Question

The same mass of copper is drawn in two wires 1 mm and 2 mm thick. Two wires are connected in series and current is passed through them. Heat produced in the wires is in the ratio.
(A). $2:1$
(B). $1:16$
(C). $4:1$
(D). $16:1$

Answer 1

Hint: In this type of question we have to take ratios of two quantities and compare each other using formulas. The resistance of wire is given by, $R=\dfrac{\rho l}{A}$ and heat resistance of the wire is given by $\text{H=}{{\text{I}}^{\text{2}}}\text{RT}$. Now, using this formula and taking the ratios of two values we will find the answer.

Formula used: $R=\dfrac{\rho l}{A}$, $\text{H=}{{\text{I}}^{\text{2}}}\text{RT}$

Complete step by step answer:
Now, in the question we are given that, two different wires of radius 1 mm and 2 mm are connected in series and then current is passed through them. So, first of all we will denote two wires as,
${{r}_{1}}=1\ mm$ and ${{r}_{2}}=2\ mm$.
Now, we know that resistance of any wire is given by the formula, $R=\dfrac{\rho l}{A}$ ……………………(i)
Now, we also know that $L=\dfrac{V}{A}$ , where $V\ =\ md$ and $A=\pi {{r}^{2}}$.
Here, V is volume, A is area of wire, r is radius of wire, m is mass, d is density of wire and R is resistance of wire.
Now, substituting the values of L, V and A in equation (1) we will get,
$R=\dfrac{\rho V}{\left( A \right)\left( A \right)}$
$\Rightarrow R=\dfrac{\rho md}{{{\pi }^{2}}{{r}^{4}}}$ …………..(ii)
Now, considering $\rho$, m, and d as constant as in equation (ii) as their values remains same, we can derive the relation between R and r as,
$R\alpha \dfrac{1}{{{r}^{4}}}$ ……………………..(iv)
Now, we know that heat produced is given by
$\text{H=}{{\text{I}}^{\text{2}}}\text{RT}$ …………………………(v)
Where, H is heat produced, I is current passed, R is resistance of wire and T is temperature produced.
Now, here also we will consider I and T as constant as their values remains same, thus we will get relation between H and R as,
$H\alpha R$ ……………………………(vi)
Substituting the value of expression (iv) in expression (vi) we will get,
$\Rightarrow H\alpha R\alpha \dfrac{1}{{{r}^{4}}}$
Thus, we can say that H is inversely proportional to radius of wires, so it is given by,
$\dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{{{r}_{2}}^{4}}{{{r}_{1}}^{4}}=\dfrac{{{2}^{4}}}{1}=\dfrac{16}{1}$
Hence, the ratio of heat produced is $\dfrac{16}{1}$.
Thus, option (d) is correct.

Note: Students must read the questions carefully while solving such questions because students might make mistakes in series and parallel and the sum may go wrong. Here, in this question it was given that wires are series so, the current was same, now if it is given that wires are parallel then current will also be there in consideration while taking ratios. Thus, students must understand the difference between them before solving such questions.