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

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

2

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)

3

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $$\omega $$. If the radius of the bottle is 2.5 cm then $$\omega $$ is close to – (density of water = 10^{3} kg/m^{3}).

A

2.50 rad s^{$$-$$1}

B

3.75 rad s^{$$-$$1}

C

5.00 rad s^{$$-$$1}

D

7.90 rad s^{$$-$$1}

Restoring force due to pressing the bottle with small
amount x,

F = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$

$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$$

$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}} $$ = 7.90 rad/s

F = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$

$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$$

$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}} $$ = 7.90 rad/s

4

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

A

$${{4\pi } \over 3}$$

B

$${3 \over 8}\pi $$

C

$${7 \over 3}\pi $$

D

$${{8\pi } \over 3}$$

$$v = \omega \sqrt {{A^2} - {x^2}} \,\,$$ . . .(1)

$$a = - {\omega ^2}x$$ . . .(2)

$$\left| v \right| = \left| a \right|$$ . . .(3)

$$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$$

$${A^2} - {x^2} = {\omega ^2}{x^2}$$

$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$$

$$ \Rightarrow \,\,\,3 = \omega \times 4$$

$$T = 2\pi /\omega $$

$$a = - {\omega ^2}x$$ . . .(2)

$$\left| v \right| = \left| a \right|$$ . . .(3)

$$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$$

$${A^2} - {x^2} = {\omega ^2}{x^2}$$

$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$$

$$ \Rightarrow \,\,\,3 = \omega \times 4$$

$$T = 2\pi /\omega $$

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*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*