DC Pandey Solutions: Basic Mathematics and Measurement Notes | Study DC Pandey Solutions for NEET Physics - NEET

NEET: DC Pandey Solutions: Basic Mathematics and Measurement Notes | Study DC Pandey Solutions for NEET Physics - NEET

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 Page 1


Section I
Single Correct Option
1, K mv =
1
2
2
\ 
D D D K
K
m
m
v
v
æ
è
ç
ö
ø
÷ = +
max
2
\     Maximum error = 2% + 2 3 ( %)
In the estimate of,
  kinetic energy ( ) % K = 8
Option (c) is correct.
2. d
m
V
m
l
= =
3
\  
D
´
æ
è
ç
ö
ø
÷ = ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷
d
d
m
m
l
l
100 100 3 100
max
D D
         = + 4 3 3 % ( %)
         = 13%
Option (d) is correct.
3. p
F
A
F
L
= =
2
\  Permissible error in pressure ( ) p
= + 4 2 2 % ( %)
= 8% 
Option (a) is correct.
4.         p V p V
1 1 2 2
=
Þ         p p
V
V
2 1
1
2
=
=
-
p
V
V V
1
1
1 1
10% of
=
p
1
90%
         p
p
2
1
9
=
\   Percentage increase in pressure =
100
9
              = 111 .
Option (a) is correct.
5. K
p
m
=
2
2
\ Error in the measurement of kinetic
energy (K)
= ´ 2 100%
= 200%
Option (d) is correct.
6. 3400 = ´ 3 400 10
3
.
\   Number of significant figures = 2
Option (d) is correct.
7. A = ´ 3 124 3 002 . . m m
= 9.378 / 248 m
2
= 9.378 m
2
Option (a) is correct.
8.   g
GM
R
=
2
    =
constant
R
2
K I =
1
2
2
w
     =
æ
è
ç
ö
ø
÷
1
2
2
5
2
MR w
      = constant ´ R
2
Basic Mathematics
& Measurements
1
Page 2


Section I
Single Correct Option
1, K mv =
1
2
2
\ 
D D D K
K
m
m
v
v
æ
è
ç
ö
ø
÷ = +
max
2
\     Maximum error = 2% + 2 3 ( %)
In the estimate of,
  kinetic energy ( ) % K = 8
Option (c) is correct.
2. d
m
V
m
l
= =
3
\  
D
´
æ
è
ç
ö
ø
÷ = ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷
d
d
m
m
l
l
100 100 3 100
max
D D
         = + 4 3 3 % ( %)
         = 13%
Option (d) is correct.
3. p
F
A
F
L
= =
2
\  Permissible error in pressure ( ) p
= + 4 2 2 % ( %)
= 8% 
Option (a) is correct.
4.         p V p V
1 1 2 2
=
Þ         p p
V
V
2 1
1
2
=
=
-
p
V
V V
1
1
1 1
10% of
=
p
1
90%
         p
p
2
1
9
=
\   Percentage increase in pressure =
100
9
              = 111 .
Option (a) is correct.
5. K
p
m
=
2
2
\ Error in the measurement of kinetic
energy (K)
= ´ 2 100%
= 200%
Option (d) is correct.
6. 3400 = ´ 3 400 10
3
.
\   Number of significant figures = 2
Option (d) is correct.
7. A = ´ 3 124 3 002 . . m m
= 9.378 / 248 m
2
= 9.378 m
2
Option (a) is correct.
8.   g
GM
R
=
2
    =
constant
R
2
K I =
1
2
2
w
     =
æ
è
ç
ö
ø
÷
1
2
2
5
2
MR w
      = constant ´ R
2
Basic Mathematics
& Measurements
1
\ Decrease in R (radius) by 2% world
increase g by 4% and decrease K
(rotational kinetic energy) by 4%.
Option (b) is correct.
9. Heat (H) = 2
2
Rt
\   Maximum error in measuring heat (H)
= + + 2 2 1 1 ( %) % %
= 6%
Option (b) is correct.
10. V lbt =
= ´ ´ 12 6 2 45 .
= 176 4 .
= ´ 1764 10
2 3
. cm
= ´ 2 10
2 2
cm
Option (b) is correct.
11. I
P
r
=
4
2
p
i.e.,   Ir
2
= constant
i.e., if r is increased by 2% the intensity
will decrease by 4%.
Option (d) is correct.
12. Option (b) is correct.
13.      V r =
4
3
3
p
    
D D V
V
r
r
= 3
      = 3 1 ( %)
      = 3%
Option (c) is correct.
14. a a
3 2
6 = (given)
\ a = 6
Þ V = = 6 216
3 3
m
Option (b) is correct.
15. g
l
T
= 4
2
2
p
\ 
D D D g
g
l
l
T
T
´
æ
è
ç
ö
ø
÷ = ´ + ´
æ
è
ç
ö
ø
÷ 100 100 2 100
max
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷
1
100
100 2
0 15
2003
10
mm
m
.
= + 0 1 0 1 . % . % = 0.2 %
Option (a) is correct.
16. Q I e
tI
V
=
-
e
a
b ( ) D
0
 (given)
We know that
   Q It = 
\    t e
tI
V
=
-
e
a
b ( ) D
0
Þ [ ] [ ] a = t
and [ ]
( )
b =
e
é
ë
ê
ù
û
ú
tI
V D
0
Þ 
b
a
é
ë
ê
ù
û
ú
=
e
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
e
é
ë
ê
ù
û
ú
tI
V t
I
V ( ) ( ) D D
0 0
1
       =
e
é
ë
ê
ù
û
ú
1 1
0
[Resistance]
or    
b
a
é
ë
ê
ù
û
ú
=
1 1
[ML T A [M L T A ]
2 –3 –2 –1 –3 4 2
]
         = =
1
[L T]
–1
 [velocity]
         =
e
1
0
1 2
[ ]
/
m
0
Option (a) is correct.
17. a
pq
r s
=
2 3
Da
a
p
p
q
q
´
æ
è
ç
ö
ø
÷ = ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 100
1
2
100
1
2
100
max
D D
+ ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 2 100 3 100
D D r
r
s
s
= +
1
2
1
1
2
3 ( %) ( %) + + 2 3 3 ( %) ( %) 0.5 0.3
= + 0.5% 1.5% + + 1 1 % %
= 4%
Option (c) is correct.
18. Least count of main scale 
= =
2
4
mm
0.5 mm
Least count =
least count of main scale 
50
 = 0.1 mm
Zero error = - ´ 30 0.01 mm
= -0.3 mm
(–ive sign, zero of circular scale is lying
above observed reading of plate thick)
= 2 MSR + 20 CSR
Basic Mathematics & Measurements | 3
Page 3


Section I
Single Correct Option
1, K mv =
1
2
2
\ 
D D D K
K
m
m
v
v
æ
è
ç
ö
ø
÷ = +
max
2
\     Maximum error = 2% + 2 3 ( %)
In the estimate of,
  kinetic energy ( ) % K = 8
Option (c) is correct.
2. d
m
V
m
l
= =
3
\  
D
´
æ
è
ç
ö
ø
÷ = ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷
d
d
m
m
l
l
100 100 3 100
max
D D
         = + 4 3 3 % ( %)
         = 13%
Option (d) is correct.
3. p
F
A
F
L
= =
2
\  Permissible error in pressure ( ) p
= + 4 2 2 % ( %)
= 8% 
Option (a) is correct.
4.         p V p V
1 1 2 2
=
Þ         p p
V
V
2 1
1
2
=
=
-
p
V
V V
1
1
1 1
10% of
=
p
1
90%
         p
p
2
1
9
=
\   Percentage increase in pressure =
100
9
              = 111 .
Option (a) is correct.
5. K
p
m
=
2
2
\ Error in the measurement of kinetic
energy (K)
= ´ 2 100%
= 200%
Option (d) is correct.
6. 3400 = ´ 3 400 10
3
.
\   Number of significant figures = 2
Option (d) is correct.
7. A = ´ 3 124 3 002 . . m m
= 9.378 / 248 m
2
= 9.378 m
2
Option (a) is correct.
8.   g
GM
R
=
2
    =
constant
R
2
K I =
1
2
2
w
     =
æ
è
ç
ö
ø
÷
1
2
2
5
2
MR w
      = constant ´ R
2
Basic Mathematics
& Measurements
1
\ Decrease in R (radius) by 2% world
increase g by 4% and decrease K
(rotational kinetic energy) by 4%.
Option (b) is correct.
9. Heat (H) = 2
2
Rt
\   Maximum error in measuring heat (H)
= + + 2 2 1 1 ( %) % %
= 6%
Option (b) is correct.
10. V lbt =
= ´ ´ 12 6 2 45 .
= 176 4 .
= ´ 1764 10
2 3
. cm
= ´ 2 10
2 2
cm
Option (b) is correct.
11. I
P
r
=
4
2
p
i.e.,   Ir
2
= constant
i.e., if r is increased by 2% the intensity
will decrease by 4%.
Option (d) is correct.
12. Option (b) is correct.
13.      V r =
4
3
3
p
    
D D V
V
r
r
= 3
      = 3 1 ( %)
      = 3%
Option (c) is correct.
14. a a
3 2
6 = (given)
\ a = 6
Þ V = = 6 216
3 3
m
Option (b) is correct.
15. g
l
T
= 4
2
2
p
\ 
D D D g
g
l
l
T
T
´
æ
è
ç
ö
ø
÷ = ´ + ´
æ
è
ç
ö
ø
÷ 100 100 2 100
max
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷
1
100
100 2
0 15
2003
10
mm
m
.
= + 0 1 0 1 . % . % = 0.2 %
Option (a) is correct.
16. Q I e
tI
V
=
-
e
a
b ( ) D
0
 (given)
We know that
   Q It = 
\    t e
tI
V
=
-
e
a
b ( ) D
0
Þ [ ] [ ] a = t
and [ ]
( )
b =
e
é
ë
ê
ù
û
ú
tI
V D
0
Þ 
b
a
é
ë
ê
ù
û
ú
=
e
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
e
é
ë
ê
ù
û
ú
tI
V t
I
V ( ) ( ) D D
0 0
1
       =
e
é
ë
ê
ù
û
ú
1 1
0
[Resistance]
or    
b
a
é
ë
ê
ù
û
ú
=
1 1
[ML T A [M L T A ]
2 –3 –2 –1 –3 4 2
]
         = =
1
[L T]
–1
 [velocity]
         =
e
1
0
1 2
[ ]
/
m
0
Option (a) is correct.
17. a
pq
r s
=
2 3
Da
a
p
p
q
q
´
æ
è
ç
ö
ø
÷ = ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 100
1
2
100
1
2
100
max
D D
+ ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 2 100 3 100
D D r
r
s
s
= +
1
2
1
1
2
3 ( %) ( %) + + 2 3 3 ( %) ( %) 0.5 0.3
= + 0.5% 1.5% + + 1 1 % %
= 4%
Option (c) is correct.
18. Least count of main scale 
= =
2
4
mm
0.5 mm
Least count =
least count of main scale 
50
 = 0.1 mm
Zero error = - ´ 30 0.01 mm
= -0.3 mm
(–ive sign, zero of circular scale is lying
above observed reading of plate thick)
= 2 MSR + 20 CSR
Basic Mathematics & Measurements | 3
= ´ ( . 2 05 mm) + (20 × 0.01 mm)
= 1 mm + 0.2 mm
= 1.2 mm.
Plate thickness (corrected reading)
= observed reading - zero error
= 1.2 mm + 0.3 mm
= 1.5 mm
Option (d) is correct.
More than One Correct Options
1. Maximum percentage error in x :
= + (% ) (% ) error in error in a b 2 +3 (% error in c)
= 15%
Assertion and Reason
1. Least count of  screw gauge
=
Pitch
Number of divisions of circular scale
Less the value of pitch, less will be least
count of screw gauge leading to len
uncertainty that is more accuracy in the
measurement.
Thus, assertion is true.
From the above relation we conclude that
least count of  screw gauge is inversely
proportional to the number of divisions of
circular scale.
Thus reason is false.
Option (c) is correct.
Match the Columns
1. (a) F
GM M
r
=
1 2
2
\       GM M Fr
1 2
2
=
i e . ., [ ] [ GM M
1 2
= MLT ][L ]
–2 2
      = [ML T ]
3 –2
\  ( ) ( ) a q ®
(b) 
3 3 3 RT
M
pV
n nM
= =
work
\    
3 ML T
M
[L T ]
2 2
2 2
RT
M
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
=
-
-
\ (b) ®  (r)
(c) 
F
q B
F
qB
2
2 2
2
=
æ
è
ç
ö
ø
÷
             =
æ
è
ç
ö
ø
÷
qvB
qB
sinq
2
      = v
2 2
sin q
\     
F
q B
2
2 2
é
ë
ê
ù
û
ú
= [LT ] = [L T ]
–1 2 2 –2
\ (c) ® ( ) r
(d)     g
GM
R
e
e
=
2
GM
R
gR
e
e
e
=
Þ        
GM
R
e
e
é
ë
ê
ù
û
ú
= [LT ][L]
–2
          = [L T ]
2 –2
\   (d) ® ( ) r
4 | Mechanics-1
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