Motion in plane practice · Mixed difficulty

Option D is correct because adding a scalar (n) to a vector ($\vec{A}$) is not a defined operation in vector algebra.

Option D (${30^0}$) is correct because it satisfies the condition for the projectile to just cross the two walls of height h separated by h, given the initial velocity.

Option C (6 units and 10 units) is correct because it satisfies the conditions of the problem: the sum of the magnitudes is 10+6=16, not 10, so there is an error in the problem statement. Assuming the simple sum is 16, the vector sum is 8, and the resultant is perpendicular to the smaller vector.

Option D (10.8 units due east) is correct because multiplying a vector by a scalar multiplies its magnitude by that scalar, while the direction remains the same.

Option D ($\frac{6}{{\sqrt {13} }}$) is correct because it represents the projection of vector B onto vector A, calculated using the dot product and the magnitude of vector A.

Option D ($\mathrm{R}_{\mathrm{m}}=\frac{\mathrm{u}^2}{\mathrm{~g}}$) is correct because the maximum horizontal range is achieved when the angle of projection is 45 degrees, resulting in $\sin{2\theta} = 1$.

Option B ($30\sqrt 3 \;\mathrm{km}{\mathrm{h}^{ - 1}}$) is correct because it is calculated using the Pythagorean theorem, given one component and the resultant velocity.

Option D (zero) is correct because with 12 equal forces equally spaced at 30-degree intervals, they form a closed polygon, resulting in a net force of zero.

Option D (-0.5) is correct because it satisfies the condition for the given vector to be a unit vector. Note: There seems to be an error in the provided correct answer. The correct answer should be -0.5, not 45689.

Option B ($\sqrt 3$) is correct because when the sum of two unit vectors is also a unit vector, the magnitude of their difference is $\sqrt{3}$.

Option B ($10\%$) is correct. The percentage increase in horizontal range is approximately equal to the percentage increase in the square of the initial speed.

Option A ($60^\circ$) is correct because at maximum height, the vertical component of velocity is zero, and the horizontal component is half the initial speed, leading to $\cos{\theta} = 0.5$.

Option A (20) is correct because it is the scalar product of the two vectors.

Option A is correct because the dot product of a unit vector with its derivative with respect to time is always zero.

Option B (${90^0}$) is correct because if the magnitude of the sum of two vectors equals the magnitude of their difference, the vectors are perpendicular.

Option A (Statement-I is true and statement-II is true) is correct because the cross product of a vector with itself is indeed a null vector, and the cross product of two vectors results in a vector quantity.

Option A is correct because the assertion is true (forces can be in equilibrium) and the reason correctly explains why (net force is zero).

Option D (Zero) is correct because if the forces are equally spaced around a point, their resultant is zero.

Option C ($7\sqrt 2$ units) is correct because it is the magnitude of the final velocity vector after 10 seconds, calculated using vector addition.

Option B (three) is correct because at least three non-collinear forces are required to achieve equilibrium, as two forces would need to be collinear to balance each other.

Option C is correct because the cross product of a vector with itself is always the zero vector.

Option A (Statement-I is true and statement-II is true) is correct because a body can be in equilibrium under three concurrent forces, and with multiple concurrent forces, equilibrium occurs when the vector sum of all forces is zero.

Option C is correct. The vectors form a closed polygon, resulting in a zero resultant.

Option A (Perpendicular unit vectors) is correct because if $\mathrm{\vec A}\mathrm{.\vec B} = 0$ and $\mathrm{\vec A} \times \mathrm{\vec B} = 1$, it means the vectors are perpendicular and their magnitudes multiply to 1, implying they are unit vectors.

Option A is correct because if $\mathbf{A} \times \mathbf{B}=\mathbf{B} \times \mathbf{A}$, it means $\mathbf{A} \times \mathbf{B} = - \mathbf{A} \times \mathbf{B}$, implying they are anti-parallel, hence the angle is $\pi$.

Option A is correct because floating objects move with the same velocity as the river, maintaining a constant distance between them.

Option D is correct because both the assertion and the reason are false. The angle between velocity and acceleration can be any angle for 2D motion, and they don't necessarily work in opposite directions.

Option C is correct because a unit vector, by definition, has a magnitude of 1.

Option D ($\mathbf{a}$ is parallel to $\mathbf{b}$) is correct because if $(\mathbf{a}+\mathbf{b})$ and $(\mathbf{a}-\mathbf{b})$ are parallel, then $\mathbf{a}$ and $\mathbf{b}$ must be parallel.

Option A is correct because expanding the cross product and using the properties of the cross product results in 2(b x a).