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

The angle of elevation of the top of a vertical tower from a point A, due east of it is 45^{o}. The angle of elevation of the top of the same tower from a point B, due south of A is 30^{o}. If the distance between A and B is $$54\sqrt 2 \,m,$$ then the height of the tower (in metres), is :

A

$$36\sqrt 3 $$

B

54

C

$$54\sqrt 3 $$

D

108

Let the height of tower = h

In triangle PQA,

tan45

$$ \Rightarrow $$ h = QA

In triangle PQB,

tan30

$$ \Rightarrow $$ BQ = $$\sqrt 3 $$h

In triangle BAQ

$$\angle $$ QAB = 90

$$ \therefore $$ QA

$$ \Rightarrow $$ h

$$ \Rightarrow $$ 2h

$$ \Rightarrow $$ $$\sqrt 2 h$$ = 54$$\sqrt 2 $$

$$ \Rightarrow $$ h = 54 m

2

An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt 3 $$ kmabove it, is obsered at an elevation of $${60^o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $${30^o}$$, then the speed (in km / hr) of the aeroplane, is :

A

1500

B

1440

C

750

D

720

For $$\Delta $$OA, A, OA_{1} = $${{\sqrt 3 } \over {\tan {{60}^o}}}$$ = 1 km

For $$\Delta $$OB_{1}, B, OB_{1} = $${{\sqrt 3 } \over {\tan {{30}^o}}}$$ = 3km

As, a distance of 3 $$-$$ 1 = 2 km is convered in 5 seconds.

Therefore the speed of the plane is

$${{2 \times 3600} \over 5}$$ = 1440 km/hr

For $$\Delta $$OB

As, a distance of 3 $$-$$ 1 = 2 km is convered in 5 seconds.

Therefore the speed of the plane is

$${{2 \times 3600} \over 5}$$ = 1440 km/hr

3

A tower T_{1} of height 60 m is located exactly opposite to a tower T_{2} of height 80 m on a straight road. Fromthe top of T_{1}, if the angle of depression of the foot of T_{2} is twice the angle of elevation of the top of T_{2}, then the width (in m) of the road between the feetof the towers T_{1} and T_{2} is :

A

$$10\sqrt 2 $$

B

$$10\sqrt 3 $$

C

$$20\sqrt 3 $$

D

$$20\sqrt 2 $$

Let the distance between T_{1} and T_{2} be x

From the figure

EA = 60 m (T_{1}) and DB = 80 m (T_{2})

$$\angle DEC = \theta $$ and $$\angle BEC = 2\theta $$

Now in $$\angle DEC$$,

$$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$$

and in $$\Delta BEC$$,

$$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$$

We know that

$$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$$

$$ \Rightarrow $$ $${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$$

$$ \Rightarrow $$ $${x^2} = 1200$$ $$ \Rightarrow $$ $$x = 20\sqrt 3 $$

From the figure

EA = 60 m (T

$$\angle DEC = \theta $$ and $$\angle BEC = 2\theta $$

Now in $$\angle DEC$$,

$$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$$

and in $$\Delta BEC$$,

$$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$$

We know that

$$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$$

$$ \Rightarrow $$ $${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$$

$$ \Rightarrow $$ $${x^2} = 1200$$ $$ \Rightarrow $$ $$x = 20\sqrt 3 $$

4

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30^{o} to 45^{o} ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

A

$$9\left( {1 + \sqrt 3 } \right)$$

B

$$18\left( {1 + \sqrt 3 } \right)$$

C

$$18\left( {\sqrt 3 - 1} \right)$$

D

$${9 \over 2}\left( {\sqrt 3 - 1} \right)$$

Assume height of tower = AB = h

From $$\Delta $$ABD,

tan45

$$ \Rightarrow $$$$\,\,\,$$ $${{AB} \over {AD}} = 1$$ [ as tan45

$$ \Rightarrow $$ $$\,\,\,$$ AB = AD

$$\therefore\,\,\,$$ AD = h

From $$\Delta $$BAC,

tan30

$$ \Rightarrow $$ $$\,\,\,$$ $${h \over {AC}} = {1 \over {\sqrt 3 }}$$

$$ \Rightarrow $$ $$\,\,\,$$ AC = h $$\sqrt 3 $$

As, AC = AD + DC

$$ \Rightarrow $$$$\,\,\,$$ DC = AC $$-$$ AD = $$\sqrt 3 h$$ $$-$$ h

Given that, time taken to reach from point C to D = 18 min.

$$\therefore\,\,\,$$ Car speed = $${{distance} \over {time}}$$ = $${{CD} \over {18}}$$ = $${{(\sqrt 3 - 1)h} \over {18}}$$

$$\therefore\,\,\,$$ Time taken to move from D to A

= $${{Distan ce\,\,of\,\,DA} \over {speed}}$$

= $${h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}$$

= $${{18} \over {\left( {\sqrt 3 - 1} \right)}}$$

= $${{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}$$

= 9 $$\left( {\sqrt 3 + 1} \right)$$ min.

Number in Brackets after Paper Name Indicates No of Questions

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