JEE Advance - Physics (2014 - Paper 1 Offline - No. 12)
Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?
4, if wires are in parallel
2, if wires are in series
1, if wires are in series
0.5, if wires are in parallel
Explanation
Resistance of initially given kettle
$$R = \rho {l \over A}$$
$$ = \rho {L \over {\pi {{(d/2)}^2}}} = {{4\rho L} \over {\pi {d^2}}}$$
Resistance of two replaced kettles
$${R_1} = {{\rho L} \over {\pi {d^2}}}$$ and
$${R_2} = {{\rho L} \over {\pi {d^2}}}$$
So, $${R_1} = {R_2} = {R \over 4}$$
If wires are in parallel then equivalent resistance
$${R_P} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {R \over 8}$$ ....... (1)
If wires are in series then equivalent resistance
$${R_S} = {R_1} + {R_2} = {R \over 2}$$ ..... (2)
Let V be the applied voltage, m = 0.5 kg be the mass of the water, and S be the specific heat of the water. Initially, the heat produced by a resistance R1 in time t1 = 4 min is V2t1/R1. This heat is used to raise the temperature of the water by $$\Delta$$T = 40 K. Thus,
$${V^2}{t_1}/{R_1} = mS\Delta T$$. ....... (3)
Let ts and tp be the time taken to raise the temperature of same amount of water by $$\Delta$$T when the resistances are connected in series and parallel. Thus,
$${V^2}{t_s}/{R_s} = mS\Delta T$$ ....... (4)
$${V^2}{t_p}/{R_p} = mS\Delta T$$ ....... (5)
Divide equation (4) by (3) and use the equation (1) to get $${t_s} = {{{R_s}} \over {{R_1}}}{t_1} = {1 \over 2} \times 4 = 2$$ min. Similarly, divide equation (5) by (3) and use the equation (2) to get $${t_p} = {{{R_p}} \over {{R_1}}}{t_1} = {1 \over 8} \times 4 = 0.5$$ min.
$$R = \rho {l \over A}$$
$$ = \rho {L \over {\pi {{(d/2)}^2}}} = {{4\rho L} \over {\pi {d^2}}}$$
Resistance of two replaced kettles
$${R_1} = {{\rho L} \over {\pi {d^2}}}$$ and
$${R_2} = {{\rho L} \over {\pi {d^2}}}$$
So, $${R_1} = {R_2} = {R \over 4}$$
If wires are in parallel then equivalent resistance
$${R_P} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {R \over 8}$$ ....... (1)
If wires are in series then equivalent resistance
$${R_S} = {R_1} + {R_2} = {R \over 2}$$ ..... (2)
Let V be the applied voltage, m = 0.5 kg be the mass of the water, and S be the specific heat of the water. Initially, the heat produced by a resistance R1 in time t1 = 4 min is V2t1/R1. This heat is used to raise the temperature of the water by $$\Delta$$T = 40 K. Thus,
$${V^2}{t_1}/{R_1} = mS\Delta T$$. ....... (3)
Let ts and tp be the time taken to raise the temperature of same amount of water by $$\Delta$$T when the resistances are connected in series and parallel. Thus,
$${V^2}{t_s}/{R_s} = mS\Delta T$$ ....... (4)
$${V^2}{t_p}/{R_p} = mS\Delta T$$ ....... (5)
Divide equation (4) by (3) and use the equation (1) to get $${t_s} = {{{R_s}} \over {{R_1}}}{t_1} = {1 \over 2} \times 4 = 2$$ min. Similarly, divide equation (5) by (3) and use the equation (2) to get $${t_p} = {{{R_p}} \over {{R_1}}}{t_1} = {1 \over 8} \times 4 = 0.5$$ min.
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