Question 161:
easyA charged particle having drift velocity of $7.5 \times 10^{-4} \text{ m s}^{-1}$ in an electric field of $3 \times 10^{-10} \text{ Vm}^{-1}$, has a mobility in $\text{m}^2 \text{ V}^{-1} \text{ s}^{-1}$ of :
(2020)
Mobility $\mu = \frac{v_d}{E}$. Substituting the given values, $\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^6 \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1}$.
Question 162:
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 combination, $R_{eq} = R_1 + R_2$. Using $R = \frac{l}{\sigma A}$, we get $\frac{2l}{\sigma_{eq} A} = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A}$. Therefore, $\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2}$, which gives $\sigma_{eq} = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$.
Question 163:
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)
The energy gained by an electron is $E_k = e E \lambda$. Therefore, the electric field is $E = \frac{E_k}{e \lambda} = \frac{2 \text{ eV}}{e \times 4 \times 10^{-8} \text{ m}} = \frac{2}{4 \times 10^{-8}} \text{ V/m} = 0.5 \times 10^8 \text{ V/m} = 5 \times 10^7 \text{ V/m}$.
Question 164:
easySpecific resistance of a conductor increases with:
(2002)
Specific resistance (resistivity) is an intrinsic property of the material and its temperature. For conductors, resistivity increases with an increase in temperature due to the increased frequency of collisions of charge carriers. It does not depend on dimensions.
Question 165:
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)
The given resistance is $47 \text{ k}\Omega \pm 10\% = 47 \times 10^3 \Omega \pm 10\%$. According to the colour code, the first digit 4 corresponds to Yellow, the second digit 7 corresponds to Violet, the multiplier $10^3$ corresponds to Orange, and the tolerance $10\%$ corresponds to Silver.
Question 166:
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)
Let the resistance of each individual wire be $R$. In parallel, the equivalent resistance is $R_{eq} = \frac{R}{4} = 0.25 \Omega$, which gives $R = 1 \Omega$. When these four wires are connected in series, the effective resistance is $R_s = 4R = 4(1) = 4 \Omega$.
Question 167:
easyA student measures the terminal potential difference ($V$) of a cell (of emf $\varepsilon$ and internal resistance $r$) as a function of the current ($I$) flowing through it. The slope, and intercept, of the graph between $V$ and $I$, then, respectively, equal:
(2009)
The terminal potential difference equation is $V = \varepsilon - Ir$, which can be rewritten as $V = -rI + \varepsilon$. Comparing this with the straight-line equation $y = mx + c$, the slope $m$ is $-r$ and the y-intercept $c$ is $\varepsilon$.
Question 168:
easyTwo cells, having the same emf, are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2$ ($r_1 > r_2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:
(2006)
Total current $I = \frac{2E}{R + r_1 + r_2}$. The potential difference across the first cell is $V_1 = E - Ir_1 = 0$, meaning $E = Ir_1$. Substituting $I$, we get $E = \frac{2E r_1}{R + r_1 + r_2}$. Simplifying gives $R + r_1 + r_2 = 2r_1$, so $R = r_1 - r_2$.
Question 169:
easyFor a cell terminal P.D. is $2.2 \text{ V}$ when circuit is open and reduces to $1.8 \text{ V}$ when cell is connected to a resistance of $R = 5 \Omega$. Determine internal resistance of cell ($r$):
(2002)
Internal resistance is given by $r = R(\frac{E}{V} - 1)$. Substituting $E = 2.2 \text{ V}$, $V = 1.8 \text{ V}$, and $R = 5 \Omega$, we get $r = 5(\frac{2.2}{1.8} - 1) = 5(\frac{0.4}{1.8}) = 5(\frac{2}{9}) = \frac{10}{9} \Omega$.
Question 170:
easyA car battery of emf $12 \text{ V}$ and internal resistance $5 \times 10^{-2} \Omega$ receives a current of $60 \text{ A}$ from external source, then terminal voltage of battery is:
(2000)
During charging, current is forced into the positive terminal, so terminal voltage $V = E + Ir$. Substituting the values, $V = 12 + (60 \times 5 \times 10^{-2}) = 12 + 3 = 15 \text{ V}$.