1

### JEE Main 2019 (Online) 10th January Morning Slot

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/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 -
A
1 : 16
B
1 : 8
C
1 : 2
D
1 : 4

## Explanation $R = {{{u^2}\sin 2\theta } \over g}$

$A = \pi \,{R^2}$

$A \propto {R^2}$

$A \propto {u^4}$

${{{A_1}} \over {{A_2}}} = {{u_1^4} \over {u_2^4}} = {\left[ {{1 \over 2}} \right]^4} = {1 \over {16}}$
2

### JEE Main 2019 (Online) 10th January Morning Slot

In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be - A
${1 \over 2}a\left( {\widehat k - \widehat i} \right)$
B
${1 \over 2}a\left( {\widehat j - \widehat i} \right)$
C
${1 \over 2}a\left( {\widehat j - \widehat k} \right)$
D
${1 \over 2}a\left( {\widehat i - \widehat k} \right)$

## Explanation

$\overrightarrow {{r_g}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$
3

### JEE Main 2019 (Online) 10th January Evening Slot

Two vectors $\overrightarrow A$ and $\overrightarrow B$ have equal magnitudes. The magnitude of $\left( {\overrightarrow A + \overrightarrow B } \right)$ is 'n' times the magnitude of $\left( {\overrightarrow A - \overrightarrow B } \right)$ . The angle between ${\overrightarrow A }$ and ${\overrightarrow B }$ is -
A
${\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$
B
${\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$
C
${\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$
D
${\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$

## Explanation

$\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2$      . . . (1)

$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$      . . . (2)

$\Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$

$\Rightarrow \,\,\,\tan {\theta \over 2} = n$
4

### JEE Main 2019 (Online) 10th January Evening Slot

A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s ? A
3 m
B
9 m
C
10 m
D
6 m

## Explanation

S = Area under graph

${1 \over 2}$ $\times$ 2$\times$2 + 2 $\times$ 2 + 3 $\times$ 1 = 9 m