Sindh Mdcat Exclusive Course March Physics Current Electricity — Solved MCQs

Question 1

An emf of 120 volts of negligible internal resistance is connected across a resistance of 1000 ohm. The current flowing through the circuit will be:

Accepted Answer

C. 120 x 10-3 A — The current(I) can be calculated using the equation V=IR(120)=(1000)(I)120/1000=II=120 x 10-3 A

Question 2

The fractional change in resistance per kelvin is known as:

Accepted Answer

B. Temperature coefficient of resistance — The fractional change in resistance per kelvin is called the temperature coefficient.The temperature coefficient of resistance is a measure of how much the electrical resistance of a material changes with temperature. It is expressed in units of ohms per ohm per degree Celsius (Ω/Ω/°C) or ohms per ohm per kelvin (Ω/Ω/K). This coefficient is denoted by the symbol α (alpha) and is an important factor in understanding and designing electronic circuits, particularly when dealing with components like resistors whose resistance can vary with temperatureResistance coefficient, abbreviated as K, a dimensionless number, is how much resistance to the flow an obstacle has. Volumetric coefficient increases in volume per unit original volume per Kelvin rise in temperature is called the coefficient of volumetric expansion.The ratio increase in length original length for the 1-degree rise in temperature is called the coefficient of linear expansion

Question 3

A uniform wire of resistance 3 ohm is folded in such a way that it forms an equilateral triangle, then what is it’s resistance across the ends of any two vertices?

Accepted Answer

C. ⅔ ohm — Since we have a uniform wire, therefore the resistance of each side of the triangle will be equal as the length of each is also equal. Three sides mean that each side will have resistance 3ohm/3 = 1 ohm, as shown below:If the resistance is to be measured across A and B, then it is clear that two of the three sides are connected in series.R’ = 1 + 1 = 2 ohmsAnd this combination is connected in parallel with the third side:1/R = 1/1 + 1/2 + 3/2 => R = 2/3 Ohm.the general formula to calculate resistance between adjacent sides of a polygon is given as (n-1/n)R where n is number of sides of the polygon

Question 4

Kirchhoff's first law is manifestation of:

Accepted Answer

D. Law of conservation of charge — Kirchhoff's first law states that the algebraic sum of current flowing into a junction is zero. This rule corresponds to conservation of charge which states that total charge in a closed system remains constant because current is the rate of flow of charge.

Question 5

The deviation of I-V graph from straight line is due to:

Accepted Answer

B. Increase in temperature and increase in resistance — The most common scenario leading to a deviation from a straight-line I-V graph is an increase in temperature and an increase in resistance. This is often observed in conductors and is known as a positive temperature coefficient of resistance. As temperature rises, the resistance increases, causing a departure from the ideal linear behavior in the I-V graph.

Question 6

A wire of resistance R is bent in the form of a circle the resistance between two points on the circumference of the wire and at the end of the diameter of circle is

Accepted Answer

A. R/4 — Solution is given below

Question 7

When a potential difference is applied across the ends of a uniform wire of length l and radius r, a current I flows in the wire. If same potential difference is applied to the ends of another wire of the same material but of length 2l and radius 2r, the current in the wire is:

Accepted Answer

B. 2I — R=ρl/AWhen radius is doubled : Area=2*pi*(2r) = 2*pi*4rHence the area has increased by 4 times.New resistance : Rx = ρ*2l/4A = ½ RV=IR where V remains same and R becomes halfI = V / ½RI become equal to 2I.

Question 8

When a conductor of a cross-sectional area 5x10-6 m2 carries a current of 6 A, the drift velocity of the conduction electrons is 1.2 x 10-4 ms1-. What is the number density (number per unit volume) of the conduction electrons?

Accepted Answer

D. 6.3x1028 m-3 — The formula for electric current relating the drift velocity is: I = neAV Where; I = Electric current n = Number of charges per unit volume e = Electron charge A = Cross sectional area V = Drift speed Solving for 'n' : n = I/eAV Here I = 6.0 Amp e = 1.6 x 10-19 C A = 5.0 x 10-6 m2 V = 1.2 x 10^-4 m/s Put all these values inabove equation and do the calculations: n = (6.0)/ (1.6 x 10-19) (5.0 x 10-6) (1.2 x 10-4) n = 6.3 × 1028 m-3 is the answer

Question 9

A thermocouple is immersed in water at 373 K and the other in ice at 273K. The emf of the thermocouple Is 90 (μ)V for each 1 K difference in temperature between junctions, and the thermocouple resistance is 6 Ω. What current will flow in the galvanometer connected in series with an internal resistance of 30 ohms?

Accepted Answer

B. 250 μA — Given details: Galvanometer resistance = 30 ohm Thermocouple couple resistance = 6 ohm So, Total resistance = 30+6 = 36 ohm Temperature difference = 373K – 273K =100K Emf of thermocouple for 1K temperature difference = 90 microvolt = 90 × 10^-6 V So, for 100K, emf = 90 × 10^-6 x 100 = 90 × 10^-4 V Solution: Current flow in galvanometer = Voltage / Resistance I = 90 × 10^-4 / 36 =0.00025 A = 250 micro Ampere

Question 10

A copper wire has resistances 10 ohms and 20 ohms at temperature 0 C and 10 C respectively. What is the temperature coefficient of the copper wire?

Accepted Answer

A. 0.1 K-1 — You can solve the question using following solution:

Question 11

A current of 4.4 amperes is following in a write. How many electrons pass a given point in the wire in one second, if the charge on an electron is 1.6x10-19 Coulomb?

Accepted Answer

B. 2.75 x1019 electrons — I=4.4A e=1.610-19C (charge on one electron)t=1sn=? (no of electrons)Solution: I=q/t As, q=ne I=ne/tRearranging n=(It)e n=(4.41)1.610-19C =2.751019C electrons Result: n = 2.751019C electrons

Question 12

Which of the following work (s) on the principle of wheatstone Bridge?

Accepted Answer

D. all of these — Slide-wire bridge,meter-bridge, and post office box all work on the principle of Wheatstone bridge.The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances is equal, and no current flows through the circuit. Under normal conditions, the bridge is in an unbalanced condition where current flows through the galvanometer.All of the given examples work on this principle.

Question 13

An ammeter reads up to 1 A. Its internal resistance is 0.81 Ω. To increase the range to 10 A, the value of the required shunt is:

Accepted Answer

D. 0.09 Ω — Attached below is the solution to the question above:

Question 14

An electric rod of 2000 watts rating boils a certain quantity of water in 10 minutes, the heat which generated for boiling this water is

Accepted Answer

B. 12 x 105 Joules — Energy=Power x time Energy=2000 x 10 x 60 Energy=12 x 105 J

Question 15

If you wish to decrease the resistance of a circuit, you will add the resistors in _.

Accepted Answer

B. Parallel — As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit decreases, and the total current of the circuit increases. Adding more resistors in parallel is equivalent to providing more branches through which charge can flow.

Question 16

If we increase the cross-section area of a conductor having a length L, the resistance R of that conductor will be:

Accepted Answer

D. Decreasing — The resistance of a conductor decreases with increase in cross-sectional area. ○If we increase the area of cross section of conductor then More electrons results in more charge crossing the cross section, thus more current. Thus if you increase the area of cross section then resistance of the wire (keeping length and resistivity same) will decrease. Resistance R = ρ L/A Result: From the above discussion and the formula of resistance that with increase in area of cross section resistance decrease and vice versa.

Question 17

The domestic electricity supply has a frequency of:

Accepted Answer

C. 50 Hz — The standard frequency for domestic electricity supply varies around the world, but it is commonly 50 Hertz (Hz). The specific frequency depends on the region and the standards established by the local electrical grid.It's important to note that the frequency of the electricity supply is a key parameter in electrical systems and devices, and appliances are designed to operate at the frequency of the local power grid.

Question 18

One kilowatt-hour is commonly termed as one commercial unit of electric energy which is equal to:

Accepted Answer

B. 3.6 x 106 J — 1 kWh can be broken down to 1000J/s multiplied by (60 x 60) seconds to give us 3.6 x 106 Joules.

Question 19

The instrument(s) which work on the principle of Wheatstone bridge is/are:

Accepted Answer

E. A, B and C — All of the above mentioned instruments work upon the principle of Wheatstone bridge to measure the resistance except calendar.The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances is equal, and no current flows through the circuit. Under normal conditions, the bridge is in an unbalanced condition where current flows through the galvanometer.All of the given examples work on this principle.

Question 20

When a wire is compressed and it’s radius becomes 2R then it’s resistance will be:

Accepted Answer

C. 1 / 16 R — Since we are aware of the change in the cross-sectional area, we don't know the change in the length hence we'll be taking volume constant.A- Cross-sectional area of wire (pi x r2)L- Length of wire.A2.L2 = A1.L1 where subscript 1 represents the conditions before the alteration of the wire and subscript 2 represents the conditions after the changes.π4r2l2=πr2l1 where π and r2 gets cancelled out giving us l2=l1/4Also, A2=4A1 (as r2 goes in the formula.)If the radius is doubled length becomes ¼, resistance is proportional to L/AThus,R2=R1/16

Sindh Mdcat Exclusive Course March Physics Current Electricity — Solved Past Paper with Answers

More Sindh / DUHS Solved Papers