Motion in plane practice · Mixed difficulty

Option C ($\mathrm{n}=\frac{2 h \mathrm{u}^2}{\mathrm{gb}^2}$) is correct because it is derived from equating the horizontal distance covered by the ball to n times the width of the step and the vertical distance to n times the height of the step.

Option B is correct because both the assertion and reason are correct, but the reason doesn't directly explain the assertion. The assertion is true because there's no external force along the x-axis, and the reason is true because mass remains constant.

Option A ($\sqrt{A^2+B^2+A B}$) is correct because the magnitude of the resultant vector is calculated using the given condition and trigonometric identities.

Option D ($\frac{4}{\sqrt{13}}$) is correct because it represents the minimum magnitude of vector $\mathbf{d}$ found by minimizing the quadratic expression obtained from the given conditions.

Option B (${90^0}$) is correct because if the dot product is 0, the vectors are perpendicular, and if the cross product is 0, they are parallel or anti-parallel. Since $\overrightarrow Q$ is perpendicular to $\overrightarrow P$ and $\overrightarrow R$ is parallel to $\overrightarrow P$, $\overrightarrow Q$ and $\overrightarrow R$ are perpendicular.

Option B (Directly proportional to $\sin ^2 \theta$) is correct because the maximum height of a projectile is proportional to the square of the sine of the launch angle.

Option A is correct because the cross product is perpendicular to the plane containing the vectors, and the sum and difference lie in that plane.

Option D is correct because if $\overrightarrow A . \overrightarrow B = 0$ and $\overrightarrow A . \overrightarrow C = 0$, then $\overrightarrow A$ is perpendicular to both $\overrightarrow B$ and $\overrightarrow C$, meaning it's parallel to their cross product $\overrightarrow B \times \overrightarrow C$.

Option D (zero) is correct because the forces are equally spaced around the body, resulting in a net force of zero.

Option A is correct because the assertion is true, and the reason provides a correct explanation for why the particle moves along a parabolic trajectory.

Option D (196.3 m/s) is correct because it correctly uses the tangent of the angle to calculate the distance traveled and then the speed.

Option C (03:01) is correct because the ratio of heights is 3:1 when the angles of elevation are 60° and 30° respectively.

Option A (Statement-I is true and statement-II is true) is correct because the cross product of two vectors is perpendicular to the plane containing those vectors, and the sum and difference of two vectors lie in the same plane.

Option B ($1: \sqrt{2}$) is correct because for the same time of flight, the vertical components of the initial velocities must be equal, leading to the given ratio.

Option A ($-\hat{i}-3 \hat{j}-3 \hat{k}$) is correct because it is the cross product of the position vector and the force vector.

Option C (7) is correct because the magnitude of the vector perpendicular to both A and B is 7 times the magnitude of the given vector.

Option A ($8 d \cdot(a+b+c)$) is correct because it correctly uses the scalar triple product properties to simplify the expression and find the value of $\lambda + \mu + v$.

Option D ($\cos ^{-1}(\frac{1}{\sqrt{3}})$) is correct because it is the angle whose cosine is the component of the vector along each axis divided by the magnitude of the vector.

Option A ($\sqrt{40}$ and $18.43^{\circ}$) is correct because the magnitude and direction of the displacement vector are calculated using the change in position coordinates.

Option A (6.5 m) is the closest to the correct answer because it is calculated by using the law of cosines to find the magnitude of the resultant displacement.

Option A (Both Assertion and Reason are true, and Reason is the correct explanation for Assertion) is correct because gravity is the only force acting on the projectile in the absence of air resistance, and it acts vertically, leaving the horizontal velocity component unchanged.

Option A ($\frac{1}{3 \sqrt{2}}(4 \hat{i}-\hat{j}-\hat{k})$) is correct because it is the unit vector in the direction of the resultant vector.

Option B (range of projectile) is correct because projectiles launched at complementary angles (60° and 30°) have the same range.

Option D is correct because by analyzing the relative velocities in both scenarios, we can determine the actual wind velocity.

Option A (${A}$ and ${B}$ are mutually perpendicular) is correct because the dot product of the two vectors is zero.

Option D ($\sqrt {{\mathrm{A}^\mathrm{2}}\mathrm{ + }{\mathrm{B}^\mathrm{2}} - \sqrt {\\mathrm{2}} \mathrm{AB}}$) is correct because it is derived from the given condition $A.B = \left| {A \times B} \right|$ which implies the angle between A and B is 45 degrees, and then applying the law of cosines.

Option B ($5 \mathrm{~ms}^{-1}$) is correct because it is calculated by comparing the given trajectory equation with the standard equation of a projectile.

Option B is correct because the magnitudes of the forces are 5N and 13N. The problem describes a right triangle where the resultant force is perpendicular to the smaller force.

Statement-I is true because the vectors form a right-angled triangle, and Statement-II is true as it represents the law of cosines.

Option D (5) is correct because the magnitude of vector $\mathrm{\vec B}$ is calculated by finding its X and Y components and then using the Pythagorean theorem.