In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction…

Year	2026	2025	2024	Total
Questions	9	10	15	34

Q jee_main_2025_04_april_morning Equilibrium of Forces

A body of mass

m

is suspended by two strings making angles

θ1

and

θ₂

with the horizontal ceiling with tensions

T₁

and

T₂

simultaneously.

T₁

and

T₂

are related by

T₁=\sqrt(3)T₂

the angles \theta_{1} and \theta_{2} are

A. $θ₁=30°, θ₂=60° with T₂=(3mg)/(4)$
B. $θ₁=60°, θ₂=30° with T₂=(mg)/(2)$
C. $θ₁=45°, θ₂=45° with T₂=(3mg)/(4)$
D. $θ₁=30°, θ₂=60° with T₂=(4mg)/(5)$

Solution

Related Formula

Horizontal equilibrium equation:

T₁ θ₁ = T₂ θ₂

Vertical balancing equation:

T₁ θ₁ + T₂ θ₂ = mg

Core Logic

Substitute $T₁ = \sqrt(3)T₂$ $T_1 = \sqrt{3}T_2$ into horizontal equilibrium:

\sqrt(3)T₂ θ₁ = T₂ θ₂ \sqrt(3) θ₁ = θ₂

Step 1: Audit Options Pattern Match

Check Option (B) where $θ₁ = 60°$ $\theta_1 = 60^{\circ}$ and $θ₂ = 30°$ $\theta_2 = 30^{\circ}$:

\sqrt(3) (60°) = \sqrt(3) \cdot (1)/(2) = \sqrt(3)2

(30°) = \sqrt(3)2

This matches the horizontal equilibrium condition $\sqrt(3) θ₁ = θ₂$ $\sqrt{3}\cos\theta_1 = \cos\theta_2$.

Step 2: Solve for T2

Substitute the angles into the vertical component balancing equation:

T₂ [\sqrt(3) (60°) + (30°)] = mg

T₂ [\sqrt(3)( \sqrt(3)2) + (1)/(2)] = mg

T₂ [(3)/(2) + (1)/(2)] = mg T₂ (2) = mg T₂ = (mg)/(2)

Pattern Recognition

Lami's Theorem or standard rectangular resolution of forces. When tension is scaled by $\sqrt(3)$ $\sqrt{3}$, it directly hints at complementary $30°--60°$ $30^{\circ}\text{--}60^{\circ}$ geometric alignments.

Evaluation Rubric / Model Answer

Option B: $θ₁=60°, θ₂=30° with T₂=(mg)/(2)$

Chapter Mix

Class 11 Physics: Laws of Motion

Q22 jee_main_2025_24_jan_evening Circular Motion

A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes

((3)/(π))

rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ____ ML.

Numerical Answer. Answer: 36 to 36

Solution

Related Formula

For a conical pendulum:

T θ = Mg

T θ = Mω² R

where $R = L θ$ $R = L \sin \theta$.

Core Logic

Substituting $R = L θ$ $R = L \sin \theta$ into the centripetal equation:

Conical pendulum string force vectors diagram Q22

T θ = M ω² (L θ)

T = M ω² L

Given rotational frequency: $f = (3)/(π) rev/s$ $f = \frac{3}{\pi}\ \mathrm{rev/s}$ Angular velocity:

ω = 2π f = 2π ((3)/(π)) = 6 rad/s

Substituting $ω$ $\omega$ back into the simplified tension equation:

T = M (6)² L = 36 ML

Pattern Recognition

In a conical pendulum, the horizontal projection of tension provides the exact centripetal force. The $θ$ $\sin\theta$ terms cancel out, making string tension independent of the semi-vertical angle.

Chapter Mix

Class 11 Physics: Laws of Motion

Q jee_main_2025_24_jan_morning Circular Motion and Banking of Roads

A car of mass 'm' moves on a banked road having radius 'r' and banking angle

θ

To avoid slipping from banked road, the maximum permissible speed of the car is

v₀.

The coefficient of friction

μ

between the wheels of the car and the banked road is :-

A. $μ= v₀²+rg~ ~θrg-v₀² ~θ$
B. $μ= v₀²+rg~ ~θrg+v₀² ~θ$
C. $μ= v₀²-rg~ ~θrg+v₀² ~θ$
D. $μ= v₀²-rg~ ~θrg-v₀² ~θ$

Solution

Related Formula

The maximum velocity limit preventing outer slide breakout on a rough banked plane profile is given by standard centrifugal force equations:

v₀ = \sqrt(rg (( θ + μ)/(1 - μ θ)))

Core Logic

By writing out the components shown in the free body layout

Circular Motion and Banking of Roads diagram for Q10 - JEE Main 2025 Morning

:

N θ + f θ = mv₀²r

N θ - f θ = mg

Step 1: Isolate Coefficient of Friction

Squaring the velocity boundary relation gives :

v₀²rg = ( θ + μ)/(1 - μ θ)

Cross multiply to isolate the variable terms :

v₀² - μ v₀² θ = rg θ + μ rg

v₀² - rg θ = μ(rg + v₀² θ)

μ = v₀² - rg θrg + v₀² θ

Pattern Recognition

Maximum limits always balance with plus signs in the numerator fraction. Simply rearrange that template expression directly to map the targeted parameter.

Chapter Mix

Class 11 Physics: Laws of Motion

Q20 jee_main_2025_28_jan_evening Newton Second Law Applications

A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be : (Take 'g' as acceleration due to gravity) [cite: 174, 175]

A. $3Ma2a - g$
B. $3 Ma2 a + g$
C. $2 Ma3 a + g$
D. $2 Ma3 a - g$

Solution

Related Formula

By Newton's second law of motion, the net upward force acting on an accelerating balloon system is given by:

Fbuoyant - mtotal g = mtotal a

Core Logic

Let $F$ $F$ be the constant buoyant force acting upward on the balloon.

Case 1 (Initial upward acceleration) :

F - M g = M a F = M(g + a)

Case 2 (After releasing mass $x$ $x$) : The new total mass becomes $(M - x)$ $(M - x)$, and its acceleration increases to $3a$ $3a$:

F - (M - x)g = (M - x)3a

Substitute the value of $F$ $F$ from Case 1 into Case 2 [cite: 836, 839]:

M(g + a) - (M - x)g = (M - x)3a

M g + M a - M g + x g = 3 M a - 3 x a

M a + x g = 3 M a - 3 x a

x(g + 3a) = 2 M a

x = (2 M a)/(3a + g)

Step 1: Visual Context

The free-body force layout for both accelerating phases is shown below:

Newton Second Law Applications free body diagrams for Q20

Pattern Recognition

Since the upward buoyant force is completely determined by the balloon's volume, it remains constant. Expressing this constant force in terms of the initial conditions allows you to quickly solve for mass changes when acceleration states vary.

Chapter Mix

Class 11 Physics: Laws of Motion

Q jee_main_2024_01_february_morning Friction

Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in

ms⁻²

is (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless):

Block and trolley tension system for Q48 - JEE Main 2024 Morning — A block and trolley mass arrangement demonstrating a horizontal kinetic interface connected over a corner pulley driven by an explicit 60N forcing function loop.

A. 3
B. 4
C. 2
D. 1.2

Solution

Related Formula

Kinetic friction force:

fk = μk N = μk m₁ g

System acceleration:

a = Fpull - fkmtotal

Core Logic

Given values: Trolley mass $m₁ = 20~kg$ $m_1 = 20\mathrm{~kg}$, total system mass component in frame $mtotal = 26~kg$ $m_{\text{total}} = 26\mathrm{~kg}$ (from solution fraction $(60-8)/(26)$ $\frac{60-8}{26}$). Applied pulling force $F = 60~N$ $F = 60\mathrm{~N}$. $μk = 0.04$ $\mu_k = 0.04$.

Calculate the kinetic friction resisting the trolley:

fk = 0.04 \times 20 \times 10 = 8~N

Step 1: Calculate Acceleration

Using the dynamic equation for connected translation systems:

a = (60 - 8)/(26) = (52)/(26) = 2~ms⁻²

Pattern Recognition

Treat connected inline systems as a single collective mass block, balancing external driving forces against collective internal friction resistance.

Chapter Mix

Class 11 Physics: Laws of Motion

Solution & Explanation

Related Formula

Core Logic

Step 1: Force Balance Equations

Pattern Recognition

Chapter Mix

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Audit Options Pattern Match

Step 2: Solve for T2

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Isolate Coefficient of Friction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Visual Context

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Acceleration

Pattern Recognition

Chapter Mix

More Laws of Motion Questions — jee_main_2026_24_january_morning

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