33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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90 NEET-AIPMT Chapterwise Topicwise Solutions Physics
38. (b) : An ideal black body is one
which absorbs all the incident
radiation without reflecting or
transmitting any part of it.
Black lamp absorbs approximately
96% of incident radiation.
An ideal black body can be realized in practice by a small
hole in the wall of a hollow body (as shown in figure)
which is at uniform temperature. Any radiation entering
the hollow body through the holes suffers a number of
reflections and ultimately gets completely absorbed. This
can be facilitated by coating the interior surface with
black so that about 96% of the radiation is absorbed
at each reflection. The portion of the interior surface
opposite to the hole is made conical to avoid the escape
of the reflected ray after one reflection.
39. (a) : Wien’s displacement law states that the
product of absolute temperature and the wavelength at
which the emissive power is maximum is constant i.e.
l max T = constant. Therefore it expresses relation between
wavelength corresponding to maximum energy and
temperature.
40. (b) : Heat flow rate dQ KA( T1−
T2)
=
= Q1
dt L
When linear dimensions are double.
A 1 ∝ r 12 , L 1 = L
A 2 ∝ 4r 12 , L 2 = 2L so Q 2 = 2Q 1
41. (b) : According to Wein’s law, l m T = constant
\ l′ = (2/3)l m
42. (a) : E = sT 4 = 20; T ′ = 2T
\ E ′ = s(2T) 4 = 16 sT 4
= 16 × 20 = 320 kcal/m 2 min
43. (b) : Temperature of black body T = 500 K
Therefore total energy emitted by the black body
E ∝ T 4 ∝ (500) 4
44. (d) : Ratio of diameters of rod = 1 : 2 and ratio of
their lengths 2 : 1.
The rate of flow of heat, (Q) = KA ∆ T A ∝ .
l l
2
Q A
Therefore,
1 l
Q
= 1 2
2 A
× 1 1
2 l
= ⎛ 1
1 ⎝ ⎜ ⎞ 2 ⎠ ⎟ × 2
= 8
or Q 1 : Q 2 = 1 : 8
45. (d) : Amount of energy radiated ∝ T 4 .
46. (b) : According to Newton’s law of cooling,
dT
= KT ( −T dt
s )
For two cases,
dT1
dT2
= KT ( 1 − Ts) and = KT ( 2 −Ts)
dt
dt
3T
+ 2T
Here, T s = T, T1
= = 25 . T
2
and dT 1 3T − 2T T
= =
dt 10 10
2T + T′
dT2
2T
− T′
T2
= and =
2 dt 10
T
So, = K(. 25T −T) ...(i)
10
and 2 T − T′ ⎛ 2
K T + T ′ ⎞
= − T
10 ⎝
⎜
2 ⎠
⎟
...(ii)
Dividing eqn. (i) by eqn. (ii), we get
T 25T
T
2T
− T′ = (. − )
or, 2 T + T′ 3
− T = ( 2T − T ′)
×
⎛ 2T
+ T′ ⎞ − T 2
2
⎝
⎜
2 ⎠
⎟
T ′ = 3(2T – T ′) or, 4T ′ = 6T; ∴ T′ =
3 T
2
47. (a) : Let T s be the temperature of the surroundings.
According to Newton’s law of cooling
T1−
T2 ⎛
K T 1+
T 2 ⎞
= −T t ⎝
⎜
2 s ⎠
⎟
For first 5 minutes,
T 1 = 70°C, T 2 = 60°C, t = 5 minutes
70 − 60 ⎛ 70 + 60 ⎞
∴ = K −
⎝
⎜ T
⎠
⎟
5 2 s = K(65 – T s ) ... (i)
For next 5 minutes,
T 1 = 60°C, T 2 = 54°C, t = 5 minutes
60 − 54 ⎛ 60 + 54 ⎞
∴ = K −
⎝
⎜ T
⎠
⎟
5 2 s
6
= K( 57 −T 5
s ) ... (ii)
Divide eqn. (i) by eqn. (ii), we get
5 65 −Ts
=
3 57 −Ts
285 – 5T s = 195 – 3T s
2T s = 90 or T s = 45°C
48. (a) : The rate of cooling is directly proportional
to the temperature difference of the body and the
surroundings. So, cooling will be fastest in the first case
and slowest in the third case.
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