A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the…

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Q37 jee_main_2024_01_february_morning Circular Motion

A ball of mass

0.5~kg

is attached to a string of length

50~cm

. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is

400~N

. The maximum possible value of angular velocity of the ball in

rad/s

is:

A. 1600
B. 40
C. 1000
D. 20

Solution

Related Formula

Centripetal force configuration for simplified horizontal rotation layout:

$T = mω² l$ $T = m\omega^2 l$

Core Logic

Given values: $m = 0.5~kg$ $m = 0.5\mathrm{~kg}$, $l = 50~cm = 0.5~m$ $l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, $Tmax = 400~N$ $T_{\text{max}} = 400\mathrm{~N}$.

Equating max tension to centripetal requirement:

400 = 0.5 \times ω² \times 0.5

Step 1: Compute Angular Velocity

400 = 0.25 ω²

ω² = (400)/(0.25) = 1600

ω = \sqrt(1600) = 40~rad/s

Pattern Recognition

Ensure units are metric ( $50~cm \to 0.5~m$ $50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics.

Chapter Mix

Class 11 Physics: Laws of Motion

Q36 jee_main_2024_29_january_evening Circular Motion and Tension

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 the string, when the stone is at the lowest point is (if

π² = 9.8

and

g = 9.8 m/s²

):

A. $97 N$ $97\text{ N}$
B. $9.8 N$ $9.8\text{ N}$
C. $8.82 N$ $8.82\text{ N}$
D. $17.8 N$ $17.8\text{ N}$

Solution

Related Formula

At the lowest point of a vertical circle, the equation of motion for a mass $m$ $m$ is:

T - mg = m r ω²

Rearranging to solve for tension $T$ $T$:

T = mg + m r ω²

Core Logic

Given data:

Mass, $m = 900 g = 0.9 kg$ $m = 900\text{ g} = 0.9\text{ kg}$
Radius, $r = 1 m$ $r = 1\text{ m}$
Frequency, $N = 10 rpm = (10)/(60) rps = (1)/(6) rps$
Angular velocity, $ω = 2π N = 2π ((1)/(6)) = (π)/(3) rad/s$

Step 1: Calculate Force Values

Now we substitute our parameters into the tension equation:

T = (0.9)(9.8) + (0.9)(1)((π)/(3))²

T = 8.82 + 0.9 \times (π²)/(9)

Since $π² = 9.8$ $\pi^2 = 9.8$:

T = 8.82 + 0.1 \times 9.8

T = 8.82 + 0.98 = 9.80 N

Free-body diagram of stone at lowest point in vertical circle for Q36

Pattern Recognition

Always convert Mass to $kg$ $\text{kg}$ and rotational speed to $rad/s$ $\text{rad/s}$ first. Using the prompt constraint $π² = 9.8$ $\pi^2 = 9.8$ yields a perfect decimal addition match.

Chapter Mix

Class 11 Physics: Laws of Motion

Q59 jee_main_2024_29_january_evening Non-uniform Circular Motion

A particle is moving in a circle of radius

50 cm

in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at

t = 0

is

4 m/s

, the time taken to complete the first revolution will be

(1)/(α)[ 1 - e-2π ] s

, where

α =

________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

For a particle in circular motion:

Normal (centripetal) acceleration: $ac = (v²)/(r)$ $a_c = \frac{v^2}{r}$
Tangential acceleration: $aₜ = (dv)/(dt)$ $a_t = \frac{dv}{dt}$

Core Logic

Given $ac = aₜ$ $a_c = a_t$:

(v²)/(r) = (dv)/(dt)

\intv₀v (dv)/(v²) = \int₀t (dt)/(r)

[ -(1)/(v) ]v₀v = (t)/(r)

-(1)/(v) + (1)/(v₀) = (t)/(r) (1)/(v) = (1)/(v₀) - (t)/(r)

v = (v₀)/(1 - (v₀ t)/(r))

Step 1: Relate Velocity to Position and Integrate

Substitute the parameters $v₀ = 4 m/s$ $v_0 = 4\text{ m/s}$ and $r = 50 cm = 0.5 m$ $r = 50\text{ cm} = 0.5\text{ m}$:

v = (4)/(1 - 8t) = (ds)/(dt)

Integrating this to find the position $s(t)$ $s(t)$:

\int₀s ds = \int₀t (4)/(1 - 8t) dt

s = 4 [ (ln(1 - 8t))/(-8) ]₀^t = -(1)/(2) ln(1 - 8t)

Step 2: Solve for Time of First Revolution

To complete the first revolution, the distance covered is:

s = 2π r = 2π (0.5) = π m

Equating the distance:

π = -(1)/(2) ln(1 - 8t)

-2π = ln(1 - 8t)

1 - 8t = e-2π

8t = 1 - e-2π t = (1)/(8) [ 1 - e-2π ] s

Comparing this to $(1)/(α)[ 1 - e-2π ] s$ , we get:

$α = 8$ $\alpha = 8$

Pattern Recognition

The condition $aₜ = ac v (dv)/(ds) = (v²)/(r) (dv)/(v) = (ds)/(r)$ . Integrating directly gives $v = v₀ es/r$ $v = v_0 e^{s/r}$. Substituting this back into $v = ds/dt$ $v = ds/dt$ makes the final time integral much more intuitive.

Chapter Mix

Class 11 Physics: Laws of Motion

Q36 jee_main_2024_27_jan_morning Banking of Tracks

A train is moving with a speed of

12 m/s

on rails which are

1.5 m

apart. To negotiate a curve of radius

400 m

, the height by which the outer rail should be raised with respect to the inner rail is (Given,

g = 10 m/s²

):

A. $6.0 cm$ $6.0\text{ cm}$
B. $5.4 cm$ $5.4\text{ cm}$
C. $4.8 cm$ $4.8\text{ cm}$
D. $4.2 cm$ $4.2\text{ cm}$

Solution

Related Formula

θ = (v²)/(Rg)

For small angles, $θ \approx θ = (h)/(d)$ , where $h$ $h$ is the raised height and $d$ $d$ is the separation between the tracks.

Core Logic

Equating the two relationships:

(h)/(d) = (v²)/(Rg)

(h)/(1.5) = (12 \times 12)/(400 \times 10)

Step 1: Compute height value

h = 1.5 \times (144)/(4000) = 1.5 \times 0.036 = 0.054 m

Converting to centimeters:

h = 0.054 \times 100 = 5.4 cm

Pattern Recognition

Whenever $θ$ $\theta$ is small, geometry permits approximating $θ$ $\tan\theta$ with $hwidth$ $\frac{h}{\text{width}}$, vastly reducing computational transcendental overhead.

Chapter Mix

Class 11 Physics: Laws of Motion

Q45 jee_main_2024_27_jan_morning Conservation of Linear Momentum

A body of mass

1000 kg

is moving horizontally with a velocity

6 m/s

. If

200 kg

extra mass is added, the final velocity (in m/s) is:

A. $6$ $6$
B. $2$ $2$
C. $3$ $3$
D. $5$ $5$

Solution

Related Formula

m₁ v₁ = m₂ v₂

Core Logic

Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.

Initial mass $m₁ = 1000 kg$ $m_1 = 1000\text{ kg}$, initial velocity $v₁ = 6 m/s$ $v_1 = 6\text{ m/s}$. Final mass $m₂ = 1000 + 200 = 1200 kg$ $m_2 = 1000 + 200 = 1200\text{ kg}$.

Step 1: Balance conservation equation

1000 \times 6 = 1200 \times v₂

6000 = 1200 v₂ v₂ = (6000)/(1200) = 5 m/s

Pattern Recognition

Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles.

Chapter Mix

Class 11 Physics: Laws of Motion

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]

Solution & Explanation

Related Formula

Core Logic

Step 1: Visual Context

Pattern Recognition

Chapter Mix

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Compute Angular Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Force Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Relate Velocity to Position and Integrate

Step 2: Solve for Time of First Revolution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute height value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Balance conservation equation

Pattern Recognition

Chapter Mix

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