NEET Practice Questions (MCQs) with Answers & Solutions

Consider east as positive x-axis, north as positive y-axis and vertically upward direction as z-axis. A helicopter first rises up to an altitude of 100 m than flies straight in north 500 m and then suddenly takes a turn towards east and travels 1000 m east. What is position vector of helicopter. (Take starting point as origin)

A force $\overrightarrow{\mathrm{F}}=6\hat{\mathrm{i}}-8\hat{\mathrm{j}}+10\hat{\mathrm{k}}$ Newton produces acceleration $1 \mathrm{m}/{\mathrm{s}}^2$ in a body. The mass of the body is (in kg)

$\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} ; \overrightarrow{\mathrm{B}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is

Given the vectors

$$\begin{gathered} \overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ \& \overrightarrow{\mathrm{C}}=\mathrm{p}\hat{\mathrm{i}}+\mathrm{p}\hat{\mathrm{j}}+2\mathrm{p}\hat{\mathrm{k}} \end{gathered}$$

Find the angle between $\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right) \& \overrightarrow{\mathrm{C}}$

If vector $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{R}}$ have magnitude 5, 12 and 13 units and $\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}=\overrightarrow{\mathrm{R}}$ the angle between $\overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{R}}$ is -

The vector having a magnitude of 10 and perpendicular to the vector $3\hat{\mathrm{i}}-4\hat{\mathrm{j}}$ is 

A force $\overrightarrow{\mathrm{F}}=3\hat{\mathrm{i}}+\mathrm{c}\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ acting on a particle causes a displacement $\overrightarrow{\mathrm{d}}=-4\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$. If the work done is 6J then the value of 'c' is

A particle moving along a straight line according to the law $\mathrm{x}=\mathrm{At}+{\mathrm{Bt}}^2+{\mathrm{Ct}}^3$, where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.

If the velocity of a particle moving on x-axis is given by $\mathrm{v}=3{\mathrm{t}}^2-12\mathrm{t}+6$. At which time is the acceleration of particle zero?

A particle moves along straight line such that at time t its position from a fixed point O on the line is $\mathrm{x}=3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is: