The masses of three wires of copper are in th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resistance of a wire is given by $R = \rho \frac{l}{A}$.Since the mass $m = A \cdot l \cdot d$ (where $d$ is density),we have $A = \frac{m}{l

Vedclass · Accepted Answer

$125 : 15 : 1$

Vedclass · Answer

(D) The resistance of a wire is given by $R = \rho \frac{l}{A}$.Since the mass $m = A \cdot l \cdot d$ (where $d$ is density),we have $A = \frac{m}{l \cdot d}$.Substituting $A$ into the resistance formula: $R = \rho \frac{l^2 \cdot d}{m}$.Since $\rho$ and $d$ are constant for copper,$R \propto \frac{l^2}{m}$.Given the ratios $m_1 : m_2 : m_3 = 1 : 3 : 5$ and $l_1 : l_2 : l_3 = 5 : 3 : 1$,we calculate the ratio of resistances:$R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5} = 25 : 3 : 0.2$.To express this in integers,multiply by $5$:$R_1 : R_2 : R_3 = 125 : 15 : 1$.

The masses of three wires of copper are in the ratio of $1 : 3 : 5$ and their lengths are in the ratio $5 : 3 : 1.$ The ratio of their electrical resistances is

Similar Questions

For which of the following does the resistance decrease upon increasing the temperature?

The resistance of the filament of an 'incandescent' bulb increases with an increase in temperature. If $R_{100}$,$R_{60}$,and $R_{40}$ are the resistances of $100 \ W$,$60 \ W$,and $40 \ W$ bulbs at room temperature,respectively,then:

$A$ resistance $R$ is stretched to four times its length. Its new resistance will be

The length of a given cylindrical wire is increased by $100\,\%$. Due to the consequent decrease in diameter,the percentage change in the resistance of the wire will be ................ $\%$.

What is conductance? Write its unit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module