Hello friends, I hope you all are doing great. In today’s tutorial, we will have a look at 2nd Year Physics Chapter 15 Solved Questions. I have started a series of tutorials related to the solution of questions given in 2nd-year physics. In the previous tutorial, I have discussed all the questions of chapter14 with the detailed.

In this post, we will have a detailed look at all the questions given in chapter 15 of FSC physics. I will explain all the questions one by one and give an answer to every question in a simple way. So let’s get started with 2nd Year Physics Chapter 15 Exercise Solved Questions.

#### 2nd Year Physics Chapter 15 Exercise Solved Questions

• So friends now we discuss all the questions one by one with the detail.

#### Question 15.1

• As we know that Faraday’s law says that induced emf depends on rate of change of flux. The formula of Faraday law is given here.

ε = -N(Δφ/Δt)

• From this equation, you can observe that emf induced depend on the flux, not the resistance of any circuit.
• As we also know that according to ohm’s law.

V=IR

I=V/R

• From this equation, you can see that current induces depend on the resistance.

#### Question 15.2

• As the field is uniform so there will be no emf induces in the loop. Due to uniform magnetic field there is no flux variation, so no emf induces.

ε = -N (Δφ/Δt)
as  (Δφ/Δt) = 0

ε = 0

#### Question 15.3

• The flux variation will induce the emf in the ring. The direction of the current will be opposite to the motion of the ring means it opposes the ring according to Lenz’s law.
• It occurs only if the field is produced due to the induced current.
• If the lower side is N pole and upper is S-pole, if we apply right-hand rule the current will be in a clockwise direction.

#### Question 15.4

• When we closed the switch current start to flow from ‘0’ to extreme value in the first winding.
• Due to current flux generated that link with the second winding. According to Lenz law direction of current generated in the second winding due to flux such that it opposes the cause that produced it.
• So the direction of current induced in the second winding is opposite to the current flowing in the first winding
•  So the current direction through the resistor R is left to right.
• If we reopen the switch the flux in the second winding will decrease and become ‘0’.
• So according to Faraday law due to induced emf current produces in second winding that will cause the current in itself in the direction in which current reduces in the first winding.
• So the current direction will be from right to left.

#### Question 15.5

• No, it is not the case, but if magnetic flux in a closed circuitry is reducing than according to Lenz’s law the direction of induced EMF is such that it will try to increases the flux.
• While it will oppose the cause that generates it.

#### Question 15.6

• If the switch is closed the current increases from ‘0’ to extreme value in very short time due to this current flux produced that linked with the ring and produced current in ring.
• The current in-ring generates its own magnetic field this field will oppose the field generated by the current of the battery.
• Due to these fields repulsion, there will force act on the metal ring due that force it will move upward.
• If the polarity of battery is reversed then the ring will also move in up direction.

#### Question 15.7

• If rotation of coil is along X-axis then the flux through the coil changed that induces the emf in the coil.
• While if coil rotates along the y and Z-axis there will be no flux variation so no em will induce in the coil.

#### Question 15.8

• If the plane of a flat loop of wire is parallel to field there is no flux variation and remains ‘0’.

φ = (B)•(A) = BAcos90
φ = 0
Δφ = 0

• As flux is zero so there will be no emf induces according to Faraday’s law.

#### Question 15.9

• The west wingtip of plane will get positively charged.
• The formula for force acting on charge q placing in field is given here.

Fm = q(V x B)

• It indicates that force Fm is at 90 degrees to both V and B. V direction is towards N (north) and the direction of B is towards downward or into paper.
• So if we use right-hand rule the direction of force Fm is towards west, so west wingtip of plane will get positively charged.

#### Question 15.10

• As we know that ‘ε’ is emf so.

Unit of ε=volt or J/C=J.C-1——(1)

The unit of ΔΦ / Δt =wb/s= T.m2/s

As we know that T=N.A-1.m-1

• so we have

ΔΦ / Δt=N.m/A.s

• As we know that N.m=J and A.s=C

ΔΦ/Δt=J/C=J.C-1——-(2)

• From equation (1) and (2) ‘ε’  and ΔΦ/Δt have same units.

#### Question 15.11

• When the motor operates its armature will rotate in the field due to that flux variation induces a voltage in the armature of the motor.
• This induced voltage is called back emf of motor that opposes the input voltage.
• Due to the generation of back emf motor can also operate as a generator.

#### Question 15.12

• Yes, it is possible to operate a DC motor as a DC generator. For this purpose there are some changing is required.
• The input of the motor connected with the load or output.
• The armature of a motor to be connected with the source of mechanical power such as a turbine.
• For field production, the permanent magnet should be used instead of an electromagnet.

#### Question 15.13

• Yes, this can be done. As we know that the formula of flux is given here.

φ = B.A

• From this formula, you can see that there should be variation in the field ‘B’ and area ‘A’.
• If we increase ‘B’ and reduces ‘A’ with the same ratio or decreases B and increases A with the same ratio.
•  Then there will be no change in flux or B.A is constant.

ε = -N Δφ/Δt

• Flux is zero so there will be no emf induces.

#### Question 15.14

• No, it is not possible.
• This scheme will not operate for a long time. As the existing mechanical power originally provided by the motor reduces in removing the friction power losses.
• A situation will occur for which there will be no mechanical power to transformed into electrical power that can be used to run the motor.

#### Question 15.15

• By putting the metallic plate below the vibrating magnet flux will change through the plate.
• This flux variation will induce the current on the surface of the plate which is known as eddy current.
• According to Lenz law, the direction of this current is such that it will oppose the cause that generated it.
• So oscillations of a magnet will be damped.

#### Question 15.16

• There are some step should be done to find the turn ratio of the transformer.
• Perform a continuity test with the AVO meter and find the terminals of primary and secondary winding.
• Find the resistance of primary and secondary windings of the transformer. If transformer is step down then the windings whose resistance is larger will e secondary and others will be primary.
• Apply some certain value of the voltage at primary and find the voltage across the secondary.
• Now use NS /NP = VS /VP to find the turn ratio of the transformer.

#### Question 15.17

(a):

• No, it is not possible for a transformer to increases power level because for an ideal transformer.

Input power =output power

VpIp = VsIs

• The transformer increases or decreases the voltage level but the power or product (VI) remains constant.

(b):

• The 2 windings of transformer are not connected with each other electrically but linked magnetically with one another.
• The flux of first coil with the second coil and according to Faraday’s law flux variation in second windings induces a voltage. In this way power transfers from one coil to other.

#### Question 15.18

• When the second winding of the transformer is open means there is no connection with it so there is no power used by the load.
• So due to absence of load current through the second winding is very less or negligible.
• As the formula of power is.

P=VI

I=P/V

• From this expression, you can see that current is less across the secondary winding.
• If the secondary winding is closed-circuit or load connected with it the load will consume power and a large amount of current will through the winding.

#### Related Posts

So friends it the detailed post about 2nd Year Physics Chapter 15 Solved Questions I tried my level best to make simple questions for you. If you have any further query ask in comments. Thanks for reading. See you in the next post. Have a good day. Soo I will upload the answer of chapter 16 exercise question.

# Free SMT Assembly Monthly

New arrival for Aluminum boards, only \$2

## New users enjoy \$54 free coupons after signing up at JLCPCB successfully

\$54 coupons can also apply to 3D printing orders, for 3D printing special offer, it gets started at \$1