# WBJEE 2015 Solved Physics Question Paper – Part 1

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 13, 2017 17:40 IST Find WBJEE 2015 Solved Physics Question Paper – Part 1 in this article. This paper consists of 5 questions (#1 to #5) 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.

1. Two particles of mass m1 and m2 ,approach each other due to their mutual gravitational attraction only. Then

(A) accelerations of both the particles are equal

(B) acceleration of the particle of mass m1 is proportional to m1

(C) acceleration of the particle of mass m1 is proportional to m2

(D) acceleration of the particle of mass m1 is inversely proportional to m1

Ans: (C)

Sol.

2. Three bodies of the same material and having masses m, m and 3m are at temperatures 400C, 500C and 600C respectively. If the bodies are brought in thermal contact, the final temperature will be

(A) 450C

(B) 540C

(C) 520C

(D) 480C

Ans: (B)

Sol.

According to law of conservation of energy,

Energy lost = Energy gained

Let the final temperature be T.

Let the specific heat capacity of the material be c.

Then,

WBJEE Sample Question Paper Set-II

3. A satellite has kinetic energy K, potential energy V and total energy E. Which of the following statements is true ?

(A) K = –V/2

(B) K = V/2

(C) E = K/2

(D) E = –K/2

Ans: (A)

Sol.

4. An object is located 4 m from the first of two thin converging lenses of focal lengths 2m and 1m respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of

(A) 8.0 m

(B) 7.5 m

(C) 6.0 m

(D) 6.5 m

Ans: (B)

Sol.

The image obtained by first lens will act as object for the second lens.

Here, for the first lens,

u = –4m,

f1 = 2m

According to lens formula,

For the second lens,

u2  =  1m

and  f2  =  1m

Again applying lens formula, we get

v2  =  0.5

Therefore distance from the object = 7 + 0.5 = 7.5 m

5. A simple pendulum of length L swings in a vertical plane. The tension of the string when it makes an angle θ with the vertical and the bob of mass m moves with a speed v is (g is the gravitational acceleration)

(A) mv2/L

(B) mg cos θ + mv2/L

(C) mg cos θmv2/L

(D) mg cos θ

Ans: (B)

Sol.

From the diagram, we can see that

mg cos θ is the component of mg.

Also, a centripetal force act on the pendulum which is mv2/L

So, the tension in the string is mg cos θ + mv2/L

WBJEE 2016 Solved Physics and Chemistry Question Paper

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