This Article is here to cover the important 2nd year Physics Questions that create the base for concepts of all other questions that appear in the MDCAT examinations related to circuits and current. there are 56 Physics Questions in the MDCAT exam. And most students find Physics Questions hard to solve.

**Q1) A 100 resistor is connected to a 220 V, 50 Hz ac supply.**

**(a) What is the rms value of current in the circuit?**

**(b) What is the net power consumed over a full cycle?**

**Ans: **Resistance of the resistor, *R* = 100

Supply voltage, *V* = 220 V

Frequency, **=** 50 Hz

**(a)** The rms value of current in the circuit is given as:

**(b)** The net power consumed over a full cycle is given as:

*P* = *VI*

= 220 × 2.2 = 484 W

**Q2) (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?**

**(b) The rms value of current in an ac circuit is 10 A. What is the peak curre**nt?

**Ans: (a)** Peak voltage of the ac supply, = 300 V

Rms voltage is given as:

**(b)** The rms value of current is given as:

*I* = 10 A

Now, peak current is given as:

**Q3) A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.**

**Ans: **Inductance of inductor, *L* = 44 mH =

Supply voltage, *V* = 220 V

Frequency, * * = 50 Hz

Angular frequency, * *=

Inductive reactance, *X*L =

Rms value of current is given as:s

Hence, the rms value of current in the circuit is 15.92 A.

**Q4) A 60 capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.**

**Ans: **Capacitance of capacitor, *C* = 60 =

Supply voltage, *V* = 110 V

Frequency, * *= 60 Hz

Angular frequency, =

Capacitive reactance

Rms value of current is given as:

Hence, the rms value of current is 2.49 A.

**Q5)** **Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?**

**Ans: **In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen () is less than the mass of incident α-particles (). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α-particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

**Q6) The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 , what is the maximum current that can be drawn from the battery?**

**Ans: **Emf of the battery, *E *= 12 V

Internal resistance of the battery,* r *= 0.4 Ω

Maximum current drawn from the battery can be calculated as:

The maximum current drawn from the given battery is 30 A.

**Q7) A battery of emf 10 V and internal resistance 3 is connected resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?**

**Ans:** Emf of the battery, E=10V

Internal resistance of the battery ,r =3

Current in the circuit, I=0.5A

Resistance of the resistor

The relation for current using Ohm’s law is,

Terminal voltage of the resistor=V

According to Ohm’s law,

V=IR

Therefore, the resistance of the resistor is 17 and the terminal voltage is 8.5v.

**Q8) Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level?**

**Ans: **Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height.

Height of the given antenna, *h* = 81 m

Radius of earth, *R* = m

For range, *d* = (2*Rh)*½, the service area of the antenna is given by the relation:

*A* =

=

=

= 3255.55

**Q9) Frequencies in the UHF range normally propagate by means of:**

**(a) Ground waves.**

**(b) Sky waves.**

**(c) Surface waves.**

**(d) Space waves.**

**Ans: (d) **Space waves

Owing to its high frequency, an ultra high frequency (UHF)wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation.

**Q10) A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of**** B at a point 2.5 m east of the wire.**

**Ans:** Current in the wire, *I* = 50 A

A point is 2.5 m away from the East of the wire.

Magnitude of the distance of the point from the wire, *r* = 2.5 m.

Magnitude of the magnetic field at that point is given by the relation, *B*

Where,

_{=} Permeability of free space =

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

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