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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in $${T \over 3}$$ seconds?
A
$${{8h} \over 9}$$ meters from the ground
B
$${{7h} \over 9}$$ meters from the ground
C
$${h \over 9}$$ meters from the ground
D
$${{7h} \over {18}}$$ meters from the ground

Explanation

We know that equation of motion, $$s = ut + {1 \over 2}g{t^2},\,\,$$

Initial speed of ball is zero and it take T second to reach the ground.

$$\therefore$$ $$h = {1 \over 2}g{T^2}$$

After $$T/3$$ second, vertical distance moved by the ball

$$h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2} $$

$$\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}$$

$$ = {h \over 9}$$

$$\therefore$$ Height from ground

$$ = h - {h \over 9} = {{8h} \over 9}$$
2

AIEEE 2003

MCQ (Single Correct Answer)
The co-ordinates of a moving particle at any time 't' are given by x = $$\alpha $$t3 and y = βt3. The speed to the particle at time 't' is given by
A
$$3t\sqrt {{\alpha ^2} + {\beta ^2}} $$
B
$$3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
C
$${t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
D
$$\sqrt {{\alpha ^2} + {\beta ^2}} $$

Explanation

Given that $$x = \alpha {t^3}\,\,\,\,$$ and $$\,\,\,\,y = \beta {t^3}$$

$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$

and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$

$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$

$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$

$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
3

AIEEE 2003

MCQ (Single Correct Answer)
A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $$30^\circ $$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $$\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]$$
A
5.20 m
B
4.33 m
C
2.60 m
D
8.66 m

Explanation



From the figure it is clear that maximum horizontal range

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

$$ = {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$$

$$ = 5\sqrt 3 $$ = 8.66 m
4

AIEEE 2003

MCQ (Single Correct Answer)
A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is
A
12 m
B
18 m
C
24 m
D
6 m

Explanation

For case 1 :
u = 50 km/hr = $${{50 \times 1000} \over {3600}}$$ m/s = $${{125} \over 9}$$ m/s, v = 0, s = 6 m, $$a$$ = ?

$$\therefore$$ 02 = u2 + 2$$a$$s

$$ \Rightarrow $$ $$a = - {{{u^2}} \over {2s}}$$

$$ \Rightarrow $$ $$a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$$ = $$-$$16 m/s2

For case 2 :
u = 100 km/hr = $${{100 \times 1000} \over {3600}}$$ m/s = $${{250} \over 9}$$ m/s, v = 0, $$a$$ = $$-$$16, s = ?

$$\therefore$$ 02 = u2 + 2$$a$$s

$$ \Rightarrow $$ $$s = - {{{u^2}} \over {2a}}$$

$$ \Rightarrow $$ $$s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$$ = 24 m

Questions Asked from Motion

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