Phy ppr1!

[tweez11]

Can sumone plz help me out abt some doubts i hav in phy ppr 1(9702/01),
Year oct/nov04,Q5!
nov 06,Q4!
june 03,Q5!
june 07,Q4!
all i need is simple explanations for these,cuz i already know the ans..jus want to know how to solve it!
would be really thankful if someone can help!

 

[hassam]

nov 06 Q4 :
SINCE y-gain setting is 5v/div so the trace will have a amplitude of from the mean/starting position.1 box vertical.
SINCE timebase setting id 10ms/div we ll convert it into sec which is .01s/div.the frequency is 50Hz which means T(TIME PERIOD) is 1/50 i.e .02s. there are total 5 div in the x-direction. so 5 * .01=.05
total no. of waves that will appear on the trace. .05/.02=2.5 waves. so the correct option is D.

 

[tweez11]

thank you so much for the help!
anybody else there to explain the rest of the qs??
thnku!

 

[beacon_of_light]

Can sumone plz help me out abt some doubts i hav in phy ppr 1(9702/01),
Year oct/nov04,Q5!
nov 06,Q4!
june 03,Q5!
june 07,Q4!
all i need is simple explanations for these,cuz i already know the ans..jus want to know how to solve it!
would be really thankful if someone can help!

June03 Q5:

3% is approximately 9.8 ... so round off the value to the nearest tens...

June07 q4:

2.5 waves in 10cm so 1 wave in 10/2.5 cm or (10/2.5) x 2.5 ms = 10ms ...
1cm = 5mV
So,
3.5cm = 3.5 x 5 is approximately equal to 17mV

Hope u get tht