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

Two particles A and B of equal mass M are moving with the same speed $$\upsilon $$ as shown in the figure. They collide completely inelastically and move as a single particle C. The angle $$\theta $$ that the path of C makes with the X-axis is given by :

A

tan $$\theta $$ = $${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$$

B

tan $$\theta $$ = $${{\sqrt 3 - \sqrt 2 } \over {1 - \sqrt 2 }}$$

C

tan $$\theta $$ = $${{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } \right)}}$$

D

tan $$\theta $$ = $${{1 - \sqrt 3 } \over {1 + \sqrt 2 }}$$

Using conservation of linear moments,

Along X-axis,

2MV'cos$$\theta $$ = MVsin30

Along Y - axis,

2Mv' sin$$\theta $$ = Mv cos30

Dividing (2) by (1), we get,

$${{\sin \theta } \over {\cos \theta }}$$ = $${{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}$$

= $${{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}$$

$$\therefore\,\,\,$$ tan$$\theta $$ = $${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$$

2

In a collinear collision, a particle with an initial speed v_{0} strikes a stationary particle of the same mass. If
the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative
velocity between the two particles, after collision, is :

A

$${{{v_0}} \over {\sqrt 2 }}$$

B

$${{v_0}} \over 4$$

C

$$\sqrt 2 {v_0}$$

D

$${{v_0}} \over 2$$

From conservation of linear momentum,

mv_{0} = mv_{1} + mv_{2}

or v_{0} = v_{1} + v_{2} ........(1)

According to the question,

K_{f} = $${3 \over 2}$$K_{i}

$$ \Rightarrow $$ $${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$$

$$ \Rightarrow $$ $$v_1^2 + v_2^2 = {3 \over 2}v_0^2$$

Using eq (1) $${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$$

$$ \Rightarrow $$ $$v_1^2 + v_2^2 + 2{v_1}{v_2}$$ = $$v_0^2$$

$$ \Rightarrow $$ $$2{v_1}{v_2}$$ = $$v_0^2 - {3 \over 2}v_0^2$$ = $$ - {1 \over 2}v_0^2$$

Now, $${\left( {{v_1} - {v_2}} \right)^2}$$ = $${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$$

= $$v_0^2 - \left( { - v_0^2} \right)$$ = $$2v_0^2$$

$$\therefore$$ $${{v_1} - {v_2}}$$ = $$\sqrt 2 {v_0}$$

mv

or v

According to the question,

K

$$ \Rightarrow $$ $${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$$

$$ \Rightarrow $$ $$v_1^2 + v_2^2 = {3 \over 2}v_0^2$$

Using eq (1) $${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$$

$$ \Rightarrow $$ $$v_1^2 + v_2^2 + 2{v_1}{v_2}$$ = $$v_0^2$$

$$ \Rightarrow $$ $$2{v_1}{v_2}$$ = $$v_0^2 - {3 \over 2}v_0^2$$ = $$ - {1 \over 2}v_0^2$$

Now, $${\left( {{v_1} - {v_2}} \right)^2}$$ = $${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$$

= $$v_0^2 - \left( { - v_0^2} \right)$$ = $$2v_0^2$$

$$\therefore$$ $${{v_1} - {v_2}}$$ = $$\sqrt 2 {v_0}$$

3

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90^{o} with respect to each other. The mass of unknown particle is :

A

$${m \over 2}$$

B

m

C

$${m \over {\sqrt 3 }}$$

D

2 m

4

Two particles of the same mass m are moving in circular orbits because of force, given by $$F\left( r \right) = {{ - 16} \over r} - {r^3}$$

The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

A

$$6 \times {10^{ - 2}}$$

B

$$3 \times {10^{ - 3}}$$

C

$${10^{ - 1}}$$

D

$$6 \times {10^{ 2}}$$

In circular motion the force required

$$\left| F \right| = {{m{v^2}} \over r}$$

$$\therefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$

$$ \Rightarrow $$ mv^{2} = 16 + r^{4}

$$\therefore\,\,\,$$ kinetic energy (K) = $${1 \over 2}$$ mv^{2} = $${1 \over 2}$$ [ 16 + r^{4}]

$$\therefore\,\,\,$$ Kinetic energy of first particle (K_{1}) = $${1 \over 2}$$ [16 + 1]

Kinetic energy of second particle (K_{2}) = $${1 \over 2}$$ [16 + 4^{4}]

$$\therefore\,\,\,\,$$ $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$$

$$\left| F \right| = {{m{v^2}} \over r}$$

$$\therefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$

$$ \Rightarrow $$ mv

$$\therefore\,\,\,$$ kinetic energy (K) = $${1 \over 2}$$ mv

$$\therefore\,\,\,$$ Kinetic energy of first particle (K

Kinetic energy of second particle (K

$$\therefore\,\,\,\,$$ $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$$

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (2) *keyboard_arrow_right*

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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*

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*

Ray & Wave Optics *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*