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BITSAT 2017 Solved Physics Practice Paper Set-I

Find BITSAT 2017 Solved Physics Practice Paper in this article. Birla Institute of Technology and Science Admission Test(BITSAT) is a computer-based admission test conducted by BITS. This year BITSAT will be held between 16th May to 30th May, 2017. Read this article to get download link of the question paper and its solution in pdf format.

May 3, 2017 18:00 IST
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Birla Institute of Technology and Science Admission Test(BITSAT) is a computer-based admission test conducted by BITS. This year BITSAT will be held between 16th May to 30th May, 2017. The Physics section of BITSAT contains 40 questions.

Few question from the sample paper are given below:

Questions

Q. In a Young's double slit experimental arrangement shown here, if a mica sheet of thickness t and refractive index μ is placed in front of the slit S1, then the optical path difference (S1P - S2P):

BITSAT 2017

(a) decreases by (μ - 1)t

(b) increases by (μ - 1)t

(c) does not change

(d) increases by μt

Q. The number of turns in the primary and secondary coils of a transformer are 1000 and 3000 respectively. If 80 volt AC is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be

(a) 240 volt                 

(b) 2400 volt              

(c) 24 volt                   

(d) 0.08 volt

Q. The double slit experiment of Young has been shown in the figure. Point Q is the position of the first bright fringe on the right side and point P is the 11th fringe on the other side as measured from point Q. If wavelength of light used is 6000 Å, S1B will be equal to

BITSAT 2017

(a) 6 × 10-6 m                          

(b) 6.6 × 10-6 m

(c) 3.138 × 10-7 m       

(d) 3.144 × 10-7 m

Hints and Solutions

Sol.       Initial optical path difference Δx = S1P - S2P

            Final optical path difference Δx' = {(S1P - t) + μt)} - S2P

                                                           = (S1P - S2P) + (μ - 1)t = Dx + (μ - 1)t

            Hence increased by (μ - 1)t.

                    (b)

Sol.       Potential difference per turn of primary and secondary coil are same and

                        = 80/1000 = 0.08 volt 

                     (d)

Sol.       P will be 10th fringe from center

            So,       S1B = 10λ = 10 × 6 × 10-7 m

                        S1B = 6 × 10-6 m

                     (a)

Download BITSAT 2017 Physics Practice Paper

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