Physics (Part-VI) (Railway)

Total Questions: 50

41. Three resistors of 9Ω each are connected in parallel. Their equivalent resistance is: [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) 3Ω
(2) 18Ω
(3) 9Ω
(4) 27Ω

Correct Answer: (1) 3Ω

Solution:ΩWhen three resistances are connected in parallel, the total resistance is given by:

• 1/Req = 1/R1 + 1/R2 + 1/R3

• So, according to the question, 1/Req=1/9+1/9+1/9=3/9

• So, Req = 3Ω

42. A proton is projected into a magnetic field as shown in the figure, the direction of force experienced by it is: [RRB Group 'D' CBT Exam, 25.08.2021 (Shift-I)]

(1) in the direction of motion of proton
(2) into the plane of paper
(3) against the direction of motion of proton
(4) outside the plane of paper

Correct Answer: (2) into the plane of paper

Solution:Protons represent the direction of current.

• We know, the direction of the force due to a magnetic field is perpendicular to the direction of motion.

• Applying Fleming's left-hand rule, if we keep the index finger in the direction of the magnetic field, and the middle finger in the direction of current, which is in the direction in which the proton is moving (since it is positively charged)

• Hence, the direction in which the force is experienced by the proton will be in the direction of the thumb, which is pointing into the plane of the paper.

43. A concave mirror forms a real, inverted and diminished Image of an object. Where is the object placed? [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) Between pole and focus
(2) Between focus and centre of curvature
(3) Beyond centre of curvature
(4) At focus

Correct Answer: (3) Beyond centre of curvature

Solution:When the object is placed beyond the centre of curvature, then a real and diminished image is formed by a concave mirror.

• In other words, when the object is placed between infinity and centre of curvature of a concave mirror. image is formed between centre of curvature & focus.

• This image is diminished, real and inverted.

44. Which of the following formulae does NOT represent the heat produced in a conductor due to flow of electric current? [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) H= V²/R t
(2) H = I²2Rt
(3) H = Vlt
(4) H = IR²t

Correct Answer: (4) H = IR²t

Solution:According to Joule's law of heating, heat produced H=I²Rt, where I is current, R is resistance, and t is time.

• Joule's law is a mathematical description of the rate at which resistance in a circuit converts electric energy into heat energy.

45. The magnetic field strength inside a current carrying solenoid is: [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) uniform and non-zero
(2) more at ends and less at centre
(3) less at ends and more at centre
(4) zero

Correct Answer: (1) uniform and non-zero

Solution:The magnetic field line is a straight line inside a solenoid. Hence, it remains uniform along its length.

• The distance from the current carrying conductor and the magnetic field are inversely proportional. Thus, towards the ends of the solenoid, the magnetic field strength reduces as they diverge.

• The magnetic field outside the solenoid is zero.

46. The refractive indices of tur-pentine oil and glass are 1.47 and 1.52 respectively. A ray of light passes from turpentine oil to glass. The refractive index of glass with respect to turpentine oils is the ray bends normal in glass. [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) 1.03, towards
(2) 0.97, away from
(3) 1.03, away from
(4) 0.97, towards

Correct Answer: (1) 1.03, towards

Solution:Refractive index of glass with respect to turpentine oil = = Refractive index of glass /Refractive index of turpentine oil 1.52/1.47= 1.034

• Now, we know greater the refractive index, the denser the medium.

• Thus, glass (1.52) is denser than turpentine oil (1.47)

• When the ray of light moves from a denser to a rarer medium, the refracted ray moves away from the normal.

• When the ray of light moves from a rarer to a denser medium, the refracted ray moves towards the normal.

47. An object is placed on the principal axis of a concave lens of focal length. 20 cm, at a distance of 10 cm. The magnification produced by the lens is_ and the image is. [RRB Group 'D' CBT Exam, 25.08.2022 (Shift-I)]

(1) more than 1, inverted
(2) less than 1, erect
(3) less than 1, inverted
(4) more than 1, erect

Correct Answer: (2) less than 1, erect

Solution:A concave lens always forms virtual, erect, and diminished image.

• Hence, the magnification is always positive, but less than 1.

48. The radius of curvature of a concave mirror is 12 cm. Following New Cartesian Sign Convention, the principal focus is located at x =_______. [RRB Group 'D' CBT Exam, 30.08.2022 (Shift-I)]

(1) 6 cm
(2)-12 cm
(3) 12 cm
(4)-6 cm

Correct Answer: (4)-6 cm

Solution:For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. It is expressed as R = 2f.

• According to the question,

- 12 =2f

∴ f= -12/2 =-6 cm

49. Consider the following statements about an electric motor: [RRB Group 'D' CBT Exam, 30.08.2022 (Shift-I)]

(a) An elecric motor uses split rings.

(b) The direction of the force on the motor's coil can be found using Fleming's right-hand rule.

Which of the above statements is/are correct?

(1) Only (b)
(2) Both (a) and (b)
(3) Only (a)
(4) Neither (a) nor (b)

Correct Answer: (3) Only (a)

Solution:In an electric motor, the direction of the force on a current carrying conductor in a magnetic field is determined 3 by Fleming's left-hand rule.

• According to Fleming's left-hand rule, when we arrange three fingers (point, middle, and thumb) as such all three are perpendicular to each other. Then,

♦ The pointer finger points the direction of the magnetic field.

♦ The middle finger points in the direction of the current.

♦ The thumb points in the direction of the magnetic thrust of force.

50. A 1100W electric toaster operates at 220V. The resistance of its coil is: [RRB Group 'D' CBT Exam, 30.08.2022 (Shift-I)]

(1) 88Ω
(2) 22Ω
(3) 44Ω
(4) 11Ω

Correct Answer: (3) 44Ω

Solution:The relationship between Power and Resistance is expressed as P = V²/R , where P is the Power in watts or W, R is the resistance measured in ohms (O), and V = the potential difference applied across the ends of the conductor.

• So, according to the question,

P = V²/R

1100 = (220)² /R

1100 = 48400/R

1100R = 48400

R = 44 Ω

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