Question 181:
easyTwo metal wires of identical dimensions are connected in series. If $\sigma_1$ and $\sigma_2$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is
(2015 Re)
In series connection, $R_{eq} = R_1 + R_2 \implies \frac{2l}{\sigma_{eq} A} = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A}$.
$\frac{2}{\sigma_{eq}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \implies \sigma_{eq} = \frac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$.
Question 182:
easyThe mean free path of electrons in a metal is $4 \times 10^{-8}\text{ m}$. The electric field which can give on an average $2\text{ eV}$ energy to an electron in the metal will be in units $\text{V/m}$.
(2009)
Energy $W = e \cdot E \cdot \lambda \implies 2\text{ eV} = e \cdot E \cdot (4 \times 10^{-8}\text{ m})$.
$E = \frac{2\text{ V}}{4 \times 10^{-8}\text{ m}} = 5 \times 10^7\text{ V/m}$.
Question 183:
easySpecific resistance of a conductor increases with:
(2002)
Specific resistance (resistivity) is an intrinsic property of a conductor that depends on temperature and material, increasing as temperature rises.
Question 184:
easyA carbon resistor of $(47 \pm 4.7)\text{ k}\Omega$ is to be marked with rings of different colours for its identification. The colour code sequence will be
(2018)
$R = 47 \times 10^3\ \Omega \pm 10\%$.
First digit 4 $\rightarrow$ Yellow, second digit 7 $\rightarrow$ Violet, multiplier $10^3 \rightarrow$ Orange, tolerance $10\% \rightarrow$ Silver.
Sequence: Yellow - Violet - Orange - Silver.
Question 185:
easyThe effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is $0.25\ \Omega$. What will be the effective resistance if they are connected in series?
(2021)
For 4 identical wires in parallel, $R_p = \frac{R}{4} = 0.25\ \Omega \implies R = 1 \Omega$.
Connected in series, $R_s = 4 R = 4 \times 1 = 4 \Omega$.
Question 186:
easyA circuit contains an ammeter, a battery of $30 \text{ V}$ and a resistance $40.8 \Omega$ all connected in series. If the ammeter has a coil of resistance $480 \Omega$ and a shunt of $20 \Omega$, the reading in the ammeter will be:
(2015 Re)
Total resistance of ammeter $R_A = \frac{480 \times 20}{480 + 20} = 19.2 \Omega$. Total resistance of circuit $R_{eq} = 40.8 + 19.2 = 60 \Omega$. Reading of ammeter $I = \frac{V}{R_{eq}} = \frac{30}{60} = 0.5 \text{ A}$.
Question 187:
easyWhen a wire of uniform cross-section $a$, length $l$ and resistance $R$ is bent into a complete circle, resistance between two of diametrically opposite points will be:
(2005)
The wire is divided into two equal semicircular parts, each having a resistance of $R/2$. Since these two parts are connected in parallel across diametrically opposite points, the equivalent resistance is $R_{eq} = \frac{(R/2) \times (R/2)}{(R/2) + (R/2)} = R/4$.
Question 188:
easyResistances $n$, each of $r \Omega$, when connected in parallel give an equivalent resistance of $R \Omega$. If these resistances were connected in series, the combination would have a resistance in ohms, equal to:
(2004)
The equivalent resistance in parallel is $R = \frac{r}{n}$, which implies $r = nR$. The equivalent resistance in series is $R_s = nr$. Substituting the value of $r$, we get $R_s = n(nR) = n^2R$.
Question 189:
easyTwo wires of the same metal have same length, but their cross-sections are in the ratio $3 : 1$. They are joined in series. The resistance of thicker wire is $10 \Omega$. The total resistance of the combination will be
(1995)
Resistance $R \propto \frac{1}{A}$. Given $\frac{A_1}{A_2} = \frac{3}{1}$ and the thicker wire's resistance $R_1 = 10 \Omega$. The thinner wire has resistance $R_2 = 3R_1 = 30 \Omega$. Total series resistance $R_s = R_1 + R_2 = 10 + 30 = 40 \Omega$.
Question 190:
easyThree resistances each of $4 \Omega$ are connected to form a triangle. The resistance between any two terminals is
(1993)
Between any two terminals of the triangle, two $4 \Omega$ resistors are in series, giving $8 \Omega$. This combination is in parallel with the third $4 \Omega$ resistor. Equivalent resistance $R_{eq} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = 8/3 \Omega$.