JEE Main Test 82

Result

JEE Main Test 82

Score
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Rank
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Weekly Quiz Competition

Question 1
4 / -1

Two point white does are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [take wavelength of light = 500 nm]

A
5 m

B
1 m

C
6 m

D
3 m

Solution
Question 2
4 / -1

A

B

C

D

Solution
Question 3
4 / -1

The angle between two diagonals of a cube is

A
cos^-1 (1/3)

B
30^o

C
cos^-1 (1/√3)

D
45^o

Solution
Question 4
4 / -1

A

B

C

D
None of these

Solution
Question 5
4 / -1

A die is thrown. If A is the event that the number obtained is greater than 3 and B is the event that the number obtained is less than 5. Then, P(A ∪ B) is

A
2/5

B
3/5

C
0

D
1

Solution
Question 6
4 / -1

A p-n junction has acceptor impurity concentration of 17 cm^-3 in the P side and donor impurity concentration of 10¹⁶ cm^-3 in the N side. What is the contact potential at the junction? (kT = thermal energy, intrinsic carrier concentration n_i = 1.4 × 10¹⁶ cm^-3 )

A

B

C

D

Solution
Question 7
4 / -1

The ionisation potential of hydrogen atom is –13.6 eV. An electron in the ground state of a hydrogen atoms absorbs a photon of energy 1275 eV. How many different spectral line can one expect when the electron make a downward transition?

A
1

B
4

C
2

D
6

Solution

Photon of energy = 12.75 eV

i.e., (-13.6+12.75)eV = -0.85eV

The photon of energy 12.25eV can excite the fourth level of H-atom.
Question 8
4 / -1

The remainder obtained when 1! + 2! + 3 ! + ..... + 11 ! is divided by 12 is

A
9

B
8

C
7

D
6

Solution
Question 9
4 / -1

The relation between half - life (T) and decay constant (λ) is

A
λT = 1

B
λT = 1/2

C
λT = log_e 2

D
λ = log.2T

Solution
Question 10
4 / -1

A polarised light of intensity I₀ is passed through another polariser whose pass axis makes an angle 60^o with the pass axis of the former. What is the intensity of emergent polarised light from second polariser?

A
I = I₀

B
I = I₀ / 6

C
I = I₀ / 5

D
I = I₀ / 4

Solution