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Physics: Post your doubts here!

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hi, this is my try:-

2) in this mcq you are given with the angle and horizontal component.use it to find the vertical comp.
Tan30 = Perpendicular / 20 , it comes 11.5N i.e C.

13) no convincing ans but the weight is making a torque of 180 Nm, how? 900 N into .20m (.20 is the perpendicular distance or the radius of the disc)
same torque will be shifted upwards in order to lift this weight i.e 180 = F into 1.20 m it comes 150 N hence B. ( i dnt knw why i took .20 as the perp distance to find the torque created by weight)

14) Power = F v , so k = P over v^3 because you need to add another v in the denominator as F = Power over velocity.

15) what i get is dat if u c in both vessels the height will decrease by half from the original height in the vessel X, means h will bcm h/4 altogether, so the answer will be B mgh/4.

18) pressure = pgh , p ang remain constant. u know that h has increased in the right side of manometer. i.e pressure has increased. there is only option D 2pgh which supports this idea. other three options shows decrease or no change in the pressure.

20) extension directly proportional to length and inversely proportional to area. length is twice so this increase the extension by 2, area has decreased as well by half so this also doubles the extension, total extension raised by 4 times. the ratio of the tension will be 4 /1. (not so clear i know)

21) strain energy = 1/2 F into x so 1/2 into 17 into 30 /100 it comes around 2.5 J but the best estimate would be less then that i.e 2 Joules which is nearest to the whole number. 3J would be very higher i.e A option

25)i dnt understand :mad: , it shud be C but its A, d sin Q = n lambda , this s what comes after moving here nd dre things
n/N = sin Q / Lambda

33) U must see that the two cells are not in one direction, cancelling each others voltage. i.e subtract these two 3V- 1.2 V = 1.8 Volt , now use this to find the current in loop 1 which has 9 ohm resistor. I = 1.8 / 9 = .20 Ampere i.e B.
Thank you so much.....:)
 
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Assalamu alaikum,
I expect you have also done the first part of this question. You get E=deltaV/delta r or E= V/r or in other words, E is the gradient of V and r graph.
You can also sketch a graph on a rough paper , and as a rough work in exam which you can erase afterwards, by taking several values of r, or x which is the same thing in this case. The graph appears to be somewhat like this.(uploaded in the picture).
You get the idea of how the graph is sketched. The graph of E against r is steeper than the graph V against r axis. In the question, you draw a graph which is steeper and a little more downwards than the V against x graph.
Note: In the graph or E against r axis and V against r axis, in the uploaded picture, value of k comes out to be one.
You might find it better to confirm this answer. I don't pretend to be an expert on graphs though.
brother i assked for QUESTION 6 part b TRANFORMER WALE K GRAPHS
 
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guyz can somebody post the answers to this attachment qs plz plz
ther topical mcqs...attachment or post or anything plz

or if not can u guyz plz plz at least post the option answers of the following Qs

for Physical quantities...ans of Qs 4, 6, 11,13, 23, 25, 27, 32, 33, 47


and for States of Matter ans for Qs 13, 17, 19, 27
thanx for reading...
 

Attachments

  • 9702_p1_properties_of_matter_all.pdf
    1.6 MB · Views: 8
  • 9702_p1_units_and_measurements_all.pdf
    2.2 MB · Views: 3
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- In the part ii) to find moment, why are we using 'sin' 18?




AOA
gain in g.p.e depends up on the height to which the object raised.

so for calculating the increase in height from the initial position we have to use the formula 61.cos(18

by this we will get the distance covered by the string which is 58cm
now how much is the string raised initially?? subract 61 - 58 U will get 3cm.

this 3cm is the increase in height from ground.

NOW calculate g.p.e = mgh (51/1000)(9.81)(3/100) = 1.5*10^-2Ans
 
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Assalamoalaikum wr wb!
I need help with Q:4c of Nov:2007 Paper:2
I don’t understand the last step…:confused:
Mark scheme
once we get
stress = force / area so area = force / stress
area = (1.9 × 103) / (9.5 × 108)
= 2.0 × 10–6 m2
now this is the minimum area the rod should have, so that it does not break.(lesser shall it be, the rod would break.) and this can only be such when the area of bubble is its maximum.
(max) area of cross-section = (3.2 – 2.0) × 10–6
= 1.2 × 10^–6
when bubble has 1.2 x 10^-6, rod has 2 x 10^-6, a total of 3.2 x 10^-6
 
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