33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Thermal Properties of Matter
Telegram @unacademyplusdiscounts
Hints & Explanations
87
7. (a) : As ρT
2
ρT
ρ
1 T1
\ 560m = 1400 or m = 2.5 g
= = ( 1 + γ ∆ T ) 1 + γ ( T2 − T1
)
Total mass of water present
= (20 + m) g = (20 + 2.5) g = 22.5 g
1. (d) :
Here, T 1 = 20°C, T 2 = 40°C
r 20 = 998 kg/m 3 , r 40 = 992 kg/m 3
∴
998
998
992 = or, 992 =
1 + γ( 40 − 20)
1 + 20 γ
998 998
\ 39°C = 39 × 2 + 39 = (78 + 39)° W
1 20
20
992 992 1 6
+ γ= or γ= − =
992
= 117° W
6 1
γ= ×
2. (c) : Mercury thermometer is based on the principle
992 20 = 3 × 10–4 /°C
of change of volume with rise of temperature and can 8. (a) : Heat required, DQ = msDT
measure temperatures ranging from –30°C to 357°C DQ = (V × r) × s DT
3. (c) : Here, F = 140°
Using F
= 4
− 32 C
=
180 100 , 140 − 32 C
3 pr3 r ⋅ s DT
= ⇒ C = 60°C
180 100
∆Q1
r1 3
r1
3 27
=
we get, fall in temperature = 40°C
∆Q2
r r
2 3 = ⎛ ⎞
15
⎝ ⎜ 2 ⎠
⎟ = = 8
4. (a) : As per question, Dl Cu = Dl Al
or, l Cu a Cu DT = l Al a Al DT
9. (d) : Thermal capacity = ms = 40 × 0.2
= 8 cal/K = 33.6 J/K.
l
−
Cuα
5
Cu 88 × 17 . × 10
lAl
= =
= 68 cm
10. (b) : Since, heat capacity of material increases with
α
−5
Al 22 . × 10
increase in temperature so, body at 100°C has more
5. (b) : Linear expansion of brass = a 1
heat capacity than body at 0°C. Hence, final common
Linear expansion of steel = a 2
temperature of the system will be closer to 100°C.
Length of brass rod = l 1 , Length of steel rod = l 2
\ T c > 50°C
On increasing the temperature of the rods by DT, new
lengths would be
11. (a) : Gravitational potential energy of a piece of ice
at a height (h) = mgh
l1′ = l1( 1+
α∆ 1 T)
... (i)
Heat absorbed by the ice to melt completely
l2′ = l2( 1+
α∆ 2 T)
... (ii) ∆Q= mgh
...(i)
4
Subtracting eqn. (i) from eqn. (ii), we get
Also, DQ = mL
...(ii)
l2′ − l1′ = ( l2 − l1) + ( l2α2 −l1α1)
∆ T
1
4L
According to question,
From eqns. (i) and (ii), mL = mgh or, h=
l2′ − l1′ = l2 −l1 (for all temperatures)
4
g
\ l 2 a 2 – l 1 a 1 = 0 or l 1 a 1 = l 2 a
Here L = 3.4 × 10 5 J kg –1 , g = 10 N kg –1
2
6. (d) : Let r 0 and r T be densities of glycerin at 0°C and ∴ = × × 5
4 34 . 10
3
h
= 4× 34× 10 = 136 km
T °C respectively. Then,
10
r T = r 0 (1 – gDT)
where g is the coefficient of volume expansion of glycerine
and DT is rise in temperature.
12. (d) : Here,
Specific heat of water, s w = 1 cal g –1 °C –1
Latent heat of steam, L s = 540 cal g –1
ρT
ρT
Heat lost by m g of steam at 100°C to change into water
= 1− γ∆T
or γ∆T
= 1−
ρ0 ρ
at 80°C is
0
Thus, ρ 0 − ρ
Q 1 = mL s + ms w DT w
T
= γ∆T
= m × 540 + m × 1 × (100 – 80)
ρ0
Here, g = 5 × 10 –4 K –1 = 540m + 20m = 560m
and DT = 40°C = 40 K
Heat gained by 20 g of water to change its temperature
\ The fractional change in the density of glycerin
from 10°C to 80°C is
ρ0
− ρT
−4 −1
= = γ∆ T = ( 5× 10 K )( 40 K) = 0020 .
Q 2 = m w s w DT w = 20 × 1 × (80 – 10) = 1400
ρ0
According to principle of calorimetry, Q 1 = Q 2