`1` meter long metallic wire is broken into two unequal parts `P and Q. P` part of the wire in uniformly extended into another wire `R`. Length of `R` is twice the length of `P` and the resistance of `R` is equal to that of `Q`. Find the ratio of the resistance `P` and `R` and also the ratio of the length `P and Q`

Explanation

Since mass or volume of wire remains same when area or length of wire is changed. For solving these problems. Keep in mind that volume of wire remains same.
`cdot` For % change, use `(("final value")/("initial value") - 1) xx 100`
`cdot` If necessary, use binomial theorem to solve
(a) (i) `A_1 l_1 = A_2 l_2`
`A_l = A_2. 2l rArr A_2 = A//2`
`R_1 = (rho l)/(A), R_2 = (rho l_2)/(A_2) = (rho. 2l)/(A//2) = (4 rho l)/(A)`
`R_2 = 4 R`
(ii) `A_1 l_1 = A_2 l_2`
`A l = 2 A l_2 rArr l_2 = l//2`
`R_1 = (rho l)/(A), R_2 = (rho l_2)/(A_2) =(rho l//2)/( 2A) = (rho l)/(4 A)`
`R_2 = ( R)/(4)`
(b) `A_1 l_1 = A_2 l_2`
`pi r^2 l = pi (2r)^2 l_2 rArr l_2 = 1//4`
`R_1 = (rho l)/(pi r^2), R_2 = (rho l_2)/(A_2) = (rhol//4)/(4 pi r^2) = (1)/(16) (rho l)/(pi r^2)`
`(R_1)/(R_2) = 16`
( c) `l_1 = l, A_1 = A, l_2 = l (1 + 0.1//100), A_2 = A'`
`A_1 l_1 = A_2 l_2 rArr Al = A' l(1 + 0.1//100)`
`A' = (A)/((1 + 0.1//100))`
`R = (rho l)/(A), R' = (rho l_2)/(A_2) = (rho l(1 + 0.1//100))/(A//(1 + 0.1//100)) = (rhol)/(A) (1 + (0.1)/(100))^2`
`(R')/(R) = (1+ (0.1)/(100))^2 = 1 + (2 xx 0.1)/(100) = 1 + (0.2)/(100)`
`(R')/(R) -1 = (0.2)/(100)`
% change `= ((R')/(R) -1) xx 100 = (0.2)/(100) xx 100 = 0.2 %`
(d) `l_1 = l, A_1 = A`
`l_1 = l(1 + (25)/(100)) = (5 l)/(4), A_2 = A'`
`A_1 l_1 = A_2 l_2`
`Al = A' (5l)/(4) rArr A' = (4A)/(5)`
`(R_2)/(R_1) = ((rho l_2)/(A_2))/((rho l_1)/(A_1)) = (l_2)/(l_1) .(A_1)/(A_2) = (5l//4)/(l).(A)/(4 A//5) = (25)/(16)`
`((R_2)/(R_1) -1) xx 100 = ((25)/(16) -1) xx 100 = (900)/(16) = 56.25 %`
(e) Let length of `P`to `l_0`
`R_p = (rho l_0)/(A), R_Q = (rho(1 - l_0))/(A)`
`R_R = (rho.2 l_0)/(A//2)`
=`4(rho l_0)/(A)` [since, `P`, is converted i nt o, `R`, by, doubli ng],[its l eng ths, hence, area, of `R`, becomes, half]
`(R_p)/(R_R) = (1)/(4)`
`R_Q = R_R rArr (1 - l_0) = 4l_0 rArr l_0 = 0.2 m`
`(l_p)/(l_Q) = (0.2)/(1-0.2) = (1)/(2)`
(f) (i) Between square faces `l = l_0, A = a xx a`
`R = (rho_l)/(A) = (rho l_0)/(a^2) = (50 xx 10^-8 xx 0.5)/(0.01 xx 0.01) = 2.5 xx 10^-3 Omega`
(ii) Between rectangular faces `l = a, A = l_0 xx a`
`R = (rho l)/(A) = (rhoa)/(l_0 a) = (rho)/(l_0) = (50 xx 10^-8)/(0.5) = 10^-6 Omega`
(g) Resistance between faces of area
(i) `(l xx b), R_1 = (rho h)/(l b)`
(ii) `(b xx h), R_2 = (rho l)/(bh) = R_(max)`
(iii) `(l xx h), R_3 = (rho b)/(l xx h) = R_(min)`
`(R_(max))/(R_(min)) = (l^2)/(b^2) = ((4b)^2)/(b^2) = 16`
(h) `m_1 = m, m_2 = 3m,m_3 = 5 m`
`l_1 = 5l, l_2 = 3l, l_3 = l`
Density and resistivity of each wire is same. Here `d` : density of wire and `rho` : resistivity of wire.
Mass = density `xx` volume = density `xx` area `xx` length
Mass `alpha ("area" xx "length")`
`m_1 = m= kA_1 l_1 = kA_1 5l rArr A_1 = (m)/(5 kl) = (lamda)/(5)`
`m_2 = 3m = kA_2 l_2 = kA_2. 3l rArr A_2 = (m)/(kl) = lamda`
`m_3 = 5m = kA_3 l_3 = kA_3. l rArr A_3 = (5m)/(kl) = 5 lamda`
`R_1 : R_2 : R_3 : = (l_1)/(A_1) : (l_2)/(A_2) :(l_3)/(A_3)`
=`(5l)/(lamda//5) : (3l)/(lamda) : (l)/(5 lamda)`
=`25 : 3 :1//5`
=`125 : 15 : 1`.