Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :

A

K_{p} < K_{e} and P_{p} = P_{e}

B

K_{p} = K_{e} and P_{p} = P_{e}

C

K_{p} < K_{e} an P_{p} < P_{e}

D

K_{p} > K_{e} and P_{p} = P_{e}

$${\lambda _p} = {h \over {{P_p}}}$$

$${\lambda _e} = {h \over {{P_e}}}$$

$$\because$$ $${\lambda _p} = {\lambda _e}$$

$$ \Rightarrow {P_p} = {P_e}$$

$${(K)_p} = {{P_p^2} \over {2{m_p}}}$$

$${(K)_e} = {{P_e^2} \over {2{m_e}}}$$

K_{p} < K_{e} as m_{p} > m_{e}

Option (a)

$${\lambda _e} = {h \over {{P_e}}}$$

$$\because$$ $${\lambda _p} = {\lambda _e}$$

$$ \Rightarrow {P_p} = {P_e}$$

$${(K)_p} = {{P_p^2} \over {2{m_p}}}$$

$${(K)_e} = {{P_e^2} \over {2{m_e}}}$$

K

Option (a)

2

MCQ (Single Correct Answer)

A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B versus time is :

(Assume that at t = 0, there are no B atoms in the sample)

(Assume that at t = 0, there are no B atoms in the sample)

A

B

C

D

A $$\to$$ B $$\to$$ C (stable)

Initially no. of atoms of B = 0 at t = 0, no. of atoms of B will starts increasing & reaches maximum value when rate of decay of B = rate of formation of B.

After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions, so best possible graph is (d).

Initially no. of atoms of B = 0 at t = 0, no. of atoms of B will starts increasing & reaches maximum value when rate of decay of B = rate of formation of B.

After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions, so best possible graph is (d).

3

MCQ (Single Correct Answer)

There are 10^{10} radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds? $$\left( {\sqrt 2 = 1.414} \right)$$

A

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

B

7 $$\times$$ 10^{9}

C

10^{5}

D

4 $$\times$$ 10^{10}

$${N \over {{N_0}}} = {\left( {{1 \over 2}} \right)^{{t \over {{t^{1/2}}}}}}$$

$${N \over {{{10}^{10}}}} = {\left( {{1 \over 2}} \right)^{{{30} \over {60}}}}$$

$$ \Rightarrow N = {10^{10}} \times {\left( {{1 \over 2}} \right)^{{1 \over 2}}} = {{{{10}^{10}}} \over {\sqrt 2 }} \approx 7 \times {10^9}$$

$${N \over {{{10}^{10}}}} = {\left( {{1 \over 2}} \right)^{{{30} \over {60}}}}$$

$$ \Rightarrow N = {10^{10}} \times {\left( {{1 \over 2}} \right)^{{1 \over 2}}} = {{{{10}^{10}}} \over {\sqrt 2 }} \approx 7 \times {10^9}$$

4

MCQ (Single Correct Answer)

At time t = 0, a material is composed of two radioactive atoms A and B, where N_{A}(0) = 2N_{B}(0). The decay constant of both kind of radioactive atoms is $$\lambda$$. However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of N_{B}(t) / N_{B}(0) with respect to time t?

[N_{A} (0) = No. of A atoms at t = 0

N_{B} (0) = No. of B atoms at t = 0]

[N

N

A

B

C

D

A $$\to$$ B, B $$\to$$ C

$${{d{N_B}} \over {dt}} = \lambda {N_A} - \lambda {N_B}$$

$${{d{N_B}} \over {dt}} = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} - \lambda {N_B}$$

$${e^{ - \lambda t}}\left( {{{d{N_B}} \over {dt}} + \lambda {N_B}} \right) = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} \times {e^{\lambda t}}$$

$${d \over {dt}}({N_B}{e^{\lambda t}}) = 2\lambda {N_{{B_0}}}$$, on integrating

$${N_B}{e^{\lambda t}} = 2\lambda t{N_{{B_0}}} + {N_{{B_0}}}$$

$${N_B} = {N_{{B_0}}}[1 + 2\lambda t]{e^{ - \lambda t}}$$

$${{d{N_B}} \over {dt}} = 0$$ at $$ - \lambda [1 + 2\lambda t){e^{ - \lambda t}} + 2\lambda {e^{ - \lambda t}} = 0$$

$${N_{{B_{\max }}}}$$ at $$t = {1 \over {2\lambda }}$$

$${{d{N_B}} \over {dt}} = \lambda {N_A} - \lambda {N_B}$$

$${{d{N_B}} \over {dt}} = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} - \lambda {N_B}$$

$${e^{ - \lambda t}}\left( {{{d{N_B}} \over {dt}} + \lambda {N_B}} \right) = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} \times {e^{\lambda t}}$$

$${d \over {dt}}({N_B}{e^{\lambda t}}) = 2\lambda {N_{{B_0}}}$$, on integrating

$${N_B}{e^{\lambda t}} = 2\lambda t{N_{{B_0}}} + {N_{{B_0}}}$$

$${N_B} = {N_{{B_0}}}[1 + 2\lambda t]{e^{ - \lambda t}}$$

$${{d{N_B}} \over {dt}} = 0$$ at $$ - \lambda [1 + 2\lambda t){e^{ - \lambda t}} + 2\lambda {e^{ - \lambda t}} = 0$$

$${N_{{B_{\max }}}}$$ at $$t = {1 \over {2\lambda }}$$

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Units & Measurements

Motion

Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Impulse & Momentum

Rotational Motion

Gravitation

Properties of Matter

Heat and Thermodynamics

Waves

Vector Algebra

Ray & Wave Optics

Electrostatics

Current Electricity

Magnetics

Alternating Current and Electromagnetic Induction

Atoms and Nuclei

Dual Nature of Radiation

Electronic Devices

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