Chemistry_Today_April_2017_vk_com_stopthepress

11. (a, d) : As, F 1 = k 1 x, F 2 = k 2 x.
1 2
1 2
Work done W1 = k1x
and W2 = k2x
2
2
W1
k1
or α= =
W2
k2
When the springs are stretched by the same force
F, the extensions in springs A and B are x 1 and x 2
respectively which are given by
F = k 1 x 1 = k 2 x 2 or x 1
k2
= …(i)
x2
k1
1 2
1 2
Work done W1′=
k1x1
and W2′ = k2x2
2
2
2
W1′
W ′ = k1
k
⋅ x1
…(ii)
2
2 2 x2
Using (i) and (ii) we get
2
W1′
β=
W ′ = k1
k
⋅ k2
k
= k2
2
2 2
k
1 1
800 − 80
12. (a, b, c, d) : Rate of heat flow H =
⎛ li
⎞ lo
⎝
⎜ KA⎟ i ⎠
+ ⎛ ⎞
⎜ ⎝ KoA⎠
⎟
which is also equal to 800 − T
. Using these two
⎛ li
⎞
⎝
⎜ KA⎠
⎟
i
720
relations we get, T = 800 −
Ki
lo
+ ⎛ . Thus
⎞ ⎛ ⎞
1
⎜ ⎝ Ko
⎠
⎟ ⎜
⎝ li
⎠
⎟
one can reduce the temperature at the interface by
any of the four options given.
13. (a, b, c, d) : Intensity, by definition, is the energy
flowing per unit area per unit time.
The intensity is related to the displacement
amplitude A of the sound wave by
2 2
I = 1 ρω v A
2
The displacement amplitude is given by A = P ,
Bk
⎛ ω⎞
where k =
⎝
⎜
v ⎠
⎟ is the propagation constant.
B
The speed is given by v = . ρ
14. (2) :
Use these relations to get the required expressions.
A
d/2 x
B 2D
Optic axis
O x d/4
1
C
3f
From similar triangles AOB and BDC
OB AO
= or x 1 d 2
= ( / )
BD CD x (
2
d / 4)
or x 1 = 2x 2
As x 1 + x 2 = 3f, 2x 2 + x 2 = 3f or x 2 = f
i.e. x 1 = 2f
n = 2.
15. (7)
16. (4) : Let the velocities of car 1 and car 2 be
v 1 m s –1 and v 2 m s –1 .
∴ Apparent frequencies of sound emitted by car 1
and car 2 as detected at end point are
υ0v
υ0v
υ1
= and υ −
2
=
v v1
v − v2
300 × 330
−1
∴ 330 =
or v1
= 30 ms
330 − v1
300 × 330
−1
and 360 =
or v2
= 55 ms
330 − v
A
2
vt 1
vt 2
1 2
B
100 m
The distance between both the cars just when the
2 n d car reaches point B(as shown in figure) is
100 m = v 2 t – v 1 t
100 100
t = = = 4 s
v − v 55 − 30
2 1
17. (5) : The capacitance of a parallel plate capacitor in
air is given by
A
C = ε 0
...(i)
d
By introducing a slab of thickness t, the new
capacitance C′ becomes
ε
C′ = 0
A
⎛ 1 ⎞
...(ii)
d′ −t⎜1
−
⎝ ⎠
⎟
K
The charge (Q = CV) remains the same in both the
cases.
Hence
ε0A
ε0A
=
d ⎛ 1 ⎞
d′ −t⎜1
−
⎝ K ⎠
⎟
⎛ 1 ⎞
or d= d′ −t
−
⎝
⎜1
K ⎠ ⎟
80 CHEMISTRY TODAY | APRIL ‘17