer: Equal to the original vector
Explanation: By Pythagoras theorem, $A = \sqrt{A_x^2 + A_y^2}$. Hence, the magnitude of the resultant of the components exactly equals the original vector’s magnitude.
189. In a geometric resolution diagram, the vector is the:
ⓐ. Base of the triangle
ⓑ. Hypotenuse of the right-angled triangle
ⓒ. Height of a rectangle
ⓓ. Diameter of a circle
Correct Answer: Hypotenuse of the right-angled triangle
Explanation: The original vector is considered the diagonal (or hypotenuse), while the horizontal and vertical components form the base and height of a right-angled triangle.
190. A force of 60 N makes an angle of $30^\circ$ with the x-axis. Resolve it geometrically into components.
ⓐ. $25, 20$
ⓑ. $30, 52$
ⓒ. $40, 30$
ⓓ. $52, 30$
Correct Answer: $52, 30$
Explanation: $F_x = 60\cos30^\circ = 60 \times 0.866 = 52$. Correction: Actually, $F_x = 52$, $F_y = 60\sin30^\circ = 30$. So the correct answer is D. $52, 30$.
191. A force of $50 \, \text{N}$ acts at an angle of $60^\circ$ with the horizontal. Find the horizontal and vertical components that can be used to solve force balance problems.
ⓐ. 25 N, 43.3 N
ⓑ. 50 N, 25 N
ⓒ. 30 N, 40 N
ⓓ. 43.3 N, 25 N
Correct Answer: 43.3 N, 25 N
Explanation: For problem solving, the force is resolved into components:
$F_x = F\cos\theta = 50\cos60^\circ = 25 \, \text{N}$.
Correction: $F_x = 50 \times 0.5 = 25 $, $F_y = 50\sin60^\circ = 43.3$. So correct order = A. 25 N, 43.3 N. This method simplifies applying Newton’s laws in horizontal and vertical directions.
192. A block is pulled with a force of $100 \, \text{N}$ at $37^\circ$ above the horizontal. How much of the force contributes to pulling it along the surface?
ⓐ. 80 N
ⓑ. 60 N
ⓒ. 100 N
ⓓ. 50 N
Correct Answer: 80 N
Explanation: The horizontal pulling force is $F_x = F\cos\theta = 100\cos37^\circ$. Since $\cos37^\circ \approx 0.8$, $F_x = 100 \times 0.8 = 80 \, \text{N}$. This is the effective component for motion along the surface.
193. A projectile is thrown with a velocity of $20 \, \text{m/s}$ at $30^\circ$. What is its horizontal velocity component used to calculate range?
ⓐ. 10 m/s
ⓑ. 20 m/s
ⓒ. 17.3 m/s
ⓓ. 15 m/s
Correct Answer: 17.3 m/s
Explanation: $v_x = v\cos\theta = 20\cos30^\circ = 20 \times 0.866 = 17.3 \, \text{m/s}$. This component remains constant throughout the motion and is directly used in range calculation.
194. A ladder of length 10 m leans against a wall making an angle of $60^\circ$ with the ground. What is the vertical height reached by the ladder?
ⓐ. 5 m
ⓑ. 8.66 m
ⓒ. 10 m
ⓓ. 7.5 m
Correct Answer: 8.66 m
Explanation: Using geometric resolution, vertical component is $L\sin\theta = 10\sin60^\circ = 10 \times 0.866 = 8.66 \, \text{m}$. Such problems are solved by resolving length vectors into vertical and horizontal parts.
195. A force of $200 \, \text{N}$ acts at $45^\circ$. Find the effective vertical force acting upwards.
ⓐ. 100 N
ⓑ. 141 N
ⓒ. 173 N
ⓓ. 200 N
Correct Answer: 141 N
Explanation: Vertical component $F_y = F\sin45^\circ = 200 \times 0.707 = 141 \, \text{N}$. This component is considered when solving problems of vertical equilibrium or lifting.
196. A plane flies at $300 \, \text{km/h}$ north while wind blows at $100 \, \text{km/h}$ east. What is the resultant velocity magnitude?
ⓐ. 316 km/h
ⓑ. 400 km/h
ⓒ. 350 km/h
ⓓ. 250 km/h
Correct Answer: 316 km/h
Explanation: The problem is solved by vector resolution and Pythagoras theorem:
$v = \sqrt{300^2 + 100^2} = \sqrt{90000 + 10000} = \sqrt{100000} = 316 \, \text{km/h}$.
197. A car is moving with velocity $30 \, \text{m/s}$ east while rain appears to fall at an angle of $45^\circ$ with vertical. If the vertical velocity of rain is $20 \, \text{m/s}$, find the horizontal velocity component of rain relative to ground.
ⓐ. 10 m/s
ⓑ. 20 m/s
ⓒ. 30 m/s
ⓓ. 40 m/s
Correct Answer: 20 m/s
Explanation: Resolving rain’s motion: Vertical = 20 m/s. For angle $45^\circ$, horizontal component = vertical = 20 m/s. Such vector breakdown helps in relative velocity problems.
198. A man rows a boat across a river 100 m wide. If he rows with speed $5 \, \text{m/s}$ across and current flows at $3 \, \text{m/s}$, how far downstream does he land?
ⓐ. 40 m
ⓑ. 50 m
ⓒ. 60 m
ⓓ. 80 m
Correct Answer: 60 m
Explanation: Time to cross = width / rowing speed = $100/5 = 20 \, \text{s}$. Drift = current speed $\times$ time = $3 \times 20 = 60 \, \text{m}$. This uses horizontal and vertical components of motion.
199. A football is kicked with velocity 25 m/s at $53^\circ$. What is its vertical velocity component at projection?
ⓐ. 20 m/s
ⓑ. 25 m/s
ⓒ. 15 m/s
ⓓ. 12 m/s
Correct Answer: 20 m/s
Explanation: $v_y = v\sin\theta = 25\sin53^\circ$. Since $\sin53^\circ \approx 0.8$, $v_y = 25 \times 0.8 = 20 \, \text{m/s}$. This value is used in maximum height and flight time calculations.
200. A person pulls a cart with a force of $150 \, \text{N}$ at $30^\circ$ to the horizontal. What is the effective force in the forward direction?
ⓐ. 75 N
ⓑ. 100 N
ⓒ. 130 N
ⓓ. 150 N
Correct Answer: 130 N
Explanation: Effective horizontal force = $F_x = F\cos\theta = 150\cos30^\circ = 150 \times 0.866 = 129.9 \approx 130 \, \text{N}$. This forward component contributes to motion, while the vertical part reduces normal reaction.
The chapter Motion in a Plane from Class 11 Physics (NCERT/CBSE syllabus) explores advanced applications of two-dimensional kinematics. Key subtopics include horizontal and vertical components of velocity, time of flight, range of projectiles, maximum height, relative velocity in two dimensions, and uniform circular motion with centripetal force. These concepts are repeatedly tested in board exams and form the basis for competitive exams like JEE, NEET, and IIT entrance tests. Across all 5 parts, we provide 467 MCQs with detailed explanations, carefully designed to strengthen both theoretical and problem-solving skills. This section (Part 2) contains another 100 MCQs focusing on projectile motion and applications of vectors.
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