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Question: A bar of iron is 10 cm at 20°C. At 19°C it will be (α of iron = 11 × 10<sup>–6</sup>/°C)...

A bar of iron is 10 cm at 20°C. At 19°C it will be (α of iron = 11 × 10–6/°C)

A

(a) 11 × 10–6 cm longer

A

(b) 11 × 10–6 cm shorter

A

(c) 11 × 10–5 cm shorter

A

(d) 11 × 10–5 cm longer

Explanation

Solution

(c)

Sol. L=L0(1+αΔθ)L = L_{0}(1 + \alpha\Delta\theta) ⇒ L1L2=1+α(Δθ)11+α(Δθ)2\frac{L_{1}}{L_{2}} = \frac{1 + \alpha(\Delta\theta)_{1}}{1 + \alpha(\Delta\theta)_{2}}

⇒ 10L2=1+11×10−6×201+11×10−6×19\frac{10}{L_{2}} = \frac{1 + 11 \times 10^{- 6} \times 20}{1 + 11 \times 10^{- 6} \times 19} ⇒ L2=9.99989L_{2} = 9.99989

⇒ Length is shorten by

10−9.99989=0.00011=11×10−5cm10 - 9.99989 = 0.00011 = 11 \times 10^{- 5}cm