Railway Science (Physics-Electric Current and its Effects) (Part-V)

Total Questions: 50

21. Which of the following instruments is used to measure electric current ? [RRB ALP Tier - I (13/08/2018) Afternoon]

Correct Answer: (b) Ammeter
Solution:

Galvanometer: This device is used on electrical circuits to measure the intensity and direction of electrical current. Voltmeter: Measures voltages of either direct or alternating electric current on a scale. Rheostat: A variable resistor which is used for controlling the flow of electric current either by increasing or decreasing the resistance.

22. The given symbol stands for a/an _______ in an electric circuit. [RRB ALP Tier - I (14/08/2018) Morning]

Correct Answer: (a) electric bulb
Solution:

Other Symbols

23. What enables an electric current to flow in a circuit? [RRB ALP Tier - I (17/08/2018) Morning]

Correct Answer: (d) The motion of electrons
Solution:

Electrical circuit is a network of electrical components with closed paths or loops in which electrons can flow. Flow of current: In solid - electrons are passed from one positively charged metallic atom to the next. In solution - the electron is carried by the ions present in the solution. A solution capable of carrying charge (electric current) is called an electrolyte.

24. If the resistance of a conductor is reduced to half its initial value, then the heating effect in the conductor will become_______. [RRB ALP Tier - I (21/08/2018) Morning]

Correct Answer: (a) half
Solution:

Heating effect of electric current: When a current is flowing in a circuit having resistance there is a heat dissipation due to the resistance. Heat dissipated (H₁)= I² R t, Where I = the current flowing in the circuit, R = the resistance of the circuit, and t = the time taken.

25. A source maintains a current in a resistor of resistance R. If V is the potential difference across the resistor, the electrical energy dissipated in the resistor in time t is given by ……. [RRB ALP Tier - I (21/08/2018) Evening]

Correct Answer: (b) VIt
Solution:

I = Current R = resistance, V = voltage, t = time during which energy dissipated According to Ohm's law, resistance (R) = 𝑉/I.
H = energy dissipated in the resistance in the form of Heat.
H = I²Rt ⇒ H = I² × 𝑉/I × t
H = Vlt.

26. What happens to the ammeter reading when the length of the wire is doubled? [RRB ALP Tier - I (29/08/2018) Morning]

Correct Answer: (a) It decreases to one half
Solution:

By Ohm's law, V = IR, where V = voltage, I = current, and R = resistance.
R = , ρ × 𝐿/A , Where, R = Resistance,  ρ = Resistivity, L = Length of material and A = cross sectional area of material. When the length (L) of the wire is doubled, Resistance (R) is also doubled and so, current (I) is halved.

27. Which of the following are bad conductors of electricity? [RRB ALP Tier - I (29/08/2018) Afternoon]

A) Mica and Quartz
B) Metals and Rubber
C) Metals and Mica

Correct Answer: (a) A only
Solution:

The materials that allow electricity to flow through them, Good conductor - Those materials which allow electricity to pass through them easily; Example - Copper, steel, gold, silver, platinum. Bad conductor - Those materials which do not allow electricity to pass through them; Example - Glass, plastic, rubber and wood.

28. In a parallel circuit, if one electrical appliance stops working due to some defect, then all the other appliances: [RRB ALP Tier - I (29/08/2018) Evening]

Correct Answer: (c) will work normally
Solution:

A circuit is said to be parallel when the electric current has multiple paths to flow through. The components that are a part of the parallel circuits will have a constant voltage across all ends. Other components will function even if one component breaks down, each has its own independent circuit. A series circuit is a circuit in which two components share a common node and the same current flows through them. If one component breaks down, the whole circuit would burn out.

29. The potential drop across the 4Ω resistor in the given circuit is: [RRC Group D 17/08/2022 (Afternoon)]

Correct Answer: (a) 2V
Solution:

The total Resistance (series combination), Rₙₑₜ = R₁ + R₂ = = 4 Ω + 6 Ω = 10 Ω.
The total Current in the circuit, I = 𝑉/R = 5 volt/10Ω = 1/2 A.
The Potential drop across resistance 4 Ω is given as V= IR = 1/2 × 4 = 2V.

30. A wire of a given material has length ‘I’ and resistance ‘R’. Another wire of the same material having three times the length and twice the area of cross section will have a resistance equal to: [RRC Group D 18/08/2022 (Morning)]

Correct Answer: (b) 3/2 R
Solution:

(Resistivity) R =  ρ  𝑙/A,
Given, R₁ = ρ 3𝑙/2A. After comparing both,
R₁ = 3/2 R