Physics P2 prob

[British Council]

Hi..... Can someone plzzzzzzz explain May' June 2009/ Physics p2 Q-5 Part 2??
Link:http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s09_qp_2.pdf
Please do reply....

 

[mista.lova.lova]

man!! ur name is real scary....

 

[beacon_of_light]

Well a waves question,

proceed step by step...

Find S2M by Pythagoras system, S2M is 128cm.

Path difference ( S2M - S1M) = 128 -100 = 28cm or 0.28m

Find two different wavelengths at different frequencies. Wavelength at 1.0 KHz is 0.33m. And wavelength at 4.0 KHz is 0.0825m.

The formula for destructive interference's path difference(since we are considering minimas' here) =( n + 1/2 lambda)

so...

we have the path difference 0.28m and two different wavelengths...

n+(0.5)(0.33) = 0.28 ....and n = 0.35..

n+(o.5)(0.0825)= 0.28...and n= 3.39...

The number of possible minima is then the number of odd numbers between 0.35 and 3.39 inclusive. So they are 1 and 3...

V get the answer...the no of possible minima are 2...

hope you understood that

 

[British Council]

@ Beacon of light: Thanks a lot dude.... If the question would have asked about the number of Maximas then we would use (n x lamda) right???

And can u please solve and explain Q.7 part (b) (iii)....??

Thanks in advance..

 

[beacon_of_light]

You'r welcome. yeah we would use n lambda when considering constructive interference.

Consider loop ABCXYZA but do not consider BY having current I3.

Now one simple rule...

If the current from the battery goes from positive to negative terminal , E would be + .....but if the current goes from negative to positive terminal, E would be negative.

Applying this rule, E1 would be positive and E2 negative...

Now make an equation using Kirchoff's second law.

E1 - E2 = I1R - I2R + I1R ( the current through resistor R in CX would be -I2 since we reverse its direction )
.......
E1 - E2 = 2I1R - I2R

 

[British Council]

Thanks dude..... U ROCK!! Can u please explain May'2008 Q.6 part (b)???? plzzzzzzzzzzzzzzzzzzzzzzzz

 

[beacon_of_light]

may 08 Q6 b...

Just remember one rule, the current ''likes'' to flow through the path with lowest resistance. here in this diagram, if switch S2 would be closed, current will flow through switch S2 but not through resistance B.

row1: power is zero since no current flows through circuit.

row2:now total resistance will be 38.4 ohms only coz current will not flow through resistance B as an alternative low resistance path by S2 is provided.

power is v^2/R .....(240)^2/ 38.4 ...power is 1.5 ohms.

Solve out other rows in the same way...

 

[1992]

nice working out