WBJEE 2014 Solved Physics Question Paper – Part 5

May 11, 2017 15:02 IST

Find WBJEE 2014 Solved Physics Question Paper – Part 5 in this article. This paper consists of 5 questions (#21 to #25) from WBJEE 2014 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.

21. A car moving with a speed of 72 km-hour–1 towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is 340 ms-1,  the difference of the two frequencies, the driver hears is

(A) 50 Hz

(B) 85 Hz

(C) 100 Hz

(D) 150 Hz

Ans: (C)

Sol:

According to Doppler’s Effect,

22. Same quantity of ice is filled in each of the two metal containers P and Q having the same size, shape and wall thickness but made of different materials. The containers are kept in identical surroundings. The ice in P melts completely in time t1 whereas that in Q takes a time t2. The ratio of thermal conductivities of the materials of P and Q is

(a) t2 : t1

(b) t1 : t2

(c) t12 : t22

(d) t22 : t12

Ans: (A)

Sol:

So, the ratio of thermal conductivities is t2 : t1

23. Three capacitors, 3μF, 6μF and 6μF are connected in series to a source of 120V. The potential difference, in volts, across the 3μF capacitor will be

(A) 24

(B) 30

(C) 40

(D) 60

Ans: (D)

Sol:

C1 = 3μF

C2 = 3μF

C3 = 3μF

24. A galvanometer having internal resistance 10Ω requires 0.01 A for a full scale defection. To convert this galvanometer to a voltmeter of full-scale deflection at 120V, we need to connect a resistance of

(A) 11990 Ω in series

(B) 11990 Ω in parallel

(C) 12010 Ω in series

(D) 12010 Ω in parallel

Ans: (A)

Sol:

We have,

V = 120

Ig = 0.01 A

(A) 1:1

(B) 2:1

(C) –1:1

(D) 3:1

Ans : (A)

Sol:

So, the ratio between α and β is 1 : 1.

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