Q46 / 256 · NTA Abhyas (NEET) 2020 · Physics · Current Electricity

A condenser of capacitance C is fully charged by a 200 V supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat 250 J kg-1 K-1 and of mass 100 g. If the temperature of the block rise by 0.4 K, then the value of C is

  • A
    300 μF
  • B
    200 μF
  • C
    400 μF
  • Correct answer
    D
    500 μF

Solution

Given that

V = 200 V

m = 100g = 0.1 kg

s=250 J kg-1K-1

∆θ=0.4 K

C =?

The energy stored in the capacitor is given by

U=12CV2

=12C×2002

=2C×104 J

The energy is used to heat up the block. Let ∆θ be the rise in temperature, then heat energy is given by

O=ms∆θ=0.1×250×0.4

= 10 J

Now,

2C×104=10 

⇒   C=102×104=5×10-4

∴   C=500 μF

Hence, the correct answer is option: 'D'.

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