Question

A battery of emf 2.1 V and internal resistance is shunted for 5 s by a wire of constant resistance 0.02 $\Omega$ ,mass 1 g and specific heat 0.1 cal/g/?C. The rise in the temperature of the wire is

• A. $10.7{}^{\circ }C$
• B. $21.4{}^{\circ }C$
• C. $107{}^{\circ }C$
• D. $214{}^{\circ }C$

Answer 1

Answer: (D)

Given, $E=2.1V,r=0.05\Omega ,t=5s,R=0.02\Omega$ $m=1g$ and $s=0.1cal/g{/}^{o}C$ $m\times s\times \Delta T=\frac{i^{2}Rt}{4.2}$ $m\times s\times \Delta T=\frac{{\left(\frac{E}{R+r}\right)}^{2}Rt}{4.2}$ $1\times 0.1\times \Delta T=\left(\frac{2.1\times 2.1}{0.07\times 0.07}\right)\times \frac{0.02\times 5}{4.2}$ $\Delta \tau =\frac{90}{0.1\times 4.2}=\frac{900}{4.2}$ $\Delta \tau =214{}^{o}C$