1

### JEE Main 2019 (Online) 9th January Evening Slot

The position co-ordinates of a particle moving in a 3-D coordinate system is given by
x = a cos$\omega$t
y = a sin$\omega$t and
z = a$\omega$t

The speed of the particle is :
A
$\sqrt 2 \,a\omega$
B
$a\omega$
C
$\sqrt 3 \,a\omega$
D
2a$\omega$

## Explanation

Given that,

x = a cos $\omega$t

y = a sin $\omega$t

z = a $\omega$t

Velocity in x-direction,

Vx = ${{dx} \over {dt}} = - a\omega \sin \omega t$

Velocity in y-direction,

Vy = ${{dy} \over {dt}}$ = a $\omega$cos $\omega$t

Velocity in z-direction,

Vz = ${{dz} \over {dt}}$ = a$\omega$

Net velocity,

$\overrightarrow V$ = Vx$\widehat i$ + Vy$\widehat j$ + Vz$\widehat k$

Speed = $\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2}$

$= \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}}$

$= \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}}$

$= \sqrt {2{a^2}{\omega ^2}}$

$= \sqrt 2 a\omega$
2

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

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

$A \propto {R^2}$

$A \propto {u^4}$

${{{A_1}} \over {{A_2}}} = {{u_1^4} \over {u_2^4}} = {\left[ {{1 \over 2}} \right]^4} = {1 \over {16}}$
3

### 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

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

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

$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$
4

### 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)

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

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