33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Thermal Properties of Matter
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89
Q =
KA( T 1 − T 2 ) t
... (i)
L
where K is the thermal conductivity of the material of
the rod.
Area of cross-section of new rod
′ = ⎛ 2 2
⎝ ⎜ R⎞
πR A
A π
⎠
⎟ = =
2 4 4
As the volume of the rod remains unchanged
\ AL = A′L′
where L′ is the length of the new rod
or L′ = L A A ′
= 4L
Now, the amount of heat flows in same time t in the new
rod with its ends maintained at the same temperatures T 1
and T 2 is given by
′ −
Q′ =
KA ( T 1 T 2)
t
L′
Substituting the values of A′ and L′
−
−
Q′ =
K ( A / 4)( T 1 T 2) t 1 KA( T1 T2)
t 1
=
= Q
4L
16 L 16
(Using (i))
26. (a) : According to the Stefan Boltzmann law, the
power radiated by the star whose outer surface radiates
as a black body at temperature T K is given by
P = s4pr 2 T 4
where, r = radius of the star, s = Stefan’s constant
The radiant power per unit area received at a distance R
from the centre of a star is
2 4 2 4
P σ4πrT
σrT
S = = =
2 2 2
4πR
4πR
R
27. (c)
28. (b) : Rate of heat radiated at (227 + 273) K
= 7 cals/(cm 2 s)
Let rate of heat radiated at (727 + 273) K
= x cals/(cm 2 s)
By Stefan’s law, 7 ∝ (500) 4 and x ∝ (1000) 4
x
∴ = 2 4 ⇒ x = 7× 2 4
2
= 112 cals/(cm s).
7
29. (c) : Similar to I = V/R
dQ kA
dt
= L ( T 1 −T
2 )
k = conductivity of the rod.
30. (a) : According to Stefan’s law, rate of energy
radiated E ∝ T 4
where T is the absolute temperature of a black body.
\ E ∝ (727 + 273) 4 or E ∝ [1000] 4 .
31. (a) : According to Wein’s displacement law,
l max T = constant
λmax
1
T2
∴ =
λ T
max2
1
λmax
× T
1 1 5000 × 1500
or λmax
= =
= 3000 Å.
2 T2
2500
32. (b) : Heat conducted
2
KA( T1−
T2) t Kπr ( T1−
T2)
t
=
=
l
l
The rod with the maximum ratio of r 2 /l will conduct
most. Here the rod with r = 2r 0 and l = l 0 will conduct
most.
33. (d) : Wein’s displacement law
l m T = constant, l m ∝ T –1
34. (a) : The slabs are in series.
Total resistance R = R 1 + R 2
l l l
⇒ = +
AKeffective AK . A2K
1 1 1 3
⇒ = + =
Keffective
K 2K 2K
∴ Keffective
=
2K
3
35. (d) : Unit of Stefan’s constant is watt/m 2 K 4 .
36. (a) :
Rate of heat loss in rod 1 = Q 1 = KA 1 1( T 1−
T 2)
l1
Rate of heat loss in rod 2 = Q 2 = K 2 A 2( T 1−
T 2)
l2
By problem, Q 1 = Q 2 .
∴
KA 1 1( T1−
T2) K2A2( T1−
T2)
=
l1
l2
\ K 1 A 1 = K 2 A 2 [Q l 1 = l 2 ]
37. (b) : Radiating power of a black body
4
= E0
= σ( T −T0 4 ) A
where s is known as the Stefan-Boltzmann constant, A is
the surface area of a black body, T is the temperature of the
black body and T 0 is the temperature of the surrounding.
\ 60 = s(1000 4 – 500 4 ) ...(i)
[T = 727°C = 727 + 273 = 1000 K, T 0 = 227°C = 500 K]
In the second case, T = 1227°C = 1500 K and let E′ be the
radiating power.
\ E′ = s(1500 4 – 500 4 ) ...(ii)
From (i) and (ii) we have
4 4 4 4
E′ 1500 − 500 15 − 5
= = =
60 4 4 4 4
1000 − 500 10 − 5
50000
∴ E′ = × 60 = 320 W.
9375
50000
9375