Electrical Resistance

A wire of resistance $160 \Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

A rectangular parallelopiped is measured as $1 \mathrm{cm}\times 1 \mathrm{cm}\times 100 \mathrm{cm}$. If its specific resistance is $3\times {10}^{-7}\mathrm{\Omega } \mathrm{m}$, then the resistance between its two opposite rectangular faces will be _____$\times {10}^{-7}\mathrm{\Omega }$.

The length of a metallic wire is increased by $20\%$ and its area of cross-section is reduced by $4\%.$ The percentage change in resistance of the metallic wire is $\_\_\_\_\_\_\_\_.$

The length of a conductor having resistance $160 \mathrm{\Omega }$, is compressed to $25\%$ of its initial value. The new resistance will be

If the length of a conductor is increased by $20$ percent and cross-sectional area is decreased by $4$ percent, then calculate the percentage change in the resistance of the conductor.

A hollow cylindrical conductor has length of $3.14 \mathrm{m}$, while its inner and outer diameters are $4 \mathrm{mm}$ and $8 \mathrm{mm}$ respectively. The resistance of the conductor is $\mathrm{n}\times {10}^{-3} \Omega$.
If the resistivity of the material is $2.4\times {10}^{-8} \Omega \mathrm{m}$. The value of $\mathrm{n}$ is _____.

In the given figure, wire used in making of frame between points $A$ and $C$ has uniform cross-section $S$ and resistivity $\rho$. The points $E,F,G$ and $H$ are mid-point of sides of square $ABCD$. If side length of square $ABCD$ is $l$, then equivalent resistance between points $A$ and $C$ is

A copper wire and an iron wire, each having an area of cross section $A$ and lengths $L_1$ and $L_2$ are joined end to end. The copper end is maintained at a potential $V_1$ and the iron end at lower potential $V_2$. If ${\sigma }_1$ and ${\sigma }_2$ are the conductivities of copper and iron respectively, the potential of the junction will be

A $1 \mathrm{m}$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of length of $X$ and$Y$.

A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

Eight copper wire of length $l$ and diameter d are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2l$ have the same resistance $\left(R\right)$ then its diameter will be _____ $d$.

The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is $31.4 \mathrm{cm}$. The diameter of wire is measured as $2.4 \mathrm{cm}$. The resistivity of the given wire is measured as $x\times {10}^{-3} \mathrm{\Omega } \mathrm{cm}$. The value of $x$ is _____ .

[Take $\pi =3.14$]

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be _____ $\%$

An aluminium wire is stretched to make its length, $0.4\%$ larger. The percentage change in resistance is

A wire has uniform resistance per unit length $\rho$ (in $\mathrm{Ohm}/\mathrm{m}$ ). Find the resistance between points $A$ and $B$ if the side of the big square is $d:$ (The inner square is made by joining mid points of the outer square)

Two wires that are made up of two different materials whose specific resistances are in the ratio $2:3$, length $3:4$ and area $4:5$. The ratio of their resistance is $\alpha$. Write value of $8\alpha$.

A graph drawn between current $I$ and voltage $V$ in a conductor is as shown in the figure. The changes in the resistance in $1^{\text{st }}$ and $3^{\text{rd }}$ parts respectively

If along a uniform rod of length $l$ carrying current $I$, the voltage $V$ changes with position $x$ along the length of the rod such that $dV/dx=-k$, where $k$ is a positive number, then the resistance of the rod is

If along a uniform rod of length $l$ carrying current $I$, the voltage $V$ changes with position $x$ along the length of the rod such that $dV/dx=-k$, where $k$ is a positive number, then the resistance of the rod is

The resistance of a $20 \mathrm{cm},$ long wire is $5 \mathrm{\Omega }.$ It is streched upto $40 \mathrm{cm}.$ The resistance of wire will now be :- (in ohm)