Question

Two identical metal balls at temperature 200 and 400 kept in air

Answer 1

Answer:

Correct option is

A

16

Using

Y=

△L/L

F/A

Y=

△L

1

/L

1

F/△

1

−(1)For bar A

Y=

△L

2

/L

2

F/△

2

−(2)For bar B

Volume is same for A and B

So,R(

2

d

1

)

2

L

1

=R(

2

d

2

)

2

L

2

d

1

2

L

1

=4d

1

2

L

2

(givend

1

=

2

1

d

2

)

L

2

L

1

=4

Dividing 1 and 2

1=

L

2

△L

1

A

1

L

1

△L

2

A

2

△L

2

△L

1

=4×4(∵

A

1

A

2

=(

d

1

d

2

)

2

=4)

△L

2

△L

1

=16

Answer 2

Explanation:

Two identical metal balls at temperature 200

∘

C and 400

∘

C kept in the air at 27

∘

C. The ratio of net heat loss by these bodies is

A

1/4

B

1/2

C

1/6

D

673

4

−300

4

473

4

−300

4

673

4

−300

4

473

4

−300

4

If the temperature of the surrounding is considered then

net loss of energy of a body by radiation

Q=Aϵσ(T4−T04)t⇒Q∝(T4−T04)⇒Q1Q2=T14−T04T24−T04Q=Aϵσ(T4−T04)t⇒Q∝(T4−T04)⇒Q1