Find WBJEE 2015 Solved Physics Question Paper – Part 5 in this article. This paper consists of 5 questions (#21 to #25) 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.
21. Particle A moves along X-axis with a uniform velocity of magnitude 10 m/s. Particle B moves with uniform velocity 20 m/s along a direction making an angle of 600 with the positive direction of X-axis as shown in the figure. The relative velocity of B with respect to that of A is
(A) 10 m/s along X-axis
(B) 10 √3 m/s along Y-axis
(C) 10 √5 m/s along bisection of velocities of A and B
(D) 30 m/s along negative X-axis
Resolving the velocity in horizontal and vertical components
The relative velocity of B with respect to that of A is given as vBA
vBA = vB - vA
vBA = (20 × cos600i + 20 sin600 j) - 10i
vBA = 10√3 j
22. When light is refracted from a surface, which of its following physical parameters does not change ?
Ans : (C)
When light is refracted from a surface, its frequency does not change.
23. A solid maintained at t10 C is kept in an evacuated chamber at temperature t20 C (t2 >> t1). The rate of heat absorbed by the body is proportional to
Since the temperature are given in Celsius, so none of the options are correct.
The rate of heat absorbed wil be proportional to (t2+ 273)4 – (t1 + 273)4
24. Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is
Resolving T in horizontal and vertical component.
Horizontal component = T cos θ
Vertical component = T sin θ
The vertical component of tension will balance the weight of block A.
Let the weight of block A be WA.
So, T sin θ = WA …(1)
The frictional force between block and the table will balance horizontal component of T.
So, T cos θ = μMg …(2)
From (1) and (2)
WA = μ W tan θ
25. The inputs to the digital circuit are shown below. The output Y is