1

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

A particle is moving with a velocity

$\overrightarrow v \, = K(y\widehat i + x\widehat j),$ where K is a constant.

The general equation for its path is :
A
y = x2 + constant
B
y2 = x + constant
C
y2 = x2 + constant
D
xy = constant

## Explanation

Given,

$\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)$

$\therefore$   Velocity in x direction,

${v_x} = {{dx} \over {dt}} = Ky\,$   . . . . . (1)

Velocity in y direction,

vy = $\,{{dy} \over {dt}}$ = Kx . . . . . . . (2)

$\therefore$   ${{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}$

$\Rightarrow$   ${{dy} \over {dx}} = {x \over y}$

$\Rightarrow$   ydy $=$ xdx

Integrating both sides we get,

$\int {ydx} = \int {xdx}$

$\Rightarrow$   ${{{y^2}} \over 2} = {{{x^2}} \over 2} + c$

$\Rightarrow$   ${y^2} = {x^2} + 2c$

$\therefore$   General equation,

y2 = x2 + constant.
2

### 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) . ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$

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

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

$\Rightarrow$  $v$ = $\sqrt {{a_1}{a_2}} .t$
3

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

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

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

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

$= \sqrt 2 a\omega$
4

### 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

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

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

$A \propto {R^2}$

$A \propto {u^4}$

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