NEET  >  DPP for NEET: Daily Practice Problems, Ch 1: Physical World, Units & Dimensions (Solutions)

# DPP for NEET: Daily Practice Problems, Ch 1: Physical World, Units & Dimensions (Solutions) - Physics Class 11

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``` Page 1

1. (d)
1
2
=
p
f
LC
æö
\
ç÷
èø
C
L
does not represent the dimension of frequency . .
2. (d) [n] = Number of particles crossing a unit area in unit
time = [L
–2
T
–1
]
[n
2
] = [n
1
] = number of particles per unit volume = [L
–3
]
[x
2
] = [x
1
] = positions
\ D =
[ ]
21
21 21
3
21
[]
[]
[]
[]
[]
--
-
-
éù
´
-
ëû
==
-
LTL
nxx
LT
nn
L
3. (d)
1 2 42
3 2 84
2 2 12
[]
3 []
--
--
--
===
X M L TA
Y M L TA
Z MTA
4. (a) In given equation,
a
q
Z
k
should be dimensionless
\
2 21
2
[]
[] []
[]
--
-
q´
a=Þa==
k MLT KK
MLT
ZL
and
2
0 20
12
[]
[] []
[]
-
--
aa éù
=Þb===
êú
b
ëû
MLT
P M LT
P
MLT
5. (c)
1/2
2
éù
=
êú
ëû
PF
v
lm
Þ
2
2
2 22
4
éù
= \µ
êú
ëû
PFF
vm
m
l lv
Þ
2
10
22
[] []
-
-
-
éù
== êú
êú
ëû
MLT
m MLT
LT
6. (d) By substituting the dimensions of mass [M], length [L]
and coefficient of rigidity [ML
–1
T
–2
] we get
T = 2p
h
M
L
is the right formula for time period of
oscillations.
7. (a) Time µ Þ=
x y z x yz
c Gh T kc Gh
Putting the dimensions in the above relation
Þ
001 1 13 2 21
[ ][ ][ ][]
- ---
=
x yz
M L T LT M LT MLT
Þ
001 322
[][]
-+ + + ---
=
y z xyz x y z
MLT M LT
Comparing the powers of M, L and T
– y + z = 0 ....(i)
x + 3y + 2z = 0 ....(ii)
– x – 2y – z = 1 ....(iii)
On solving equations (i) and (ii) and (iii)
51
,
22
-
= == x yz
Hence, dimension of time are [G
1/2
h
1/2
c
–5/2
].
8. (a) Let radius of gyration [ ] [ ][] [] µ
x yz
k h cG
By substituting the dimension of
[k] = [L]
211
[ ] [ ],[ ] [ ], h ML T c LT
--
==
132
[][] G M LT
--
=
and by comparing the power of both sides
we can get x = 1/2, y = – 3/2, z = 1/2
So dimension of radius of gyration are
[h]
1/2
[c]
–3/2
[G]
1/2
9. (b) Because magnitude is absolute.
10. (c) Stefan's law is
4
4
() =s Þs=
E
ET
T
where, E =
2
Energy Watt
Area Time
m
=
´
2
24
4
Wattm
WattmK
K
-
--
-
s= =-
11. (d) ct
2
must have dimensions of L
Þ c must have dimensions of L/T
2
= L T T
–2
12. (b) 6 × 10
–5
= 60 × 10
–6
= 60 microns
13. (d)
12
2
0
1
4
qq
F
r
=Þ
pÎ
2 21 12
0
2
1
4
qq
CmN
Fr
--
Î==
p
14. (b) According to the defition.
15. (b) Pyrometer is used for measurement of temperature.
16. (d) x = Ay + B tan Cz,
From the dimensional homogenity
[ ] [ ] [ ] []
éù éù
= = Þ ==
êú êú
ëû ëû
xB
xAyBy
AA
[Cz] = [M
0
L
0
T
0
] = Dimensionless
x and B; C and z
–1
; y and
B
A
have the same dimension
but x and A have the different dimensions.
17. (a) Let µr
x yz
T Sr
by substituting the dimension of
[T] = [T],
23
[][ ],[][],[][]
--
= = r= S MT r L ML
and by comparing the power of both the sides
x = –1/2, y = 3/2, z = 1/2
so
3
3
/
r
µ r Þ=
r
T rS Tk
S
18. (c)
22 1 21
[ ][][][][]
- --
= Þ = Þ= E hv MLT h T h MLT
19. (c) P AB =+
u r u r ur
Q A–B =
u r u r ur
Page 2

1. (d)
1
2
=
p
f
LC
æö
\
ç÷
èø
C
L
does not represent the dimension of frequency . .
2. (d) [n] = Number of particles crossing a unit area in unit
time = [L
–2
T
–1
]
[n
2
] = [n
1
] = number of particles per unit volume = [L
–3
]
[x
2
] = [x
1
] = positions
\ D =
[ ]
21
21 21
3
21
[]
[]
[]
[]
[]
--
-
-
éù
´
-
ëû
==
-
LTL
nxx
LT
nn
L
3. (d)
1 2 42
3 2 84
2 2 12
[]
3 []
--
--
--
===
X M L TA
Y M L TA
Z MTA
4. (a) In given equation,
a
q
Z
k
should be dimensionless
\
2 21
2
[]
[] []
[]
--
-
q´
a=Þa==
k MLT KK
MLT
ZL
and
2
0 20
12
[]
[] []
[]
-
--
aa éù
=Þb===
êú
b
ëû
MLT
P M LT
P
MLT
5. (c)
1/2
2
éù
=
êú
ëû
PF
v
lm
Þ
2
2
2 22
4
éù
= \µ
êú
ëû
PFF
vm
m
l lv
Þ
2
10
22
[] []
-
-
-
éù
== êú
êú
ëû
MLT
m MLT
LT
6. (d) By substituting the dimensions of mass [M], length [L]
and coefficient of rigidity [ML
–1
T
–2
] we get
T = 2p
h
M
L
is the right formula for time period of
oscillations.
7. (a) Time µ Þ=
x y z x yz
c Gh T kc Gh
Putting the dimensions in the above relation
Þ
001 1 13 2 21
[ ][ ][ ][]
- ---
=
x yz
M L T LT M LT MLT
Þ
001 322
[][]
-+ + + ---
=
y z xyz x y z
MLT M LT
Comparing the powers of M, L and T
– y + z = 0 ....(i)
x + 3y + 2z = 0 ....(ii)
– x – 2y – z = 1 ....(iii)
On solving equations (i) and (ii) and (iii)
51
,
22
-
= == x yz
Hence, dimension of time are [G
1/2
h
1/2
c
–5/2
].
8. (a) Let radius of gyration [ ] [ ][] [] µ
x yz
k h cG
By substituting the dimension of
[k] = [L]
211
[ ] [ ],[ ] [ ], h ML T c LT
--
==
132
[][] G M LT
--
=
and by comparing the power of both sides
we can get x = 1/2, y = – 3/2, z = 1/2
So dimension of radius of gyration are
[h]
1/2
[c]
–3/2
[G]
1/2
9. (b) Because magnitude is absolute.
10. (c) Stefan's law is
4
4
() =s Þs=
E
ET
T
where, E =
2
Energy Watt
Area Time
m
=
´
2
24
4
Wattm
WattmK
K
-
--
-
s= =-
11. (d) ct
2
must have dimensions of L
Þ c must have dimensions of L/T
2
= L T T
–2
12. (b) 6 × 10
–5
= 60 × 10
–6
= 60 microns
13. (d)
12
2
0
1
4
qq
F
r
=Þ
pÎ
2 21 12
0
2
1
4
qq
CmN
Fr
--
Î==
p
14. (b) According to the defition.
15. (b) Pyrometer is used for measurement of temperature.
16. (d) x = Ay + B tan Cz,
From the dimensional homogenity
[ ] [ ] [ ] []
éù éù
= = Þ ==
êú êú
ëû ëû
xB
xAyBy
AA
[Cz] = [M
0
L
0
T
0
] = Dimensionless
x and B; C and z
–1
; y and
B
A
have the same dimension
but x and A have the different dimensions.
17. (a) Let µr
x yz
T Sr
by substituting the dimension of
[T] = [T],
23
[][ ],[][],[][]
--
= = r= S MT r L ML
and by comparing the power of both the sides
x = –1/2, y = 3/2, z = 1/2
so
3
3
/
r
µ r Þ=
r
T rS Tk
S
18. (c)
22 1 21
[ ][][][][]
- --
= Þ = Þ= E hv MLT h T h MLT
19. (c) P AB =+
u r u r ur
Q A–B =
u r u r ur
2
DPP/ P 01
P.Q0 =
u r ur
Þ (A B).(A – B) 0 +=
u r ur u r ur
Þ A
2
– B
2
= 0 Þ |A| |B| =
u r ur
PQ ^ Q
20. (a) By substituting dimension of each quantitity in R.H.S.
of option (a) we get
2
1
11
[ ].
-
-
--
éù
éù ´
== êú
êú
h
´ ëûêú
ëû
mg M LT
LT
r
MLTL
This option gives the dimension of velocity.
21. (a) By principle of dimensional homogeneity
2
[]
éù
=
êú
ëû
a
P
V
\
2 1 2 6 52
[ ][ ][ ] [ ] [ ][ ]
---
= = ´= a P V ML T L ML T
22. (c) f = c m
x
k
y
;
Spring constant k = force/length.
[M
0
L
0
T
–1
] = [M
x
(MT
–2
)
y
]

=[ M
x + y
T
–2y
]
Þ
2
1
y or 1 y 2 , 0 y x = - = - = +
Therefore,
2
1
– x =
23. (a) Try out the given alternatives.
When x = 1, y = –1, z = 1
S
PC
C S P C S P
1 1 1 z y x
= =
-

–1 –2 –1
0 00
2 –22
[ML T ] [LT ]
[M LT]
[MLT /LT]
==
24. (c) Dimensions of angular momentum, []
21
L MLT
-
éù
=
ëû
Dimensions of work, []
22
W MLT
-
éù
=
ëû
Dimensions of torque, []
22
MLT
-
éù
t=
ëû
Dimensions of energy , []
22
E MLT
-
éù
=
ëû
Dimensions of Y oung¢s modulus,
[]
12
Y MLT
--
éù
=
ëû
Dimensions of light year = [L]
Dimension of wavelength = [L]
25. (d),      26.  (b)
5
hG
c
=
2 1 3 12
55
kgm s m kg s
m /s
- --
´
=
2
s = s
Putting the values of h, G and c in above relation
Planck time = 1.3 × 10
– 43
s.
27. (a)
348
19
7
6.6 10 3 10
4.95 10 J
4 10
hc
E
-
-
-
´ ´´
= = =´
l
´
= 3.0 eV
28. (a) Both statement -1 and statement -2 are correct and
statement -1 follows from statement -2
29. (c) Let us write the dimension of various quantities on
two sides of the given relation.
L.H.S. = T = [T],
R.H.S. =
2
1
2 / []
-
-
p ==
LT
glT
L
(\ 2p has no dimension).
As dimension of L.H.S. is not equal to dimension of
R.H.S. therefore according to principle of homogeneity
the relation.
2 /. =p T gl
30. (a) Unit of quantity (L/R) is Henry/ohm.
As Henry = ohm × sec,
hence unit of L/R is sec
i.e. [L/R] = [T].
Similarly , unit of product CR is farad × ohm
or
Coulomb Volt
×
Volt Amp
or
Sec×Amp
Amp
= second
i.e. [CR] = [T]
therefore,  [L/R] and [CR] both have the same
dimension.
```

## FAQs on DPP for NEET: Daily Practice Problems, Ch 1: Physical World, Units & Dimensions (Solutions) - Physics Class 11

 1. What is the importance of daily practice problems for NEET preparation?
Ans. Daily practice problems are crucial for NEET preparation as they help students understand and apply the concepts they have learned. Regular practice not only reinforces the knowledge but also improves problem-solving skills and time management, which are essential for NEET success.
 2. How can solving practice problems in Chapter 1: Physical World, Units & Dimensions help in NEET preparation?
Ans. Solving practice problems in Chapter 1: Physical World, Units & Dimensions helps in NEET preparation by familiarizing students with the fundamental concepts and principles related to physical world, units, and dimensions. It allows students to gain a strong foundation in these topics, which are essential for understanding more complex concepts in later chapters.
 3. How can I effectively incorporate daily practice problems into my NEET study schedule?
Ans. To effectively incorporate daily practice problems into your NEET study schedule, allocate a specific time for solving problems each day. Start with a manageable number of problems and gradually increase the difficulty level. Review the solutions and identify areas where you need improvement. Additionally, you can join online NEET practice platforms that provide a variety of practice problems for each chapter.
 4. Are the solutions provided for the practice problems in the article reliable?
Ans. Yes, the solutions provided for the practice problems in the article are reliable. They are designed to help students understand the step-by-step approach to solving the problems accurately. However, it is always recommended to cross-check the solutions with other trusted sources or consult a teacher or mentor if you have any doubts.
 5. Can solving daily practice problems in Chapter 1 alone guarantee success in the NEET exam?
Ans. No, solving daily practice problems in Chapter 1 alone cannot guarantee success in the NEET exam. While Chapter 1 is important, NEET is a comprehensive exam that covers multiple subjects and chapters. It is essential to have a well-rounded preparation strategy that includes studying all subjects, practicing problems from various chapters, and regularly evaluating your progress through mock tests and previous years' question papers.

## Physics Class 11

130 videos|483 docs|210 tests

## Physics Class 11

130 videos|483 docs|210 tests

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