Physics MCQs

[yasser37]

Hi all

so I did the pastpapers and this is what I came up with

Nov. 08
21
June 07
33
20
13
Nov. 06
31
18
2
June 06
33
22
9


Will come up with more questions tomorrow

Thanks

[yasser37]

No one??

[cashem'up]

Hi all

so I did the pastpapers and this is what I came up with

Nov. 08
21
June 07
33
20
13
Nov. 06
31
18
2
June 06
33
22
9


Will come up with more questions tomorrow

Thanks

for q21
NOW there is something very important you should know
1. when there is a wire or spring of length l extension will be x but when length is 2l extension will also be 2x for same load
2. when springs are hung together on same load parallel L will be distributed amongst them

now for this question lets compare this on the basis of a singe spring extension as x
so A will have x/2 as load will be divided
B will have x/2 + x/2 so x as load acts on both sets but is also distributed in both
C it will be x for lower + x/2 for upper
and D x + X/3

so overall C gives greatest of 1.5x

[cashem'up]

june 07 q13
now the definition of torgue is Force into PERPENDICULAR distance between  both forces
therefore first its imporant to get this perpendicular distance
so lets find perpendicular length that is
0.6cos 30(by trigo)
Ans x 8 gives 4.15 rounded to 4.2 so B
QUESTION 20 EVEN I DONT KNOW CAN SOMEONE PLEASE HELP 

FOR 33
first  take notice that voltage in parallel is same...
now first find V1
2/10 x 5 = 1
and V2
2/5 x 3 = 1.2

and now V1-V2=1-1.2 = -0.2
hope u got it Thanks

[yasser37]

OH THANK YOU GUYS

you really helped me out understanding the way to do those questions

hope that someone can answer the rest so that I can put the rest of my questions

[yasser37]

Bump please

[nid404]

Nov 06

2) You know Energy=Work done= Force X displacement

displacement=velocityX time

Energy=Fvt hence B

18) for this one you need to find the power input.

since it's efficiency is 80%
80% of the input power= 4X10^6 W
Input power= 5X10^6W
P=VI
I=P/V
 =5X10^6W/ 25000V
 =200A

31) current= charge/time
        I=Q/t
t=1/f
therefore I=Qf

the charge is 4Q
therefore I=4Qf

[yasser37]

Thanks m8

this is all left that I didn't know

Hi guys

so I've done past papers and I have some doubts


Nov 01
37
June 02
14
18
23
35
40

NOV 02
24
34
36
40

June 03
5
22

Nov 03
13
24
31
40

June 04
18
26


Nov 04

11
18 ( what is the equation used here? )
20
37


June 05
4
13

Nov 05
34

[hesho21]

Guys plz number,10,13,22 in the attached paper

[astarmathsandphysics]

10 Force*time=change in momentum
-60*0.5=mv-,u
-30=30v-90
v=2
13 tension =torque/radius=3/0.1=30
torque at P=force*eadius=30*0.15=4.5Nm
22 Y=Fl/Ae
e/l=F/AY=20/(5.1*10^-43.14*(2.5*10^-4)^2*2*10^11)=0.051%

[hesho21]

Thanks man that really helped 

[halosh92]

10 Force*time=change in momentum
-60*0.5=mv-,u
-30=30v-90
v=2
13 tension =torque/radius=3/0.1=30
torque at P=force*eadius=30*0.15=4.5Nm
22 Y=Fl/Ae
e/l=F/AY=20/(5.1*10^-43.14*(2.5*10^-4)^2*2*10^11)=0.051%

for q13  how come we dont use the diameter here?

[hesho21]

for q13  how come we dont use the diameter here?

No u actually use the diameter, its just a small mistake anyway number 22 I still didnt get it !

[halosh92]

No u actually use the diameter, its just a small mistake anyway number 22 I still didnt get it !

but then again the answer would be different, if u use the diamter instead of radiius right 

[hesho21]

but then again the answer would be different, if u use the diamter instead of radiius right 

Yeh of course u should use the diameter because its the perpindicular distance between the two forces.If u use the radius u would get the wrong answer
if u can explain the method for 22 plz i still have difficulty with it thanx