Respuesta :
Answer:decreases by 30.55%
Explanation:
Given
length of wire is increased by 20 % keeping volume constant
Let the length of wire be L and its area of cross section be A
Thus new length=1.2 L
Volume is constant
[tex]AL=1.2 L\times A'[/tex]
A'=0.833 A
and resistance is given by
[tex]R=\frac{\rho L}{A}[/tex]
where [tex]\rho [/tex]=resistivity
New resistance [tex]R'=\frac{\rho\times 1.2L}{0.833A}[/tex]
R'=1.44 R
heat produced for same potential
[tex]H_1=\frac{V^2t}{R}[/tex]
[tex]H_2=\frac{V^2t}{1.44R}=0.694H_1[/tex]
% change in heat
[tex]\frac{H_2-H_1}{H_1}\times 100[/tex]
[tex]=\frac{0.694-1}{1}[/tex]
=30.55 decreases
Answer:
30.55 %
Explanation:
Assumptions:
- l = initial length of the wire
- L = final length of the wire
- v = initial volume of the wire
- V = final volume of the wire
- a = initial cross sectional area of the wire
- A = final cross sectional area of the wire
- h = initial heat of generated by the wire
- H = final heat generated by the wire
- P = potential difference across the wire
- t = time for which the potential difference is created across the wire
- r = initial resistance of the wire
- R = final resistance of the wire
- [tex]\Delta H[/tex] = change in heat produced
According to the question, we have
[tex]L = l + 20\ \% l = \dfrac{120l}{100}\\V=v\\\Rightarrow LA=la\\\Rightarrow A= \dfrac{la}{L}\\\Rightarrow A= \dfrac{la}{\dfrac{120l}{100}}\\\Rightarrow A= \dfrac{100a}{120}[/tex]
Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have
[tex]r = \dfrac{\rho l}{a}\\R=\dfrac{\rho L}{A}=\dfrac{\rho\times \dfrac{120l}{100} }{\dfrac{100a}{120}}=(\dfrac{120}{100})^2\dfrac{\rho l}{a}= 1.44r\\[/tex]
Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have
[tex]h = \dfrac{P^2}{r}t\\H = \dfrac{P^2}{R}t= \dfrac{P^2}{1.44r}t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = \dfrac{P^2}{r}t-\dfrac{P^2}{1.44r}t\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(-\dfrac{1}{1.44})\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(\dfrac{0.44}{1.44})\\\therefore \textrm{Percentage change in the heat produced}= \dfrac{\Delta H}{h}\times 100\ \%= \left (\dfrac{\dfrac{P^2t}{r}(\dfrac{0.44}{1.44})}{\dfrac{P^2}{r}t} \right )\times 100\ \% = 30.55\ \%[/tex]
Hence, the percentage change in the heat produced in the wire is 30.55 %.
