Question

If the length of a wire is increased by 20% keeping its volume constant. what will be the % change in heat produced when connected across same potential difference. please explain properly!!

Answer 1

30.55 %

Step-by-step 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
• 
  `\Delta H` = change in heat produced

According to the question, we have

`L = l + 20\ \% l = (120l)/(100)\\V=v\\\Rightarrow LA=la\\\Rightarrow A= (la)/(L)\\\Rightarrow A= (la)/((120l)/(100))\\\Rightarrow A= (100a)/(120)`

Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have

`r =  (\rho l)/(a)\\R=(\rho L)/(A)=(\rho* (120l)/(100) )/((100a)/(120))=((120)/(100))^2(\rho l)/(a)= 1.44r\\`

Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have

`h = (P^2)/(r)t\\H = (P^2)/(R)t= (P^2)/(1.44r)t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = (P^2)/(r)t-(P^2)/(1.44r)t\\\Rightarrow \Delta H = (P^2t)/(r)(-(1)/(1.44))\\\Rightarrow \Delta H = (P^2t)/(r)((0.44)/(1.44))\\\therefore \textrm{Percentage change in the heat produced}= (\Delta H)/(h)* 100\ \%= \left (((P^2t)/(r)((0.44)/(1.44)))/((P^2)/(r)t)  \right )* 100\ \% = 30.55\ \%`

Hence, the percentage change in the heat produced in the wire is 30.55 %.

Answer 2

Answer:decreases by 30.55%

Step-by-step 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

`AL=1.2 L* A'`

A'=0.833 A

and resistance is given by

`R=(\rho L)/(A)`

where
`\rho`=resistivity

New resistance
`R'=(\rho* 1.2L)/(0.833A)`

R'=1.44 R

heat produced for same potential

`H_1=(V^2t)/(R)`

`H_2=(V^2t)/(1.44R)=0.694H_1`

% change in heat

`(H_2-H_1)/(H_1)* 100`

`=(0.694-1)/(1)`

=30.55 decreases