UPSC SCRA Question Papers

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 SCRA-UPSC-Physical Science-2011 solved question papers

 1. Two a-particles, each of 13.2 MeV energy are produced when a  ₃Li⁶ nucleus is bombarded with a 4.4 MeV deuteron. What is the Q-value of the reaction ?

(a) 22 MeV (Ans)

(b) 30.8 Mev

(c) 22 J

(d) 30.8 J

Explanations :  Q-value of reaction = (2 x 13.2 - 4.4)

= (26.4 - 4.4) = 22 MeV

Magnetic field at the centre.

 2. A current is flowing in a circular coil of radius r and the magnetic field at the centre is B₀. At what distance from the centre on the axis of the coil is the magnetic field B₀/27?

(a) √3 r

(b) 2√2 r(Ans)

(c) √2 r

(d) 2 r

Explanations :  B₀ = µ₀ Ni/2r

At axis,        B_axis = µ₀/2 Nir²/(x² + r²)^3/2

⇒               B₀/2₇ = µ₀/2 Nir²/(x² + r²)^3/2

∴   µ₀Ni/2 * r * 27 = µ₀/2 Nir²/(x² + r²)^3/2

or                   27 = (1 + x²/r²)^3/2

                    (3)² = (1 + x²/r²)

                       9 =  (1 + x²/r²)

                     8r² = x²

                       x = 2√2r

 Alternative

The ratio of magnetic field at the centre of circular coil and on its axis is given by

    B_centre/B_axis   =  (1 + x²/r²)^3/2

⇒     B₀/ B₀/27 =  (1 + x²/r²)^3/2

or               27 = (1 + x²/r²)^3/2

                    x = 2√2r

 3. In the circuit given below, the ammeter reading is zero. What is the value of the resistance R?

(a) 50 Ω

(b) 100 Ω (Ans)

(c) 200 Ω

(d) 400 Ω

Explanations : The terminal potential difference across R due to 12 V battery should be equal to 2 V which is the emf of the cell in the loop containing the ammeter.

12 - 12/500 + R * 500 = 2

10 = 12 * 500/500 + R

500 + R = 60 or R = 100 Ω

 4.  In the above circuit, in which of the following cases, the reading of ammeter will change if the ammeter resistance G is changed ?

(a) G = 50 Ω

(b) G = 100 Ω

(c) G = 500 Ω

(d) None of these (Ans)

Explanations : Reading will remain zero, whatever may be the value of ammeter resistance.