A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6× 10- 5mm . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (linear coefficient of expansion for steel is given as .α=12 × 10-6 ° C -1)

Question

A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6× 10- 5mm . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (linear coefficient of expansion for steel is given as .α=12 × 10-6 ° C -1) (A) 4.0° C (B) 4.5° C (C) 5.0° C (D) 5.5° C

Tardigrade Admin · Accepted Answer

We know that thermal extension is given by, Δ l=l0α Δ T 6× 10- 5=1× 12× 10- 6× Δ ⊤ Δ T=(6 × 10- 5/12 × 10- 6) Δ T=5° C

$4. 0^{\circ} C$

$4. 5^{\circ} C$

$5. 0^{\circ} C$

$5. 5^{\circ} C$

We know that thermal extension is given by,
$Δ l = l_{0} α Δ T$
$6 \times 1 0^{- 5} = 1 \times 12 \times 1 0^{- 6} \times Δ⊤$
$Δ T = \frac{6 \times 1 0 ^{- 5}}{12 \times 1 0 ^{- 6}}$ $Δ T = 5^{\circ} C$

4.0∘C

4.5∘C

5.0∘C

5.5∘C

We know that thermal extension is given by, Δl=l0​αΔT 6×10−5=1×12×10−6×Δ⊤ ΔT=12×10−66×10−5​ ΔT=5∘C

$4. 0^{\circ} C$

$4. 5^{\circ} C$

$5. 0^{\circ} C$

$5. 5^{\circ} C$

We know that thermal extension is given by,
$Δ l = l_{0} α Δ T$
$6 \times 1 0^{- 5} = 1 \times 12 \times 1 0^{- 6} \times Δ⊤$
$Δ T = \frac{6 \times 1 0 ^{- 5}}{12 \times 1 0 ^{- 6}}$ $Δ T = 5^{\circ} C$