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Question

A fuse is made of tin whose melting point is 23C. The ambient temperature is 25C. 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 10C and there’s a short circuit that allows 4A into the device. What’s the delay in melting of the fuse?


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

Solution

The correct option is D 0.14s
Given 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×(TTo)=m×0.2×(23025)=41mNow,32R=41mRm=4132
Now given that the temperature on a particular day is
10C
TTo=220C
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=t2s=2.142=0.14s.

Physics

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