33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Units and Measurements
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5
unit area per sec. The non zero integers x, y, z such
that P x S y c z is dimensionless are
(a) x = 1, y = 1, z = 1
(b) x = –1, y = 1, z = 1
(c) x = 1, y = –1, z = 1
(d) x = 1, y = 1, z = –1 (1992)
49. The frequency of vibration f of a mass m suspended
from a spring of spring constant k is given by a
relation f = am x k y , where a is a dimensionless
constant. The values of x and y are
1 1
1 1
(a) x= , y=
(b) x=− , y=−
2 2
2 2
1 1
1 1
(c) x= , y=−
(d) x=− , y=
2 2
2 2
(1990)
50. If x = at + bt 2 , where x is the distance travelled by
the body in kilometers while t is the time in seconds,
then the units of b is
(a) km/s
(b) km s
(c) km/s 2 (d) km s 2 (1989)
ANSWER KEY
1. (a) 2. (c) 3. (c) 4. (c) 5. (c) 6. (c) 7. (d) 8. (c) 9. (b) 10. (d)
11. (d) 12. (a) 13. (b) 14. (b) 15. (d) 16. (c) 17. (c) 18. (b) 19. (d) 20. (a)
21. (c) 22. (b) 23. (a) 24. (c) 25. (b) 26. (b) 27. (a) 28. (c) 29. (b) 30. (d)
31. (c) 32. (a) 33. (d) 34. (c) 35. (a) 36. (a) 37. (d) 38. (d) 39. (a) 40. (c)
41. (d) 42. (c) 43. (b) 44. (c) 45. (b) 46. (d) 47. (b) 48. (c) 49. (d) 50. (c)
Hints & Explanations
Qx
1. (a) : K =
AT ( −T ) t
, where Q is the amount of heat
1 2
flow, x is the thickness of the slab, A is the area of crosssection,
and t is the time taken.
K =
Jm
− −
= W
1 1 = Wm
1 K
1
2
m Ks mK
2. (c) : Damping force, F ∝ v or F = kv
where k is the constant of proportionality
−2
F N kg ms −1
∴ k = = = = kg s
v −1
−1
ms ms
3. (c) : Force between two charges
2
2
q
q
F = 1
4 r
⇒ = 1
2 0
4 Fr
= 2 2
ε
C /N-m
πε
2
0
π
4. (c) : Given : least count = 0.01 and number of
circular scale divisions = 50.
\ Pitch = L.C × No. of circular scale division
= 0.01 × 50 = 0.5 mm.
5. (c) : X = AB 2 1/
2
13 / 3
C D
Maximum percentage error in X
⎛ dX ⎞ ⎛ dA 1 dB 1 dC dD ⎞
⎝
⎜
X ⎠
⎟ × 100 = + + +
⎝
⎜2
3
A B C D ⎠
⎟ × 100
2 3
= 2 1 1 2 2 1 3 3 3 4
× + × + × + × = 16%
6. (c) : If n divisions of vernier scale coincides with
(n – 1) divisions of main scale.
Therefore, n VSD = (n – 1) MSD
⇒ 1 VSD = ( n − 1 ) MSD
n
( n − 1)
\ Least count = 1 MSD – 1 VSD = 1 MSD−
MSD
n
1
= 1MSD− 1MSD + MSD = 1
n n MSD
1 1 1 ⎡
1 ⎤
= × = cm =
⎣
⎢
1 MSD cm
n n 2
⎦
⎥
n n
7. (d) : Diameter of the ball
= MSR + CSR × (Least count) – Zero error
= 5 mm + 25 × 0.001 cm – (–0.004) cm
= 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm.
8. (c) : As P ab 3 2
=
% error in P is cd
∆P
⎡ ⎛ ∆a
⎞ ∆b
∆c
∆d
× =
P ⎝
⎜
a ⎠
⎟ + ⎛ ⎝ ⎜ ⎞
b ⎠
⎟ + c
+ ⎤
100 ⎢3 2
⎣
d
⎥
⎦
× 100
= [3 × 1% + 2 × 2% + 3% + 4%] = 14%
9. (b) : From the relation, h= ut + 1 2
gt
2
1 2 2h
h= gt ⇒ g = (Q body initially at rest)
2
2
t
Taking natural logarithm on both sides, we get
ln g = ln h – 2 ln t
Differentiating, ∆g
∆ h ∆ t
= −2
g h t