DC Pandey Solutions: Basic Mathematics and Measurement

# DC Pandey Solutions: Basic Mathematics and Measurement | DC Pandey Solutions for NEET Physics PDF Download

``` 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.
= 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
```

122 docs

## FAQs on DC Pandey Solutions: Basic Mathematics and Measurement - DC Pandey Solutions for NEET Physics

 1. What are the basic mathematical concepts covered in DC Pandey Solutions: Basic Mathematics and Measurement?
Ans. DC Pandey Solutions: Basic Mathematics and Measurement covers various fundamental mathematical concepts such as arithmetic, algebra, geometry, and statistics. It includes topics like numbers, fractions, decimals, equations, inequalities, angles, triangles, circles, data interpretation, and more.
 2. How can DC Pandey Solutions: Basic Mathematics and Measurement help in exam preparation?
Ans. DC Pandey Solutions: Basic Mathematics and Measurement is a comprehensive study material that provides detailed explanations, solved examples, and practice questions for each topic. It helps in building a strong foundation in mathematics and measurement, which is essential for scoring well in exams. By understanding the concepts and practicing the questions, students can improve their problem-solving skills and gain confidence in tackling mathematical problems.
 3. Are there any measurement concepts covered in DC Pandey Solutions: Basic Mathematics and Measurement?
Ans. Yes, DC Pandey Solutions: Basic Mathematics and Measurement includes measurement concepts such as units and conversions, length, area, volume, time, speed, and temperature. It helps students understand the principles and techniques of measurement and apply them in solving real-life problems.
 4. How can I access DC Pandey Solutions: Basic Mathematics and Measurement study material?
Ans. DC Pandey Solutions: Basic Mathematics and Measurement study material is available in both online and offline formats. It can be accessed through the official website of DC Pandey or purchased from authorized bookstores. The online version may require a subscription or login credentials, while the offline version can be obtained as a physical book.
 5. Is DC Pandey Solutions: Basic Mathematics and Measurement suitable for competitive exams?
Ans. Yes, DC Pandey Solutions: Basic Mathematics and Measurement is suitable for various competitive exams that have a mathematics section. It covers the essential mathematical concepts required for exams like engineering entrance exams, bank exams, and other competitive exams. By studying this material, students can strengthen their mathematical skills and improve their chances of scoring well in these exams.

## DC Pandey Solutions for NEET Physics

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