An iron ball has a diameter of 12cm and is 0.02 cm too large to pass through a hole in a brass plate when the ball and plate are at the same temperature of 25°C. At what temperature the same for both ball and plate would the ball just pass through the hole. (linear expansivties for iron and brass are 1.32×10^-5 K ^-1 and 2.0×10^-5 K^-1​

270°C

Explanation:

Since αB > αI

We want LB - LI = 0.02 cm at a given temperature t

But

ΔLB= LBαBΔt

ΔLI = LIαIΔt

Where;

LI = length of the iron

LB= length of he brass

αB = Linear expansivity of brass

αI = Linear expansivity of iron

Δt = Change in temperature

ΔLB - ΔLI =   (LBαB - LIαI)Δt = 0.02

LB = LI = 12 cm

Δt = 0.02/12( 2.0×10^-5 -  1.32×10^-5)

Δt = 0.02/12(0.68 ×10^-5)

Δt = 245°C

25°C + 245°C = 270°C

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s