33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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78 NEET-AIPMT Chapterwise Topicwise Solutions Physics
ANSWER KEY
1. (b) 2. (d) 3. (d) 4. (a) 5. (c) 6. (d) 7. (a) 8. (b) 9. (c) 10. (b)
Hints & Explanations
1. (b)
2. (d) : Given : initial length = L, area of cross section = A
New length after mass M is suspended on the wire = L
\ Change in length, DL = L 1 – L.
Now Young’s modulus, Y = Stress
= mg
A
L MgL
∆L
or AL ( 1 − L)
3. (d) : Stress = F Mg
=
A A
,
Strain = ∆L L l L l
= + − =
L L L
Energy stored in the wire is,
Strain = F
A
×
L
∆L
U = 1 × Stress × Strain × Volume
2
= 1 2 × Mg
× l
A L × A×
L = 1 2 Mgl
Fl
4. (a) : Young’s modulus, Y =
A ∆ l
Since initial volume of wires are same and their areas
of cross sections are A and 3A so lengths are 3l and l
respectively.
For wire 1,
F
∆l = ⎛ l
⎝ ⎜ ⎞
AY ⎠
⎟ 3
...(i)
For wire 2, let F ′ force is applied
F′ =
A Y ∆ l
3 l
⎛ F′
⎞
⇒ ∆l =
⎝
⎜
3 AY ⎠
⎟ l
From eqns (i) and (ii),
⎛ F ⎞ ⎛ F′
⎞
l
⎝
⎜
AY ⎠
⎟ 3 =
⎝
⎜
AY ⎠
⎟ l ⇒ F ′ = 9
3
F
...(ii)
5. (c) : Bulk modulus B is given as
pV
B = − ...(i)
∆ V
The volume of a spherical object of radius r is given as
4 4
V = πr 3 , ∆V = π( 3r 2 ) ∆r
3 3
4 3
πr
V
∴ − =
3
V r
or − =−
∆V
4 ∆V
∆r
r ∆r
3 3 2
3
π
pr
Put this value in eqn. (i), we get B =− 3∆r
∆r p
Fractional decrease in radius is − =
r 3B
6. (d) : Let L and A be length
and area of cross section of each
wire. In order to have the lower
ends of the wires to be at the
same level (i.e. same elongation
is produced in both wires), let
weights W s and W b are added to steel and brass wires
respectively. Then, by definition of Young’s modulus, the
elongation produced in the steel wire is
WL
⎛
s
∆Ls
= as Y W / A ⎞
=
YA
⎝
⎜
LL⎠
⎟
∆ /
s
WL b
and that in the brass wire is ∆Lb
=
YA b
But DL s = DL b (given)
WL s WL b Ws
Ys
∴ = or =
YA s YA b Wb
Yb
As Y s
Y
= 2 ; ∴ Ws
= 2
b Wb
1
7. (a) : Depth of ocean, d = 2700 m
Density of water, r = 10 3 kg m –3
Compressibility of water, K = 45.4 × 10 –11 Pa –1
∆V
V = ?
Excess pressure at the bottom, DP = rgd
= 10 3 × 10 × 2700 = 27 × 10 6 Pa
∆P
We know,
B =
( ∆VV
/ )
⎛ ∆V
⎞ ∆P
K ∆P K
⎝
⎜
V ⎠
⎟ = B
= ⎛
= 1 ⎞
.
⎝
⎜
B ⎠
⎟
= 45.4 × 10 –11 × 27 × 10 6 = 1.2 × 10 –2
8. (b) : As V = Al ... (i)
where A is the area of cross-section of the wire.
( F/ A)
Fl
Young’s modulus, Y = =
( ∆l/ l)
A∆l
2
Fl Fl
∆l = YA
=
(Using (i))
YV
Dl ∝ l 2
Hence, the graph between Dl and l 2 is a straight line.