First slide

Elastic Modulie

Question

On loading a metal wire of cross section 10^–6 m² and length 2 m by a mass of 210 kg it extends by 16 mm and suddenly broke from the point of support. If density of that metal is 8000 kgm^–3 and its specific heat is 420 Jkg^–1K^–1, the rise in temperature of wire is (g = 10 ms^-2)

Moderate

A

2.5 ⁰C
B

5 ⁰C
C

6 ⁰C
D

10 ⁰C

Solution

Gravitational P.E. lost by the mass (M = 210 kg) = Mg.e 50% of this lost energy is converted into heat. so
$\large \frac 12Mge=ms.\Delta T$
. where m = mass of wire.

$\large \rho = \frac{m}{V},m = \rho V \Rightarrow m = \rho A\ell$

$\large \frac{1}{2}mge = \rho A\ell s\Delta T \Rightarrow \frac{1}{2} \times 210 \times 10 \times 16 \times {10^{ - 3}}$

$\large = 8000 \times {10^{ - 6}} \times 2 \times 420 \times \Delta T$

$\large \Delta T = \frac{{{\textstyle{1 \over 2}} \times 210 \times 10 \times 16 \times {{10}^{ - 3}}}}{{8000 \times {{10}^{ - 6}} \times 2 \times 420}} = {2.5^0}C$

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