1proper solution:
R/R'=l/l'
=(\rho l/\pi r^2)/(\rho (2l)/\pi (r/2)^2)
=1/8=l/l'
therfore l'=8l.............
hence solved.....;-))))
a metre bridge is used to determine d resistance of an unknown wire by measuring d balance point L.if d wire is replaced by another wire of same material with double d length n half d thickness d balancin point is expected to be...........................................
pls help me....
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28 Answers
1
62ramkumar.. i think what you are doing is to extend the meter wire..
but i think the question has some flaws and i dont seehow it can be solved.
1The question falters somewhere......
No conditions giv an answer from the options[92][79]
1since material being same d resistivity also remains constant....
R is PROPORTIONAL TO l ....
so it shld change similarly.................
62ramkumar..
resistance becomein 8R does not mean that balancing length changes similarly.. can you prove this claim?
1R/R'=l/l'
R=\rhol/A
R==\rhol/pi r2
l'=2l
r'=r/2
on substitutin d values we get
R'=8R
THEREFORE THE NEW BALANCIN LENGTH IS EXPECTED TO BE 8l
11Even i didn't understand my answer bhaiya......i thot i had understood the question but it seems i didn't afterall!!!
62I am not convinced with that answer Ani..
I guess either I am not able to understand the question or the question is not able to udnerstand me :D
I am very close to assuming that there was a mistake inteh VITEEE2008 paper!
11half, the thickness means....1/2 d.ohhhhhhhh
SORRY MACHAN
So answer is L/8??
1The balance point would be obtained at 2L as the thickness of wire is uniform, (whatever be the width it is uniform). As the resistance remains same, and length of wire( say d) doubles.
L/(d-L) = l/(2d-l)
→l=2L
Is that right??
11Ya i noticed.....but,Bhaiya i think he meant (\frac{1}{4})L
1but ani u made a mistake in calculatin d new resistance.........
11Let original resistance per unit length be δ
Since thickness (diameter) is halved, area will be 1/4 original area....
SO the new resistance will be =Ï(2l)/4A =R/2
Hence new resistance per unit length will be \frac{R}{2(2l)}=R/4l= δ/4
Thus for the same R, new balancing length must be L/4
1practically thinkin ll d balancin length ll be lesser or greater fr d above conditions???????????????????????
62you have to make a balanced wheatstone
so the ratio of the resistances should be equal....
L/(1-L)= r/R
new r'=8r (double length and half thickness)
r'/R=8r/R = x/(1-x)
so 8L/(1-L)=x/(1-x)
8L-8Lx=x-Lx
8L=1x+7Lx
x=8L/(1+7L)
I dont know how the above answer came!
Correction after Ani's post!)
1NO....i m bit confused .... weak in solvin experimental prob...
i ll give u d option..
a)1/8L
b)1/4L
c)8L
d)16L