WBJEE 2015 Solved Physics Question Paper – Part 7

Find WBJEE 2015 Solved Physics Question Paper – Part 7 in this article. This paper consists of 5 questions (#31 to #35) 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.

31. A cylinder of height h is filled with water and is kept on a block of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1,2,3 and 4 are at the side of the cylinder and at heights 0, h/4 and 3h/4 respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole no.

(A) 1

(B) 2

(C) 3

(D) 4

Ans: (B)

Sol.

For maximum range,

 32. The pressure p, volume v and temperature T for a certain gas are related by

where A and B are constants. The work done by the gas when the temperature changes from T1   to T2 while the pressure remains constant, is given by

Ans: (C)

Sol.

We have,

WBJEE 2016 Solved Physics and Chemistry Question Paper

33. In the circuit shown below, the switch is kept in position ‘a’ for a long time and is then thrown to position ‘b’. The amplitude of the resulting oscillating current is given by

Ans: (D)

Sol.

The charge through the capacitor is given as:

qo = CE

where, C is capacitance and E is potential difference

Now by conservation of energy,

Energy stored by capacitor = Energy stored by inductor

 34. A charge q is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero. The flux through any one of the other three faces is

Ans: (D)

Sol.

Let’s use Gauss's law and symmetry to solve this problem. On placing 7 more identical cubes around the original, the charge is covered symmetrically. So, according to Gauss's law

35. Two cells A and B of e.m.f 2V and 1.5V respectively, are connected as shown in figure through an external resistance 10Ω. The internal resistance of each cell is 5 Ω. The potential difference EA and EB across the terminals of the cells A and B respectively are

Ans: (C)

Sol.

Let i be the current flowing through the circuit

On applying KVL,

WBJEE Sample Question Paper Set-II

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