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1

### JEE Main 2019 (Online) 12th January Morning Slot

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is :
A
$${{\sqrt 3 } \over 2}$$v
B
$${{2v} \over {\sqrt 3 }}$$
C
v
D
$${v \over 2}$$

## Explanation

AB = VP $$\times$$ t

BC = Vt

cos60o = $${{AB} \over {BC}}$$

$${1 \over 2} = {{{V_P} \times t} \over {Vt}}$$

VP = $${V \over 2}$$
2

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

A particle moves from the point $$\left( {2.0\widehat i + 4.0\widehat j} \right)$$ m, at t = 0, with an initial velocity $$\left( {5.0\widehat i + 4.0\widehat j} \right)$$ ms$$-$$1. It is acted upon by a constant force which produces a constant acceleration $$\left( {4.0\widehat i + 4.0\widehat j} \right)$$ ms$$-$$2. What is the distance of the particle from the origin at time 2 s?
A
15 m
B
$$20\sqrt 2$$ m
C
$$10\sqrt 2$$ m
D
5 m

## Explanation

$$\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4$$

$$= 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$$

$$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$$

$$\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j$$

$$\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2$$
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) 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

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

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

$$A \propto {R^2}$$

$$A \propto {u^4}$$

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

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