33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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34 NEET-AIPMT Chapterwise Topicwise Solutions Physics
40. (d) : Given u = V, final velocity = 0.
Using v = u + at
0
\ 0 = V – at or, − a = − V V
=−
t t
Force of friction, f = mR =mmg
V V
Retardation, a = mg ∴ t = =
a µ g
.
41. (d) :
the weight of hanging part must be balanced by the force
of friction on the portion of the table.
( L– y)
Free body diagram of two masses is
W = f L
But from figure
.....(i)
W = M L yg and R= W′ = M L ( L−
y)
g
We get equations
T + m A a = f or T = mN A (for a = 0)
and T = m B a + m B g or T = m B g (for a = 0)
∴ µ NA = mBg ⇒ mB = µ mA = 0.2 × 2 = 0.4 kg
42. (b) : m = 10 kg, R = mg
\ Frictional force = f k
= m k R = m k mg
= 0.5 × 10 × 10 = 50 N
\ Net force acting on the body, F = P – f k
= 100 – 50 = 50 N
\ Acceleration of the block, a = F/m
= 50/10 = 5 m/s 2
43. (a) :
f rL = m s N = m s × mg = 0.6 × 1 × 10 = 6 N.
where f rL is the force of limiting friction.
Pseudo force = ma = 1 × 5 ; F = 5 N
If F < f rL block does not move. So static friction is present.
Static friction = applied force .
\ f r = 5 N.
44. (d) : The acceleration is nullified by force of kinetic
friction on the block.
mg sinq is force downwards.
m k is the coefficient of kinetic friction.
m k mg cosq is friction force acting upwards.
\ mg sinq – m k mg cosq = mass × acceleration
acceleration = 0 as v is constant
\ m k = tanq
45. (b)
46. (a) : Let M is the mass of the chain of length L. If y is
the maximum length of chain which can hang outside the
table without sliding, then for equilibrium of the chain,
So that f L = mR = µ M L ( L−
y)
g
Substituting these values of W and f L in eqn.(i), we get
µ M L ( L− y)
g = M
L yg
µ L 025 . L L
or m(L – y) = y or y = = =
µ + 1 125 . 5
y
or
L = 1
= 1
5
× 100
5 % = 20 %
47. (b) : The various forces
acting on the body have been
shown in the figure. The
force on the body down the
inclined plane in presence of
friction is
F = mgsinq – f = mgsinq – mN = ma
or a = gsinq – mgcosq.
Since block is at rest thus initial velocity u = 0
\ Time taken to slide down the plane
2s
2s
t1
= =
a gsinθ − µ gcosθ
In absence of friction time taken will be
2s
t2
=
g sinθ
Given : t 1 = 2t 2 .
2 2
2s
2s
× 4
∴ t1
= 4t2
or
=
g(sinθ − µ cos θ) g(sin θ)
3
or sinq = 4sinq – 4mcosq or µ = tan θ=
075 .
4
48. (d) : To keep the block stationary,
Frictional force ≥ Weight
mN ≥ Mg
Here, N = Mw 2 r
r = 1 m, m = 0.1
For minimum w, mMw 2 r = Mg
g
w =
µ r
= 10
= 10 rad s –1
01 . × 1
N
Mg
f