Question

3
When the temperature of a metal wire is increased
from 0°C to 10°C, its length increases by 0.02%. The
percentage change in its mass density will be closest
[JEE Main-2020 (September)]
(1) 2.3
(2) 0.06
(3) 0.8
(4) 0.008
to​

Answer 1

Answer:

Answer is (2)0.06

Explanation:

∆l / l = ∝∆t

[latex]\alpha =\frac{\Delta l}{l\Delta t}=\frac{0.02}{100\times 10}=2\times 10^{-5}[/latex] [latex]\gamma =3\alpha =6\times 10^{-5}[/latex]

ℓ = m / v

∆ ℓ / ℓ = ∆m / m – ∆v / v

Here, ∆ m / m = 0

∆ ℓ / ℓ = – ∆v / v

Now, [latex]\frac{\Delta v}{v}\times 100=\gamma \Delta t.100=6\times 10^{-5}\times 10\times 100[/latex]

=0.06

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Answer 2

Answer:

3) 0.8

Explanation:

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