Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under
`Kettle A: `
specific heat capacity `= 1680 Jkg^(-1)K^(-1)`
Mass = 200 g
Cost = Rs. 400
Kettle B:
Specific heat capacity = `2450 Jkg^(-1) K^(-1)`
Mass = 400 g
Cost = Rs. 400
When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as
(Energy used for liquid heating)/(Total energy supplied)
They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)`
Efficiency of kettle A is

Options

A. `63.34%`
B. `83.34%`
C. `93.34%`
D. `73.34%`

Answer

B

Explanation

b. `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 4.8(theta-theta_1)) xx 100 = 250//3 = 83.34%`
`eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%`
`H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)`
`= 420 xx [0.8 +4] (theta - theta_1)`
`H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)`
`= 420 xx (2.4 + 4)(theta-theta_1)` .
`H_(A)/H_(B) = 4.8/6.4 = 3/4`
`H_(A) = V^2/R_(A) xx 6`
`H_(B) = V^2/R_(B) xx 8`
`H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 `
Then `R_(B) = R_(A)`