Question

Two copper wires have length in the ratio 1:2 and masses in the ratio 3:4. The ratio of their
resistances is
fa) 1:3
(b) 3:16
(C) 16:3
(d)8:3​

Answer 1

Answer:

Option (a) 1: 3 is the correct answer

Explanation:

Given,  ratio of length of copper is 1 : 2 and

ratio of mass of copper is 3: 4

Let $m_1 = 3 s and \ m_2 = 4 s$ and

let $l_1 = 1kand \ l_2 = 2k$   ,where s and k are constant .

And we have

R  ∝ $\frac{l}{A}$

⇒R = $p\frac{l}{A}$    = $p \frac{l.l}{A.l } = p \frac{l^{2} }{V}$ = $p\frac{l^{2} }{m} d$    

[∴A × l = volume and V = $\frac{density }{mass}$]    

Therefore , we get ,

$R_1= p \frac{l_1^{2} }{m_1} d \ and R _2 = \ p \frac{l_2^{2} }{m_2}d$

Now ratio of resistances is ,

$\frac{R_1}{R_2}$ = $p \frac{l_1^{2} }{m_1} d$ × $\frac{m_2}{p.l_2^2.d}$ = $\frac{l_1^2}{m_1}$ ×$\frac{m_2}{l_2^2}$          [where  p and d are cancel out ]

⇒$\frac{R_1}{R_2 } = (\frac{l_1}{l_2} )^2$×$\frac{m_2}{m_1}$

⇒$\frac{R_1}{R_2 } = (\frac{k}{2k} )^2$ ×$\frac{4s}{3s}$ = $\frac{1}{4}$ ×$\frac{4}{3}$ = $\frac{1}{3}$

Hence, the ratio of resistance of the given two copper is  1: 3 .

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