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Physics: Post your doubts here!

Tkp

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May june 11(12)-29,34,16,8
 

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Tkp

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May june 11(13)-27,14
 

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Tkp

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well what is the ans?d
if its d then stress =f/a
and the radius is1/4 and the area is 1/16
so f/1/16=16f
and the orginal 1s radius is 1/2 and area is 1/4 so stress is 4f so it increases to 4
and the strain=extension/original length
so diameter changes doesnt affect strain
 
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Thank u soo much.....could u please explain these questions 2...
M j 2012 paper 11 question 12,18,23
M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
May ALLAH bless u.

12) you need to know mechanics m1 for this method: 120g-t=120a and ( T stands for tension in rope which is constant)
t-80g=80a , add both equations t gets cancelled off and find a using g=10ms-2, use u=0 s=9m because at half way only the man's head is at the bottom of the barrel as acceleration is constant for both of them, find v using v^2=u^2+2as

q18) output energy / input energy
output is in power convert it into energy using p=E/t, pxt=E, and obviously input is 40Mj x 20
 
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Just the use of a basic principle .
Resistance = resistivity x length/Area.
Remember if length decreases then area increases . by the same factor .
Here the length is halved so the area will become twice of that as before .



For dia 1 : R1= l/A

For dia 2 :
R2=0.5l/2A

if we rearrange the eq for dia 1 then we get 4R2 = l/A

Substitute l/A to R1 : 4R2=R1

R1 = 8
4R2=8
R2=8/4 = 2 Ohms

Hope you got it :)
 
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