TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body,
⇒ Rate of heat transfer ∝ Surface area of body ⇒‌‌‌

Q2

Q1

=‌

s2

S1

⇒Q2=Q1×‌

s2

S1

Now, volume remains same in stretching, so
πr12l1‌‌=πr22l2
⇒‌‌&‌‌l2=‌

r12

r22

⋅l1=‌

(2.5)2

(0.5)2

×5=125cm
So, area before stretching, S1=2πr1(l1+r1)
and area after stretching, S2=2πr2(l2+r2)
Hence by Eq. (i), we get
Q2=Q1×‌

S2

S1

=Q1×‌

2πr2(l2+r2)

2πr1(l1+r1)

=Q1×‌

r2(l2+r2)

r1(l1+r1)

=‌

1×0.5×10−2(125+0.5)×10−2

2.5×10−2(5+2.5)×10−2

=3.346=335W