The `2.0 Omega` resistor show in figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is `2000 J K^(-1)`. (a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water? (b) Suppose the `6.0 Omega` resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes? (Figure)

Answer

A, B, C

Explanation

`R_(eff)=((12)/(8)) + 1`
`=((3)/(2)) + 1 = (5)/(2)`
`I = (12)/(5) amp`
Here `I'xx6 = (i -i)^2`
`rArr I' 6 = ((12)/(5)) xx 2-2i'`
`rArr 8i = (24)/(5)`
` rArr I' = (24)/((5xx8)) = (3)/(5) amp. So, `
`ixxj' = (12)/( 5) - (3)/(5) = (9)/(5) amp. (a) Heat `
`= i^2Rt`
`=((9)/(5)) xx ((9)/(5)) xx2 xx 15 xx60`
`= 81xx36xx2 = 5832`
` 2000J of heat raises the temp. by 1K 5832 J raises the temp. by 2.916K (b) When `
`6Omega`
` resistor gets burnt `
` R_(eff) = 1 + 2 = 3 Omega`
`i=(6)/(3) = 2 amp`
`= 2xx2xx2xx15xx60 = 7200 J`
` 2000 J raises the temp. by 1K 7200 J raises the temp. by 3.6 K`