Masses of three wires of copper are in the ra | vedclass

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

Question

(B) The resistance $R$ of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$,we can write $A = \frac{V}{l}$.Substituting this int

Vedclass · Accepted Answer

$\sqrt{125}: 15: 1$

Vedclass · Answer

(B) The resistance $R$ of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = \frac{ho l^2}{V}$.Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$.Thus,$R = \frac{ho l^2 d}{m}$.For wires of the same material,$ho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.Given the ratios $m_1:m_2:m_3 = 1:3:5$ and $l_1:l_2:l_3 = 5:3:1$,the ratio of resistances is:$R_1: R_2: R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$$R_1: R_2: R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$$R_1: R_2: R_3 = 25 : 3 : 0.2$Multiplying by $5$ to clear the fraction:$R_1: R_2: R_3 = 125 : 15 : 1$.

Masses of three wires of copper are 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 . . . . . . .

Similar Questions

If the length of a wire is increased by $0.1\%$,what is the percentage increase in its resistance (in $\%$)?

The resistance of a wire at $300 \, K$ is found to be $0.3 \, \Omega$. If the temperature coefficient of resistance of the wire is $1.5 \times 10^{-3} \, K^{-1}$, the temperature at which the resistance becomes $0.6 \, \Omega$ is: (in $K$)

The specific resistance of a wire is $\rho$,its volume is $3\,m^3$ and its resistance is $3\,\Omega$. Then its length will be:

The resistance of a conductor increases with:

Arrange the following materials in increasing order of their resistivity: Nichrome,Copper,Germanium,Silicon.

Vedclass Test Series

Exam Paper Generator

Online Exam Module