Physics (Part-VIII) (Railway)

Total Questions: 50

41. If the focal length of a spherical mirror increases to double, then the radius of curvature of the mirror will [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) increase to double
(2) remain the same
(3) increase to four times
(4) decrease to half

Correct Answer: (1) increase to double

Solution:Focal length of a spherical mirror is equal to half of the radius of curvature spherical mirror.

• It is expressed as f = R/2, where f is the focal length and R is the Radius of curvature.

• Hence if radius of curvature of a mirror is doubled, its focal length is double.

42. Suppose in a circuit, a silver wire of length L and cross - sectional area A is replaced by an aluminium wire of length 5L and cross-sectional area 9A. The resistance of the circuit will_________. [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(Given that Psilver = 1.6 x 10-⁸ Ω m and P aluminium = 2.6 x 10-⁸ Ωm)

(1) increase to 1.1 times of it-self
(3) decrease to 0.9 times of itself
(3) decrease to 0.9 times of itself
(4) decrease to 0.8 times of itself

Correct Answer: (3) decrease to 0.9 times of itself

Solution:R = PL/A

R¹/R = P¹L¹/PL * A/A¹

= 2.6 x 10-⁸ * 5 L / 1.6 x 10-⁸ * L * A/9A

2.6 * 5/1.6 * 9 = 0.90

Hence, R¹ = 0.90 R

∴ The resistance of the circuil will decrease to 9.0 times of itself.

43. When a car is moving through a dusty road during night, a path of beam from the head light is clearly visible due ________ to effect. [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) scattering
(2) dispersion
(3) refraction
(4) immersion

Correct Answer: (1) scattering

Solution:When a car is moving through a dusty road during night, a path of beam from the headlight is clearly visible due to scattering.

• The tiny dust particles present in the atmosphere scatter the beam of light all around. This is called Tyndall effect.

44. Consider a wire of resistance 10 Ω. If there is another wire of the same material and the same length as the previous one, but the cross-sectional area is double, then the resistance of the new wire is: [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) 7.5 Ω
(2) 5 Ω
(3) 20 Ω
(4) 10 Ω

Correct Answer: (2) 5 Ω

Solution:Resistance of a wire is directly proportional to the length of the conductor and inversely proportional to the area of the cross-section.

• So, when the cross-sectional area is doubled, then the resistance of the new wire will be halved.

45. A house is supplied electricity through a 15 A fuse. The number of 100 W lamps that can be used simultaneously along with a 2KW AC is: (The AC and lamps both are rated for 220 V Supply) [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) 13
(2) 23
(3) 18
(4) 15

Correct Answer: (1) 13

Solution:1 kw = 1000 W.

∴ 2 kw = 2000 W

Power of Ac = 2000 W

Power of bulb = 100 W

We known, Power, P = VI.

The current drawn by the Ac,

I = P ac/V = 2000/220 = 0.09 A

maximum electric current, Which can pass through bulb

= 15 - 9.09

= 5.90 A

Now, Current With draw by a single bulb, I

46. In order to obtain an inverted and magnified image of an object by a spherical mirror of radius of curvature 40 cm, the object should be placed: [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) between 20 cm and 40 cm from a concave mirror
(2) at infinity from a Convex mirror
(3) between 20 cm and 40 cm from a convex mirror
(4) 20 cm from a convex mirror

Correct Answer: (1) between 20 cm and 40 cm from a concave mirror

Solution:between 20 cm and 40 cm from a concave mirror

• Radius of curvature is observed to be equal to twice the focal length for spherical mirrors. Hence R = 2 f

• So, according to the question, f = R / 2 = 40/2 = 20 cm

• When an object is placed between the centre of curvature and the principal focus of a concave mirror, its image is formed beyond the centre of curvature.

• The image formed is real. inverted and magnified.

47. The resistance of a metal wire of length 1 m is 50 Ω at 30° C. Suppose the diameter of the wire is 0.3 mm. Now another wire has the same material land the same length (at 30° C), but the diameter is increased to 0.6 mm. The resistance of the new wire is: [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) 12.5 Ω
(2) 20.5 Ω
(3) 2.5 Ω
(4) 120.5 Ω

Correct Answer: (1) 12.5 Ω

Solution:Resistance is inversely proportional to the cross Dectional area.

• If a certain wire has resistance R Then the resistance of another wire identical with the first except having twice its diameter is D.25 R.

⇒ 50 * 0.25 = 12.5Ω

48. The strength of magnetic field inside a long current-carrying straight solenoid is: [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) uniform at all points inside the solenoid.
(2) the minimum at the centre.
(3) more at the end than the centre.
(4) the maximum at the middle.

Correct Answer: (1) uniform at all points inside the solenoid.

Solution:The magnetic field inside a long straight solenoid-carrying current is the same at all points.

• One end of the solenoid basically acts as a magnetic north pole whereas the other acts as a magnetic south pole.

• The field lines inside the solenoid are known to be parallel and straight.

49. A light ray enters from medium A to medium B and as a result, it bends away from the normal in the medium B. The refractive index of medium B relative to medium A is: [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-I)]

(1) greater than unity
(2) equal to unity
(3) less than unity
(4) equal to 2

Correct Answer: (3) less than unity

Solution:According to the question, ray of light bends away from the normal, when it goes from medium A to medium B.

• This means that the medium A is denser and medium B is rarer.

• This means that the refractive index of medium B with respect to refractive index of medium A will be less than 1.

50. The magnification of an image is +1.5 and the object distance is 30cm from a spherical mirror. The image is formed at_______. [RRB Group 'D' CBT Exam, 20.09.2022 (Shift-II)]

(1) 45 cm behind the mirror
(2) 45 cm in front of the mirгог
(3) 20 cm in front of the mirгог
(4) 20 cm behind the mirror

Correct Answer: (1) 45 cm behind the mirror

Solution:We know, Magnification, M=v/u, where v is the image distance and u is the object distance

• According to the question, 1.5

= v / u = v / 30

• v = 30 * 1.5 = 45 cm

• Image formed at 45 cm behind the mirror.

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