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Question

Physics Question on Current electricity

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

A

10.7C10.7{}^\circ C

B

21.4C21.4{}^\circ C

C

107C107{}^\circ C

D

214C214{}^\circ C

Answer

214C214{}^\circ C

Explanation

Solution

Given, E=2.1V,r=0.05Ω,t=5s,R=0.02ΩE=2.1V,\,r=0.05\,\Omega ,\,t=5s,\,R=0.02\Omega m=1gm=1\,g and s=0.1cal/g/oCs=0.1\,cal/g{{/}^{o}}C m×s×ΔT=i2Rt4.2m\times s\times \Delta T=\frac{{{i}^{2}}Rt}{4.2} m×s×ΔT=(ER+r)2Rt4.2m\times s\times \Delta T=\frac{{{\left( \frac{E}{R+r} \right)}^{2}}Rt}{4.2} 1×0.1×ΔT=(2.1×2.10.07×0.07)×0.02×54.21\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} Δτ=900.1×4.2=9004.2\Delta \tau =\frac{90}{0.1\times 4.2}=\frac{900}{4.2} Δτ=214oC\Delta \tau =214{{\,}^{o}}C