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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 isA. `63.34%`B. `83.34%`C. `93.34%`D. `73.34%` |
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Answer» Correct Answer - B 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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