A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in…

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More Laws of Motion Previous-Year Questions — Page 7

Q31 jee_main_2024_31_jan_evening Pulley Systems and Tension

A light string passing over a smooth light fixed pulley connects two blocks of masses

m₁

and

m₂

. If the acceleration of the system is

g/8

, then the ratio of masses is

Pulley Systems and Tension diagram for Q31 - JEE Main 2024 Evening — The image displays two masses suspended over a single fixed smooth pulley.

A. $(9)/(7)$ $\frac{9}{7}$
B. $(8)/(1)$ $\frac{8}{1}$
C. $(4)/(3)$ $\frac{4}{3}$
D. $(5)/(3)$ $\frac{5}{3}$

Solution

Related Formula

a = ((m₁ - m₂)g)/((m₁ + m₂))

Core Logic

Assuming $m₁ > m₂$ $m_1 > m_2$, the net pulling force is $(m₁ - m₂)g$ $(m_1 - m_2)g$ and the total mass to be accelerated is $(m₁ + m₂)$ $(m_1 + m_2)$.

Given that the acceleration of the system is $a = (g)/(8)$ $a = \frac{g}{8}$.

Step 1: Algebraic Manipulation

(g)/(8) = ((m₁ - m₂)g)/((m₁ + m₂))

m₁ + m₂ = 8m₁ - 8m₂

8m₂ + m₂ = 8m₁ - m₁

$9m₂ = 7m₁$ $9m_2 = 7m_1$

(m₁)/(m₂) = (9)/(7)

Pattern Recognition

For standard Atwood machines, $a = g \times Difference in massSum of mass$ . If $a/g = 1/8$ $a/g = 1/8$, then $(m₁-m₂)/(m₁+m₂) = 1/8$ $(m_1-m_2)/(m_1+m_2) = 1/8$, which can be solved using componendo and dividendo: $m₁/m₂ = (8+1)/(8-1) = 9/7$ $m_1/m_2 = (8+1)/(8-1) = 9/7$.

Chapter Mix

Class 11 Physics: Laws of Motion

Q42 jee_main_2024_31_jan_evening Friction on an Inclined Plane

A block of mass

5 kg

is placed on a rough inclined surface as shown in the figure.

Friction on an Inclined Plane diagram for Q42 - JEE Main 2024 Evening — The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.

F₁

is the force required to just move the block up the inclined plane and

F₂

is the force required to just prevent the block from sliding down, then the value of

| F₁| - | F₂|

is: [Use

g = 10 m/s²

]

A. $25\sqrt(3) N$ $25\sqrt{3}\text{ N}$
B. $50\sqrt(3) N$ $50\sqrt{3}\text{ N}$
C. $5 \sqrt(3)2 N$ $\frac{5 \sqrt{3}}{2} \text{ N}$
D. $10 N$ $10 \text{ N}$

Solution

Related Formula

fk = μ mg θ

F₁ = mg θ + fk

F₂ = mg θ - fk

Core Logic

To move the block up, the applied force $F₁$ $F_1$ must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force $F₂$ $F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.

Step 1: Calculate Friction

fk = μ mg θ

fk = 0.1 \times 5 \times 10 \times (30°)

fk = 5 \times \sqrt(3)2 = 2.5\sqrt(3) N

Step 2: Force Equations

Moving up:

F₁ = mg θ + fk = 50 (30°) + 2.5\sqrt(3) = 25 + 2.5\sqrt(3)

Preventing slip down:

F₂ = mg θ - fk = 50 (30°) - 2.5\sqrt(3) = 25 - 2.5\sqrt(3)

Step 3: Difference calculation

|F₁| - |F₂| = (25 + 2.5\sqrt(3)) - (25 - 2.5\sqrt(3))

|F₁| - |F₂| = 2 \times 2.5\sqrt(3) = 5\sqrt(3) N

Note: The official options had an anomaly where $5\sqrt(3) N$ $5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as $50\sqrt(3)$ $50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.

Pattern Recognition

The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely $2 fk$ $2 f_k$ ( $2 μ mg θ$ $2 \mu mg \cos \theta$). Bypass calculating the $mg θ$ $mg \sin \theta$ terms entirely.

Chapter Mix

Class 11 Physics: Laws of Motion

Q jee_main_2024_31_jan_morning Pulley And Incline Friction

In the given arrangement of a doubly inclined plane two blocks of masses

M

and

m

are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is

0.25

. The value of

m

, for which

M = 10 kg

will move down with an acceleration of

2 m/s²

is : (take

g = 10 m/s²

and

37° = 3 / 4

)

Pulley And Incline Friction diagram for Q43 - JEE Main 2024 Morning — Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.

A. $9 kg$ $9\mathrm{\ kg}$
B. $4.5 kg$ $4.5\mathrm{\ kg}$
C. $6.5 kg$ $6.5\mathrm{\ kg}$
D. $2.25 kg$ $2.25\mathrm{\ kg}$

Solution

Related Formula

Σ F = ma

fk = μk N = μk mg θ

Core Logic

Since block $M$ $M$ moves down the incline, kinetic friction opposes its motion (acts upwards). Block $m$ $m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards).

For $M$ $M$ block ( $53^°$ $53^\circ$ slope):

Mg 53° - μ Mg 53° - T = Ma

Step 1: Tension Calculation

Given $M = 10 kg$ $M = 10\mathrm{\,kg}$, $a = 2 m/s²$ $a = 2\mathrm{\,m/s^2}$, $μ = 0.25$ $\mu = 0.25$, $g = 10 m/s²$ $g = 10\mathrm{\,m/s^2}$. $53^° = 4/5 = 0.8$ $\sin 53^\circ = 4/5 = 0.8$, $53^° = 3/5 = 0.6$ $\cos 53^\circ = 3/5 = 0.6$.

10(10)(0.8) - 0.25(10)(10)(0.6) - T = 10(2)

80 - 15 - T = 20

65 - T = 20 \Rightarrow T = 45 N

Step 2: Evaluate mass m

For $m$ $m$ block ( $37^°$ $37^\circ$ slope) moving upward:

T - mg 37° - μ mg 37° = ma

$37^° = 3/5 = 0.6$ $\sin 37^\circ = 3/5 = 0.6$, $37^° = 4/5 = 0.8$ $\cos 37^\circ = 4/5 = 0.8$.

45 - m(10)(0.6) - 0.25(m)(10)(0.8) = m(2)

45 - 6m - 2m = 2m

$45 - 8m = 2m$ $45 - 8m = 2m$

10m = 45 \Rightarrow m = 4.5 kg

Chapter Mix

Class 11 Physics: Laws Of Motion

Q jee_main_2024_31_jan_morning Circular Motion Friction

A coin is placed on a disc. The coefficient of friction between the coin and the disc is

μ

. If the distance of the coin from the center of the disc is

r

, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:

A. $(μ g)/(r)$ $\frac{\mu g}{r}$
B. $\sqrt((r)/(μ g))$ $\sqrt{\frac{r}{\mu g}}$
C. $\sqrt((μ g)/(r))$ $\sqrt{\frac{\mu g}{r}}$
D. $μ\sqrt(rg)$ $\frac{\mu}{\sqrt{rg}}$

Solution

Related Formula

fₛ \leq μₛ N

Fc = mrω²

Core Logic

Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning

To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. $f = mω² r$ $f = m\omega^2 r$

The normal force on the flat disc is $N = mg$ $N = mg$. The maximum static friction is $fmax = μ N = μ mg$ $f_{\text{max}} = \mu N = \mu mg$.

For no slipping:

m r ω² \leq μ mg

ω² \leq (μ g)/(r)

ωmax = \sqrt((μ g)/(r))

Chapter Mix

Class 11 Physics: Laws Of Motion

A stone of mass 900 g$900\text{ g}$ is tied to a string and moved in a vertical circle of radius 1 m$1\text{ m}$ making 10 rpm$10\text{ rpm}$. The tension in the string, when the stone is at the lowest point is (if π² = 9.8$\pi^2 = 9.8$ and g = 9.8 m/s²$g = 9.8\text{ m/s}^2$):

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Force Values

Pattern Recognition

Chapter Mix

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Algebraic Manipulation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Friction

Step 2: Force Equations

Step 3: Difference calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Tension Calculation

Step 2: Evaluate mass m

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

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A stone of mass $900 g$ $900\text{ g}$ is tied to a string and moved in a vertical circle of radius $1 m$ $1\text{ m}$ making $10 rpm$ $10\text{ rpm}$. The tension in the string, when the stone is at the lowest point is (if $π² = 9.8$ $\pi^2 = 9.8$ and $g = 9.8 m/s²$ $g = 9.8\text{ m/s}^2$):