WBJEE 2014 Solved Physics Question Paper – Part 8
In this article find 5 questions (#36 to #40) from WBJEE 2014 Solved Physics Question Paper. This solved paper will help students in their final level of preparation for WBJEE Exam. 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.
Find WBJEE 2014 Solved Mathematics Question Paper – Part 8 in this article. This paper consists of 5 questions (#36 to #40) from WBJEE 2014 Mathematics 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.
37. A uniform solid spherical ball is rolling down a smooth inclined plane from a height h. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity v, the maximum height to which the ball will rise is
38. Two coherent monochromatic beams of intensities I and 4I respectively are superposed. The maximum and minimum intensities in the resulting pattern are
(A) 5I and 3I
(B) 9I and 3I
(C) 4I and I
(D) 9I and I
Ans : (D)
39. If the bandgap between valence band and conduction band in a material is 0 eV, then the material is
(B) good conductor
If there is no band gap, then the material is semi conductor.
40. Consider a blackbody radiation in a cubical box at absolute temperature T. If the length of each side of the box is doubled and the temperature of the walls of the box and that of the radiation is halved, then the total energy
(D) remains the same
Assuming temperature of the body and cubica box is same initially i.e. T and finally it becomes T/2. Because temperature of body and surrounding remains same Hence there is no net loss of radiation through the body. Thus total energy remains constant.