1

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

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed '$\upsilon$' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then '$\upsilon$' is equal to :
A
${{2{a_1}{a_2}} \over {{a_1} + {a_2}}}t$
B
$\sqrt {2{a_1}{a_2}} t$
C
$\sqrt {{a_1}{a_2}} t$
D
${{{a_1} + {a_2}} \over 2}t$

## Explanation

For both car initial speed ($\mu$) = 0

Let the acceleration of car A and car B is $a$1 and $a$2 respectively.

Also let the time taken to reach the finishing point for car A is t1 and for car B is t2.

Let at finishing point speed of car A is $v$1 and speed of car B is $v$2

According to the question,

t2 $-$ t1 = t

and   $v$1 $-$ $v$2 = $v$

$\Rightarrow$  $a$1t1 $-$ $a$2t2 = $v$

$\Rightarrow$  $a$1t1 $-$ $a$2(t + t1) = $v$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So,   xA = xB

$\Rightarrow$  ${1 \over 2}{a_1}t_1^2 = {1 \over 2}{a_2}t_2^2$

$\Rightarrow$  $a$1t$_1^2$ = $a$2 (t + t1)2

$\Rightarrow$  $\sqrt {{a_1}} .{t_1}$ = $\sqrt {{a_2}}$ . (t + t1)

$\Rightarrow$  $\sqrt {{a_1}} .{t_1} - \sqrt {{a_2}} .{t_1} = \sqrt {{a_2}} .t$

$\Rightarrow$  t1 = ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)$

Now put the value of t1 in equation (2),

($a$1 $-$ $a$2) t1 $-$ $a$2t = $v$

$\Rightarrow$  (a1 $-$ a2) . ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$

$\Rightarrow$  $\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\sqrt {{a_2}} .t - {a_2}t = v$

$\Rightarrow$  $\sqrt {{a_1}{a_2}} .t + {a_2}.t - {a_2}t = v$

$\Rightarrow$  $v$ = $\sqrt {{a_1}{a_2}} .t$
2

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

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

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

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

$= \sqrt 2 a\omega$
3

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

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

$A \propto {R^2}$

$A \propto {u^4}$

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

### 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

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

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

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