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

An oscillator of mass M is at rest in its equilibrium position in a potential

V = $${1 \over 2}$$ k(x $$-$$ X)^{2}. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)

V = $${1 \over 2}$$ k(x $$-$$ X)

A

$${1 \over {\sqrt 3 }}$$

B

$${1 \over 2}$$

C

$${2 \over 3}$$

D

$${3 \over {\sqrt 5 }}$$

2

A particle executes simple harmonic motion and is located at x = a, b and c at times t_{0}, 2t_{0} and 3t_{0} respectively. The freqquency of the oscillation is :

A

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$

B

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$

C

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$

D

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$

In general equation of simple harmonic motion, y = A sin $$\omega $$t

$$\therefore\,\,\,$$ a = A sin $$\omega $$t_{0}

$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t_{0}

$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t_{0}

a + c = A[sin $$\omega $$t_{0} + sin 3$$\omega $$t_{0}]

= 2A sin 2$$\omega $$t_{0} cos$$\omega $$t_{0}

$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t_{0}

$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t_{0}

$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos^{$$-$$1} $$\left( {{{a + c} \over {2b}}} \right)$$

$$\therefore\,\,\,$$ f = $${\omega \over {2\pi }}$$

= $${1 \over {2\pi {t_0}}}$$ cos^{$$-$$1} $$\left( {{{a + c} \over {2b}}} \right)$$

$$\therefore\,\,\,$$ a = A sin $$\omega $$t

$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t

$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t

a + c = A[sin $$\omega $$t

= 2A sin 2$$\omega $$t

$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t

$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t

$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos

$$\therefore\,\,\,$$ f = $${\omega \over {2\pi }}$$

= $${1 \over {2\pi {t_0}}}$$ cos

3

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :

A

$${A \over 2}$$

B

$${A \over {2\sqrt 2 }}$$

C

$${A \over {\sqrt 2 }}$$

D

A

Total energy of particle = $${1 \over 2}k{A^2}$$

Potential energy (v) = $${1 \over 2}$$ kx^{2}

Kinetic energy (K) = $${1 \over 2}$$ kA^{2} $$-$$ $${1 \over 2}$$kx^{2}

According to the question,

Potential energy = Kinetic energy

$$ \therefore $$ $${1 \over 2}$$kx^{2} = $${1 \over 2}$$kA^{2} $$-$$ $${1 \over 2}$$ kx^{2}

$$ \Rightarrow $$ kx^{2} = $${1 \over 2}$$ kA^{2}

$$ \Rightarrow $$ x = $$ \pm $$ $${A \over {\sqrt 2 }}$$

Potential energy (v) = $${1 \over 2}$$ kx

Kinetic energy (K) = $${1 \over 2}$$ kA

According to the question,

Potential energy = Kinetic energy

$$ \therefore $$ $${1 \over 2}$$kx

$$ \Rightarrow $$ kx

$$ \Rightarrow $$ x = $$ \pm $$ $${A \over {\sqrt 2 }}$$

4

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)

A

4

B

7

C

6

D

5

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency.

$$ \therefore $$ (2n + 1) f_{0} $$ \le $$ 20,000

(f_{0} is fundamental frequency = 1.5 KHz)

$$ \therefore $$ n = 6

$$ \therefore $$ Total number of overtone that can be heared is 7. (0 to 6)

$$ \therefore $$ (2n + 1) f

(f

$$ \therefore $$ n = 6

$$ \therefore $$ Total number of overtone that can be heared is 7. (0 to 6)

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*