Solved past paper MCQs for Vectors & Equilibrium from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
Boards Included:PMDC
Live Exam Simulation
Want to test these 50 questions against a real-time exam countdown with anti-cheat Leaderboard scoring?
PRINT EDITION
100% Toner-Friendly • 0% Heavy Black Blocks
Download Solved Vectors & Equilibrium Past Papers (Print Edition)
Official printable A4 booklet containing all 50 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
A single non-zero vector can never have a zero resultant because there is no other force to cancel its magnitude.
Two vectors can only yield a zero resultant if they have equal magnitudes and point in exactly opposite directions. If they are of unequal magnitude, they cannot cancel each other out.
Three coplanar vectors of unequal magnitude can be arranged such that they form a closed triangle when joined head-to-tail, which makes their vector sum exactly zero.
Why other options are incorrect:
A is incorrect because two unequal vectors will always leave a non-zero net force in the direction of the larger vector.
C and D are incorrect because while 4 or 5 unequal vectors can sum to zero, they do not represent the minimum number required.
If a vector \(\vec{A}\) has components \(A_x = 1.5\text{ cm}\) and \(A_y = -1.0\text{ cm}\), in which quadrant of the Cartesian coordinate system does the vector point?
A
Quadrant I
B
Quadrant II
C
Quadrant III
D
Quadrant IV
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Cartesian Coordinate Quadrants and the signs of rectangular vector components.
Solution: Let's look at the algebraic signs of the given rectangular components:
The x-component is positive: \(A_x > 0\) (points to the right).
The y-component is negative: \(A_y < 0\) (points downwards).
Any point with a positive horizontal coordinate and a negative vertical coordinate lies in the fourth quadrant (Quadrant IV).
Why other options are incorrect:
Quadrant I requires both components to be positive (\(A_x > 0, A_y > 0\)).
Quadrant II requires \(A_x < 0, A_y > 0\).
Quadrant III requires both components to be negative (\(A_x < 0, A_y < 0\)).
Two forces of magnitude 20 N and 50 N act simultaneously on a body. Which of the following forces cannot be a resultant of these two forces?
A
40 N
B
30 N
C
20 N
D
70 N
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: The range of possible values for the resultant of two vectors.
Formula: $$|F_1 - F_2| \le R \le F_1 + F_2$$
Solution: Given \(F_1 = 50\text{ N}\) and \(F_2 = 20\text{ N}\):
The minimum possible resultant magnitude occurs when the two forces act in opposite directions (\(\theta = 180^\circ\)): $$\text{Min Resultant} = 50\text{ N} - 20\text{ N} = 30\text{ N}$$
The maximum possible resultant magnitude occurs when they act in the same direction (\(\theta = 0^\circ\)): $$\text{Max Resultant} = 50\text{ N} + 20\text{ N} = 70\text{ N}$$
The magnitude of the resultant force must fall within the range \([30\text{ N}, 70\text{ N}]\). Since 20 N is less than 30 N, it cannot be a possible resultant force.
The sum of the magnitudes of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to the smaller force, then the magnitudes of the individual forces are:
A
6 N and 10 N
B
8 N and 8 N
C
4 N and 12 N
D
2 N and 14 N
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: Vector addition using right-triangle properties when the resultant is perpendicular to one of the component forces.
Solution: Let the smaller force be \(F_1\) and the larger force be \(F_2\). Thus, \(F_2 = 16 - F_1\). Since the resultant \(R = 8\text{ N}\) is perpendicular to \(F_1\), these vectors form a right-angled triangle where \(F_2\) is the hypotenuse: $$F_1^2 + 8^2 = (16 - F_1)^2$$ $$F_1^2 + 64 = 256 - 32F_1 + F_1^2$$ $$32F_1 = 256 - 64$$ $$32F_1 = 192 \implies F_1 = 6\text{ N}$$ Calculating the larger force: $$F_2 = 16 - 6 = 10\text{ N}$$
Why other options are incorrect:
8 N and 8 N is incorrect because if both are 8 N, their sum is 16 N but they cannot form a right triangle with a perpendicular resultant of 8 N.
4 N and 12 N, and 2 N and 14 N are incorrect because these pairs do not satisfy the Pythagorean relation with a third side of 8 N. For example, \(4^2 + 8^2 = 80 \neq 12^2\).
For which angle between two non-zero vectors \(\vec{A}\) and \(\vec{B}\) is the relation \(|\vec{A} \cdot \vec{B}| = |\vec{A} \times \vec{B}|\) satisfied?
A
\(30^\circ\)
B
\(45^\circ\)
C
\(60^\circ\)
D
\(90^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Equating the definitions of the scalar (dot) product and vector (cross) product magnitudes.
Formula: $$|\vec{A} \cdot \vec{B}| = AB \cos\theta$$ $$|\vec{A} \times \vec{B}| = AB \sin\theta$$
Solution: Set the two magnitudes equal to each other: $$AB \cos\theta = AB \sin\theta$$ Since the vectors are non-zero (\(A \neq 0, B \neq 0\)), we can divide both sides by \(AB \cos\theta\): $$\frac{\sin\theta}{\cos\theta} = 1 \implies \tan\theta = 1$$ $$\theta = \arctan(1) = 45^\circ$$
Why other options are incorrect:
\(30^\circ\): \(\sin(30^\circ) = 0.5\) while \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\) (they are not equal).
\(60^\circ\): \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\) while \(\cos(60^\circ) = 0.5\) (they are not equal).
\(90^\circ\): The dot product is zero, and the cross product is at its maximum magnitude.
What is the scalar (dot) product of the orthogonal unit vectors \(\hat{i}\) and \(\hat{j}\)?
A
0
B
1
C
-1
D
\(\hat{k}\)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: The scalar product of orthogonal vectors.
Formula: $$\vec{A} \cdot \vec{B} = AB \cos\theta$$
Solution: The unit vectors \(\hat{i}\) and \(\hat{j}\) lie along the positive x-axis and y-axis respectively. The angle \(\theta\) between them is exactly \(90^\circ\): $$\hat{i} \cdot \hat{j} = (1)(1) \cos(90^\circ) = 1 \cdot 0 = 0$$
Why other options are incorrect:
1 and -1 are incorrect because the dot product of any two distinct, perpendicular unit vectors is always zero, not 1 or -1.
\(\hat{k}\) is incorrect because the scalar product yields a scalar, not a vector (which is the result of the cross product \(\hat{i} \times \hat{j}\)).
The direction of a vector product (cross product) is determined by which of the following rules?
A
Head-to-tail rule
B
Right-hand rule
C
Left-hand rule
D
Fleming's left-hand rule
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Finding the orientation of the cross-product vector.
Solution: The direction of \(\vec{C} = \vec{A} \times \vec{B}\) is perpendicular to the plane containing vectors \(\vec{A}\) and \(\vec{B}\). This direction is uniquely determined by the Right-hand rule: curl the fingers of your right hand from the first vector \(\vec{A}\) to the second vector \(\vec{B}\) through the smaller angle; your extended thumb will point in the direction of the product vector \(\vec{C}\).
Why other options are incorrect:
A is incorrect because the head-to-tail rule is used for vector addition, not multiplication.
C and D are incorrect because left-hand rules do not define standard vector cross-product conventions in mathematics.
What is the cross product of the unit vectors \(\hat{i} \times \hat{j}\)?
A
0
B
-\(\hat{j}\)
C
-\(\hat{k}\)
D
\(\hat{k}\)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Cyclic permutation of Cartesian unit vectors in cross products.
Solution: For the orthogonal unit vectors, the cross product of any two yields the third. The standard positive direction follows the cyclic order: \(\hat{i} \rightarrow \hat{j} \rightarrow \hat{k} \rightarrow \hat{i}\). Therefore: $$\hat{i} \times \hat{j} = \hat{k}$$
Why other options are incorrect:
A is incorrect because \(\hat{i}\) and \(\hat{j}\) are orthogonal, meaning their cross product is at maximum magnitude (1), not zero.
B and C are incorrect because they represent the wrong directions. Only a positive \(\hat{k}\) points in the direction determined by the right-hand rule.
Solution: For a body to remain in complete equilibrium, it must not undergo translational acceleration nor rotational acceleration. This requires both the net external force and the net external torque about any axis to be zero.
Why other options are incorrect:
A is incorrect because satisfying only the first condition allows the body to undergo angular acceleration.
B is incorrect because satisfying only the second condition allows the body to undergo linear translation.
D is incorrect because the horizontal and vertical forces being equal does not ensure they sum to zero.
The unit vector \(\hat{a}\) in the direction of vector \(\vec{a}\) is mathematically defined as:
A
\(|\vec{a}|\)
B
\(\vec{a} \cdot \vec{a}\)
C
\(\vec{a} \times \vec{a}\)
D
\(\vec{a} / |\vec{a}|\)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Mathematical definition of a unit vector.
Formula: $$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$
Solution: A unit vector is a vector that has a magnitude of exactly 1 and points in the same direction as the original vector. It is obtained by dividing the vector \(\vec{a}\) by its scalar magnitude \(|\vec{a}|\).
Why other options are incorrect:
A is incorrect because \(|\vec{a}|\) is a scalar magnitude, not a vector.
B is incorrect because the dot product \(\vec{a} \cdot \vec{a}\) yields a scalar equal to \(|\vec{a}|^2\).
C is incorrect because the cross product of any vector with itself is always the null vector.
Which mathematical condition must be satisfied to guarantee that an object is in rotational equilibrium?
A
\(\Sigma \vec{\tau} = 0\)
B
\(\Sigma \vec{F} = 0\)
C
\(\Sigma \vec{F} = 0\) and \(\Sigma \vec{\tau} = 0\)
D
\(\Sigma \vec{F} > 0\)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: The specific condition for rotational equilibrium.
Formula: $$\Sigma \vec{\tau} = 0$$
Solution: Rotational equilibrium means that the angular acceleration of the body is zero. According to Newton's second law for rotation (\(\Sigma \vec{\tau} = I\vec{\alpha}\)), this state is achieved if and only if the vector sum of all external torques acting on the body is zero.
Why other options are incorrect:
B is incorrect because it defines translational equilibrium, not rotational.
C is incorrect because while it defines complete equilibrium, the question specifically asks for the condition of rotational equilibrium.
D is incorrect because a non-zero net force has no direct bearing on rotational equilibrium unless it creates a torque.
If a force vector is given by \(\vec{F} = 2\hat{i} + 3\hat{j} - 4\hat{k}\) and the displacement vector is \(\vec{D} = \hat{i} - 2\hat{j} - 5\hat{k}\), calculate the work done by the force.
A
28 J
B
-24 J
C
-8 J
D
16 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Work defined as the scalar (dot) product of force and displacement.
Solution: Substitute the components of the vectors into the formula: $$W = (2)(1) + (3)(-2) + (-4)(-5)$$ $$W = 2 - 6 + 20$$ $$W = 16\text{ J}$$
Why other options are incorrect:
A, B, and C result from algebraic sign errors during multiplication (e.g., writing \((-4) \times (-5) = -20\) or incorrectly adding/subtracting the components).
The sum and the difference of two perpendicular vectors of the same magnitude are always:
A
Parallel to each other
B
Perpendicular to each other
C
At an acute angle with each other
D
At an obtuse angle with each other
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Geometric properties of vector addition and subtraction.
Formula: Let the perpendicular vectors be \(\vec{P}\) and \(\vec{Q}\) with \(|\vec{P}| = |\vec{Q}| = a\) and \(\vec{P} \cdot \vec{Q} = 0\). $$\vec{S} = \vec{P} + \vec{Q}$$ $$\vec{D} = \vec{P} - \vec{Q}$$
Solution: Take the scalar product of the sum \(\vec{S}\) and the difference \(\vec{D}\): $$\vec{S} \cdot \vec{D} = (\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q})$$ $$\vec{S} \cdot \vec{D} = \vec{P} \cdot \vec{P} - \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{P} - \vec{Q} \cdot \vec{Q}$$ Since \(\vec{P} \cdot \vec{Q} = 0\) (perpendicular) and \(\vec{P} \cdot \vec{P} = |\vec{P}|^2 = a^2\): $$\vec{S} \cdot \vec{D} = a^2 - 0 + 0 - a^2 = 0$$ Since their dot product is zero, the sum vector and difference vector are perpendicular to each other.
Why other options are incorrect:
A is incorrect because parallel vectors would require a non-zero dot product equal to the product of their magnitudes.
C and D are incorrect because the dot product is exactly zero, which uniquely corresponds to an angle of \(90^\circ\).
Zero magnitude and a specific direction along the x-axis
B
Unit magnitude and an arbitrary direction
C
Zero magnitude and no specific direction
D
Infinite magnitude and a random direction
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: The properties of the null (zero) vector.
Solution: A null vector (represented as \(\vec{0}\)) is a vector with a magnitude of exactly zero. Because its length is zero, it does not point in any specific direction, and its direction is mathematically defined as arbitrary or undefined.
Why other options are incorrect:
A is incorrect because a null vector cannot have any specified direction.
B and D are incorrect because they assign non-zero magnitudes (1 and infinity) to the zero vector.
The net external torque acting on a rigid body determines its:
A
Linear acceleration
B
Angular acceleration
C
Linear momentum
D
Moment of inertia
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Newton's second law for rotational motion.
Formula: $$\vec{\tau} = I \vec{\alpha}$$
Solution: Just as net force determines the linear acceleration of a mass (\(\vec{F} = m\vec{a}\)), the net external torque \(\vec{\tau}\) acting on a body determines its angular acceleration \(\vec{\alpha}\).
Why other options are incorrect:
A and C are incorrect because linear acceleration and linear momentum are governed by translational forces, not torques.
D is incorrect because the moment of inertia is an intrinsic geometric property of the mass distribution, independent of the applied torque.
Solution: To minimize \(R\), we need to minimize \(\cos\theta\). The minimum value of the cosine function is \(-1\), which occurs when: $$\theta = 180^\circ$$ In this case: $$R = \sqrt{F_1^2 + F_2^2 - 2F_1F_2} = \sqrt{(F_1 - F_2)^2} = |F_1 - F_2|$$ This matches the scenario where the forces point in opposite directions.
Why other options are incorrect:
\(0^\circ\) yields the maximum possible resultant (\(F_1 + F_2\)).
\(90^\circ\) yields \(\sqrt{F_1^2 + F_2^2}\), which is greater than the minimum difference.
Which of the following pairs contains exactly one vector quantity and one scalar quantity?
A
Displacement, Acceleration
B
Force, Kinetic Energy
C
Power, Speed
D
Momentum, Velocity
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Identifying vectors and scalars within physical quantity pairings.
Solution: Let's analyze each pair:
Force is a vector quantity (magnitude and direction), whereas Kinetic Energy is a scalar quantity (only magnitude). This satisfies the requirement of exactly one vector and one scalar.
Why other options are incorrect:
A: Displacement and acceleration are both vectors.
Which of the following objects has every point on its surface equidistant from its center of gravity?
A
An egg
B
A cubic box
C
A triangular prism
D
A table tennis ball
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Uniform mass distribution, geometric centers, and centers of gravity of regular shapes.
Solution: A table tennis ball is highly spherical and has a uniform mass distribution. For a uniform sphere, the center of gravity coincides with its geometric center. Because every point on the surface of a sphere is equidistant from its center, the table tennis ball perfectly satisfies this property.
Why other options are incorrect:
An egg is ellipsoidal, so surface points at the poles are further from the center of gravity than points at the equator.
A cubic box and a triangular prism have vertices and flat faces, meaning corners are much further from the center of gravity than the center of the faces.
A unit vector is primarily used in physics to specify which of the following properties?
A
The direction of a vector
B
The magnitude of a vector
C
The angle of a vector with the origin
D
The position of a vector in space
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: The purpose of unit vectors.
Solution: Unit vectors have a magnitude of exactly 1. Multiplying any scalar value by a unit vector assigns a direction to that value without changing its magnitude. Thus, they are used to describe direction.
Why other options are incorrect:
B is incorrect because the magnitude of a unit vector is always fixed at 1, so it cannot describe the variable magnitude of other vectors.
C and D are incorrect because a unit vector alone does not specify position in space or reference angles.
If the rectangular x-component \(R_x\) of a vector is positive and its y-component \(R_y\) is negative, in which quadrant does the resultant vector lie?
A
First quadrant
B
Second quadrant
C
Third quadrant
D
Fourth quadrant
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Quadrant determination using component signs.
Solution: The signs of the components determine the direction:
\(R_x > 0\) (points along the positive x-axis, to the right)
\(R_y < 0\) (points along the negative y-axis, downwards)
The right-down direction corresponds to the fourth quadrant.
Why other options are incorrect:
First quadrant: Requires both components to be positive.
Second quadrant: Requires negative x-component and positive y-component.
Third quadrant: Requires both components to be negative.
Torque is also commonly referred to in physics as the:
A
Moment arm
B
Moment of force
C
Moment of inertia
D
Impulse of force
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Scientific nomenclature for rotational force effects.
Solution: Torque measures the tendency of a force to rotate an object about an axis. In classical mechanics, this rotational effect is also called the moment of force.
Why other options are incorrect:
Moment arm is the perpendicular distance from the axis of rotation to the line of action of the force.
Moment of inertia is a measure of an object's resistance to rotational acceleration.
Impulse is the product of force and the time interval over which it acts.
By standard sign convention, an anticlockwise (counterclockwise) torque is considered:
A
Positive
B
Negative
C
Zero
D
Imaginary
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: Standard rotational sign conventions.
Solution: By standard mathematical convention (the right-hand rule applied to the coordinate plane), rotations and torques in the anticlockwise direction are defined as positive, whereas clockwise rotations are defined as negative.
Why other options are incorrect:
Negative is the sign convention reserved for clockwise torques.
Zero and Imaginary are incorrect because active torques have non-zero real values.
Two equal and opposite forces acting on a body along different lines of action form a:
A
Resultant vector
B
Moment arm
C
Couple
D
Linear impulse
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Definition and properties of a force couple.
Solution: A couple is defined as two parallel forces that are equal in magnitude, opposite in direction, and do not share a common line of action. A couple produces a pure torque with zero net translational force.
Why other options are incorrect:
Resultant vector: The net vector sum of these two forces is actually zero, so they do not produce a net translational force.
Moment arm: This is a distance measurement, not a system of forces.
Linear impulse: This is the change in linear momentum, which is zero since the net force of a couple is zero.
When a rigid body satisfies the first condition of equilibrium (\(\Sigma \vec{F} = 0\)), it is guaranteed to be in:
A
Translational equilibrium
B
Rotational equilibrium
C
Complete static equilibrium
D
Dynamic rotational equilibrium
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: The first condition of equilibrium and its physical consequence.
Solution: The first condition of equilibrium states that the vector sum of all forces acting on a body must be zero (\(\Sigma \vec{F} = 0\)). According to Newton's second law (\(\Sigma \vec{F} = m\vec{a}\)), this ensures that the translational acceleration of the body's center of mass is zero. Therefore, the body is in translational equilibrium.
Why other options are incorrect:
Rotational equilibrium requires the sum of all torques to be zero (the second condition).
Complete static equilibrium requires both translational and rotational equilibrium to be satisfied, and the body must be at rest.
A force of magnitude 10 N acts at an angle of 30° relative to the y-axis. What is the magnitude of its rectangular x-component?
A
5.0 N
B
8.66 N
C
10.0 N
D
0.0 N
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: Resolving a vector using an angle defined relative to the vertical (y-axis).
Formula: $$F_x = F \sin\theta_y$$
Solution: When the angle \(\theta = 30^\circ\) is measured relative to the y-axis, the horizontal (x) component is associated with the sine of that angle: $$F_x = F \sin(30^\circ)$$ $$F_x = 10\text{ N} \times 0.5 = 5.0\text{ N}$$
Why other options are incorrect:
8.66 N is the vertical (y) component, calculated using \(10 \cos(30^\circ) = 10 \times 0.866 = 8.66\text{ N}\).
The rectangular components of a vector in a 2D plane are equal in magnitude if the angle \(\theta\) of the vector relative to the x-axis is:
A
\(30^\circ\)
B
\(45^\circ\)
C
\(60^\circ\)
D
\(90^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Equality of rectangular components.
Formula: $$A_x = A \cos\theta$$ $$A_y = A \sin\theta$$
Solution: For the magnitudes to be equal (\(|A_x| = |A_y|\)): $$A \cos\theta = A \sin\theta \implies \tan\theta = 1$$ For angles in the first quadrant, this yields: $$\theta = 45^\circ$$
Why other options are incorrect:
\(30^\circ\): The horizontal component is larger (\(\cos 30^\circ \approx 0.866 > \sin 30^\circ = 0.5\)).
\(60^\circ\): The vertical component is larger (\(\sin 60^\circ \approx 0.866 > \cos 60^\circ = 0.5\)).
\(90^\circ\): The horizontal component is zero, and the vertical component is equal to the vector's full magnitude.
If the position vector \(\vec{r}\) of a point of application of force and the force vector \(\vec{F}\) are collinear (pointing in the same direction), what is the resulting torque?
A
Maximum
B
Minimum but non-zero
C
Zero
D
Negative
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Calculating torque when the force is parallel to the displacement vector.
Solution: If the vectors point in the same direction, the angle \(\theta\) between them is \(0^\circ\). $$\tau = rF \sin(0^\circ) = rF(0) = 0$$ Therefore, the torque is exactly zero.
Why other options are incorrect:
A is incorrect because the maximum torque occurs when the force is perpendicular to the position vector (\(\theta = 90^\circ\)).
B and D are incorrect because the sine of \(0^\circ\) is zero, making the torque zero.
If a rigid body is either at rest or rotating with a constant angular velocity, the net external torque acting on the system must be:
A
Maximum
B
Negative
C
Zero
D
Positive
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Rotational equilibrium and its relation to torque.
Formula: $$\vec{\tau}_{\text{net}} = I \vec{\alpha}$$
Solution: If a body is at rest or rotating with a constant angular velocity, its angular acceleration \(\vec{\alpha}\) is zero: $$\vec{\alpha} = \frac{d\vec{\omega}}{dt} = 0$$ Substituting this into Newton's rotational second law yields: $$\vec{\tau}_{\text{net}} = I(0) = 0$$
Why other options are incorrect:
A, B, and D are incorrect because any non-zero net torque (whether positive, negative, or maximum) would produce an angular acceleration, causing the angular velocity to change over time.
The area of a parallelogram formed by two vectors \(\vec{A}\) and \(\vec{B}\) as its adjacent sides is equal to:
A
AB
B
\(AB \cos\theta\)
C
\(AB \sin\theta\)
D
\(AB \tan\theta\)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Geometric interpretation of the magnitude of the vector cross product.
Formula: $$\text{Area} = |\vec{A} \times \vec{B}| = AB \sin\theta$$
Solution: The base of the parallelogram is \(A\). The height is the perpendicular component of vector \(\vec{B}\), which is given by \(B \sin\theta\). $$\text{Area} = \text{base} \times \text{height} = A(B \sin\theta) = AB \sin\theta$$ This is exactly equal to the magnitude of the cross product of the two vectors.
Why other options are incorrect:
\(AB\) is the area of a rectangle with sides of length \(A\) and \(B\).
\(AB \cos\theta\) is the scalar dot product of the two vectors.
\(AB \tan\theta\) does not correspond to any standard geometric area for these vectors.
The magnitude of the dot product of two vectors is \(6\sqrt{3}\) and the magnitude of their cross product is 6. What is the angle between the two vectors?
A
\(0^\circ\)
B
\(30^\circ\)
C
\(45^\circ\)
D
\(60^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Finding the angle between vectors using the ratio of cross and dot products.
Formula: $$\vec{A} \cdot \vec{B} = AB \cos\theta = 6\sqrt{3}$$ $$|\vec{A} \times \vec{B}| = AB \sin\theta = 6$$
Solution: Divide the magnitude of the cross product by the dot product: $$\frac{AB \sin\theta}{AB \cos\theta} = \frac{6}{6\sqrt{3}}$$ $$\tan\theta = \frac{1}{\sqrt{3}}$$ $$\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$$
Why other options are incorrect:
\(0^\circ\): The cross product would be zero.
\(45^\circ\): The dot product and cross product would be equal in magnitude.
\(60^\circ\): This would swap the two values, making the dot product 6 and the cross product \(6\sqrt{3}\).
If the cross product of two non-zero vectors \(\vec{a}\) and \(\vec{b}\) is directed along the z-axis, then both vectors \(\vec{a}\) and \(\vec{b}\) must lie in the:
A
yz-plane
B
zx-plane
C
xy-plane
D
3D space of any orientation
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: The perpendicular nature of the vector cross product.
Solution: The vector cross product produces a vector that is always perpendicular to the plane containing the original vectors. If the resulting vector points along the z-axis, then both \(\vec{a}\) and \(\vec{b}\) must be perpendicular to the z-axis. This means they must lie within the flat plane perpendicular to the z-axis, which is the xy-plane.
Why other options are incorrect:
yz-plane contains vectors whose cross products must point along the x-axis.
zx-plane contains vectors whose cross products must point along the y-axis.
Given vectors \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = 4\hat{i} + 2\hat{j} - 2\hat{k}\), find a vector \(\vec{x}\) that is parallel to \(\vec{A}\) but has a magnitude equal to that of \(\vec{B}\).
A particle undergoes a displacement from position vector \(\vec{r}_1 = 3\hat{i} + 2\hat{j} - 6\hat{k}\) to position vector \(\vec{r}_2 = 14\hat{i} + 13\hat{j} + 9\hat{k}\) under the action of a constant force \(\vec{F} = 8\hat{i} + 2\hat{j} + 6\hat{k}\). Find the work done by the force.
A
50 J
B
125 J
C
155 J
D
200 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept: Work done calculated using the displacement vector and the force vector.
A, B, and C are incorrect values that result from mathematical errors when calculating the displacement coordinates (e.g. evaluating \(9 - 6 = 3\) instead of \(9 - (-6) = 15\)).
A vector (such as the velocity of a body undergoing uniform translational motion) that can be displaced parallel to itself without altering its effect is known as a:
A
Unit vector
B
Position vector
C
Free vector
D
Null vector
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Categorization of vectors based on localization.
Solution: A free vector is a vector whose point of application is not restricted to a specific point in space. It can be translated parallel to itself to any location without changing its physical meaning or mathematical representation (such as velocity or displacement).
Why other options are incorrect:
Unit vector is defined strictly by having a magnitude of one.
Position vector is tied to a specific point relative to a chosen coordinate origin.
Null vector has a magnitude of zero and no direction.
Two position vectors \(\vec{R}_1\) and \(\vec{R}_2\) make angles of 30° and 90° with the positive x-axis, respectively. If \(|\vec{R}_1| = 4\text{ cm}\) and \(|\vec{R}_2| = 3\text{ cm}\), find the magnitude of their vector product.
A
\(12\sqrt{3}\text{ cm}^2\)
B
\(6\sqrt{3}\text{ cm}^2\)
C
\(6\sqrt{12}\text{ cm}^2\)
D
\(3\sqrt{6}\text{ cm}^2\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Calculating the magnitude of a cross product using the angle between two vectors.
Solution: First, determine the angle \(\theta\) between the two vectors: $$\theta = \theta_2 - \theta_1 = 90^\circ - 30^\circ = 60^\circ$$ Now substitute the magnitudes and the angle into the formula: $$|\vec{R}_1 \times \vec{R}_2| = (4\text{ cm})(3\text{ cm}) \sin(60^\circ)$$ $$|\vec{R}_1 \times \vec{R}_2| = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\text{ cm}^2$$
Why other options are incorrect:
A is incorrect because it omits the factor of \(1/2\) from the sine value (using 1 instead of \(\sin 60^\circ\)).
C and D are mathematically incorrect results from applying the wrong trigonometric functions or angles.
Find the work done in moving an object along a straight line from position \( (3, 2, -1) \) to position \( (2, -1, 4) \) under a force field given by \(\vec{F} = 4\hat{i} - 3\hat{j} + 2\hat{k}\).
A
10 J
B
12 J
C
15 J
D
17 J
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Calculating work done between two spatial coordinates.
What is the vector projection of \(\vec{E} = 2\hat{i} - 3\hat{j} + 6\hat{k}\) onto the direction of vector \(\vec{F} = \hat{i} + 2\hat{j} + 2\hat{k}\)?
A
1/2
B
8/3
C
3/5
D
5/7
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Finding the scalar projection of one vector onto another.
Two vectors \(\vec{S}\) and \(\vec{R}\) have magnitudes of 4 and 6, respectively. If their scalar product is \(\vec{S} \cdot \vec{R} = 13.5\), find the angle between the two vectors.
A
\(22.99^\circ\)
B
\(55.77^\circ\)
C
\(77.81^\circ\)
D
\(99.18^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Calculating the angle between vectors using the definition of the dot product.
Solution: Substitute the given values into the equation: $$\cos\theta = \frac{13.5}{4 \times 6} = \frac{13.5}{24} = 0.5625$$ $$\theta = \arccos(0.5625) \approx 55.77^\circ$$
Why other options are incorrect:
A, C, and D are incorrect angles that result from arithmetic calculation errors, such as dividing by the sum of the magnitudes (10) instead of their product (24).
The resultant of two forces has a magnitude of 20 N. If one of the forces has a magnitude of \(20\sqrt{3}\text{ N}\) and makes an angle of 30° with the resultant force, what is the magnitude of the other force?
A
\(10\sqrt{3}\text{ N}\)
B
20 N
C
\(20\sqrt{3}\text{ N}\)
D
10 N
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Solving for a component vector magnitude using the law of cosines.
Formula: $$B^2 = A^2 + R^2 - 2AR \cos\phi$$
Solution: Let the resultant force be \(R = 20\text{ N}\) and the known force be \(A = 20\sqrt{3}\text{ N}\). The angle between them is \(\phi = 30^\circ\). Using the law of cosines on the vector triangle: $$B^2 = (20\sqrt{3})^2 + (20)^2 - 2(20\sqrt{3})(20) \cos(30^\circ)$$ $$B^2 = (400 \times 3) + 400 - 800\sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$$ $$B^2 = 1200 + 400 - 1200$$ $$B^2 = 400 \implies B = 20\text{ N}$$
Why other options are incorrect:
A, C, and D are mathematically incorrect and do not satisfy the law of cosines for this vector geometry.
If \(\vec{A} = \vec{B} + \vec{C}\) and the magnitudes of vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) are 5, 4, and 3 units respectively, what is the angle between vector \(\vec{A}\) and vector \(\vec{C}\)?
A
\(\cos^{-1}(3/5)\)
B
\(\cos^{-1}(4/5)\)
C
\(\sin^{-1}(3/4)\)
D
90°
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept: Using right-angled triangle properties to find the angle between vector components.
Solution: The magnitudes of the vectors (5, 4, and 3) satisfy the Pythagorean theorem: $$5^2 = 4^2 + 3^2 \implies 25 = 16 + 9$$ This means vectors \(\vec{B}\) and \(\vec{C}\) are perpendicular to each other, and \(\vec{A}\) forms the hypotenuse of a right-angled triangle. The angle \(\theta\) between \(\vec{A}\) (the hypotenuse) and \(\vec{C}\) (the adjacent side) is given by: $$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{C}{A} = \frac{3}{5}$$ $$\theta = \cos^{-1}(3/5)$$
Why other options are incorrect:
B is incorrect because it defines the angle between the hypotenuse \(\vec{A}\) and the vector \(\vec{B}\).
C and D are mathematically incorrect and do not match the geometric configuration of this right-angled triangle.
If the expression \(0.8\hat{i} + 0.5\hat{j} + c\hat{k}\) represents a unit vector, find the value of the constant \(c\).
A
\(\sqrt{0.64}\)
B
0.20
C
\(\sqrt{0.11}\)
D
1.00
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: The magnitude of a unit vector is exactly equal to one.
Formula: $$\sqrt{x^2 + y^2 + z^2} = 1$$
Solution: Set the magnitude of the given vector to 1: $$\sqrt{(0.8)^2 + (0.5)^2 + c^2} = 1$$ Square both sides of the equation: $$0.64 + 0.25 + c^2 = 1$$ $$0.89 + c^2 = 1$$ $$c^2 = 1 - 0.89 = 0.11 \implies c = \sqrt{0.11}$$
Why other options are incorrect:
A is incorrect because it equals 0.8, which is the value of the x-component.
B and D are mathematically incorrect and do not satisfy the unit magnitude condition.
Solution: Let's define East as the positive x-direction and West as the negative x-direction. Each trip occurs over a duration of \(t = 1\text{ hour}\). Let's calculate the net displacement for the journeys described in Option B:
Car 1: Travels +40 km, then -20 km: $$\vec{d}_{\text{net1}} = 40\text{ km} - 20\text{ km} = +20\text{ km (East)}$$ $$\vec{v}_{\text{avg1}} = \frac{20\text{ km}}{1\text{ hr}} = 20\text{ km/hr East}$$
Car 2: Travels +20 km: $$\vec{d}_{\text{net2}} = +20\text{ km (East)}$$ $$\vec{v}_{\text{avg2}} = \frac{20\text{ km}}{1\text{ hr}} = 20\text{ km/hr East}$$
Both cars have the exact same net displacement vector over the same time interval, so they have identical average velocities.
Why other options are incorrect:
A, C, and D contain pairs of journeys with unequal net displacements (for example, in D, Car 1 has a net displacement of 0 km, whereas Car 2 has a net displacement of 20 km).
A uniform 100 cm meter rod is balanced at its center of gravity (the 50 cm mark). A downward force of 5 N is applied at the 0 cm mark. Where must a downward force of 10 N be applied to keep the rod in balance?
A
80 cm mark
B
75 cm mark
C
70 cm mark
D
65 cm mark
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept: Using rotational equilibrium to balance torque about a pivot.
Solution: The pivot is at the 50 cm mark. Let's calculate the moment arms:
The 5 N force is at the 0 cm mark. Its moment arm is: $$d_1 = 50\text{ cm} - 0\text{ cm} = 50\text{ cm}$$
This force produces an anticlockwise torque: $$\tau_{\text{anticlockwise}} = 5\text{ N} \times 50\text{ cm} = 250\text{ N}\cdot\text{cm}$$
The 10 N force must produce an equal clockwise torque of \(250\text{ N}\cdot\text{cm}\) on the opposite side of the pivot: $$\tau_{\text{clockwise}} = 10\text{ N} \times d_2 = 250\text{ N}\cdot\text{cm}$$ $$d_2 = 25\text{ cm}$$
Since the clockwise torque must be on the right side of the pivot, the position of this force is: $$\text{Position} = 50\text{ cm} + 25\text{ cm} = 75\text{ cm mark}$$
Why other options are incorrect:
A, C, and D do not satisfy the condition of rotational equilibrium. If placed at these marks, the clockwise and anticlockwise torques would be unbalanced, causing the rod to tilt.
3, 7, and 10 are incorrect magnitudes that result from mathematical errors (such as using \(3 + 4 = 7\) instead of the square root of the sum of squares, or failing to divide by 2).
A force couple is formed by two equal and opposite forces that do not share a common line of action. Which of the following statements about a force couple is correct?
A
Its torque depends on the chosen coordinate origin.
B
It produces a non-zero net translational force.
C
Its net torque is independent of the coordinate origin.
D
It can cause translational acceleration of the body.
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept: Properties of a force couple.
Solution: A couple consists of two forces of equal magnitude pointing in opposite directions (\(\vec{F}_1 = -\vec{F}_2\)). The net force is exactly zero: $$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = 0$$ Because the net force is zero, the torque produced by a couple is constant and has the same value about any point in space, making it completely independent of the choice of coordinate origin.
Why other options are incorrect:
A is incorrect because the origin does not affect the torque calculation for a couple.
B and D are incorrect because the net force is zero, which means there is no translational acceleration.
Join thousands of pre-med students utilizing BeambePrep's full combat suite: Swarm Mode timed challenges, Normal Grind Mode, Propolis Ward mistake notebooks, Infinite Full-Length Practice (FLP) Mocks, and FSRS Spaced Repetition.
After 42 compilation builds with zero physical Apple hardware and months of relentless engineering, the native BeambePrep iOS app is officially approved and live on the Apple App Store for iPhone and iPad.
42 Production Builds
800+ Active Bees
+2,300% Monthly Growth
100% Free & Community Driven