33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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4 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Hints & Explanations
1. (c) : In case I : Q r – Q
A
B
Q
F =− 1 2
4πε
2
0 r
In Case II : Q Q Q Q Q Q
A = − , B =− +
4 4
1
∴ F′ =
4πε
0
⎛ ⎞
−
⎝
⎜
⎠
⎟ ⎛
⎝
⎜ − + ⎞
Q Q Q Q
⎠
⎟
4 4
r
2
....(i)
⎛ 3 ⎞ ⎛ −3
⎞
⎝
⎜ Q
⎠
⎟
⎝
⎜ Q
1
⎠
⎟
2
4 4 1 9 Q
...(ii)
=
=−
4πε
2
0
4πε0
16 2
r
r
From equations (i) and (ii), F′= 9 F
16
2. (b) : A hydrogen atom consists of an electron and a
proton.
\ Charge on one hydrogen atom
= q e + q p = –e + (e + De) = De
Since a hydrogen atom carries a net charge De
e
\ Electrostatic force, Fe
= 1 2
( ∆ )
...(i)
4πε
2
o d
will act between two hydrogen atoms.
The gravitational force between two hydrogen atoms is
given as
Gmhmh
Fg
=
...(ii)
2
d
Since, the net force on the system is zero, F e = F g
Using eqns. (i) and (ii), we get
2 2
( ∆e)
Gmh
= or, 2
2
( ∆e)
= 4πε
Gm
2 2
0 h
4πεod
d
= 6.67 × 10 –11 × (1.67 × 10 –27 ) 2 [1/(9 × 10 9 )]
De ≈ 10 –37 C
3. (a) :
From figure, T cos q = mg...(i)
2
kq
T sinθ=
...(ii)
2
x
From eqns. (i) and (ii), tanθ= kq 2
2
xmg
Sinceθissmall, ∴ tanθ≈ sinθ
= x 2l
2
x kq
2 3
∴ = ⇒ q = x mg
32 /
or q∝
x
2l
2
xmg
2lk
dq 3
⇒ ∝ x dx 3
= xv
dt 2 dt 2
dq
Since,
constant
dt = 1
; ∴ v ∝
x
4. (d) : Let m be mass of each ball and q be charge on
each ball. Force of repulsion,
q
F = 1 2
4πε
2
0 r
In equilibrium
Tcosq = mg
Tsinq = F
...(i)
...(ii)
Divide (ii) by (i), we get, tanθ
From figure (a),
1 q
r/2
4 0 r
= πε 2
y mg
...(iii)
For figure (b)
1
2
q
1
4πε
2
0 r′
r′
/ 2 4πε
tan θ′ = ⇒
0
=
mg y/
2
F
= =
mg
r
mg
2
q
2
′
1 q
4πε0 r
2
mg
r
...(iv)
Divide (iv) by (iii), we get
2
3
2r′ r 3 r
r
= ⇒ r′ = ⇒ r′ =
r ′
2 2 3
r
2
5. (c) : According to Coulomb’s law, the force of
repulsion between the two positive ions each of charge q,
separated by a distance d is given by
q q
F = 1
2
( )( )
q
or F =
4πε
2
0 d
4πε 0d
2
q 2 = 4pe 0 Fd 2 or q= 2
4πε 0Fd
...(i)
Since, q = ne
where, n = number of electrons missing from each ion,
e = magnitude of charge on electron