# WBJEE 2015 Solved Physics Question Paper – Part 3

Find WBJEE Solved Physics Question Paper for the year 2015. This solved paper will help students in their final level of preparation for WBJEE Exam.

Created On: Feb 15, 2017 14:13 IST
Modified On: Feb 22, 2017 15:16 IST Find WBJEE 2015 Solved Physics Question Paper – Part 3 in this article. This paper consists of 5 questions (#11 to #15) 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.

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.

11. Assume that each diode shown in the figure has a forward bias resistance of 50Ω and an infinite reverse bias resistance. The current through the resistance 150 Ω is

(A) 0.66 A

(B) 0.05 A

(C) Zero

(D) 0.04 A

Ans : (D)

Sol.

Under the given condition, one of the diodes is reverse biased and other one is forward biased. In reverse biased mode the diode act as open circuit and in reverse biased mode the diode act as short circuit. So, the equivalent diagram is as below:

WBJEE 2016 Solved Physics and Chemistry Question Paper

12. The r.m.s speed of oxygen is v at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the r.m.s speed becomes

(A) v

(B) √2v

(C) 2v

(D) 4v

Ans : (C)

Sol.

The rms speed of any molecule is given as:

13. Two particles, A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R1 and R2 respectively. The ratio of the masses of A and B is

Ans: (C)

Sol.

The radius of circular path, r is given as,

14. A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the center of mass of the system from the origin is

(A)  = R

(B)  ≤ R

(C)  > R

(D) ≥ R

Ans : (B)

Sol.

If OAB is a sector of a circle, radius r, with angle 2α at the centre then the centre of mass lies on line of symmetry such that,

If, arc AB tends to zero then the centre of mass is at distance R from the origin. And as the arc length AB increases, centre of mass starts moving down.

15. A straight conductor 0.1m long moves in a uniform magnetic field 0.1T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The e.m.f. induced between the two ends of the conductor is

(A) 0.10 V

(B) 0.15 V

(C) 1.50 V

(D) 15.00 V

Ans : (B)

Sol.

The e.m.f. induced between the two ends of the conductor is given as:

E = Blv = 0.1×0.1×15 = 0.15V

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