A wire when connected to 220 V mains supply has power dissipation `P_1`. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is `P_2`. Then `P_2: P_1` is

Options

A. `1`
B. `4`
C. `2`
D. `3`

Answer

B

Explanation

Using the `P=(V^(2))/R....(i)`
Where `R` is resistance of wire, `V` is voltage across wire and `P` is power is dissipation in wire and `R=(rhol)/A...(ii)`
From `(i)` and `(ii)`
`P_(1)=(V^(2))/(rhol//A)=(V^(2))/(rhol).A`
`P_(1)=(V^(2))/(rhol).A....(iii)`
In `2nd` case
Let `R_(2)` is net resistance
`R_(2)=(RxxR)/(R+R)=R/2`
where, `R` is the resistance of half wire.
`:. R_(2)=(rho(l/2))/(A.2)=(rhol)/(4A)`
`:. P_(2)=(V^(2))/(rhol).4A....(iv)`
Hence, from Eqs. `(iii)` and `(iv)`
`(P_(1))/(P_(2))=1/4implies(P_(2))/(P_(1))=4/1`