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 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

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

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

2

A body of mass m starts moving from rest along x-axis so that its velocity varies as $$\upsilon = a\sqrt s $$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :

A

$${1 \over 8}\,$$ m a^{4} t^{2}

B

8 m a^{4} t^{2}

C

4 m a^{4} t^{2}

D

$${1 \over 4}\,$$ m a^{4} t^{2}

Given,

$$\upsilon $$ = a $$\sqrt s $$

$$ \Rightarrow $$$$\,\,\,$$ $${{ds} \over {dt}} = a\sqrt s $$

$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt} $$

$$ \Rightarrow $$$$\,\,\,$$ 2$$\sqrt s $$ = at

$$ \Rightarrow $$$$\,\,\,$$ s = $${{{a^2}{t^2}} \over 4}$$

= $${1 \over 2}.{{{a^2}} \over 2}.{t^2}$$

$$\therefore\,\,\,\,$$ acceleration = $${{{a^2}} \over 2}$$

$$\therefore\,\,\,$$ Force (F) = m $$ \times $$ $${{{a^2}} \over 2}$$

$$\therefore\,\,\,\,$$ Work done = F. S

= $${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$$

= $${{m{a^4}{t^2}} \over 8}$$

$$\upsilon $$ = a $$\sqrt s $$

$$ \Rightarrow $$$$\,\,\,$$ $${{ds} \over {dt}} = a\sqrt s $$

$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt} $$

$$ \Rightarrow $$$$\,\,\,$$ 2$$\sqrt s $$ = at

$$ \Rightarrow $$$$\,\,\,$$ s = $${{{a^2}{t^2}} \over 4}$$

= $${1 \over 2}.{{{a^2}} \over 2}.{t^2}$$

$$\therefore\,\,\,\,$$ acceleration = $${{{a^2}} \over 2}$$

$$\therefore\,\,\,$$ Force (F) = m $$ \times $$ $${{{a^2}} \over 2}$$

$$\therefore\,\,\,\,$$ Work done = F. S

= $${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$$

= $${{m{a^4}{t^2}} \over 8}$$

3

A block of mass m, lying on a smooth horizontal surface, is attached to a sring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is :

A

$${{2F} \over {\sqrt {mk} }}$$

B

$${F \over {\pi \sqrt {mk} }}$$

C

$${{\pi F} \over {\sqrt {mk} }}$$

D

$${F \over {\sqrt {mk} }}$$

When block is pulled x distance with F force, then it is distance with F force, then it is at maximum position

$$ \therefore $$ F = Kx

$$ \Rightarrow $$ x = $${F \over K}$$

Applying work energy theorem, work done by all the forces = change in Kinetic energy.

$$ \Rightarrow $$ W

$$ \Rightarrow $$ Fx $$-$$ $${1 \over 2}$$Kx

$$ \Rightarrow $$ $$F$$$$\left( {{F \over K}} \right) - {1 \over 2}K{\left( {{F \over K}} \right)^2}$$ = $${1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $${{{F^2}} \over K} - {{F{}^2} \over {2K}}$$ = $${1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $${{F{}^2} \over {2K}}$$ = $${1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $$v$$ = $${F \over {\sqrt {mK} }}$$

4

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed $$\upsilon $$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically $${5 \over 6}$$th of the initial kinetic energy is lost in whole process. What is value of M/m ?

A

5

B

2

C

4

D

3

As in elastic or in elastic clollision, momentum is conserved.

$$ \therefore $$ P_{i} = P_{f}

P_{i} = Initial momentum

P_{f} = Final Momentum

mv = (2m + m) V_{f}

$$ \Rightarrow $$ V_{f} = $${{mv} \over {2m + M}}$$

Here due to collision $${5 \over 6}$$th of kinetic energy is lost.

$$ \therefore $$ Remaining kinetic energy,

K_{f} = $${1 \over 6}$$ Ki

$$ \Rightarrow $$ $${1 \over 2}$$(2m + M) $$ \times $$ $${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$$ = $${1 \over 6} \times {1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $${{{m^2}{v^2}} \over {2m + M}}$$ = $${1 \over 6}m{v^2}$$

$$ \Rightarrow $$ $${m \over {2m + M}}$$ = $${1 \over 6}$$

$$ \Rightarrow $$ 6m = 2m + M

$$ \Rightarrow $$ M = 4m

$$ \Rightarrow $$ $${M \over m}$$ = 4

$$ \therefore $$ P

P

P

mv = (2m + m) V

$$ \Rightarrow $$ V

Here due to collision $${5 \over 6}$$th of kinetic energy is lost.

$$ \therefore $$ Remaining kinetic energy,

K

$$ \Rightarrow $$ $${1 \over 2}$$(2m + M) $$ \times $$ $${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$$ = $${1 \over 6} \times {1 \over 2}m{v^2}$$

$$ \Rightarrow $$ $${{{m^2}{v^2}} \over {2m + M}}$$ = $${1 \over 6}m{v^2}$$

$$ \Rightarrow $$ $${m \over {2m + M}}$$ = $${1 \over 6}$$

$$ \Rightarrow $$ 6m = 2m + M

$$ \Rightarrow $$ M = 4m

$$ \Rightarrow $$ $${M \over m}$$ = 4

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Ray & Wave Optics *keyboard_arrow_right*

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