Practical Physics- 2 Practice Questions - DPP for NEET

 Page 1


1. (b) Voltmeter measures voltage across two points so it is
connected in parallel and ammeter measures current
so it is connected in series.
2. (a)
PR
QS
= so by changing the gap resistance of copper
strip gets cancelled.
3. (b) Potential gradient  = Potential drop/unit length
V = IR = I 
A
r l
\ 
VI
A
r
=
l
]
4. (d) Refractive index is the property of the material, hence
it does not depend on angle of the prism.
5. (a) A meter-bridge is a device which is based on the
principle of Wheatstone bridge.
6. (a) A potentiometer is device which is used to compare
e.m.f.'s of two cells as well as to determine the internal
resistance of a cell. It is based on the principle that when
a current flows through a wire of uniform thickness and
material, potential difference between its two points is
directly proportional to the length of the wire between
the two points.
7. (c) When a p-n junction diode is connected in reverse
biased mode, a reverse current flows.
8. (b) A Zener diode is a heavily doped p-n junction diode
which operates on reverse bias beyond breakdown
voltage.
9. (c) In a transfer characteristics  V
i
 is plotted along x-axis
and V
0
 along y-axis.
10. (a) Current gain in CE configuration is b = 
C
B
I
I
?
?
.
I
c
 >> I
B
, hence it is maximum.
11. (c) V = IR = I
VI
k
AA
l
l
rr
\ << ,
increase in I, will increase k, so it will decrease
sensitivity.
12. (c)
13. (c) According to the figure the voltmeter and the resistor
are connected in parallel.
14. (b) Here I = 4A
c c
30
30
180 6
´pp æö
q=°==
ç÷
èø
Now, 
I 4 467 267
k
22 11
6
´´ ´´
== ==
p q æö
ç÷
èø
             
84
7.6 A / rad
11
==
15. (c) In reverse-bias mode, a reverse current flows.
Therefore, (c) represents the form.
16. (c) Larger the length, lesser will be the potential gradient,
so more balancing length will be required.
17. (a) LED is a p-n junction diode which always operates on
forward bias.
18. (b) Emitter current is the sum of base and collector current
by Kirchhoff's 1st law.
19. (b) V = IR = 
I
A
rl
 \ potential gradient  = k
when A is decreased, k will increase.
20. (d) Positive terminal is at lower potential
(0V) and negative terminal is at higher potential + 5V .
21. (d)
22. (d) V = 30.0,  I = 0.020 A,   
V 30.0
R 1.50k
I 0.020
== =W
Error : As 
V R VI
R
I R VI
D DD
= \ =+
Þ  
VI
RR
VI
DD æö
D=+
ç÷
èø
= 1.50 × 10
3 
0.1 0.001
0.080 k
30.0 0.020
æö
+ =W
ç÷
èø
23. (a) u = – 0.30 m, v = – 0.60m
By mirror formula,
1 11
v uf
+=
Þ 
1 11
f 0.30 0.60
-
=- Þ  
3.0
f f 0.20m
0.60
-
= Þ=
Þ 
2 22
1 1 1 df dv du
f vu
f vu
- --
=+Þ = -
Þ df = (0.20)
2
 
22
0.01 0.01
(0.60) (0.30)
éù
+
êú
êú
ëû
Þ df = 0.0055 » 0.01m
Þ Focal length f = (0.20 ± 0.01) m
24. (d) As shown in the figure.
P
,P
Q 100 100
=µ
--
ll
ll
P(unknown)Q
G
l
B
100l