A uniform wire of length l and radius r has resistance 100 Ω . It is recasted into a thin wire of (i) length 2l (ii) radius (r/2.) The resistance of the new wire in each case will be :

Question

A uniform wire of length l and radius r has resistance 100 Ω . It is recasted into a thin wire of (i) length 2l (ii) radius (r/2.) The resistance of the new wire in each case will be : (A) 200 Ω 400 Ω  (B) 100 Ω 200 Ω  (C) 400 Ω 160 Ω  (D) 400 Ω 1600 Ω

Tardigrade Admin · Accepted Answer

(i) Since, given wire of fixed mass is recanted to twice the length

R \propto l^{2}

.....(i)
Let R' be the new resistance

R^{'} \propto l^{'}^{2}

...(2)
Hence,

\frac{R ^{'}}{R} = (\frac{l ^{'}}{l})^{2}

or

R^{'} = (\frac{l _{1} ^{'}}{l})^{2} R

Here,

R = 100Ω l^{'} = 2 l

So,

R^{'} = (\frac{2 l}{l})^{2} \times 100 = 400Ω

(ii)
Now in terms of radius

R \propto \frac{1}{r ^{4}}

Here,

r^{'} = \frac{r}{2}

Hence,

\frac{R ^{'}}{R} = (\frac{r}{r ^{'}})^{4}

or

R^{'} = (\frac{r}{r ^{'}})^{4} \times R

So,

R^{'} = (\frac{\frac{r}{r}}{2})^{4} R = 16 R

= 16 \times 100 = 1600Ω

Q. A uniform wire of length $l$ and radius $r$ has resistance $100 Ω$ . It is recasted into a thin wire of (i) length 2l (ii) radius $\frac{r}{2.}$ The resistance of the new wire in each case will be :

200 $Ω$ 400 $Ω$

100 $Ω$ 200 $Ω$

400 $Ω$ 160 $Ω$

400 $Ω$ 1600 $Ω$

Q. A uniform wire of length l and radius r has resistance 100Ω . It is recasted into a thin wire of (i) length 2l (ii) radius 2.r​ The resistance of the new wire in each case will be :

200 Ω 400 Ω

100 Ω 200 Ω

400 Ω 160 Ω

400 Ω 1600 Ω

Q. A uniform wire of length $l$ and radius $r$ has resistance $100 Ω$ . It is recasted into a thin wire of (i) length 2l (ii) radius $\frac{r}{2.}$ The resistance of the new wire in each case will be :

200 $Ω$ 400 $Ω$

100 $Ω$ 200 $Ω$

400 $Ω$ 160 $Ω$

400 $Ω$ 1600 $Ω$