Physical World, Units & Dimensions Practice Questions - DPP for NEET

 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.