Question

A steel rod is 1.5 cm in radius at 25°C. A brass
ring has an interior diameter of 2.992 c, at
25°C. The common temperature at which the
ring just slides on to the rod is.
1) 273°C 2) 473 °C 3) 373 °C 4) 173 °C
​

Answer 1

Given info : A steel rod is 1.5 cm in radius at 25°C. A brass ring has an interior diameter of 2.992 cm, at 25°C.

To find : The common temperature at which the

ring just slides on to the rod is....

solution : radius of steel rod , R = 1.5 cm at 25°C

interior radius of brass ring , R' = 2.992/2 = 1.496 cm at 25°C

from experimental data,

coefficient of linear expansion of steel and brass is 11 × 10¯⁶/°C and 19 × 10¯⁶/°C respectively.

let at temperature T, ring just slides onto the rod.

so, radius of steel rod at T = interior radius of brass ring at T.

⇒R[1 + α∆T] = R'[1 + α'∆T]

⇒1.5 [1 + 11 × 10¯⁶ × ∆T] = 1.496 [ 1 + 19 × 10¯⁶ × ∆T]

⇒1.5 - 1.496 = (1.496 × 19 - 1.5 × 11) × 10¯⁶ ∆T

⇒0.004 × 10⁶ = 11.9 ∆T

⇒4000/11.9 = ∆T

⇒∆T = 336°C

⇒(T - 25°C) = 336°C

⇒T = 336 + 25 = 361°C

Therefore option must be (3) 373°C

[ note : you didn't mention coefficient of linear expansion that's why we couldn't get exact value of temperature. but it is the nearest value of 373°C . so definitely option (c) is correct one ]