Question 1:
easy2. A flow of $10^{7}$ electrons per second in a conducting wire constitutes a current of (1994)
Current $I = \frac{q}{t} = \frac{ne}{t}$. Given $\frac{n}{t} = 10^{7} \text{ s}^{-1}$. Thus, $I = 10^{7} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-12} \text{ A}$.
Question 2:
easy3. The velocity of charge carriers of current (about $1 \text{ ampere}$) in a metal under normal conditions is of the order of (1991)
The drift velocity of electrons in a typical metallic conductor under normal conditions is extremely small, typically on the order of $10^{-4} \text{ m/s}$ or a fraction of a $\text{mm/sec}$.
Question 3:
easyTwo solid conductors are made up of same material have same length and same resistance. One of them has a circular cross section of area $A_{1}$ and the other one has a square cross section of area $A_{2}$. The ratio $A_{1}/A_{2}$ is
(2020-Covid)
Resistance is given by $R = \rho \frac{l}{A}$. Since the material (hence $\rho$), length $l$, and resistance $R$ are the same for both conductors, their cross-sectional areas must be equal. Therefore, $A_{1} = A_{2}$ and the ratio is $1$.
Question 4:
easyThe resistance of a wire is ‘$R$’ ohm. If it is melted and stretched to ‘$n$’ times its original length, its new resistance will be:
(2017-Delhi)
The volume of the wire remains constant during stretching. Resistance $R = \rho \frac{l}{A} = \rho \frac{l^{2}}{V}$. If the length becomes $nl$, the new resistance is $R' = \rho \frac{(nl)^{2}}{V} = n^{2}R$.
Question 5:
easyA wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire would be:
(2013)
When a wire is stretched, its new resistance becomes $R' = n^{2}R$. Given $n=2$ and $R=4 \Omega$, we have $R' = (2)^{2} \times 4 = 16 \Omega$.
Question 6:
easyA wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively:
(2008)
Specific resistance is a property of the material and remains the same. When length increases by $10\%$, $l' = 1.1l$. Since $R \propto l^{2}$ (volume constant), new resistance $R' = (1.1)^{2}R = 1.21R$.
Question 7:
easyThe electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then:
(2004)
Specific resistance depends only on the material, so it remains unchanged. Resistance $R = \rho \frac{l}{\pi r^{2}}$. If $l \rightarrow 2l$ and $r \rightarrow 2r$, $R' = \rho \frac{2l}{\pi (2r)^{2}} = \frac{1}{2} \rho \frac{l}{\pi r^{2}} = \frac{R}{2}$.
Question 8:
easyCopper and silicon is cooled from $300 \text{ K}$ to $60 \text{ K}$, the specific resistance:
(2001)
Copper is a metal, and its specific resistance decreases with a decrease in temperature. Silicon is a semiconductor, and its specific resistance increases as temperature decreases.
Question 9:
easyThree copper wires of lengths and cross-sectional areas are $(l, A)$, $(2l, A/2)$ and $(l/2, 2A)$. Resistance is minimum in
(1997)
Resistance $R = \rho \frac{l}{A}$. For the three wires: $R_{1} = \rho \frac{l}{A}$, $R_{2} = \rho \frac{2l}{A/2} = 4\rho \frac{l}{A}$, and $R_{3} = \rho \frac{l/2}{2A} = \frac{1}{4}\rho \frac{l}{A}$. The minimum resistance is for the wire with area $2A$.
Question 10:
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 $\sigma_{eq} = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$.