CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If masses of three wires of copper are in the ratio of 1 : 3 : 5 and there lengths are in the ratio of 5 : 3 : 1, then ratio of their electrical resistance is


A
1 : 3 : 5
loader
B
5 : 3 : 1
loader
C
1 : 15 : 125
loader
D
125 : 15 : 1
loader

Solution

The correct option is D 125 : 15 : 1
Let the masses be
$$m_1=m$$
$$m_2=3m$$
$$m_3=5m$$
$$\Rightarrow$$ Volume
$$v_1=v$$
$$v_2=3v$$
$$v_3=5v$$
Given lengths
$$l_1=5l$$
$$l_2=3l$$
$$l_3=l$$
Areas $$(A=v/l)$$
$$A_1=\dfrac{v}{5l}$$

$$A_2=\dfrac{3v}{3l}$$

$$A_3=\dfrac{5v}{l}$$

Resistance $$\propto \dfrac{l}{A}$$
$$\Rightarrow R_1=\rho \dfrac{25l^2}{v}$$

$$R_2=\rho\dfrac{3l^2}{v}$$

$$R_3=\rho\dfrac{1}{5}\dfrac{l^2}{v}$$

 $$Ratio=R_1:R_2:R_3\ =\ \ 25:3:\dfrac15$$
multiple with 5

$$\therefore$$$$Ratio =R_1:R_2:R_3 \ = \ \ 125:15:1$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image