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1.

Match list I (electromagnetic wave type)  with List II (Its association/application  ) and select  the correct option from the choices given below the lists

A) A:4,B:3,C:2,D:1

B) A:1,B:2,C:4,D:3

C) A:3,B:2,C:1,D:4

D) A:1,B:2,C:3, D:4

2.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

A) A meter scale

B) Avernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm.

C) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

D) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm

3.

Hydrogen (1H1), deuterium (1H2), singly ionised helium   $(_{2}He^{4})^{+}$   and doubly ionised lithium$(_{3}Li^{8})^{++}$ all have one electron around the nucleus. Consider an electron transition from n=2 to n=1 . If the wavelength s of emitted radiation are   $\lambda_{1},\lambda_{2},\lambda_{3}$ and   $\lambda_{4}$, respectively for four elements, then approximately which one of the following is correct?

A) $4\lambda_{1}=2\lambda_{2}=2 \lambda_{3}=\lambda_{4}$

B) $\lambda_{1}=2\lambda_{2}=2 \lambda_{3}=\lambda_{4}$

C) $\lambda_{1}=\lambda_{2}=4\lambda_{3}=9\lambda_{4}$

D) $\lambda_{1}=2\lambda_{2}=3\lambda_{3}=4\lambda_{4}$

4.

The radiation corresponding to 3→ 2 transition of hydrogen atom falls on a metal surface to produce photo electrons . These electrons are made toi enter a magnetic field of  $3\times 10^{-4}T$ . If the radius of the largest circular path followed by these electrons is 10.0mm. the work function of the metal is close to

A) 1.8eV

B) 1.1eV

C) 0.8 eV

D) 1.6 eV

5.

Two beams, A and B , of plane  polarised light with mutually perpendicular planes of polarisation are seen through a polarised. From the position  when the beam  A has  maximum intensity (and beam B has  zero intensity)  , a rotation of polarised through 30° makes the two beams appear equally bright. If the initial intensities  of the  two beams are IA and IB  respectively, then IA/IB equals

A) 3

B) 3/2

C) 1

D) 1/3

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