AIPMT Examination
AIPMT Solved / Unsolved Papers
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AIPMT 2007 Physics Screening
Q. 41. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio 
will be proportional to
- R
- 1/R
- 1/R2
- R2
Sol . Radius of path 
Correct choice: (3)
Q. 42. If the nucleus 
13 has a nuclear radius of about 3.6 fm, then 
Te
would have its radius approximately as
- 4.8 fm
- 6.0 fm
- 9.6 fm
- 12.0 fm
Sol. 
Correct choice: (2)
Q. 43. In radioactive decay process, the negatively charged emitted 
particles are
- the electrons orbiting around the nucleus
- the electrons present inside the nucleus
- the electrons produced as a result of the decay of neutrons inside the nucleus
- the electrons produced as a result of collisions between atoms
Sol . When neutron decays as given 
Correct choice: (3)
Q. 44. A nucleus 
has nucleus A X
Z has mass represented by M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and B.E the binding energy in Me V then
Sol. 
Correct choice: (3)
Q. 45. Two radioactive substances A and B have decay constants 
respectively. At t = 0 they have the same number of
nuclei. The ratio of number of nuclei of A to those of B will be 
after a time interval
Sol. 
Correct choice: (1)
Q. 46. The total energy of electron in the ground state of hydrogen atom is –13.6 eV. The kinetic energy of an electron in the firstm excited state is
- 1.7 eV
- 3.4 eV
- 6.8 eV
- 13.6 eV
Sol. Total energy in 1st excited state 
Correct choice: (2)
Q. 47. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is
- an n-type semiconductor
- a p-type semiconductor
- an insulator
- a metal
Sol. In diagram acceptor level is near valance band which will happen in p-type semiconductor. Correct choice: (2)
Q. 48. A common emitter amplifier has a voltage gain of 50, an input impedance of 
and an output impedance of 
. The
power gain of the amplifier is
- 100
- 500
- 1000
- 1250
Sol. 
Correct choice: (4)
Q. 49. In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table
| (i) | A | B | Y |
| 0 | 0 | 0 | |
| 0 | 1 | 1 | |
| 1 | 1 | 1 |
| (ii) | A | B | Y |
| 0 | 1 | 0 | |
| 1 | 0 | 0 | |
| 1 | 1 | 1 |
| (iii) | A | B | Y |
| 0 | 0 | 1 | |
| 0 | 1 | 1 | |
| 1 | 0 | 1 | |
| 1 | 1 | 0 |
| (iv) | A | B | Y |
| 0 | 0 | 1 | |
| 0 | 1 | 0 | |
| 1 | 0 | 0 | |
| 1 | 1 | 0 |
Q. 50. For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct?
Sol. the cubic system is the most symmetric out of all seven crystal systems. All edges for the unit cell are same i.e., a = b = c and all angles are right angles i.e., 
Correct choice: (1)