Physics (Part-V) (Railway)

Total Questions: 50

11. Planets do not twinkle because- [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(1) they scatter less light
(2) they act as extended source of light
(3) they act as a point source of light
(4) they scatter more light

Correct Answer: (2) they act as extended source of light

Solution:Planets do not seem to twinkle like stars, although light coming from them also passes through the variable refractive index of the atmosphere.

• This is because stars appear as point objects to us due to their enormous distance from the earth.

• On the other hand, planets are quite close to the earth, therefore, they are more like extended objects and can be considered as a collection of a large number of point-sized sources of light.

12. Which of the following is NOT a unit of heat energy? [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(1) Calorie
(2) Watt second
(3) Kilowatt
(4) Joule

Correct Answer: (3) Kilowatt

Solution:A kilowatt (Kw) is a unit of electrical energy.

• It is simply a measure of how much power an electric appliance consumes.

• It is equal to 1000 Watts, and one Watt is equal to 1 Joule per second. So, a kilowatt is 1000 Joules per second.

13. The property/properties that must be possesed by a material to be chosen for making heating element of heating devices is/are: [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(i) high melting point

(ii) high resisitivety

(iii) low resistance

(1) Only (i) and (ii)
(2) Only (1) and (iii)
(3) (i), (ii), (iii)
(4) Only (ii) and (iii)

Correct Answer: (1) Only (i) and (ii)

Solution:A heating wire should be such that it produces more heat when current is passed through it and also does not melt.

• It will be so if it has high specific resistance and high melting point.

14. The magnification produced by a spherical mirror is -0.5. The image formed by the mirror is: [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(1) real, inverted and diminished
(2) real, inverted and enlarged
(3) virtual, erect and diminished
(4) virtual, erect and enlarged

Correct Answer: (1) real, inverted and diminished

Solution:The magnification is the ratio of the height of the image to the height of the obJect.

• Magnification is -0.5. The negative sign of magnification Indicates that the image is real while 0.5 indicates that the image is diminished.

• A convex mirror only forms a real and diminished image of an object.

15. Which of the following line(s) 2 Act as a normal to a spherical mirror? [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(i) Line joining the pole and centre of curvature

(ii) Line joining the centre of curvature and point of incidence

(iii) Line joining focus and. point of incidence

(1) Both (i) and (iii)
(2) (i), (ii) and (iii)
(3) Both (1) and (ii)
(4) Both (ii) and (iii)

Correct Answer: (3) Both (1) and (ii)

Solution:The normal at any point of the spherical mirror is the straight line obtained by joining that point with the centre of the mirror.

• By definition, the centre of curvature of spherical mirrors is the centre of the (imaginary) sphere of which they are a part. Hence, every ray that is incident along the normal of a spherical mirror passes through its centre of curvature.

• A line joining any point of the spherical mirror to its centre of curvature is always normal to the mirror at that point.

16. The current across the 3 ohm resistance in the given circuit is: [RRB Group 'D' CBT Exam, 18.08.2022 (Shift-III)]

(1) 7/5 A
(2) 5/3 A
(3) 3/5 A
(4) 5/7 A

Correct Answer: (4) 5/7 A

Solution:We are provided with a circuit which has resistances and voltage connected to it as an electrical component. 3Ω and 4Ω are connected in series with a voltage of 5V.

• Total resistance in Series = 3Ω +4Ω =7Ω

• So, the current flowing through circuit I=V/R =5/7A

17. The resistance of a conductor is NOT dependent on: [RRB Group 'D' CBT Exam, 22.08.2022 (Shift-I)]

(1) area of cross section of the conductor
(2) current flowing through the conductor
(3) temperature of the conductor
(4) length of the conductor

Correct Answer: (2) current flowing through the conductor

Solution:The resistance of a conductor depends on the cross sectional area of the conductor, the length of the conductor, and its resistivity.

• It does also depend on the flow of current through the conductor or other physical factors like temperature, etc.

18. A potential difference of 5V when applied across a conductor produces a current of 2.5mA. The resistance of the conductor (in W) is: [RRB Group 'D' CBT Exam, 22.08.2022 (Shift-I)]

(1) 2000
(2) 200
(3) 2
(4) 20

Correct Answer: (1) 2000

Solution:The relationship between voltage, current, and resistance is described by Ohm's law. This equation, I = V/R. So, R = V/I

• Now, we know 1 mA (milliampere) = 0.001 A (Ampere), So, 2.5 mA = 0.0025 ampere

∴ R=5/0.0025 = 2000 Ω

19. The magnification 'm' produced by a convex lens when the object is placed at a distance 2f from the lens is given by: [RRB Group 'D' CBT Exam, 22.08.2022 (Shift-I)]

(1) m = +1
(2) m = -1
(3) m-2
(4) m = +2

Correct Answer: (2) m = -1

Solution:

u = -2f , 1/f = 1/v - 1/u

1/f = 1/v + 1/2f = 1/v = 1/f - 1/2f

1/v = 1/f [1/1 - 1/2f ] =

1/v = 1/f [2-1/2]

1/v = 1/f (1/2) = 1/2f

v = 2f

Magnification (m),

= v/u = 2f / -2 j = -1

∴ m = -1

20. The refractive index of a given medium is 1.5 what will be the speed of light in that medium. [RRB Group 'D' CBT Exam, 22.08.2022 (Shift-I)]

(2) 4.5 * 10⁸ m/sec
(1) 2 * 10⁸ m/sec
(3) 3 * 10⁸ m/sec
(4) 0.5 * 10⁸ m/sec

Correct Answer: (1) 2 * 10⁸ m/sec

Solution:The refractive index of a medium (n) is equal to the speed of light (c) divided by the velocity of light through the medium (v). In other words, n = c/v

• Rearranging the equation allows us to see the relationship regarding v. so, v = c/n

• According to the question, n= 1.5

• So, v = (3 x 10⁸ m/sec)/1.5 = 2 x 10⁸ m/sec

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