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Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

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Reinforced Cement Concrete

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Engineering Mathematics

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General Aptitude

1

Imagine that a reactor converts all given mass into energy and that it operates at a power level of 10^{9} watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is
3×10^{8} m/s)

A

0.96 gm

B

0.8 gm

C

4 $$ \times $$ 10^{$$-$$2} gm

D

6.6 $$ \times $$ 10^{$$-$$5} gm

2

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let
$${\lambda _n}$$, $${\lambda _g}$$ be the de Broglie wavelength of the electron in the n^{th} state and the ground state respectively. Let
$${\Lambda _n}$$ be the wavelength of the emitted photon in the transition from the n^{th} state to the ground state. For large n, (A, B are constants)

A

$${\Lambda _n} \approx A + {B \over {\lambda _n^2}}$$

B

$${\Lambda _n} \approx A + B{\lambda _n}$$

C

$$\Lambda _n^2 \approx A + B\lambda _n^2$$

D

$$\Lambda _n^2 \approx \lambda$$

We know,

Wavelength of emitted photon from n_{2} state to n_{1} state is

$${1 \over \lambda }$$ = RZ^{2} $$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$

Here electron comes from n^{th} state to ground state (n = 1),

then the wavelength of photon is ,

$${1 \over {{\Lambda _n}}}$$ = RZ^{2} $$\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$$

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$_{n} = $${1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}}$$

As n is very large, so using binomial theorem

$$\Lambda $$_{n} = $${1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} \right)$$

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$_{n} = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} \right)$$

We know,

$$\lambda $$_{n} = $${{2\pi r} \over n}$$

= 2$$\pi $$ $${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$$

$$\therefore\,\,\,$$ $$\lambda $$_{n} $$ \propto $$ n

$$ \Rightarrow $$$$\,\,\,$$ n = K $$\lambda $$_{n}

$$\therefore\,\,\,$$ $$\Lambda $$_{n} = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right)$$

Let A = $${1 \over {R{Z^2}}}$$ and B = $${1 \over {{K^2}R{Z^2}}}$$

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$_{n} = A + $${B \over {\lambda _n^2}}$$

Wavelength of emitted photon from n

$${1 \over \lambda }$$ = RZ

Here electron comes from n

then the wavelength of photon is ,

$${1 \over {{\Lambda _n}}}$$ = RZ

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$

As n is very large, so using binomial theorem

$$\Lambda $$

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$

We know,

$$\lambda $$

= 2$$\pi $$ $${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$$

$$\therefore\,\,\,$$ $$\lambda $$

$$ \Rightarrow $$$$\,\,\,$$ n = K $$\lambda $$

$$\therefore\,\,\,$$ $$\Lambda $$

Let A = $${1 \over {R{Z^2}}}$$ and B = $${1 \over {{K^2}R{Z^2}}}$$

$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$

3

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its
energy is p_{d}; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_{c}. The values of p_{d} and p_{c} are respectively :

A

(0, 1)

B

(0.89, 0.28)

C

(0.28, 0.89)

D

(0, 0)

Applying conservation of momentum :

mv + 0 = mv

$$ \Rightarrow $$$$\,\,\,$$ v = v

As collision is elastic,

So, coefficient of restitution, e = 1

$$\therefore\,\,\,$$ e = 1 = $${{velocity\,\,of\,\,separation} \over {Velocity\,\,of\,\,approach}}$$

$$ \Rightarrow $$$$\,\,\,$$ 1 = $${{{v_2} - {v_1}} \over {v - 0}}$$

$$ \Rightarrow $$$$\,\,\,$$ v = v

Add (1) and (2),

2v = 3v

$$ \Rightarrow $$$$\,\,\,$$ v

put value of v

v

= v $$-$$ $${{4v} \over 3}$$

= $$-$$ $${v \over 3}$$

$$\therefore\,\,\,$$ Fractional loss of energy of neutron.

P

= $${{{1 \over 2}m{v^2} - {1 \over 2}mv_1^2} \over {{1 \over 2}m{v^2}}}$$

= $${{{v^2} - {{{v^2}} \over 9}} \over {{v^2}}}$$

= $${8 \over 9}$$

= 0.89

Applying momentum of conservation,

mv + 0 = mv

$$ \Rightarrow $$$$\,\,\,$$ v = v

Here also e = 1

$$\therefore\,\,\,$$ e = 1 = $${{{v_2} - v{}_1} \over {v - 0}}$$

$$ \Rightarrow $$$$\,\,\,$$ v = v

adding (3) and (4), we get

2v = 13v

$$ \Rightarrow $$$$\,\,\,$$ v

put this v

v

= $$-$$ $${{11v} \over {13}}$$

$$\therefore\,\,\,$$ Frictional loss

p

= $${{48} \over {169}}$$

= 0.28

4

If the series limit frequency of the Lyman series is $${\nu _L}$$, then the series limit frequency of the Pfund series is:

A

$${\nu _L}/25$$

B

$$25{\nu _L}$$

C

$$16{\nu _L}$$

D

$${\nu _L}/16$$

(1) In Lyman Series, transition happens in n = 1 state

from n = 2, 3, . . . . . $$ \propto $$

(2) In Balmer Series, transition happens in n = 2 state

from n = 3, 4, . . . . . $$ \propto $$

(3) In Paschen Series, transition happens in n = 3 state

from n = 4, 5, . . . . . $$ \propto $$

(4) In Bracktt Series, transition happens in n = 4 state

from n = 5, 6 . . . . . . $$ \propto $$

(5) In Pfund Series, transition happens in n = 5 state

from n = 6, 7, . . . . $$ \propto $$

We know,

$${1 \over \lambda }$$ = RZ

Series limit means transition happens

from n = $$ \propto $$ to n = 1, for Lyman Series.

In series limit for Lyman series,

$${1 \over {{\lambda _L}}}$$ = RZ

$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {{\lambda _L}}}$$ = RZ

We know,

E = $${{hc} \over \lambda }$$ = h$$\gamma $$

$$ \Rightarrow $$$$\,\,\,$$ $$\gamma $$ = $${c \over \lambda }$$

So, frequency in Lyman Series,

$$\gamma $$

In Pfund series,

n

$$\therefore\,\,\,$$ $${1 \over {{\lambda _P}}}$$ = RZ

$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {{\lambda _P}}}$$ = $${{R{Z^2}} \over {25}}$$

$$\therefore\,\,\,$$ $${\gamma _P}$$ = $${c \over {{\lambda _P}}}$$ = c $$ \times $$ $${{R{Z^2}} \over {25}}$$

$$\therefore\,\,\,$$ $$\gamma $$

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