Question

A hole is drilled in a copper sheet. The diameter of the hole 4.24cm at 27°C. Diameter of the hole when it is heated to 35°C is
1. less than 4.24cm
2.equal to 4.24cm
3. more than 4.24cm
4 data insufficient ​

Answer 1

Answer:

it will be more than 4.24cm

Explanation:

on heating or incresing in temperature metal expand so hole diameter will be more

Answer 2

Answer:

Explanation:

Here, diameter of hole ( D1) = 4.24 cm

So, initial area of hole (Ao)= πr² = 22/7 ( 4.24/2)²

= 4.494π cm²

intial temperature ( T1) = 27°C = 27+ 273 = 300 K

Final temperature ( T2) = 227°C = 227+ 273 = 500K

coefficient of linear expansion (a) = 1.7 × 10^-5/°C

coefficient of superficial expansion (b) = 2× linear expansion

= 2 × 1.7 × 10^-5/°C

= 3.4 × 10^-5 /°C

Use formula,

A = Ao( 1 + b∆T)

A = 4.494π [1 + 3.4 × 10^-5 × (500-300)]

= 4.494π[ 1 + 3.4 × 10^-5 × 200]

= 4.494π [ 1 + 6.8 × 10^-3 ]

= 4.494π [ 1 + 0.0068]

= 4.494π × 1.00068

= 4.525π cm² = πD2²/4

D2² = 4.525 × 4

D2 = 4.2544 cm

Change in diameter (∆D) = D2 - D1

= 4.2544 - 4.24

= 0.0144 cm