Physics (Part-VI) (Railway)

Total Questions: 50

31. According to Fleming's right-hand rule, the index finger and the central finger of the right hand represent directions of and геspectively. [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-II)]

(1) induced current, magnetic field
(2) motion of the conductor, induced current
(3) motion of the conductor, magnetic field
(4) magnetic field, induced current

Correct Answer: (4) magnetic field, induced current

Solution:Fleming's right-hand rule is used to determine the direction of the induced current.

• Fleming's Right Hand Rule states that if we arrange our thumb, forefinger and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

32. A current of 5 A flows through a conductor having resistance 2Ω. The potential difference (in volt) across the ends of the conductor is: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) 7
(2) 10
(3) 2.5
(4) 3

Correct Answer: (2) 10

Solution:ΩElectric potential (V), current (I) and resistance (R) are related as V = IR

• So, according to the question, V= 5 x 2 = 10V

33. Fleming's right hand rule can be used to determine the direction of induced current when the angle between the magnetic field and the motion of the conductor is: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) 30°
(2) 0°
(3) 180°
(4) 90°

Correct Answer: (4) 90°

Solution:Fleming's right-hand rule is used to determine the direction of the induced current.

• It states that if we arrange our thumb, forefinger and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

34. The laws of reflection hold true for which of the following mirror(s)? [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(A) Concave mirror

(B) Convex mirror

(1) Only A and B
(2) Only A and C
(3) Only B and C
(4) A, B and C

Correct Answer: (4) A, B and C

Solution:The laws of reflection hold true for all the reflecting surfaces either spherical or plane.

• Therefore, the plane mirror, concave mirror and the convex mirror all will follow the laws of reflection.

• According to the law of reflection

• During the phenomenon of reflection, the angle of incidence and the angle of reflection remain equal.

• The law of reflection states that the incident ray. The reflected ray and the normal to the Surface of the mirror all lies in same Plane.

35. Identify the correct relation between radius of curvature 'R', object distance 'u' and image distance 'v' for a spherical mirror [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) R = 2(u + v)/uv
(2) R = 2uv/u + v
(3) R = (u + v)/2uv
(4) R = uv/2(u + v)

Correct Answer: (2) R = 2uv/u + v

Solution:For a spherical mirroг, u is distance of object from pole, v is distance of image from pole and f is focal length of mirror.

From mirror formula: 1/v + 1/u = 1/f

Or, 1/f=u+v/uv

So, f = uv/u+v

Since R = 2f = 2uv/u+v

36. Two resistors R₁ and R₂ with resistances 2Ω and 3Ω, respectively, are connected in series to a 15V battery source. The current across R (in A) is: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(2) 7.5
(3) 15
(1) 3
(4) 5

Correct Answer: (1) 3

Solution:The total resistance of a series circuit is equal to the sum of individual resistances.

Or, R = R1 + R2 = 2Ω + 3Ω = 5Ω Electric potential (V), current (I) and resistance (R) are related as V = IR

Or, I = V/R = 15/5 = 3A

37. A ray of light is Incident on an interface separating two media at an angle of incidence equal to 45°, for which the angle of refraction is 30°. The refractive index of the second medium with respect to first, is equal to: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) 1/√2
(2) √3/2
(3) √2
(4) 1/2

Correct Answer: (3) √2

Solution:Sin 45°√1/2, sin 30° 1/2 From snell's law,

μ sin i = μ sin i

μ2/μ1 = sin i/sin r

= sin 45°/sin 30° = 1/√2 * 2

μ2

= √2

μ1

38. The magnetic field lines inside a bar magnet are directed from: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) south pole to north pole
(2) north pole to south pole
(3) Centre to south pole
(4) centre to north pole

Correct Answer: (1) south pole to north pole

Solution:Magnetic field lines are known to travel from its North Pole to the South Pole. But this is for the outside of the magnet.

• For the inside, in order for the magnetic field lines to form a closed loop, they have to be directed from the south to the north.

• Therefore, we found that the magnetic lines of force inside a bar magnet are directed from the South Pole to the North Pole of the magnet.

39. Identify the correct statement (s) from among the following. [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

a. Resistivity of a material is dependent on the nature of the material. .

b. Resistivity of a material is independent of the area of cross section.

c. Resistivity of a pure metal is more than its alloy.

(1) Both (a) and (b)
(2) Both (b) and (c)
(3) Only (a)
(4) Only (b)

Correct Answer: (1) Both (a) and (b)

Solution:The resistivity of alloys is generally higher than that of pure metals and increases with impurity contamination.

40. A light ray is travelling from air medium to water medium (refractive index = 1.3) such that angle of incidence is x degree and angle of refraction is y degree. The value of ratio (sin y)/(sin x) is: [RRB Group 'D' CBT Exam, 24.08.2022 (Shift-III)]

(1) 0.3
(2) 1.3
(3) 1
(4) 1/1.3

Correct Answer: (4) 1/1.3

Solution:n₁ sin ( α ₁) = n₂ sin ( α ₁)

Sin (i) = 1.3 sin (r)

sin (x) = 1.3 sin (y)

⇒ (sin y)/(sin x) = 1/1.3

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