33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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100 NEET-AIPMT Chapterwise Topicwise Solutions Physics
50. (c) : Efficiency of a Carnot engine
η= 1−
T2
T1
where T 1
is the temperature of source and T 2
is the
temperature of sink respectively.
T
For engine A, η A = 1−
T
For engine B, η B
1
T2
= 1−
T
As per question, h A
= h B
T T2
T T2
∴ 1− = 1− ⇒ = or T =
T1
T T1
T
51. (c) : Efficiency of an engine, η= 1−
TT 1 2
T2
T1
where T 1
is the temperature of the source and T 2
is the
temperature of the sink.
1
∴ = − =
6 1 T2
T2
5
or, ...(i)
T1
T1
6
When the temperature of the sink is decreased by 62°C
(or 62 K), efficiency becomes double.
Since, the temperature of the source remains unchanged
−
−
∴ 2 × 1 = −
= −
6 1 ( T2
62)
1
3 1 ( T2
62)
or,
T1
T1
or,
2
3
T2
− 62
=
T1
or, 2T1 = 3T2
−186
or, 2T
3 5 1 = ⎡ T1
186
⎣ ⎢ ⎤ 6⎦ ⎥ − [using (i)]
∴
⎡5
⎤
⎢ − ⎥ = =
⎣2 2 ⎦
186 T1
T1
or, 186
2
or, T 1
= 372 K = 99°C
T
52. (d) : Efficiency of a Carnot engine, 2
η= 1−
T
T1
2
or,
T
= 1 − 40 3
η = 1 −
1
100
= 5
5 5
∴ T1 = × T2
= × 300 = 500 K.
3 3
Increase in efficiency = 50% of 40% = 20%
New efficiency, h′ = 40% + 20% = 60%
T2
60 2
∴ = 1− =
T1
′ 100 5
T′ 5 5
1 = × T2
= × 300 = 750 K.
2 2
Increase in temperature of source = T 1
′ – T 1
= 750 – 500 = 250 K
vvv
T
53. (d) :
2 Q2
1−
= 1−
⇒ 1− 400 = −
T1
Q 1 500 1 Q2
6 × 10
4 Q2
⇒ =
5 6 ×
4
⇒ Q
10
2
= 4.8 × 10 4 cal
Net heat converted into work
= 6.0 × 10 4 – 4.8 × 10 4 = 1.2 × 10 4 cal
54. (d) : Efficiency of Carnot engine
W T2
= = 1 −
Q T
1
1
W 400
1
W
6
= − 1 6
500
= 5
⇒ = 5 = 12 . kcal.
55. (c) : Efficiency (h) of a carnot engine is given by
T2
η= 1−
T , where T 1
is the temperature of the source and
1
T 2
is the temperature of the sink.
Here, T 2
= 500 K
∴ 05 . = 1− 500 ⇒ T1
= 1000 K .
T1
T ′
2
Now, η= 06 . = 1−
(T 1000 2
′ is the new sink temperature)
⇒ T 2
′ = 400 K.
56. (a) : Efficiency is maximum in Carnot engine which
is an ideal engine.
400 − 300
η= × 100% = 25%
400
\ efficiency 26% is impossible for his heat engine.
57. (b) : 1 6 1 T2
5 T2
= − or =
T1
6 T1
Now, 1 3 1 T2
− 62 5 62
= − = 1 − +
T1 6 T1
T 1
= 62 × 6 = 372 K = 99°C and T 2
= 310 K = 37°C
58. (b) : Reservoir temperature (T 1
) = 100°C = 373 K
and T 2
= – 23°C = 250 K.
The efficiency of a Carnot engine
T1−
T2
373 − 250
η= = .
T1
373
59. (c) : Efficiency of Carnot engine (h 1
) = 40% = 0.4
New efficiency (h 2
) = 50 % = 0.5.
T
The efficiency (h) =
2 T2
1−
or = 1−
η
T1
T1
For first case, T 2
500 = 1− 04 . or T 2
= 300 K
For second case, 300 1 05
T
= − . or T 1 = 600 K
1
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