Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

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

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

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

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

2

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

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

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

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

$$ = \sqrt 2 a\omega $$

3

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

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

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

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

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

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

4

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

$${1 \over 2}a\left( {\widehat i - \widehat k} \right)$$

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

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

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

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

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

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Atoms and Nuclei *keyboard_arrow_right*

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