33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
36 NEET-AIPMT Chapterwise Topicwise Solutions Physics
1 1 1
= + Req
= 18 5
Req
( 4+
2) ( 6+
3) ; / Ω
V V
V = iReq
⇒ i= = 5 Req
18 .
98. (c) : Equivalent circuit of given circuit is shown in
figure (i).
F
Also this is equivalent to a balanced
Wheatstone bridge C and D are at
equal potential, no current will flow
in this resistance therefore this
resistance can be neglected.
Thus equivalent resistance of this remaining
circuit shown in figure (ii) is R.
Then current in AFCEB branch is
V 2R
V
i1
= × =
R 2R+
2R
2R
99. (c) : In balance Wheatstone bridge,
the galvanometer arm can be neglected.
So equivalent resistance will be
2R×
2R
4R
= = R
2R+
2R
4R
100. (a)
2
101. (b) : This is a balanced Wheatstone’s bridge so no
current flows through the 7 W resistor.
1 1 1
14
∴ = + or R =
4 + 3 6 +
eq Ω
8 3
R e q
102. (d) : The circuit is equivalent to a balanced Wheatstone
bridge. Therefore resistance between A and B is 2 W.
103. (b) : Unknown is X, R = 10 W.
Here, l 1 3
= ; X l
= 1 l2
2 R l2
⇒ X = 3 × 10 ⇒ X = 15 W
2
Thus, 1.5 m length has resistance 15 W hence, length of
1 W of the resistance wire = 15 .
= 01 . m = 10 . × 10 −1
m
15
104. (d) : Yes, the bridge will work. For a balanced
condition, the current drawn from the battery will be
zero. Also, P ∝ l 1 and Q ∝ l 2 Therefore, the condition
A
X
l 1
G
l 2
R
B
P l
= 1 will remain same after interchanging the cell and
Q l2
galvanometer.
105. (b) : In the first case,
At balance point
5 l1
=
R 100 − l
...(i)
1
In the second case,
At balance point
5 16 . l1
= ...(ii)
( R / 2)
100−16
. l
Divide eqn. (i) by eqn. (ii), we get
1 100 −16
. l1
=
2 16100 . ( − l )
1
1
or 160 – 1.6l 1 = 200 – 3.2l 1
40
16 . l1 = 40 or l1
= = 25 cm
16 .
Substituting this value in eqn. (i), we get
5 25
R =
375
R = Ω=
15 Ω
75 or 25
106. (d) : Metre bridge works on the principle of
Wheatstone’s bridge.
P l
l
∴ = or, P = Q
. .
Q 100 − l 100 − l
× = 20
80 × 1=
0 25 Ω
107. (b) : A potentiometer is an accurate and versatile
device to make electrical measurement of emf because
the method involves a condition of no current flow
through the galvanometer. The device can be used to
measure potential difference, internal resistance of a cell
and compare emf ’s of two sources.
108. (b) : Suppose two cells have emfs e 1 and e 2 (also e 1 > e 2 ).
Potential difference per unit length of the potentiometer
wire = k (say)
When e 1 and e 2 are in series and support each other then
e 1 + e 2 = 50 × k
...(i)
When e 1 and e 2 are in opposite direction
e 1 – e 2 = 10 × k
...(ii)
On adding eqn. (i) and eqn. (ii)