JEE (Main + Advanced) 2025
GUIDED REVISION ENTHUSIAST & LEADER COURSE
PHASE : ALL
SOLUTION GR_ROTATIONAL DYNAMICS
PHYSICS 2. Ans ( A )
SECTION-I (i)
1. Ans ( D )
For the semicricular lamina of mass m, the moment
of inertia about an axis through C is IC = . Torque acting about the point B = (mg)
Let ICM = moment of inertia about an axis through If I is the moment of inertia of the stick about B and a
its centre of mass. is the angular acceleration then
IC = ICM + md2 ⇒ ICM = – m = I α = (mg) ⇒ α =
Acceleration of centre of mass a = α
N
Moment of inertia of this stick about its edge I =
E ⇒ a=
OR
For angular motion of the stick
LL
τ = mg =I α
moment of intertia of stick about B is I =
⇒ mg α ⇒ α=
A
Acceleration of centre of mass = α
3. Ans ( C )
For toppling
KTJAPHELGR25004 HS-1/6
Target:JEE (Main + Advanced) 2025
4. Ans ( A ) PHYSICS
= +
SECTION-I (ii)
= ↓
8. Ans ( B,C )
= Rω2 = For (B) : 2mv0 ℓ = 6m ℓ 2 ω
= =
5. Ans ( C )
Only translation is possible if force is applied at
center of mass
N
∴ ω=
T = 2m ω 2 ℓ =
For (C) : IA = 2m(3 ℓ )2; IB = m(3 ℓ )2
E 9. Ans ( B,C )
cm = = l Balancing torque about support,
6. Ans ( C ) 85g × 1.5 + 15 g × 0.5 = T × 0.5
⇒ T = 2700 N
Balancing force in vertical direction,
LL
N = 15g + 85g + 2700 = 3700
10. Ans ( A,B,C,D )
... (i)
... (ii)
Ry = m = 15 N
A
7. Ans ( C ) =4
= So accelerations of A & B is
= ×
= Mvl
HS-2/6 KTJAPHELGR25004
Phase-ENTHUSIAST & LEADER COURSE
11. Ans ( A,B ) 13. Ans ( A,B )
mghA = KB = (Ktrans)B + (Krot)B
... (i)
, where
v= ;ω=
... (ii) ⇒t=
N
From (i) & (ii)
hA > hC ; KB > KC s = ut + = =
and hA > hC ; KC > KA
12. Ans ( A,B,C ) PHYSICS
E SECTION-I (iii)
For (A) a = =
15. Ans ( A )
For (B) V2 = 0 + 2 × l ⇒ V =
(i)
LL
⇒ ω= =
For (C & D) At toppling cm is just a box base line.
For pure roling a = α R
=
⇒ Mgsina xc = =
A
– f = 2f ycm =
⇒f= tan (90° – α ) = ⇒ tan α =
μ mg cos α = ⇒μ=
(ii) for it to not slide 3mg sin α ≤ 3m mg cos α
⇒ μ ≥ 0.66
(iii) an = g sin α
⇒a= ⇒
KTJAPHELGR25004 HS-3/6
Target:JEE (Main + Advanced) 2025
16. Ans ( B ) 22. Ans ( A )
Let ball leaves with velocity u & disc rotates with
(i) at toppling cm is just a box ω clockwise direction
Angular moment conservation about centre of disc
base line.
O = muR –
xc = ω= ... (i)
Energy conservation
mv2 = mu2 + I ω 2 ... (ii)
=
u= ,ω=
ycm =
N
tan (90° –
α) =
E ⇒ tan α =
(ii) for it to not slide 3mg sin α ≤ 3m mg cos α
LL
⇒ μ ≥ 0.66
(iii) an = g sin α
⇒ a= ⇒
17. Ans ( C )
A
… (i) … (ii)
… (iii)
For pure rolling ...(iv)
On solving , ,
∴
HS-4/6 KTJAPHELGR25004
Phase-ENTHUSIAST & LEADER COURSE
26. Ans ( B ) PHYSICS
SECTION-II
1. Ans ( 10 )
F = ma
a = 5m/s2
N1 + N2 = mg = 200 .... (i)
Case-1 : F(h – 0.9) + N1(0.3) – N2 (0.3) = 0
N
N2 – N1 = .... (ii)
From (i) & (ii)
E Taking torque about axis OO' & putting it equal to
zero
For not tipping, N1 > 0
h < 0.9 + 0.6
LL
.... (i)
h < 1.5 m
.... (ii)
Case-2 : F (0.9 – h) + N1(0.3) – N2(0.3) = 0 .... From (i) & (ii), putting the values we get
(iii) v = 10 m/s
From (i) & (iii)
2. Ans ( 36.00 )
For not tipping, N1 > 0
A
I0 =
h > 0.9 – 0.6
h > 0.3 m Now CM = d =
∴ ICM = I0 – md2 =
m=
KTJAPHELGR25004 HS-5/6
Target:JEE (Main + Advanced) 2025
PHYSICS PHYSICS
SECTION-III SECTION-IV
1. Ans ( 8 ) 1. Ans ( A->Q,B->P,C->S,D->R )
τ = Ι α ; τ = mg ; , (A)
IO = MR2
2. Ans ( 8 )
N
For the given sphere :
mg – fs = m (acm)rel ...(1)
N= ; fs × R = mR2 α ; a = α R
3. Ans ( 9 ) E
(i) Li = Lf; = MV × (C) Ibase = =
⇒ V= (D) I = I(+) + I( – )
LL
(ii) Pi = Pf; 0 = MV –
mV1 ⇒ V = ⇒
(iii) e =1= ⇒ V + V1 = ;
= ; =3
A
⇒ =2 ⇒ M= = 9 kg.
4. Ans ( 4 )
mv0 · 2R = ;ω= =4
rad/s
5. Ans ( 2 )
mg sin θ – T cos θ = ma ∴ a = 2m/s2
HS-6/6 KTJAPHELGR25004
