Question

Two thin metal strips, one of brass and the
other of iron are fastened together parallel to
each other. Thickness of each strip is 1 mm. If
the strips are of equal length at 0°C. The radius
of the are formed by the bimetallic strip when
heated to 80°C is (Coefficient of linear
expansion of brass = 19x10-7°C & of iron =
12x10-6/°C).​

Answer 1

Two thin metal strips, one of brass and the other of iron are fastened together parallel to each other. thickness of each strip is 1 mm. if the strips are of equal length at 0°C.

To find : the radius formed by the bimetallic strip when heated to 80°C.

solution : using formula,

$R=\frac{2t}{(\alpha_B-\alpha_I)\Delta T}$

here t is the width of each metal strip.

$\alpha_B$ is coefficient of linear expansion of brass.

$\alpha_I$ is coefficient of linear expansion of iron.

∆T is change in temperature.

here , t = 1mm = 10¯³ m , ∆T = (80°C - 0°C) = 80°C

$\alpha_B$ = 19 × 10¯⁶/ °C

$\alpha_I$ = 12 × 10¯⁶/°C

so R = (2 × 10¯³)/[(19 × 10¯⁶ - 12 × 10¯⁶) × 80]

= 2 × 10³/(7 × 80)

= 2000/560

= 3.57 m

Therefore the radius formed by bimetallic strip when heated at 80°C.