GUIDED REVISION

JEE (Advanced) 2024 2024

JEE (Advanced)
ENTHUSIAST COURSE
ENTHUSIAST COURSE
STAR BATCH
STAR BATCH

PHYSICS GR # THERMAL CONDUCTION AND RADIATION

SOLUTION
SECTION-I
Single Correct Answer Type 8 Q. [3 M (–1)]
1. Ans. (C)
2. Ans. (B)
Sol. Thermal resistance of a shell of radius x and thickness dx will be
dx 1 xn 1
dRth = = dx = x n -2dx .....(i)
k4px 2
r.4p x 2
r.4 p
Radial heat current (which is constant) will be
dT
= H ( cons tan t )
dR th

4prdT
=H
x n -2dx
dT
= a constant if n = 2
dx
\ n0 = 2
3. Ans. (C)
Sol. 100 = kms (50 – 20) = k(100) (30)
100 1
k= =
3000 30
dQ 1
= ´ 100 ( T - 20 ) = 10 ( T - 20 )
dt loss 30 3
dT dQ
ms dt = Pheater - dt
loss

dT 10
100 = 200 - ( T - 20 )
dt 3
dT 1 8 T 1
=2- ( T - 20 ) = - = ( 80 - T )
dt 30 3 30 30
T 30
dT 1
ò50 80 - T = ò 30 dt
0

4. Ans. (C)
T T0

Sol. r

dQ
H= = s4pr2(T4 – T04)
dt

Physics / GR # Thermal Conduction and radiation E-1/8

 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

T'
T0
R T

H' = s4pr2(T4 – T'4) = s4pR2(T'4 – T04)

r 2 T4 + R 2T04
T'4 =
r 2 + R2
H' T4 - T04 R2
= 4 =
H T - T'4 R 2 + r 2
5. Ans. (A)
Sol. E = esAT4
Þ 100 = 0.3 × 5.62 × 10–8 × A × (2780)4
\ Surface area or bulb
100 -4 2
= 4
= 0.98 ´ 10 m
101.46 ´ 10
6. Ans. (B)
Sol. (P – PL)t = mSDT
(P – PL( × 120 = 1.8 × 4200 × 5
P – PL = 9 × 35
mSDT
PL =
60
4200 ´ 1.8 ´ 1
= = 126 W
60
P = 315 + 126 = 441 W
7. Ans. (A)
dQ 17
Sol. = 1 ´1 ´ ´ 10 -8 (625 – 81) × 108
dt 3
544 ´ 17
=
3
dQ kA
= ´ ( 500 - 300 )
dt d
544 ´ 17
Þk= ´ 10-3 = 1.54 × 10–2W/mk
3 ´ 200
8. Ans. (B)
Sol. We must take into account here that the heat transferred per unit time is proportional to the temperature
difference. Let us introduce the following notation: T out1, Tout2 and Tr1 , Tr2 are the temperatures outdoors
and in the room in the first and second cases respectively. The thermal power dissipated by the radiator
in the room is k1(T – Tr), where k1 is a certain coefficient. The thermal power dissipated from the room
is k2(Tr – Tout), where k2 is another coefficient. In thermal equilibrium, the power dissipated by the
radiator is equal to the power dissipated from the room.
Therefore, we can write
E-2/8 Physics / GR # Thermal Conduction and radiation
 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

k1 (T - Tr1 ) = k 2 (Tr1 - Tout1 )

Similarly, in the second case,
k1 (T - Tr2 ) = k 2 (Tr2 - Tout2 )
Dividing the first equation by the second, we obtain
T - Tr1 Tr1 - Tout1
=
T - Tr2 Tr2 - Tout2
Hence we can determine T:
Tr2 Tout1 - Tr1 Tout2
T= = 60o C
Tr2 + Tout1 - Tout2 - Tr1
Multiple Correct Answer Type 6 Q. [4 M (–1)]
9. Ans. (C,D)
dT l dT 1
Sol. = at steady state µ
dx kA dx k
10. Ans. (A,C)
11. Ans. (A,C,D)
12. Ans. (A, B)
kA
Sol. P = CA(360 – 300) + (360 - 340)
t
kA
(360 - 340) = CA(340 - 300)
t
P = 100 CA
2CA
P = CA(Tt - 300) + (Tt - Tb )
2
kA
(Tt - Tb ) = CA(Tb - 300)
2t
TT – Tb = Tb – 300
100 CA = CA (2Tt – 300 – Tb)
100 = 4Tb – 600 – 300 – Tb
1000
= Tb
3
2000 1100
TT = - 300 =
3 3
13. Ans. (A, C)
Sol. Cylinder absorbs energy from left face and radiate from both, So
P
(1 – cos60°) = sAT4 + sAT’4
2
Þ P = 68 W
Heat current = 1W
kADT
=1
l

Physics / GR # Thermal Conduction and radiation E-3/8

 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

Þ k = 0.057
14. Ans. (A, C, D)
dH As(6004 - 4004 )
= = As ´ 3004
Sol. dt æ 1 1 ö
ç 1 + e - 1÷
è ø

3004 81
Þe= 4 4
=
600 - 400 81 ´16 - 256
81 81 81
= = =
16[81 - 16] 16 ´ 65 1040
17
P =1´ ´ 10-8 ´ 81 ´ 108 = 459 W
3
k ´1
= -3
(100) Þ h = 1836 ´ 10-5
4 ´10
Matching List Type (4 × 4) 1Q. [3 M (–1)]
15. Ans. (A)
Sol. (P) CD is like wheatstone bridge heat how through AB = BE = EF there temp of B = 140ºC
kA
( 200 - 140 ) = Q
l
kA 60
= Q half of it flows through BC
l
Q kA
= ( K10 - TC )
2 l
TC = 110ºC
(Q) heatBD = heatBA + heatBC
3kA 2kA kA
(100 - T ) = (T - 50 ) + (T - 0º )
l l l
(R)heat lost = heat gained
(S) Using Newton's law of cooling
SECTION-II
Numerical Answer Type Question 6 Q. [3M(0)]
(upto second decimal place)
16. Ans. 105.00
25T
100°C 0°C
Sol.
80W
kA kA
80 = ( T - 100 ) + (T - 0)
0.25 0.75
100 T
80 = = T - 100 +
4 ´ 50 3

E-4/8 Physics / GR # Thermal Conduction and radiation

 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

4T
140 =
3
420
ÞT = = 105°C
4
17. Ans. 25.00
x2 Kqt
Sol. We havestudied in growth of ice on pond = rL
2 f

d12 Kq t 1
= .......(i)
2 rL f

(d 2 ) 2 Kq t 2
= .......(ii)
2 rL f

éæ d ö 2 ù
êç 2 ÷ ú
\ t2 = êç d ÷ ú t1 = 25 days
ëè 1 ø û
18. Ans. 35
Sol. PH = rate of heat loss to surrounding
L1 L2
Net thermal resistance RNet = K A + K A
1 2

2 ´10 -2 10 ´ 10- 2
= +
0.2 ´ 50 0.5 ´ 50
= 2 × 10–3 + 4 × 10–3
= 6 × 10–3
DT 15 - ( -10) 25
PH = = =
6 × 10
3
Rnet 6 ´10 -3
Watts
12500
PH = Watts
3
Ti - ( -15)
2PH =
R
25000 T + 15
=
3 6 ´10-3
2 × 25 = T + 15 Þ T = 35°C
19. Ans. 1.40 to 1.50
dH
Sol. dt ´ p R 2 = 4p R 2s T 4
4p r 2
E E E

TE r = Const.

Physics / GR # Thermal Conduction and radiation E-5/8

 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

288 ´ r = T ¢ ´ 1.01r
288
T¢ = = 288(1 + 0.01)-1/2
1.01
æ 0.01 ö 2.88
= 288 ç 1 - ÷ = 288 -
è 2 ø 2
DT » 1.445
20. Ans. 3.12 to 3.13
Sol. 500 Tsun = 400 Tstar
Tsun 4
=
Tstar 5
dH
= 4p R 2s T 4
dt
dH
sun 2
æ Rsun ö æ Tsun ö
4
dt =4=ç
dH ÷ ç ÷
star è Rstar ø è Tstar ø
dt
2
256 æ Rsun ö
4= ç ÷
625 è Rstar ø

Rsun 50 25
= = = 3.125
Rstar 16 8
21. Ans. 200

Sol.

dH
1.7pl =
dt
(
= 2pr1ls T14 - T24 )
= 2pr2lsT24

r2 4
Þ T14 - T24 - T
r1 2
1/4
ær +r ö
T1 = ç 1 2 ÷ T2 = 2T
è r1 ø
2

1 17 15
Þ 1.7p = 2p ´ ´ ´10-8 T14 ´
100 3 16
T14= 16 ´ 108

E-6/8 Physics / GR # Thermal Conduction and radiation

 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
T1 = 200 K
SECTION-III
Numerical Grid Type (Ranging from 0 to 9) 5 Q. [4 M (0)]
22. Ans. 9
dm 3.35 ´ p ´ 0.22
Sol. ´ 3.35 ´ 105 = ´ (100 - 0 )
dt 10 -3
dm
= 0.04p
dt
m
= Vi
Pi

m
= Vf
Pw

ADx = D V =
m m
- =m
(10 - 9 )
Pi Pw 9000

dx dm
=
dt dt ´ 9 ´ 103 ´ A
0.04p
= m/s
0.04p ´ 9 ´ 103
23. Ans. 3
l
Sol. R =
KA
When heat is transferred from first vessel to second, temperature of first vessel decreases while that of
second vessel increases. Due to both there reasons, difference between temperature of vessels decreases.
Let at an instant t, the temperature difference between two vessels be q.
q KAq
H= =
R l
KAq
dQ = Hdt = dt ….(i)
l
Since gases are contained in two vessels, therefore, processes on gases in two vessels are isochoric.
Hence, decrease in temperature of gas in first vessel,
dQ dQ dQ
= =
Dq1 = nCv 2 ´ 5 R 5R
2
Increase in temperature of gas in second vessel is
dQ dQ
=
Dq2 = 4 ´ 3R 6 R
2
\ Decrease in temperature difference
(–dq) = Dq1 + Dq2
dQ 11
– dq = ´
R 30
Physics / GR # Thermal Conduction and radiation E-7/8
 GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH

dq KA ´ 11
25 t

or – ò q = 30lR ò dt
50 0

t = 3 seconds.
24. Ans. 7
dx
Sol. dR = 60(10 + x) ; lTB = b; TB = 103 K

1 é æ 3öù
R = 60 êln çè 2 ÷ø ú
ë û
(TA - TB )
= eA s éë TB4 - (300)4 ùû
R
TA » 1400 K
25. Ans. 2
heat energy
Sol. Pab =
time
msDT
(Intensity) (cross section area) =
t
(4 ´ 10 -3 )(4200)(5)
t= = 600 s = 10 min.
(1000)(1.4 ´ 10 -4 )
26. Ans. 3

E-8/8 Physics / GR # Thermal Conduction and radiation