A wire of length L and 3 identical cells of n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $R$ be the resistance and $m$ be the mass of the first wire. The second wire has length $2L$,so its resistance is $2R$ and its mass is $2m$.Le

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $R$ be the resistance and $m$ be the mass of the first wire. The second wire has length $2L$,so its resistance is $2R$ and its mass is $2m$.Let $E$ be the $EMF$ of each cell and $S$ be the specific heat capacity of the material.For the first wire,the current is $i_1 = \frac{3E}{R}$. The heat produced is $H_1 = i_1^2 R t = m S \Delta T$.Substituting $i_1$,we get $(\frac{3E}{R})^2 R t = m S \Delta T \implies \frac{9E^2 t}{R} = m S \Delta T$.For the second wire,the current is $i_2 = \frac{NE}{2R}$. The heat produced is $H_2 = i_2^2 (2R) t = (2m) S \Delta T$.Substituting $i_2$,we get $(\frac{NE}{2R})^2 (2R) t = 2m S \Delta T \implies \frac{N^2 E^2 t}{4R^2} \cdot 2R = 2m S \Delta T \implies \frac{N^2 E^2 t}{2R} = 2m S \Delta T$.Dividing the two heat equations: $\frac{9E^2 t / R}{N^2 E^2 t / 2R} = \frac{m S \Delta T}{2m S \Delta T} \implies \frac{18}{N^2} = \frac{1}{2} \implies N^2 = 36 \implies N = 6$.

Similar Questions

$A$ thermocouple produces $200\,\mu V$ between $0\,^oC$ and $100\,^oC$. If it produces $64\,\mu V$ and $76\,\mu V$ for temperature ranges $(0\,^oC - 32\,^oC)$ and $(32\,^oC - 70\,^oC)$ respectively,then the thermo $emf$ produced between $70\,^oC$ and $100\,^oC$ will be ........... $\mu V$.

Two electric bulbs marked $25\ W - 220\ V$ and $100\ W - 220\ V$ are connected in series to a $440\ V$ supply. Which of the bulbs will fuse?

The supply voltage to a room is $120\ V$. The resistance of the lead wires is $6\,\Omega$. $A$ $60\ W$ bulb is already switched on. What is the decrease in voltage across the bulb when a $240\ W$ heater is switched on in parallel to the bulb? ............. $V$

The thermo-emf of a hypothetical thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700^{\circ}C$. If the cold junction is kept at $0^{\circ}C$,then the neutral temperature is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module