262@swaraj - we have to consider that there is no viscous force .
the block comes to rest momentarily at the maximum displacement (as rishab said)
21Q 8)
Fing the initial potential as kQ2/2r i.e the P.E. of charges in a metallic shell.
Now after introduction of charge q, the charges redistribute as -q in the inner surface and q+Q on outer surface.
Now total P.E.= PE of shell with radius r and charge -q + PE of shell with radius 2r and charge (q+Q) + the 3 combinations with the central charge q.
1@swaraj can u post a bit more detailed solution fr Q 8 ? "5" other potentials ?
21Vivek:I didn't get exactly what you are trying to say.
In my approach The work done in bringing the extra charge=Change in potential energy of the system(including the change of potential energy caused to to redistribution of charges in the shell)
71Sorry Swaraj, But 1 is incorrect. As I already said it is a very good sum. The formula W = q ΔV doesn't works here.
The reason for it is that ' Your approach doesn't takes into account the additional work done by the Electric forces in changing the charge configuration upon the shell '
AS the point charge q is being taken out, the charges on surface of shell redistribute and this requires some energy.
21In Q 8 initially find Potential.
When charge is inserted the charges redistribute and the final Potential is the sum of 5 other potentials.
Work done=Diff. in P.E.
262@swaraj - thnx for Q10 .
tell me whether you applied conservation of energy in Q8
218)I got ans as 1.
4)@rishab:The Q says that at t=0 the switch pos. is changed.In the meantime it is possible that the current value decreases from max.It never mentions that the switch is changed instantaneously.
1i guess Q4 is wrong.
because the value of 'I' at t=0 and t=∞ should be same. because no matter what the inductance is , the max current will be E/R.
718. It is a very very good question. Infact I'll try to find out the original detailed question.
2110) magnetic field is proportional to q(VxR)
Hence magnetic field at a pt will be zero when velocity and displacement of the charge from that pt. are in same direction.
Simply draw 2 tangents from that pt.(at pt of tangency VxR=0)
Angle subtended by arc between pts of tangency=120
Hence min. time=9/3=3(ans)
71Why should there be no viscous force ?.. See the question clearly mentions it, ' in a viscous liquid .. .'
It's like damped oscillation.
215)A planet revolves around the sun in an elliptical path because of the influence of other bodies.
Hence the energy equation wont hold valid.
We can use keepers law,that states a planet sweeps equal area's in equal time.
Consider time dt and path traced out by the planet as a sector.
1/2*(a-ae)Vp=1/2*(a+ae)Va
Vp/Va=1+e/1-e ...option (A)
17) the block when pulled slightly from it's mean position would undergo SHM and the moment it comes to rest is when it is at it's amplitude,
=> kx0 + mg/2 = mg
=> x0 = mg/2k
so 'x' when the block will come to rest instantaneously = 2x0 = mg/k ..(1)
now applying work energy theorum,
ΔK.E = work done by all the forces
=> 0 = Wspring + mgx - mgx/2
=> Wspring = - mgx/2 = - m2g22k
21Aditya for 7 work done will be both by viscosity and upthrust(as it does not oscillate but comes to rest)
262change in PE of spring + change in PE of block + Work done by Buoyant force = 0
71Still my answer for 4 remains same.
@Aditya.. In question 7 , what have you done is second line.. could you tell please.
I had tried the same.. ie., PE lost = PE Stored+Work done by liquid .. Is it same? then why didn't I get the answer.!! :(
Infact it has nothing to do with the final answer..
1057we know that v=√Tm
so we are getting v =70m/s
now the consecutive frequencies are in the ratio 3:5:7..
now it is surely a closed pipe.....
the fundamental frequency(f) is 105/3=35 Hz...
wavelength=vf=7035=2m
262@ketan - Ans is 2m but i am getting 1m . plz post ur soln.
4. Ans is 3
7. Buoyant force = FB = Mg/2
at rest position (v=0) , 1/2*kx2 +FBx = Mgx
x= Mg/k
work done by liquid = FB*net displacement = - FB*x = - m2g22k
717. If the spring is relaxed i.e, net force on it is zero, then why would it move when left. IT has to be displaced a bit. Now how you have 8 as answer is that you have assumed that total initial PE stored in the spring (m2g28k) is lost. This is not so because when the block comes to rest, the spring will still have some potential energy.
2627) u got - 8
i got - 4
given - 2
btw, i think relaxed means no compression in the spring.
217)I am getting n=8.
Consider the fact that it says that the block is released when spring was relaxed(at rest).i.e it was already extended by x=mg/k when block was released in liquid.
