Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let $$\overrightarrow A $$ = $$\left( {\widehat i + \widehat j} \right)$$ and, $$\overrightarrow B = \left( {2\widehat i - \widehat j} \right).$$ The magnitude of a coplanar vector $$\overrightarrow C $$ such that $$\overrightarrow A .\overrightarrow C = \overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B ,$$ is given by :

A

$$\sqrt {{{10} \over 9}} $$

B

$$\sqrt {{{5} \over 9}} $$

C

$$\sqrt {{{20} \over 9}} $$

D

$$\sqrt {{{9} \over 12}} $$

Let $$\overrightarrow C $$ = a$$\widehat i$$ + b$$\widehat j$$

Given, $$\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B $$

$$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 2 $$-$$ 1

$$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 1 . . . . .(1)

also given

$$\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B $$

$$ \Rightarrow $$ $$\,\,\,\,$$ 2a $$-$$ b = 1 . . . . (2)

Solving (1) and (2), we get,

a = $${1 \over 3}$$ and b = $${2 \over 3}$$

$$\therefore\,\,\,\,$$ $$\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$$

$$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $$

= $$\sqrt {{5 \over 9}} $$

Given, $$\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B $$

$$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 2 $$-$$ 1

$$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 1 . . . . .(1)

also given

$$\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B $$

$$ \Rightarrow $$ $$\,\,\,\,$$ 2a $$-$$ b = 1 . . . . (2)

Solving (1) and (2), we get,

a = $${1 \over 3}$$ and b = $${2 \over 3}$$

$$\therefore\,\,\,\,$$ $$\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$$

$$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $$

= $$\sqrt {{5 \over 9}} $$

2

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 :

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

The general equation for its path is :

A

y = x^{2} + constant

B

y^{2} = x + constant

C

y^{2} = x^{2} + constant

D

xy = constant

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,

v_{y} = $$\,{{dy} \over {dt}}$$ = 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,

y^{2} = x^{2} + constant.

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

$$ \therefore $$ Velocity in x direction,

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

Velocity in y direction,

v

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

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

$$ \Rightarrow $$ 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,

y

3

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 a_{1} and a_{2} 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

$${{{a_1} + {a_2}} \over 2}t$$

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 t_{1} and for car B is t_{2}.

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

According to the question,

t_{2} $$-$$ t_{1} = t

and $$v$$_{1} $$-$$ $$v$$_{2} = $$v$$

$$ \Rightarrow $$ $$a$$_{1}t_{1} $$-$$ $$a$$_{2}t_{2} = $$v$$

$$ \Rightarrow $$ $$a$$_{1}t_{1} $$-$$ $$a$$_{2}(t + t_{1}) = $$v$$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So, x_{A} = x_{B}

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

$$ \Rightarrow $$ $$a$$_{1}t$$_1^2$$ = $$a$$_{2} (t + t_{1})^{2}

$$ \Rightarrow $$ $$\sqrt {{a_1}} .{t_1}$$ = $$\sqrt {{a_2}} $$ . (t + t_{1})

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

$$ \Rightarrow $$ t_{1} = $${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)$$

Now put the value of t_{1} in equation (2),

($$a$$_{1} $$-$$ $$a$$_{2}) t_{1} $$-$$ $$a$$_{2}t = $$v$$

$$ \Rightarrow $$ (a_{1} $$-$$ a_{2}) . $${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$$

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

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

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

Let the acceleration of car A and car B is $$a$$

Also let the time taken to reach the finishing point for car A is t

Let at finishing point speed of car A is $$v$$

According to the question,

t

and $$v$$

$$ \Rightarrow $$ $$a$$

$$ \Rightarrow $$ $$a$$

As, Total distance covered by both car is equal.

So, x

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

$$ \Rightarrow $$ $$a$$

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

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

$$ \Rightarrow $$ t

Now put the value of t

($$a$$

$$ \Rightarrow $$ (a

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

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

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

4

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 :

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$$\omega $$

Given that,

x = a cos $$\omega $$t

y = a sin $$\omega $$t

z = a $$\omega $$t

Velocity in x-direction,

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

Velocity in y-direction,

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

Velocity in z-direction,

V_{z}_{} = $${{dz} \over {dt}}$$ = a$$\omega $$

Net velocity,

$$\overrightarrow V $$ = V_{x}$$\widehat i$$ + V_{y}$$\widehat j$$ + V_{z}$$\widehat k$$

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

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

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

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

$$ = \sqrt 2 a\omega $$

x = a cos $$\omega $$t

y = a sin $$\omega $$t

z = a $$\omega $$t

Velocity in x-direction,

V

Velocity in y-direction,

V

Velocity in z-direction,

V

Net velocity,

$$\overrightarrow V $$ = V

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

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

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

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

$$ = \sqrt 2 a\omega $$

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (3) *keyboard_arrow_right*

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Simple Harmonic Motion *keyboard_arrow_right*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*