1. Xtra Edge February 2012 - Career Point

1. For the process A → B
1 3 1
U ∝ ⇒ RT ∝
ρ 2 M
V
or T ∝ V
So AB is isobaric process (pressure in constant)
3. TV
8U
0 ⎡ 3 ⎤
T A = ⎢As
4U
0 = RTA
3R
⎥
⎣ 2 ⎦
M
VA
=
2ρ
0
16U
so
0ρ
0
M
PA
= and VB
=
3M
4ρ
0
so WAB = P[VB - VA]
2.
16U
0ρ
0 ⎡ M M ⎤ 16U
0ρ
0 M
⇒ ⎢ − ⎥ = − ×
3M
⎣4ρ
0 2ρ
0 ⎦ 3M
4ρ0
Option [D] is correct
P
The slope of the graph is
0
m = −
2V0
The equation of this graph is given by
P0V
3P0
P = − +
2V0
2
nRT −P0
V 3P0
⇒ = +
V 2V0
2
−P0
V 2 3P0
⇒ T = − V + V
2nRV0
2nR
dT −2P0V0
3P0
= + = 0
dV 2nRV0
2nR
P0V ⇒
nRV0
2P0
3V0
= ⇒ V =
2nR
2
3V
At 0
V = , temp will be maximum [n = 1]
2
2
P0
9V0
3P0
3V0
Tmax
= − × + ×
2RV0
4 2R
2
9 P0V0
9 P0V0
⇒ − +
8 R 4 R
9 P0V0
⇒
8 R
Option [C] is correct
-3 PV
= k − 3
⇒ V = k
nR
for this polytrophic process, PV -2 = constant
so x = -2
nR(
T2
− T1
) nR(
T2
− T1
) 2
w =
=
= nRT0
1−
x
3 3
∆U = nCV∆T = n 3/2 R(3T0 – T0) = 3nRT0
⎛ 2 ⎞ 11
∆ Q = ∆U
= W = ⎜ + 3⎟nRT0
= nRT0
⎝ 3 ⎠ 3
∆Q
11 nRT 11
∆Q = nC∆T 0
⇒ C = = = R
n∆T
3 n(
2T0
) 6
Option [A,C,D] is correct
y
4.
R
ωt
F
Fx
Fy
x
F = mω 2 R
Fy = F sin ωt
Fy = mω 2 R sin (ωt)
x
* In figure as cos ωt = ⇒ x = R cos ωt
R
* For one complete circle, θ = ωt
y
*
R
ωt
vx
x vx = – ωR sin ωt
v = ωR
Option [A,C] is correct
5. From 1 st 8
Solution
Set # 9
Physics Challenging Problems
Questions were Published in January Issue
law,
∆Q = ∆U + W
nCαT + nCvαT + pαV
⇒ nCαT = nCVαT + nRT/V αV
α V
αV
nRT αV
⇒ nαT0e
αV
= C VαT0
e αV
+ T0e
αV
V
⇒ αC = αCV + R/V ⇒ C = CV + R/αV
Option [B] is correct
6. Option [C] is correct
7. Option [B] is correct
8. Option [D] is correct
XtraEdge for IIT-JEE 16 FEBRUARY 2012