JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Current Electricity

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 0.5kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length

L

and diameter

2 d

. 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

(A)4 if wires are in parallel
(B)2 if wires are in series✓ Correct
(C)1 if wires are in series
(D)0.5 if wires are in parallel✓ Correct

Answer: (B), (D)

Solution

Heat required:

Δ Q = m s Δ T

is constant. Power

P = \frac{V ^{2}}{R}

, so time

t = \frac{Δ Q}{P} = \frac{Δ Q \cdot R}{V ^{2}}

.

Original wire:

R = \frac{ρ L}{π d ^{2} /4}

. Time:

t_{1} = 4

min.

New wire (diameter 2d):

R^{'} = \frac{ρ L}{π ( 2 d ) ^{2} /4} = \frac{R}{4}

.

Series connection:

R_{e q} = R^{'} + R^{'} = \frac{R}{2}

. Time:

t = \frac{Δ Q \cdot R /2}{V ^{2}} = \frac{4}{2} = 2

min. So (B) is correct.

Parallel connection:

R_{e q} = \frac{R ^{'} \cdot R ^{'}}{R ^{'} + R ^{'}} = \frac{R ^{'}}{2} = \frac{R}{8}

. Time:

t = \frac{Δ Q \cdot R /8}{V ^{2}} = \frac{4}{8} = 0.5

min. So (D) is correct.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Current Electricity

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