Find WBJEE 2015 Solved Physics Question Paper – Part 4 in this article. This paper consists of 5 questions (#16 to #20) from WBJEE 2015 Physics paper. Detailed solution of these questions has been provided so that students can match their solutions.
Importance of Previous Years’ Paper:
Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.
About WBJEE Exam
WBJEE is a common entrance examinations held at state level for admission to the Undergraduate Level Engineering and Medical courses in the State of West Bengal. The Physics section of WBJEE 2015 engineering entrance exam consists of 40 questions.
16. A ray of light is incident at an angle i on a glass slab of refractive index μ. The angle between reflected and refracted light is 90. Then the relationship between i and μ is
By Snell’s law:
17. Two particles A and B are moving as shown in the figure. Their total angular momentum about the point O is
(A) 9.8 kg m2/s
(C) 52.7 kg m2/s
(D) 37.9 kg m2/s
Ans : (A)
Angular momentum(L) = mv × r
Now, since v and r are perpendicular to each other so,
mv × r = mvr sin90
L = (6.5) (2.2) (1.5)(-k) + (2.8) (3.1)(3.6)(k) = 9.8k kgm2/s
18. A 20 cm long capillary tube is dipped vertically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length of water column in the tube will be
(A) 5 cm
(B) 10 cm
(C) 15 cm
(D) 20 cm
Ans : (D)
Under free fall condition, geff = 0
In gravity less condition the entire tube fills with water.
So, the length of water column in the tube is 20 cm.
19. A train is moving with a uniform speed of 33 m/s and an observer is approaching the train with the same speed. If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s then the apparent frequency of the sound that the observer hears is
(A) 1220 Hz
(B) 1099 Hz
(C) 1110 Hz
(D) 1200 Hz
Ans : (A)
According to Doppler effect,
When both observer and source are moving, then apparent frequency is given as:
fapp = apparent frequency
v = velocity of sound in air
v0 = velocity of observer
vs = velocity of source
On putting the given values in above equation, we get
20. A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct? (Consider, Planck’s constant = 4 × 10–15 eVs, velocity of light = 3 × 108 m/s, ionization energy of hydrogen = 13.6eV)
(A) Electron will be knocked out of the atom
(B) Electron will go to any excited state of the atom
(C) Electron will go only to first excited state of the atom
(D) Electron will keep orbiting in the ground state of atom
Ans : (D)
The energy contained in the photon of given wavelength is given by,
Putting the values given in question,
λ = 300 × 10-9
h = 4 × 10–15 eVs
This energy is less than first excitation energy of hydrogen i.e. 10.2 eV.
So, electron will keep orbiting in its ground state.