Solution

KEYSTONE 6.3 11TH

NEET-UG - Physics

1. (a) r ⃗ × F ⃗
Explanation:
⃗
τ ⃗ = r⃗ × F

2.
(d) 14^i − 38^j + 16k
^

Explanation:
^ ^ ^
−21 i + 3 j + 5k

∣ ^ ^ ^∣
i j k
∣ ∣
⃗
τ ⃗ = r⃗ × F = ∣ 7 3 1∣
∣ ∣
∣ −3 1 5∣

^ ^ ^
= 14 i − 38 j + 16k

3.
(c) it will be remain equal to ω
Explanation:
it will be remain equal to ω

4.
(c) both angular acceleration and linear acceleration
Explanation:
The body will experience both linear and angular accelerations.

5.
(b) 6 Hrs
Explanation:
As the Moment of inertia of earth considered as the sphere is I = 2

5
MR2, thus according to the law of conservation of angular
momentum as the radius contracts to half, thus a new moment of inertia of earth will be I

4
, thus the angular velocity will
increase 4 times and making the length of the day to 6 hrs.

6.
(c) vf = vr
Explanation:
Speeds of the topmost points of both wheels will be equal and equal to that of CM of the bicycle.

2
Ml ω
7. (a) 3t

Explanation:
As Torque(τ ) is equal to the product of Moment of Inertia (I) and Angular acceleration (α)
τ = Iα
Δω
τ = I
Δt
2
M(2l) ω
τ = [ ][ ]
12 t

2
Ml ω
τ =
3t

8.
(b) purely rotational motion
Explanation:

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 The resultant of the two forces forming a couple is zero. So a couple produces a rotational motion, not a linear motion.

9.
2
4πMR
(d) 5T

Explanation:
2

L = Iω = 2

5
MR
2
⋅
2π

T
=
4πMR

5T

10.
dL
(b) dt
= τext

Explanation:
Newton’s second law for rotational motion
τ =
dL
= τ ext
dt

11.
(b) all of these
Explanation:
⃗

Torque, τ ⃗ = dL

dt
⃗
= A × L
⃗

Clearly, dL

dt
is perpendicular to L⃗ After time Δt the new angular momentum vector L + ΔL has the same magnitude as the
initial angular momentum vector L⃗ but has a different direction. Thus the magnitude of L⃗ remains constant while its direction
changes. dL⃗ has no component in the direction of A. So the component of L⃗ in the direction of A⃗ does not change with time.
Hence all given options are true.

12.
(d) t = 1 s
Explanation:
θ (t) = 2 t3 - 6t2
ω =
dθ

dt
= 6t2 - 12t
α =
dω

dt
= 12t - 12 = 0
⇒ t = 1s

13. (a) FN( L

2
) cos θ
Explanation:
τ = FN × perpendicular distance from CM

= FN × L

2
sin(90
∘
− θ)

= FN ( L

2
) cos θ

14. (a) −7^i − 4^j − 8k

^

Explanation:

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 ⃗ ^ ^ ^
F = 4 i + 5 j − 6k

^
⃗ = (2 − 2)^
r ⃗ − r0
^
i + (0 + 2) j + (−3 + 2)k

^ ^ ^
= 0 i + 2j − k

⃗ )× F⃗
τ ⃗ = (r ⃗ − r 0

∣^ ^ ^ ∣
i j k
∣ ∣
^ ^ ^
= ∣0 2 −1 ∣ = −7 i − 4 j − 8k
∣ ∣
∣4 5 −6 ∣

15. (a) 8
Explanation:

⃗ 1
ΔA = (r ⃗ × Δr )
⃗
2
1
= [r ⃗ × (vΔt)]
⃗
2
⃗ ⃗
ΔA 1 P
= (r ⃗ × )
Δt 2 m

⃗
ΔA

Δt
=
1
L
⃗
... (L⃗ = r ⃗ × P⃗)
2 m

⃗
⃗ ΔA
L = 2 m
Δt

⃗
|L| =2×2×2
2

= 8 kg m

min

16. (a) purely rotational motion

Explanation:
A couple produces purely rotational motion.
17. (a) remains constant
Explanation:

Mathematically speaking* angular momentum is moment of momentum about the origin. The angle goes on decreasing from
90°. But it is the perpendicular distance to line of motion x mv, which is angular momentum. This is a constant. Therefore, the
answer expected is that the angular momentum is a constant. (For rectilinear motion, can one discuss angular momentum? But
let us take it only as a mathematical exercise)
18.
(d) 9.8 kg m2/s
Explanation:

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 Total angular momentum about O is
2.8 × 3.1 × 3.6 - 6.5 × 2.2 × 1.5 = 9.8 kg m2/s

19.
−
−−−
(d) 2√2gC
Explanation:
y = 4Cx2 ⇒
dy

dx
= tan θ = 8Cx
At P, tan θ = 8 Ca

For steady circular motion

mω a cos θ = mg sinθ
2

−−−−−
g tan θ
⇒ ω = √
a
−−−−−
g×8aC −−−
−
∴ ω = √ = 2√2gC
a

20.
(d) L

Explanation:
Angular momentum, L = I ω
Rotational kinetic energy, K = 1

2
Iω
2

L Iω 2
= =
K 1 ω
2
Iω
2

or L = 2K

ω
′ ′
L K ω K ω
∴ = × = ( )( )
L K ω′ 2K 2ω

′ L
L =
4

21.
(c) 531 Nm
Explanation:
P = 100 kW = 105 W
rad s-1
rev
ω = 1800 = 30rps = 30 × 2π
min
5
P 10
τ =
ω
=
60π
= 531 Nm

22.
(c) -48 k
^

Explanation:
Position o f particle is, r = 2t^i - 3t2^j
where, t is instantaneous time.
Velocity of particle is v = dr

dt
^ ^
= 2 i − 6t j

Now, angular momentum of particle with respect to origin is given by L = m(r × v)

= m{(2t^i - 3t2^j ) × (2^i - 6t^j )}
= m (-12t2 (^i × ^j ) − 6t 2 ^ ^
( j × i ))

As, ^i × ^i = ^j × ^j = 0
⇒ L = m(-12t2k
^
+ 6t2k
^
)
As, ^i × ^j = k
^
and j × ^i = −k
^

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 ⇒ L = -m (6t2) k
^

So, angular momentum of particle o f mass 2 kg at time t = 2 s is L = (-2 × 6 × 22) k

^
= -48 k
^

23.
(c) (F1 + F2)
Explanation:
It is seen that perpendicular distance of each line of action of force from centre is same as r.
Now taking momentum about O
F1 × r + F2 × r - F3 × r = 0
Hence, F3 = F1 + F2

24.
(b) remains constant
Explanation:
|v| = v = constant and |r| = r (say)
Angular momentum o f the particle about origin O will be given by L = r × p = m (r × v)
or |L| = L = mrv sin θ = mv (rsin θ) = mvh
Now, m.v and h all are constants.

Therefore, angular momentum of particle about origin will remain constant. The direction of r × v also remains the same
(negative z).

25.
(c) F (^i + ^j )
Explanation:
Torque, τ ⃗ = r ⃗ × F ⃗
⃗ ^
F = −F k and r ⃗ = ^i − ^j
∣^ ^ ^ ∣
i j k
∣ ∣
⃗
∴ r⃗ × F = ∣ 1 −1 0 ∣
∣ ∣
∣0 0 −F ∣

^ ^ ^ ^
= i (F ) − j (−F ) = F ( i + j )

26.
(d) -80 k
^

Explanation:
Angular momentum L = m (v × r)
= 2 kg ( dr

dt
× r) = 2 kg (4t ^i × 5^i - 2t2^j )

= 2 kg (-20 t k
^
) = 2 kg × (-20) × 2 m-2 s-1 k
^

= -80 k
^

27.
(d) x-axis
Explanation:

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 ∣^ ^ ^ ∣
i j k
∣ ∣
⃗
L = r⃗ × p⃗ = ∣ 1 2 −1 ∣
∣ ∣
∣3 4 −2 ∣

⃗ ^ ^ ^
L = r ⃗ × p ⃗ = i (−4 + 4) − j (−2 + 3) + k(4 − 6)

^ ^ ^
= 0 i − 1 j − 2k

L
⃗
has components along -y-axis and -z-axis but it has no components in the x-axis. The angular momentum is in y-z plane, i.e.,
perpendicular to x-axis.

28.
(c) zero
Explanation:
Torque on the stone,
⃗
τ ⃗ = r ⃗ × F = rF sin θ

= rF sin 0° = 0

29.
(c) Pattern A is more sturdy
Explanation:
Pattern A is more sturdy because the moment of the tension about the fulcrum is maximum in A to counterbalance the moment
of mg acting from the centre of mass of the rod.

30.
(d) the body will rotate about its centre of mass
Explanation:
Net force on centre of mass is zero, i.e., the centre of mass cannot move at all. Hence, the body will rotate about its centre of
mass only.

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