JEE Advanced 2014Paper 1
- Exam
- JEE Advanced
- Year
- 2014
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Current Electricity
JEE Advanced 2014 · Paper 1 · Physics · Current Electricity
Heater of an electric kettle is made of a wire of length L and...
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 40K?
Options
- A4 if wires are in parallel
- B2 if wires are in series
- C1 if wires are in series
- D0.5 if wires are in parallel.
Correct answer
B. 2 if wires are in series
Step-by-step solution
Joules law,
(\mathrm{H}=^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}})
Also,
(\mathrm{R}=\rho \frac{1}{\mathrm{ A}} \Rightarrow \mathrm{R} \propto \frac{1}{\mathrm{ A}} \Rightarrow \mathrm{R} \propto \frac{1}{\mathrm{r}^{2}})
(\therefore R^{\prime}=\frac{R}{4})
parallel (\Rightarrow \mathrm{R}_{\mathrm{p}}=\left(\frac{1}{\mathrm{R} / 4}+\frac{1}{\mathrm{R} / 4}\right)^{-1}=\frac{\mathrm{R}}{8})
(\text{ series } \Rightarrow \mathrm{R}_{\mathrm{s}}=\frac{\mathrm{R}}{4}+\frac{\mathrm{R}}{4}=\frac{\mathrm{R}}{2})
(\therefore \frac{\mathrm{V}^{2} 4}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{t}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{V}^{2} \mathrm{t}_{\mathrm{s}}}{\mathrm{R}_\mathrm{s}})
(\Rightarrow \mathrm{t}_{\mathrm{p}}=0.5 \mathrm{ min}, \mathrm{t}_{\mathrm{s}}=2 \mathrm{ min})