metre bridge...

@byah.. frm viteee 2008............

proper 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.....;-))))

byah.... wats d conclusion??????????

The question falters somewhere......

No conditions giv an answer from the options[92][79]

since material being same d resistivity also remains constant....

R is PROPORTIONAL TO l ....

so it shld change similarly.................

R/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

Even i didn't understand my answer bhaiya......i thot i had understood the question but it seems i didn't afterall!!!

For wire::
R_initital=ρl/Ï€(d/2)²
R _final=2ρl/Ï€(d/4)²

half, the thickness means....1/2 d.ohhhhhhhh

SORRY MACHAN

So answer is L/8??

The 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??

wat calc. mistake??

Ya i noticed.....but,Bhaiya i think he meant (\frac{1}{4})L

but ani u made a mistake in calculatin d new resistance.........

I am also weak in this part,,,,, so don't trust my answer

Let 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 =ρ(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

practically thinkin ll d balancin length ll be lesser or greater fr d above conditions???????????????????????