1
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

To get an output Y = 1 in given circuit which of the following input will be correct ?

A
A $$ \to $$ 1,  B $$ \to $$ 0,  C $$ \to $$ 0
B
A $$ \to $$ 1,  B $$ \to $$ 0,  C $$ \to $$ 1
C
A $$ \to $$ 1,  B $$ \to $$ 1,  C $$ \to $$ 0
D
A $$ \to $$ 0,  B $$ \to $$ 1,  C $$ \to $$ 0

Explanation



The Boolen expression for the given combination is

output Y = (A + B) ยท C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1
2
MCQ (Single Correct Answer)

AIPMT 2009

Sodium has body centred packing. Distance between two nearest atoms is 3.7 $$\mathop A\limits^ \circ $$. The lattice parameter is
A
4.3 $$\mathop A\limits^ \circ $$
B
3.0 $$\mathop A\limits^ \circ $$
C
8.6 $$\mathop A\limits^ \circ $$
D
6.8 $$\mathop A\limits^ \circ $$

Explanation

Distance between nearest atoms in body centred cubic lattice (bcc),

d = $${{\sqrt 3 } \over 2}a$$

Given d = 3.7 $$\mathop A\limits^ \circ $$

$$ \therefore $$ $$a = {{3.7 \times 2} \over {\sqrt 3 }}$$ = 4.3 $$\mathop A\limits^ \circ $$
3
MCQ (Single Correct Answer)

AIPMT 2009

A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
A
4000 nm
B
6000 nm
C
4000 $$\mathop A\limits^ \circ $$
D
6000 $$\mathop A\limits^ \circ $$

Explanation

$$\lambda $$max = $${{hc} \over E}$$

= $${{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2.5 \times 1.6 \times {{10}^{ - 19}}}}$$

= 5000 $$\mathop A\limits^o $$

Now the wavelength detected by photodiode be less than $$\lambda $$max, hence it can detect a signal of wave length 4000 $$\mathop A\limits^o $$.
4
MCQ (Single Correct Answer)

AIPMT 2009

A transistor is operated in common-emitter configuration at VC = 2V such that a change in the base current from 100 $$\mu $$A to 200 $$\mu $$A produces a change in the collector current from 5 mA to 10 mA. The current gain is
A
100
B
150
C
50
D
75

Explanation

Current gain, $$\beta $$ = $${{\Delta {I_C}} \over {\Delta {I_B}}}$$

= $${{\left( {10 - 5} \right) \times {{10}^{ - 3}}} \over {\left( {200 - 100} \right) \times {{10}^{ - 6}}}}$$ = 50

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