Question 171:
easyThe terminal potential difference of a cell is greater than its emf when:
(1998)
Terminal potential difference exceeds emf ($V > E$) only when the cell is being charged. This occurs when a battery of higher emf is connected in series (opposing) to drive current into the positive terminal of the cell.
Question 172:
easy$n$ equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?
(1989)
Maximum resistance is obtained in a series combination ($R_s = nR$). Minimum resistance is obtained in a parallel combination ($R_p = R/n$). The ratio is $\frac{R_s}{R_p} = \frac{nR}{R/n} = n^2$.
Question 173:
easyA set of ‘$n$’ equal resistors, of value ‘$R$’ each, are connected in series to a battery of emf ‘$E$’ and internal resistance ‘$R$’. The current drawn is $I$. Now, the ‘$n$’ resistors are connected in parallel to the same battery. Then the current drawn from battery becomes $10 I$. The value of ‘$n$’ is
(2018)
Current in series: $I = \frac{E}{nR + R}$. Current in parallel: $10I = \frac{E}{R/n + R}$. Dividing the two equations gives $10 = \frac{nR + R}{R/n + R} = \frac{n(n+1)R}{(n+1)R} = n$. Thus, $n = 10$.
Question 174:
easyThe internal resistance of a $2.1 \text{ V}$ cell which gives a current of $0.2 \text{ A}$ through a resistance of $10 \Omega$ is:
(2013)
Using the relation $I = \frac{E}{R + r}$, we substitute the given values: $0.2 = \frac{2.1}{10 + r}$. Solving for $r$, we get $10 + r = \frac{2.1}{0.2} = 10.5$, which yields $r = 0.5 \Omega$.
Question 175:
easyA wire $50\text{ cm}$ long and $1\text{ mm}^2$ in cross-section carries a current of $4\text{ A}$ when connected to a $2\text{ V}$ battery. The resistivity of the wire is:
(1994)
Resistance $R = \frac{V}{I} = \frac{2}{4} = 0.5\ \Omega$.
Resistivity $\rho = \frac{R A}{L} = \frac{0.5 \times 10^{-6}}{0.5} = 10^{-6}\ \Omega \cdot \text{m} = 1 \times 10^{-6}\ \Omega \cdot \text{m}$.
Question 176:
easyIdentify the set in which all the three materials are good conductors of electricity
(1994)
Copper ($\text{Cu}$), Silver ($\text{Ag}$), and Gold ($\text{Au}$) are all metals with high free-electron density, making them excellent conductors of electricity.
Question 177:
easyThe masses of the wires of copper is in the ratio of $1 : 3 : 5$ and their lengths are in the ratio of $5 : 3 : 1$. The ratio of their electrical resistance is:
(1988)
Resistance $R = \rho \frac{L}{A} = \frac{\rho d L^2}{m} \propto \frac{L^2}{m}$.
$R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5} = 25 : 3 : 0.2 = 125 : 15 : 1$.
Question 178:
easyA copper wire of length $10\text{ m}$ and radius $(10^{-2}/\sqrt{\pi})\text{ m}$ has electrical resistance of $10\ \Omega$. The current density in the wire for an electric field strength of $10\text{ V/m}$ is:
(2022)
Area $A = \pi r^2 = \pi \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2 = 10^{-4}\text{ m}^2$.
Resistivity $\rho = \frac{R A}{L} = \frac{10 \times 10^{-4}}{10} = 10^{-4}\ \Omega \cdot \text{m}$.
Current density $J = \frac{E}{\rho} = \frac{10}{10^{-4}} = 10^5\text{ A/m}^2$.
Question 179:
easyColumn-I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations. (2021)
Column-I:
(A) Drift Velocity
(B) Electrical Resistivity
(C) Relaxation Period
(D) Current Density
Column-II:
(P) $\frac{m}{ne^2\rho}$
(Q) $ne v_d$
(R) $\frac{e E}{m}\tau$
(S) $\frac{E}{J}$
Drift velocity $v_d = \frac{e E}{m}\tau \rightarrow \text{(R)}$.
Resistivity $\rho = \frac{E}{J} \rightarrow \text{(S)}$.
Relaxation period $\tau = \frac{m}{n e^2 \rho} \rightarrow \text{(P)}$.
Current density $J = n e v_d \rightarrow \text{(Q)}$.
Question 180:
easyA charged particle having drift velocity of $7.5 \times 10^{-4}\text{ m s}^{-1}$ in an electric field of $3 \times 10^{-10}\text{ V m}^{-1}$, has a mobility in $\text{m}^2\text{ V}^{-1}\text{ S}^{-1}$ of: (2020)
Mobility $\mu = \frac{v_d}{E} = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^6\text{ m}^2\text{ V}^{-1}\text{ s}^{-1}$.