  Question

A fuse is made of tin whose melting point is 23∘C. The ambient temperature is 25∘C. The specific heat constant of tin is 0.2J/g/K. The fuse rating is 4A and it melts in 2s. On an unfortunate day, the temperature drops to 10∘C and there’s a short circuit that allows 4A into the device. What’s the delay in melting of the fuse?

A
0.2s  B
0.14s  C
0.4s  D
Can’t say  Solution

The correct option is D 0.14sGiven c = 0.2J/g/K The power across the fuse is I2R when the fuse melts P=42R=16R Given that the fuse melts in 2s, the energy required to meet the melting point of the fuse is given by.P×t=16R×2=32R Now, heat energy required to reach the melting point of fuse is m×c×(T−To)=m×0.2×(230−25)=41mNow,32R=41mRm=4132 Now given that the temperature on a particular day is 10∘C T−To=220∘C If the time taken for the fuse to melt is t P=16RH=16Rt16Rt=m(0.2)(220)16×4132×mt=m(44)t=8841=2.14sDelay=t−2s=2.14−2=0.14s.Physics

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