WBJEE 2015 Solved Physics Question Paper – Part 2

Find WBJEE 2015 Solved Physics Question Paper – Part 2 in this article. This paper consists of 5 questions (#6 to #10) 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.

6. The length of a metal wire is L1 when the tension is T1 and L2 when the tension is T2. The unstretched length of the wire is

Ans: (C)

Sol.

Let the unstretched length of the wire be l.

WBJEE 2016 Solved Physics and Chemistry Question Paper

7. The line AA’ is on a charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge σ and B is a ball of mass m with a like charge of magnitude q. B is connected by a string from a point on the line AA’. The tangent of the angle (θ) formed between the line AA’ and the string is

8. The current I in the circuit shown is

(A) 1.33 A

(B) Zero

(C)  2.00 A

(D) 1.00 A

Ans: (A)

Sol.

9.  A hollow sphere of external radius R and thickness t (<< R) is made of a metal of density Ρ, sphere will float in water if

Ans: (B)

Sol.

If the sphere has to float in water, then the weight of water displaced by liquid must be greater than or equal to weight of sphere.

10. A metal wire of circular cross-section has a resistance R1. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R2, then R2 : R1 is

(A) 1 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Ans : (C)

Sol.

Let the original length of wire be l and original cross-sectional area be A.

From (1) and (2)

WBJEE 2016 Solved Mathematics Question Paper

Next Page

Previous Page

Advertisement

Related Categories