1.

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

Answer»

`63.34%`
`83.34%`
`93.34%`
`73.34%`

Solution :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)`


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