Practical Physics - 1 Practice Questions - DPP for NEET

 Page 1


1. (b)
20 division of vernier scale = 8 div. of main scale
Þ 1 V .S.D. = 
82
M.S.D. M.S.D.
205
æ ö æö
=
ç ÷ ç÷
è ø èø
Least count
= 1 M.S.D – 1 V .S.D. = 1 M.S.D. – 
2
5
æö
ç÷
èø
 M.S.D
= 
2
1 M.S.D.
5
æö
-
ç÷
èø
 = 
33
M.S.D. 0.1 cm. 0.06 cm.
55
æö
=´=
ç÷
èø
(Q 1 M.S.D. = 
1
10
 cm. = 0.1 cm.)
Directly we can use
ba
L.C.MVM
b
- æö
= -=
ç÷
èø
         = 
20813
cm. cm. 0.06 cm.
20 10 50
- æ öæö
==
ç ÷ç÷
è øèø
2. (c) Within elastic limit it obeys Hooke's Law i.e., stress µ
strain.
3. (c) Least count 
111
cm
N 10 10N
=´=
4. (b) 5th division of vernier scale coincides with a main scale
division. L.C. = 
1
0.1mm
10
=
\ Zero error = – 5 × 0.1 = – 0.5 mm
This error is to be subtracted from the reading taken for
measurement. Also, zero correction = + 0.5 mm.
5. (b) If Y = Young's modulus of wire, M = mass of wire,
g = acceleration due to gravity , x = extension in the wire,
A= Area of cross-section of the wire,
l = length of the wire.
Mgx Y M xA
Y
A Y MxA
DDD D D
= Þ = + ++
l
ll
Þ 
Y 0.01 0.01 2 0.001 0.001
0.065
Y 3.00 0.87 0.041 2.820
D´
= + + +=
or 
Y
100 6.5%
Y
D
´ =±
6. (b) The instrument has negative zero error.
7. (c) If A, B and C be the points corresponding to the
impressions made by the legs of a spherometer then
3
a bc
c
++
=
A
C B
b d
a
If h is the depression or elevation then the radius of
curvature is given by 
2
62
=+
ch
r
h
8. (b) L.C. = 
Pitch
No. of circular divisions
9. (b) The specific heat of a solid is determined by the method
of mixture.
10. (a)
11. (d) Least count of screw gauge = 
0.5
mm 0.01mm
50
=
\ Reading = [Main scale reading  + circular scale
                            reading × L.C] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
12. (d) 30 Divisions of vernier scale coincide with 29 divisions
of main scales
Therefore 1 V .S.D = 
29
30
MSD
Least count = 1 MSD – 1VSD
= 1 MSD 
29
30
-
 MSD
= 
1
MSD
30
= 
1
0.5
30
´° = 1 minute.
13. (c) Least count = 
0.5
0.01mm
50
=
Zero error = 5 × L.C
           = 5 × 0.01 mm
           = 0.05 mm
Diameter of ball = [Reading on main scale] + [Reading
on circular scale × L . C] – Zero error
= 0.5 × 2 + 25 × 0.01 – 0.05
= 1.20 mm
14. (d) 2
gT
gT
D DD
=+
l
l
Dl and DT are least and number of readings are
maximum in option (d), therefore the measurement of
g is most accurate with data used in this option.
Page 2


1. (b)
20 division of vernier scale = 8 div. of main scale
Þ 1 V .S.D. = 
82
M.S.D. M.S.D.
205
æ ö æö
=
ç ÷ ç÷
è ø èø
Least count
= 1 M.S.D – 1 V .S.D. = 1 M.S.D. – 
2
5
æö
ç÷
èø
 M.S.D
= 
2
1 M.S.D.
5
æö
-
ç÷
èø
 = 
33
M.S.D. 0.1 cm. 0.06 cm.
55
æö
=´=
ç÷
èø
(Q 1 M.S.D. = 
1
10
 cm. = 0.1 cm.)
Directly we can use
ba
L.C.MVM
b
- æö
= -=
ç÷
èø
         = 
20813
cm. cm. 0.06 cm.
20 10 50
- æ öæö
==
ç ÷ç÷
è øèø
2. (c) Within elastic limit it obeys Hooke's Law i.e., stress µ
strain.
3. (c) Least count 
111
cm
N 10 10N
=´=
4. (b) 5th division of vernier scale coincides with a main scale
division. L.C. = 
1
0.1mm
10
=
\ Zero error = – 5 × 0.1 = – 0.5 mm
This error is to be subtracted from the reading taken for
measurement. Also, zero correction = + 0.5 mm.
5. (b) If Y = Young's modulus of wire, M = mass of wire,
g = acceleration due to gravity , x = extension in the wire,
A= Area of cross-section of the wire,
l = length of the wire.
Mgx Y M xA
Y
A Y MxA
DDD D D
= Þ = + ++
l
ll
Þ 
Y 0.01 0.01 2 0.001 0.001
0.065
Y 3.00 0.87 0.041 2.820
D´
= + + +=
or 
Y
100 6.5%
Y
D
´ =±
6. (b) The instrument has negative zero error.
7. (c) If A, B and C be the points corresponding to the
impressions made by the legs of a spherometer then
3
a bc
c
++
=
A
C B
b d
a
If h is the depression or elevation then the radius of
curvature is given by 
2
62
=+
ch
r
h
8. (b) L.C. = 
Pitch
No. of circular divisions
9. (b) The specific heat of a solid is determined by the method
of mixture.
10. (a)
11. (d) Least count of screw gauge = 
0.5
mm 0.01mm
50
=
\ Reading = [Main scale reading  + circular scale
                            reading × L.C] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
12. (d) 30 Divisions of vernier scale coincide with 29 divisions
of main scales
Therefore 1 V .S.D = 
29
30
MSD
Least count = 1 MSD – 1VSD
= 1 MSD 
29
30
-
 MSD
= 
1
MSD
30
= 
1
0.5
30
´° = 1 minute.
13. (c) Least count = 
0.5
0.01mm
50
=
Zero error = 5 × L.C
           = 5 × 0.01 mm
           = 0.05 mm
Diameter of ball = [Reading on main scale] + [Reading
on circular scale × L . C] – Zero error
= 0.5 × 2 + 25 × 0.01 – 0.05
= 1.20 mm
14. (d) 2
gT
gT
D DD
=+
l
l
Dl and DT are least and number of readings are
maximum in option (d), therefore the measurement of
g is most accurate with data used in this option.
86
DPP/ P 31
15. (b) We know that Y = 
2
4
mgL
D
´
p
l
Þ
( )( )
22
33
4 419.8 2
0.4 10 0.8 10
mgL
Y
D --
´´´
==
p
p ´ ´´
l
112
2.0 10 N/m =´
Now
2 YD
YD
D DD
=+
l
l
[Q the value of m, g and L are exact]
= 
0.01 0.05
2
0.4 0.8
´+ = 2 × 0.025 + 0.0625
= 0.05 + 0.0625 = 0.1125
Þ
DY = 2 × 10
11
 × 0.1125 = 0.225 × 10
11
112
0.2 10 N/m =´
16. (b) The time period of a simple pendulum is given by
2 T
g
=p
l
\
22
4 T
g
=p
l
 Þ 
2
2
4 g
T
=p
l
Þ 100 100 2 100
gT
gT
D DD
´ = ´ +´
l
l
Case (i)
Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s
100 0.3125
g
g
D
\ ´=
Case (ii)
Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s
100 0.46875
g
g
D
\ ´=
Case (iii)
Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s
100 1.055
g
g
D
\ ´=
Clearly , the value of 100
g
g
D
´ will be least in case (i)
17. (c) Terminal velocity , 
2
12
2()
9
-
=
h
T
rd dg
v
2
2
(10.5 1.5) 9
0.2
0.2 (19.5 1.5) 18
-
= Þ =´
-
T
T
v
v
2
0.1/ \=
T
v ms
2 2
2
0
ìü
ïï æö
- - +=
íý ç÷
èø
ïï
îþ
d y dy dy
x By B By
dx dx
dx
Þ 
2 2
2
0
æö
+ -=
ç÷
èø
d y dy dy
xy xy
dx dx
dx
18. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
F
v
B=Vg r
2 
W=V    g
1
r
2
12 t
V g V g kv \ r = r+     
11
() r -r
\=
t
Vg
v
k
19. (d) From the figure it is clear that liquid 1 floats on liquid
2.  The lighter liquid floats over heavier liquid.
Therefore we can conclude that 
12
r <r
Also r
3
 < r
2
 otherwise the ball would have sink to the
bottom of the jar.
Also r
3
 > r
1
 otherwise the ball would have floated in
liquid 1. From the above discussion we conclude that
r
1
 < r
3
 < r
2
.
20. (c) In case of water, the meniscus shape is concave
upwards. Also according to ascent formula
2 cosq
=
r
T
h
rg
The surface tension (T) of  soap solution is less than
water. Therefore rise of soap solution in the capillary
tube is less as compared to water. As in the case of
water, the meniscus shape of soap solution is also
concave upwards.
21. (c)
Y
l
A
Wire (1)
Wire (2)
3A
Y
l/3
As shown in the figure, the wires will have the same
Y oung’s modulus (same material) and the length of the
wire of area of cross-section 3A will be  l/3 (same
volume as wire 1).
Page 3


1. (b)
20 division of vernier scale = 8 div. of main scale
Þ 1 V .S.D. = 
82
M.S.D. M.S.D.
205
æ ö æö
=
ç ÷ ç÷
è ø èø
Least count
= 1 M.S.D – 1 V .S.D. = 1 M.S.D. – 
2
5
æö
ç÷
èø
 M.S.D
= 
2
1 M.S.D.
5
æö
-
ç÷
èø
 = 
33
M.S.D. 0.1 cm. 0.06 cm.
55
æö
=´=
ç÷
èø
(Q 1 M.S.D. = 
1
10
 cm. = 0.1 cm.)
Directly we can use
ba
L.C.MVM
b
- æö
= -=
ç÷
èø
         = 
20813
cm. cm. 0.06 cm.
20 10 50
- æ öæö
==
ç ÷ç÷
è øèø
2. (c) Within elastic limit it obeys Hooke's Law i.e., stress µ
strain.
3. (c) Least count 
111
cm
N 10 10N
=´=
4. (b) 5th division of vernier scale coincides with a main scale
division. L.C. = 
1
0.1mm
10
=
\ Zero error = – 5 × 0.1 = – 0.5 mm
This error is to be subtracted from the reading taken for
measurement. Also, zero correction = + 0.5 mm.
5. (b) If Y = Young's modulus of wire, M = mass of wire,
g = acceleration due to gravity , x = extension in the wire,
A= Area of cross-section of the wire,
l = length of the wire.
Mgx Y M xA
Y
A Y MxA
DDD D D
= Þ = + ++
l
ll
Þ 
Y 0.01 0.01 2 0.001 0.001
0.065
Y 3.00 0.87 0.041 2.820
D´
= + + +=
or 
Y
100 6.5%
Y
D
´ =±
6. (b) The instrument has negative zero error.
7. (c) If A, B and C be the points corresponding to the
impressions made by the legs of a spherometer then
3
a bc
c
++
=
A
C B
b d
a
If h is the depression or elevation then the radius of
curvature is given by 
2
62
=+
ch
r
h
8. (b) L.C. = 
Pitch
No. of circular divisions
9. (b) The specific heat of a solid is determined by the method
of mixture.
10. (a)
11. (d) Least count of screw gauge = 
0.5
mm 0.01mm
50
=
\ Reading = [Main scale reading  + circular scale
                            reading × L.C] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
12. (d) 30 Divisions of vernier scale coincide with 29 divisions
of main scales
Therefore 1 V .S.D = 
29
30
MSD
Least count = 1 MSD – 1VSD
= 1 MSD 
29
30
-
 MSD
= 
1
MSD
30
= 
1
0.5
30
´° = 1 minute.
13. (c) Least count = 
0.5
0.01mm
50
=
Zero error = 5 × L.C
           = 5 × 0.01 mm
           = 0.05 mm
Diameter of ball = [Reading on main scale] + [Reading
on circular scale × L . C] – Zero error
= 0.5 × 2 + 25 × 0.01 – 0.05
= 1.20 mm
14. (d) 2
gT
gT
D DD
=+
l
l
Dl and DT are least and number of readings are
maximum in option (d), therefore the measurement of
g is most accurate with data used in this option.
86
DPP/ P 31
15. (b) We know that Y = 
2
4
mgL
D
´
p
l
Þ
( )( )
22
33
4 419.8 2
0.4 10 0.8 10
mgL
Y
D --
´´´
==
p
p ´ ´´
l
112
2.0 10 N/m =´
Now
2 YD
YD
D DD
=+
l
l
[Q the value of m, g and L are exact]
= 
0.01 0.05
2
0.4 0.8
´+ = 2 × 0.025 + 0.0625
= 0.05 + 0.0625 = 0.1125
Þ
DY = 2 × 10
11
 × 0.1125 = 0.225 × 10
11
112
0.2 10 N/m =´
16. (b) The time period of a simple pendulum is given by
2 T
g
=p
l
\
22
4 T
g
=p
l
 Þ 
2
2
4 g
T
=p
l
Þ 100 100 2 100
gT
gT
D DD
´ = ´ +´
l
l
Case (i)
Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s
100 0.3125
g
g
D
\ ´=
Case (ii)
Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s
100 0.46875
g
g
D
\ ´=
Case (iii)
Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s
100 1.055
g
g
D
\ ´=
Clearly , the value of 100
g
g
D
´ will be least in case (i)
17. (c) Terminal velocity , 
2
12
2()
9
-
=
h
T
rd dg
v
2
2
(10.5 1.5) 9
0.2
0.2 (19.5 1.5) 18
-
= Þ =´
-
T
T
v
v
2
0.1/ \=
T
v ms
2 2
2
0
ìü
ïï æö
- - +=
íý ç÷
èø
ïï
îþ
d y dy dy
x By B By
dx dx
dx
Þ 
2 2
2
0
æö
+ -=
ç÷
èø
d y dy dy
xy xy
dx dx
dx
18. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
F
v
B=Vg r
2 
W=V    g
1
r
2
12 t
V g V g kv \ r = r+     
11
() r -r
\=
t
Vg
v
k
19. (d) From the figure it is clear that liquid 1 floats on liquid
2.  The lighter liquid floats over heavier liquid.
Therefore we can conclude that 
12
r <r
Also r
3
 < r
2
 otherwise the ball would have sink to the
bottom of the jar.
Also r
3
 > r
1
 otherwise the ball would have floated in
liquid 1. From the above discussion we conclude that
r
1
 < r
3
 < r
2
.
20. (c) In case of water, the meniscus shape is concave
upwards. Also according to ascent formula
2 cosq
=
r
T
h
rg
The surface tension (T) of  soap solution is less than
water. Therefore rise of soap solution in the capillary
tube is less as compared to water. As in the case of
water, the meniscus shape of soap solution is also
concave upwards.
21. (c)
Y
l
A
Wire (1)
Wire (2)
3A
Y
l/3
As shown in the figure, the wires will have the same
Y oung’s modulus (same material) and the length of the
wire of area of cross-section 3A will be  l/3 (same
volume as wire 1).
DPP/ P 31
87
For wire 1,
                
/
/
=
D l
FA
Y
x
...(i)
For wire 2 ,
                     
'/3
/( / 3)
FA
Y
x
=
D l
...(ii)
From (i) and (ii) ,
'
33
´ =´
DD
ll FF
A x Ax
 
'9 Þ= FF
22. (b) Lower the vernier constant, more accurate measurement
is possible by it.
23. (b) Effective length = MC = MN + NC = 
l
d
+
2
C
d /2
N
l
M
24. (c) Here, original length (L) = y ,
Extension (l) = x, Force applied (F) = p
Area of cross-section (A) = q
Now, Young's modulus (Y) = 
FL
AL
Þ=
yp
Y
xq
25. (a) Screw gauge is used to measure the diameter (d) of the
wire so that the area of cross-section is calculated by the
formula
2
4
p
=
d
A
26. (b) Both the statements (1) & (2)are precautions to be taken
during the experiment.
27. (a) The liquid cools faster first and slowly later on when its
temperature gets close to surrounding temperature.
28. (a) Maximum percentage error in measurement of e, as given
by Reyleigh’s formula.
(Given error is measurement of radius is 0.1 cm)
De = 0.6 DR = 0.6 × 0.1 = 0.06 cm.
Percentage error is
e 0.06
100 100 3.33%
e 0.63
D
´= ´=
´
29. (b) Speed of sound at the room temperature.
l
1 
= 4.6 cm,  l
2 
= 14.0 cm.,
l = 2 (l
2
 – l
1
) = 2 (14.0 – 4.6) = 18.8 cm.
v = f l = 
18.8
2000 376 m / s
100
´=
30. (c) End correction obtained in the experiment.
21
3 14.0 3 4.6
e 0.1 cm.
22
- -´
= ==
ll