**Question:**

**Two guns $A$ and $B$ can fire bullets at speeds $1 \mathrm{~km} / \mathrm{s}$ and $2 \mathrm{~km} / \mathrm{s}$ respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is:**

$1: 16$

$1: 2$

$1: 4$

$1: 8$

Question of from chapter.

JEE Main Previous Year 10 Jan. 2019 I

Correct Option: 1

**Solution:**

### Related Questions

Let $\left|\overrightarrow{A_{1}}\right|=3,\left|\overrightarrow{A_{2}}\right|=5$ and $\left|\overrightarrow{A_{1}}+\overrightarrow{A_{2}}\right|=5$. The value of $\left(2 \overrightarrow{\mathrm{A}_{1}}+3 \overrightarrow{\mathrm{A}_{2}}\right) \cdot\left(3 \overrightarrow{\mathrm{A}_{1}}-2 \overrightarrow{\mathrm{A}_{2}}\right)$ is :

In the cube of side ‘a’ shown in the figure, the vector from the central point of the face $\mathrm{ABOD}$ to the central point of the face BEFO will be:

Two forces $P$ and $Q$, of magnitude $2 F$ and $3 F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is:

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitude of $(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})$ is ‘ $n$ ‘ times the magnitude of $(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})$.

The angle between $\vec{A}$ and $\vec{B}$ is:

Let $\overrightarrow{\mathrm{A}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})$ and $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})$. The magnitude of a coplanar vector $\overrightarrow{\mathrm{C}}$ such that $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}$ is given by

A vector $\vec{A}$ is rotated by a small angle $\Delta \theta \operatorname{radian}(\Delta \theta<<1)$

to get a new vector $\vec{B}$. In that case $|\vec{B}-\vec{A}|$ is :

If $\vec{A} \times \vec{B}=\vec{B} \times \vec{A}$, then the angle between $\mathrm{A}$ and $\mathrm{B}$ is

A balloon is moving up in air vertically above a point A on the ground. When it is at a height $h_{1}$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h_{2}$, it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point $\mathrm{C}$ ). Then the height $h_{2}$ is (given $\tan 30^{\circ}=0.5774$ ):

Starting from the origin at time $t=0$, with initial velocity

$5 \hat{j} \mathrm{~ms}^{-1}$, a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{i}+4 \hat{j}) \mathrm{ms}^{-2}$. At time $t$, its coordiantes are $\left(20 \mathrm{~m}, y_{0} \mathrm{~m}\right)$. The values of $t$ and $y_{0}$ are, respectively:

The position vector of a particle changes with time according to the relation $\vec{r}(\mathrm{t})=15 \mathrm{t}^{2} \hat{i}+\left(4-20 \mathrm{t}^{2}\right) \hat{j}$. What is the magnitude of the acceleration at $t=1 ?$