33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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60 NEET-AIPMT Chapterwise Topicwise Solutions Physics
32. (c) : Load W = Mg = 75 × 10 = 750 N
Effort P = 250 N
\ Mechanical advantage
load W 750
= = = = 3
effort P 250
Velocity ratio
distance travelled by effort 12
= = = 4
distance travelled by load 3
Mechanical advantage
Efficiency, η=
Velocity ratio
= (3/4) × 100 = 75%.
33. (d)
1 2
E Isωs
2
Sphere
Is
s
34. (b) : =
2 ω
=
ECylinder
1 2
2
I I
cωc
cωc
2
2 2 1 2
Here, Is = MR , Ic = MR ,ωc = 2ωs
5 2
2 2 2
E
mR × ω
Sphere
s
=
5
4 1 1
= × =
ECylinder
1 2 2 5 4 5
mR × ( 2ωs)
2
35. (a) : Here, l 1 + l 2 = l
Centre of mass of the system,
m1× 0 + m2
× l ml 2
l1
=
=
m1+
m2
m1+
m2
ml
l 2 = l – l 1 =
1
m1+
m2
Required moment of inertia of the system,
I = m 1 l 2 2
1 + m 2 l 2
l
= ( mm 1 2 2 + mm 2 1 2 2
)
( m1+
m2 )
2
mm 1 2( m1+
m2)
l
2 mm 1 2
=
=
m + m m + m l 2
( 1 2)
2 1 2
36. (a) : Mass of the disc = 9M
Mass of removed portion of disc = M
The moment of inertia of the complete
disc about an axis passing through
its centre O and perpendicular to its
9
plane is
2
I1
= MR
2
Now, the moment of inertia of the removed portion of
the disc
I M R 2
1 ⎛ ⎞ 1 2
2 =
MR
2 ⎝
⎜
3⎠
⎟ =
18
Therefore, moment of inertia of the remaining portion of
disc about O is
2 2 2
MR MR 40MR
I = I 1 – I 2 = 9 − =
2 18 9 2
2 M⋅
R
37. (d) : M.I. of a circular disc, Mk =
2
M.I. of a circular ring = MR 2 .
1
\ Ratio of their radius of gyration =
2 : 1=
1:
2
38. (c) : KE . .= 1 2
Iω
2
1 1
∴ Iω1 2 = ⋅
2 2 2 Iω 2 2 ω
;
1 2 ω1
ω ω
2 2 = 2
2
1
⇒ = 2 1
.
39. (c) : The moment of inertia of the system
= m A r 2 A + m B r 2 2
B + m C r C
= m A (0) 2 + m(l) 2 + m(lsin30°) 2
= ml 2 + ml 2 × (1/4) = (5/4) ml 2
40. (a) : A circular disc may be divided into a large
number of circular rings. Moment of inertia of the disc
will be the summation of the moments of inertia of these
rings about the geometrical axis. Now, moment of inertia
of a circular ring about its geometrical axis is MR 2 , where
M is the mass and R is the radius of the ring.
Since the density (mass per unit volume) for iron is more
than that of aluminium, the proposed rings made of iron
should be placed at a higher radius to get more value of
MR 2 . Hence to get maximum moment of inertia for the
circular disc, aluminium should be placed at interior and
iron at the exterior position.
41. (b) : As effective distance of mass from BC is greater
than the effective distance of mass from AB, therefore
I 2 > I 1 .
42. (d) : The intersection of medians is the centre of
mass of the triangle. Since the distances of centre of mass
from the sides is related as x BC < x AB < x AC .
Therefore I BC > I AB > I AC or I BC > I AB .
43. (d) : The moment of inertia is minimum about EG
because mass distribution is at minimum distance from
EG.
44. (c) : K.E. = 1 2
Iω
2
2K.E.
2
I = = × 360
= 08 . kg m 2
2
ω 30 × 30
45. (a) : Kinetic energy = 1 2
Iω , and for ring I = mr 2 .
2
Hence KE = 1 2 2
mr ω
2
M
46. (d) : Mass per unit area of disc =
Mass of removed portion of disc,
πR
2
′ = × ⎛ 2
M
⎝ ⎜ R ⎞ M
M π
⎠
⎟ =
2
πR
2 4
Moment of inertia of removed portion about an axis
passing through centre of disc O and perpendicular to
the plane of disc,
2
I′ = I + Md ′
O
O′
= × × ⎛ 2
⎝ ⎜ ⎞
⎠
⎟ + × ⎛ 2
1 M R M
⎝ ⎜ R ⎞
⎠
⎟
2 4 2 4 2
2 2 2
MR MR 3MR
= + =
32 16 32