NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIPMT 2008

If $$a$$ stands for the edge length of the cubic systems: simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively
A
$${1 \over 2}a:{{\sqrt 3 } \over 2}a:{{\sqrt 2 } \over 2}a$$
B
$$1a:\sqrt 3 a:\sqrt 2 a$$
C
$${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$$
D
$${1 \over 2}a:\sqrt 3 a:{1 \over {\sqrt 2 }}a$$

## Explanation

For Simple cubic : r+ + r = $${a \over 2}$$

For Body centred : r+ + r = $${{a\sqrt 2 } \over 4}$$

For Face centered: r+ + r = $${a \over {2\sqrt 2 }}$$

$$\therefore$$ Ratio of radii of the three will be

= $${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$$
2

### AIPMT 2008

With which one of the following elements silicon should be doped so as to give p-type of semiconductor?
A
Selenium
B
Boron
C
Germanium
D
Arsenic

## Explanation

The semiconductors formed by the introduction of impurity atoms containing one elecron less than the parent atoms of insulators are termed as p-type semiconductors. Therefore silicon containing 14 electrons has to be doped with boron containing 13 electrons to give a p-type semi-conductor.
3

### AIPMT 2007

If NaCl is doped with 10$$-$$4 mol% of SrCl2, the concentration of cation vacancies will be (NA = 6.02 $$\times$$ 1023 mol$$-$$1)
A
6.02 $$\times$$ 1016 mol$$-$$1
B
6.02 $$\times$$ 1017 mol$$-$$1
C
6.02 $$\times$$ 1014 mol$$-$$1
D
6.02 $$\times$$ 1015 mol$$-$$1

## Explanation

As each Sr2+ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl2 added.

$$\therefore$$ Concentration of cation vacancies

= 10–4 mole %

= $${{{{10}^{ - 4}}} \over {100}} \times 6.023 \times {10^{23}}$$

= $$6.023 \times {10^{23}}$$ $$\times$$ 10-6

= $$6.023 \times {10^{17}}$$
4

### AIPMT 2007

The fraction of total volume occupied by the atoms present in a simple cube is
A
$${\pi \over {3\sqrt 2 }}$$
B
$${\pi \over {4\sqrt 2 }}$$
C
$${\pi \over 4}$$
D
$${\pi \over 6}$$

## Explanation

The maximum properties of the available volume which may be filled by hard sphere in simple cubic arrangement is $${\pi \over 6}$$ or 0.52.

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12