JEE (Main + Advanced) 2025
GUIDED REVISION ENTHUSIAST & LEADER
PHASE : ALL
SOLUTION GR_Fluid Statics & Dynamics
PHYSICS 2. Ans ( A )
SECTION-I (i)
1. Ans ( A )
For equilibrium ⇒ FBD ⇒
N
FB = mg(Water) + mg(container)
For the section of length dx,
E
dT = ρ A dx ω 2x
∴ T at centre =
This shows that if ρ C < 0.5
LL
So Ans. (A)
= =
3. Ans ( B )
=
∴ Pressure = =
A
Writing Bernoulli's equation along flow line at
point A & B
Ptank + P0 + 0 +
[VA << VB as area of nozzle is very less as
compared to area of tank]
KTJAPHELGR25007 HS-1/6
Target:JEE (Main + Advanced) 2025
4. Ans ( C ) PHYSICS
As the height decreases, the rate of flow with which
the water is coming out decreases. SECTION-I (ii)
5. Ans ( B ) 12. Ans ( B,C )
where
Fv = (M + m)g
= V = volume/second.
6. Ans ( A )
Let y be the height of liquid at some instant.
Then = constant (given)
From equation of continuity, 13. Ans ( B,C,D )
Pnew = ρ (g + ) h & Bnew = Pnew × V
( π x2) (a = area of hole)
14. Ans ( A,B )
N
Here, , a and g are constants
Hence, squaring the equation, we get y = kx4.
10. Ans ( A )
E
Pressure inside the film is less than outside by an
amount, , where r1 and r2 are the
radii of curvature of the meniscus. Here r1 = t/2 and
r2 = , then the force required to separate the two A1 = 4 × 10 – 3 m2
A2 = 8 × 10 – 3 m2
LL
glass-plates, between which a liquid film is
enclosed (figure) is, v1 = 1 m/sec
ρ W = 103 kg/m3
F=P×A= ,
Continuity equation at P & Q
where t is the thickness of the film, A = area of A1 v1 = A2 v2
film. (4 × 10 – 3) (1) = (8 × 10 – 3) (v2)
Work done per unit volume by pressure = Δ P =
A
pressure difference
Applying Bernoulli's theorem between P & Q
F = =
= 45 N
11. Ans ( A ) = 29625 J/m3
By force balance on the wire, Work done per unit volume by gravity force =
2T cos θ ℓ = λ ℓ g
So, work done/volume = ρ g (h1 – h2)
= 1000 × 10 [2 – 5]
= – 30000 J/m3
HS-2/6 KTJAPHELGR25007
Phase-ENTHUSIAST & LEADER
15. Ans ( B,C ) 18. Ans ( B )
Surface energy of drop after detachment = T ×
surface area
= T × 4 π R2 = 0.11 × 4 × π × (1.4 × 10 – 3)2
≈ 2.7 × 10 – 6 Joule
19. Ans ( D )
Force exerted by liquid = (mass of liquid displaced)
×
= ρ ωv ×
20. Ans ( B )
N
PHYSICS
SECTION-I (iii)
16. Ans ( C )
R – r = ℓ sin θ .....(i)
E ℓ – h = ℓ cos θ
m' ω 2r = T sin θ
......(ii)
m' = ( ρ ω – ρ body)v & m'g = T cos θ
.....(iv)
from (i), (ii), (iii) & (iv)
.....(iii)
LL
FT = T·2 π r
FV = FTsin θ
17. Ans ( A )
A
Drop detaches from dropper
When :
FV = mg
≈ 1.4 × 10 – 3 m
KTJAPHELGR25007 HS-3/6
Target:JEE (Main + Advanced) 2025
21. Ans ( A ) PHYSICS
h1 = = 0.26 ; h2 = = 0.195
SECTION-II (i)
Plug opens when level of water reaches 'h'.
• FBD of plug
1. Ans ( 0.1 )
Y = 10 – 5/°C, Δ T = 100°C, h1 = 300 cm
h2 – h1 = 0.1 cm ans.
FW = T ⇒ P. π R2 = T 2. Ans ( 4.5 )
T = 25 π h
N
• FBD of float Pf =
FB = mg + T
E [h = 4.5 m]
ρ WgxA = 3 × 10 + T
PHYSICS
10000x π × = 30 + 25 π h SECTION-II (ii)
100 π x = 30 + 25 π h 3. Ans ( 100 )
LL
100 π (h – 0.1) = 30 + 25 π h
75 π h = 30 + 10 π
∴ Water level its be lowered = 2 cm
∴ Area occupied by water level at the top = 100
• To again close the plug
cm2
mg = FB (for the float) → [When water has risen
∴ 2t = 100 × 2 ⇒ t = 100 sec
to height y for the float]
4. Ans ( 110 )
A
where V = pr2 ℓ
FB = mg so = 35 π = 110 rad/sec2
ρ WgyA = 3g
1000 × 10 × y × π ×
100py = 3g
So h = 0.1 m +
HS-4/6 KTJAPHELGR25007
Phase-ENTHUSIAST & LEADER
5. Ans ( 2500 ) 2. Ans ( 1 )
A1V1 = A2V2
10 × 1 = 5 V2
V2 = 2 m/s
P2 = 2500 If final pressure of gas in tube is P2 then
6. Ans ( 40 ) P2 – = P0 (as levels inside and outside are
...(i) same)
i.e. P2 = P0 +
...(ii) but as temperature remains constant
∴ P1 V 1 = P 2 V 2
From equation (i) & (ii)
N
; P0AL = A(L – h)
On solving, h = 0.01 m
E PHYSICS 3. Ans ( 4 )
SECTION-III
p= =
1. Ans ( 5 ) V = π r3
dV = 4 π r2dr
LL
=B×
dr = = = × 10 – 8 =
4Å
4. Ans ( 6 )
= 1.5 × 4 – 2y = 1.5 × 4 = 6
For minimum velocity at orifice
A
PHYSICS
SECTION-IV
This will give us minimum height for this velocity
1. Ans ( A->PQ,B->PQ,C->PRST,D->PRST )
If dgh1 > P ⇒ height of liquid will not cross h1
v1 = & v2 = – dh/dt dgh1 < P ⇒ height of liquid will cross h1
By equation of continuity av1 =Av2
⇒ t = 125 sec = (5)3 = (a)3 ⇒ α = 5
KTJAPHELGR25007 HS-5/6
Target:JEE (Main + Advanced) 2025
2. Ans ( A->PQRST,B->PQT,C->RS,D->Q )
(A) θ < θ R
(C)
mg sin θ – f = macm .......(i)
2
f(R) = ( MR2 ) α .......(ii)
5
Block will not slide acm
α= ......(iii)
Block in equilibrium R
On solving
Force exerted by A = Mg 2
f= mg sin θ
N = mg cos θ 7
f = mg sin θ F A = √f 2 + N 2 =
FA = 2
2
mg cos θ 2 + ( mg sin θ ) < mg
N
√
work done by FA = 0 (displacement is zero) 7
Hence, mechanical energy as well as KE remains velocity of pt. P = 0
const. Hence, →s P = 0
work done by FA = 0.
(B)
E
Body falling with terminal velocity
(D)
LL
KE is const.
Hence, Fnet = 0 m1 < m2
FA = mg m2g – T = m2a
If density of ball increased then mass will increase T – m1g = m1a
and it will start accelrating since it (m2 − m1 ) g
a=
acquire new terminal velocity. m1 + m2
2m m
T = ( m +1 m2 ) g ≠ mg
1 2
If density of body is changed, then mass will
A
change and so will acceleration.
HS-6/6 KTJAPHELGR25007
