1

### 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

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

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

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

### 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)

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

$\Rightarrow \,\,\,\tan {\theta \over 2} = n$
3

### 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
4

### JEE Main 2019 (Online) 11th January Morning Slot

A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 60o with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the value of R is :
A
2.8 m
B
5.1 m
C
2.5 m
D
10.3 m

## Explanation vx = 10cos60o = 5 m/s vy = 10cos30o = $5\sqrt 3$ m/s

velocity after t = 1 sec.

vx = 5 m/s

vy = $\left| {\left( {5\sqrt 3 - 10} \right)} \right|$ m/s = 10 $-$ 5$\sqrt 3$

an = ${{{v^2}} \over R} \Rightarrow \,R\,$ = ${{v_x^2 + v_y^2} \over {{a_n}}}$ = ${{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos \theta }}$

tan$\theta$ = ${{10 - 5\sqrt 3 } \over 5}$ = 2 $-$ ${\sqrt 3 }$ $\Rightarrow$  $\theta$ = 15o

R = ${{100\left( {2 - \sqrt 3 } \right)} \over {10\cos 15}} = 2.8m$