Single Choice

The resistance if a wire is '$$R$$' ohm. If it is melted and stretched to '$$n$$' times its original length, its new resistance will be:

A$$\dfrac{R}{n^2}$$
B$$nR$$
C$$\dfrac{R}{n}$$
D$$n^2R$$
Correct Answer

Solution

$$R_1=\dfrac{\rho l_1}{A_1}$$ $$R_2=\dfrac{\rho l_2}{A_2}$$ $$=\dfrac{\rho n^2l_1}{A_1}$$ $$l_1A_1=l_2A_2 \rightarrow $$ (volume constant) $$l_1A_1 = nl_1A_2$$ $$A_2=\dfrac{A_1}{n}$$ Hence $$R_2=n^2R_1$$

SIMILAR QUESTIONS

Current Electricity

A copper wire is stretched to make it $$0.5\%$$ longer. The percentage change in its electrical resistance if its volume remains unchanged is:

Current Electricity

A uniform wire of length $$l$$ and radius $$r$$ has a resistance of $$100\Omega$$. It is recast into a wire of radius $$\dfrac{r}{2}$$. The resistance of new 2 wire will be :

Current Electricity

When $$5V$$ potential difference is applied across a wire length $$0.1\ m$$, the drift speed of electron is 2.5 $$\times 10^{-4} ms^{-1}$$. If the electron density in the wire is 8 $$\times 10^{28} m^{-3}$$, the resistivity of the material is close to

Current Electricity

The resistance of a wire is $$R$$. It is bent at the middle by $$180^{\circ}$$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is

Current Electricity

The resistance of a 20 cm long wire is 5 $$\Omega$$. The wire is stretched to form a uniform wire of 40 cm length. The resistance now will be :

Current Electricity

If a copper wire is stretched to make its radius decrease by $$0.1$$%, then the percentage increase in its resistance is approximately:

Current Electricity

If a rod has resistance $$4\Omega$$ and if rod is turned as half circle, then the resistance along diameter is

Current Electricity

A wire of resistance $$4\:\Omega$$ is stretched to double its original length. The resistance of the stretched wire would be

Current Electricity

In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $$2.7 \times 10^{-8} \Omega$$ m and $$1.0 \times 10^{-7} \Omega$$ m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

Current Electricity

If the length and area of cross-section of a conductor are doubled, then its resistance will be

Contact Details