33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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6 NEET-AIPMT Chapterwise Topicwise Solutions Physics
For maximum permissible error,
⎛ ∆g
⎞ ⎛ ∆h
⎞ ∆t
or ×
⎝
⎜ g ⎠
⎟ = ×
⎝
⎜
h ⎠
⎟ + × ⎛
⎝
⎜
t
× ⎞
100 100 2 100⎟
⎠
max
According to problem
∆h
∆t
× 100 = e1 and × 100 =e2
h
t
⎛ ∆g ⎞
Therefore, × e e
⎝
⎜ 100
g ⎠
⎟ = 1 + 2 2
max
4
10. (d) :
⎛ 4 ⎞
V = πR 3 ;lnV = ln R
3
⎝
⎜ π
3 ⎠
⎟ + ln
3
Differentiating, dV dR
= 3
V R
Error in the determination of the volume
= 3 × 2% = 6%
⎛
11. (d) : Maximum error in mass ∆m ⎞
⎝
⎜
m ⎠
⎟ = 3 % = 3
100
⎛ ∆l ⎞ 2
and maximum error in length
⎝
⎜
l ⎠
⎟ = 2 % =
100
Maximum error in the measurement of density
∆ρ
∆ ∆
ρ = m
+ ⎛
⎝
⎜ × l ⎞
⎠
⎟ = 3
+ ⎛
⎝
⎜ × 2 ⎞
3
3
m l 100 ⎠
⎟ = 3
100 100
+ 6
100
9
= = 9%
100
12. (a) : Percentage error in mass = 2% = 2
100 and
3
percentage error in speed = 3 % =
100
KE . .= 1 2
mv
2
Therefore the error in measurement of kinetic energy
∆ KE . . ∆m
∆v
2
= + 2 × = + × =
KE . . m v 100 2 3 8
100 100 = 8%
13. (b) : Density ρ= mass m
volume V .....(i)
Take logarithm to take base e on the both sides of
eqn (i), we get
lnr = lnm – lnV.....(ii)
Differentiate eqn (ii), on both sides, we get
∆ρ
∆ ∆
ρ = m
m
− V
V
Errors are always added.
Error in the density r will be
⎡∆m
∆V
⎤
= ⎢ + ⎥
⎣ m V ⎦
× 100%
⎡ 001 . 01 . ⎤
= ⎢ + ⎥
⎣22 42 47⎦
× 100% = 2%
. .
14. (b) : 9.99 – 0.0099 = 9.9801 m
Least number of significant figure are 3. Hence, required
answer will be 9.98 m.
15. (d) : We know, stress = Force
Area
Dimensions of force is [M 1 L 1 T –2 ] and that of area is [L 2 ].
\ Dimensions of stress = [ 1 1 −2
MLT ] = [M 1 L –1 T –2 ].
2
[ L ]
16. (c) : Impulse = Force × time
= [MLT –2 ][T] = [MLT –1 ]
− 2
Surface tension = Force [MLT ] 0 −2
= = [ ML T ]
length [L]
Angular momentum
= Moment of inertia × angular velocity
= [ML 2 ][T –1 ] = [ML 2 T –1 ]
Work = Force × distance = [MLT –2 ][L] = [ML 2 T –2 ]
Energy = [ML 2 T –2 ]
Torque = Force × distance = [MLT –2 ][L] = [ML 2 T –2 ]
Young’s modulus
Force/
Area
=
Change in length /original length
− 2 2
[ MLT ]/[ L ] −1 −2
= = [ ML T ]
[ L]/[ L]
Hence, among the given pair of physical quantities work
and torque have the same dimensions [ML 2 T –2 ].
17. (c) : The speed of the light in vacuum is
1
−12
/
c = = ( µε 0 0)
µε 0 0
\ [(m 0 e 0 ) –1/2 ] = [c] = [LT –1 ]
18. (b) : Energy density of an electric field E is
uE = 1 ε E
2 0 2
where e 0 is permittivity of free space
2 −2
Energy ML T
u E = =
Volume 3
L
= ML–1 T –2
Hence, the dimension of 1 2 0 2 ε E is ML –1 T –2
19. (d) : Pressure,
force mass × acceleration
P = =
area
area
1 −2
MLT 1 −1 −2
∴ [ P] = = [ ML T ] = M
2
a L b T c .
L
\ a = 1, b = –1, c = –2.
Work done
20. (a) : [ Energy density]
= ⎡ ⎣ ⎢ ⎤
⎥
Volume ⎦
−2
MLT ⋅L
=
3 = [ML–1 T –2 ]
L
Force []
[ Young’s modulus] = [ Y ] = ⎡ ⎣ ⎢ ⎤
Area
⎥
⎦
× l
[ ∆l]
−2
MLT L −1 −2
= ⋅ = [ ML T ]
2
L L
The dimensions of 1 and 4 are the same.
21. (c) : According to Ohm’s law,
V = RI or R=
V I
Dimensions of V = W 2 −2 q = [ML T ]
[IT]