Physics P1 Questions

[ds29]

Hey I've got some doubts on this Physics MCQ paper if someone could clear it up for me, I'd be grateful
All are from ON 09 Paper 12

Q 8 - I don't how to do it
Q 12 - Don't understand this question
Q 20 -

Q 21 - I get the right answer but I'm not sure if my method is right.

What I did was Y = (FL) / (eA)
5e5 = (20*L) / (e*A) where A is you divide the diameter by 2 square it and multiply it by Pi.
Simplifying you'll get 625Pi = (L/e)
Since they wan't e/L you take the inverse of it and you get 5.09*10^-4 and multiply that by 100 to get the percentage.

and Q 25.

Thanks

 

[ds29]

anyone?

 

[sse2010]

for Q8
first remember dat da momentum needz to be conservd n the momentum b4 collision = 2mu-mu=mu
so da momentum after collision shud be mu. this is in option A n C .( option D is wrng as shws da result of inelastic colision)
now remember dat reletive velocity of aproach = reltive velocity of seperation.
therefor reletive velocit before= u+u= 2u
da reletive velocity is 2u only in A. so A is corect....hope u get da answer

 

[sse2010]

FOR q20
assume dat the line till P from origin is a straight line... since strain energy stored is= the area under the force extension graph
the area below the straight line ( strain energy)= 1/2 x 2x10^-3 x100= 0.1 J (convert mm into m)
now since we have underestimated da total area below da curve by taking da straight line, the total area wud be a lil more dan 0.1J, so C...obviously it cannot be 0.2 J cuz dat representz da area under da square ie 100 x 2 x10^-3=0.2 J

 

[sse2010]

for q 21.
wat da q asks is to find extension or compression/original lengh x100%
since extension or compression / lentgh = strain...therefore we first have to find strain.
calculate da area = pi d^2/4 ....then calculate stress ie force/area
now calculate strain = stress/yung modulus
now da percentage of l dat contractz = strain x 100%

 

[sse2010]

for q 25
since da formula for difraction grating=> a sin@=nxwavelntgh(a is seperation)
a=1/300000 = 3.33x10^-6
now when ever we have to find da totlal number of orders of difraction "n" we put @=90'
=> 3.33x10^-6 x sin90 = n x 450x10^-9
=>n=7.4....therfore total orders of difraction wud be 7....dis means dat 7 maxima wud be above da central maxima n 7 maxima below da central maxima.. since da q asks for total no. of maxima...dont forgot to count da central maxima which alwayz exists in a difration gratin... therfore total wud be 7 above+ 7 below+ central maxima= 15.....
hope u gpot da answer!!

 

[ds29]

Thanks a lot for the help!
If you could also answer this one.
MJ 09 Q 15

 

[PlanetMaster]

May/June 2009 Paper 1 MCQ 18:
Change in height = h+h
Increase in pressure = (h+h)ρg
Increase in pressure = 2hρg

 

[ds29]

May/June 2009 Paper 1 MCQ 18:
Change in height = h+h
Increase in pressure = (h+h)ρg
Increase in pressure = 2hρg

I think you saw the wrong question admin. I asked for help in Q 15
Thanks anyways

 

[PlanetMaster]

Yeah sorry about that!

May/June 09 Paper 01 MCQ 15:
After tap is opened, water in Y has height h/2 and mass m/2.
So potential energy lost from X = Potential energy gained in Y
= (m/2) x g x (h/2)
= (mgh)/4

 

[mz23]

can anyone please help in question number 21, november 06? please be quick!

 

[mz23]

nobody???

 

[ds29]

Ok P = h$g where P = Pressure h = height $ = density and g = gravitational

We know total Pressure is 17.5 * 10^6 Pa

We first consider Pressure by Oil
P = x * 830 * 9.81
P = 8142.3x

Next we consider Pressure by water
Here we have to consider height as (2000-x) since we are given total length is 2000m

P = (2000-x) * 1000 * 9.81
= 1.962 * 10^7 - 9810x

Now both of these pressures = Total Pressure
Therefore
8142.3X - 9810X = 17.5*10^6 - 1.962 * 10^7
-1667.7X = -2.12* 10^6
X = 1271m
Approximate equal to 1270 and therefore answer is D


Also thanks Admin for answering my question. I would have not figured that one out.

 

[ds29]

Can someone explain O/N 04 P1 Q20
Thanks

 

[intel1993]

O/N 2004.............

as the tubes are open frm both sides so preesure si same on both sides due to atm....................

so Pgh= Qgh

g cancels

p(2)= Q(1)

P/Q= 1/2

so the ans is A........

 

[ds29]

Thanks