Newtonian Mechanics MCQ Quiz in मराठी - Objective Question with Answer for Newtonian Mechanics - मोफत PDF डाउनलोड करा

Concept:

• For a frictionless pulley system, the net acceleration can be calculated using Newton's second law.
• The effective force acting on each block is determined by its weight component along the incline.
• For each block:
  • Block 1: Downward force = m₁g × sin 37°
  • Block 2: Upward force = m₂g × sin 37°
  • Block 3: Downward force = m₃g × sin 53°
• The net force (F_net) and total mass (M_total) are used to find acceleration:
• a = F_net / M_total

Calculation:

Masses of blocks: m1 = 2 kg, m2 = 4 kg, m3 = 6 kg

Angles: sin 37° = 3/5, sin 53° = 4/5

System is frictionless, and we need to find the acceleration (a).

Net force:

⇒ T₁ - (2mg × sin 37°) =2ma                    (1)

⇒ T₂  +(4mg sin 53°) -T₁=4ma                 (2)

⇒ 6mgsin 53°-T₂ = 6ma                             (3)

Adding 1,2 and 3

⇒ (2m+4m+6m)a =6mgsin 53°+4mg sin 53°-2mg × sin 37°

⇒ a = (34g/5) / 12

⇒ a = (17g) / (30)

∴ The acceleration of the system is (17/30)g.

Explanation:

Given:

Drag coefficient \(b=3\ kgs^{-1}\)

mass of spring system is 0.5 kg.

spring constant \(2 \ Nm^{-1}\)

Solution:
\(b^2 = 4mk \Rightarrow m = \frac{b^2}{4k} = \frac{3^2}{4 \times 2} = \frac{9}{8} = 1.12 \, \text{kg}\)

Excess mass:
\(m_{\text{excess}} = 1.12 - 0.5 = 0.62 \, \text{kg}\)

The correct answer is 0.62 .

Calculation:

To calculate the power needed to keep the conveyor belt running, we use the formula for power:

\(P = F \times v\)

Where:

• P:  is the power in watts.
• F: is the force applied.
• v:  is the speed of the conveyor belt.

The force required to move the sand is the rate of change of momentum, which is given by:

\(F = \frac{dm}{dt} \times v\)

Where:

\(\frac{dm}{dt} \) is the rate at which sand is falling \(1.5 \, \text{kg/s}\).

\(v \) is the velocity of the belt \(7\text{m/s}\).

Now, substituting the values:

\(F = 1.5 \, \text{kg/s} \times 7 \, \text{m/s} = 10.5 \, \text{N}\)

Now calculate the power:

\(P = 10.5 \, \text{N} \times 7 \, \text{m/s} = 73.5 \, \text{W}\)

So, the power required to keep the conveyor belt running is \(\boxed{73.5 \, \text{W}} \).

Calculation:

 Time Period of Motion:

given:

\(\text{mass of the body }m=1\text{kg}\)

\(\text{semi-major axis }a=1000m\)

semi-minor axis b=100m

Orbital angular momentum, \(L=100kgm^2s^{-1}\)

The Eccentricity of the orbit is:

\(\begin{align*} e &= \sqrt{1 - \frac{b^2}{a^2}} \\ &= \sqrt{1 - \frac{100^2}{1000^2}} \\ &= \sqrt{1 - 0.01} \\ &= \sqrt{0.99} \end{align*}\)

The angular momentum for an elliptical orbit is:

\(\begin{align*} L &= m \sqrt{GM \, a \, (1 - e^2)} \end{align*}\)

Substitute \(L = 100, m = 1, a = 1000, \text{and } 1 - e^2 = 0.01 \)

\(\begin{align*} 100 &= \sqrt{GM \cdot 1000 \cdot 0.01} \end{align*}\)

Square both sides:

\(\begin{align*} 10000 &= GM \cdot 10 \end{align*}\)

\(\begin{align*} GM &= 1000 \, \text{m}^3 \cdot \text{s}^{-2} \end{align*}\)

Step 3: Time Period of the Orbit

Using Kepler's Third Law:

\(\begin{align*} T &= 2\pi \sqrt{\frac{a^3}{GM}} \end{align*}\)

Substitute \(a = 1000 ,\text{and }GM = 1000\)

\(\begin{align*} T &= 2\pi \sqrt{\frac{1000^3}{1000}} \\ &= 2\pi \sqrt{1000^2} \\ &= 2\pi \cdot 1000 \end{align*}\)

\(\begin{align*} T &= 6283.185 \, \text{s} \end{align*}\)

Convert to hours:

\(\begin{align*} T &= \frac{6283.185}{3600} \\ &\approx 1.746 \, \text{hours} \end{align*}\)

The time period of motion of the body is:

\(\begin{align*} \boxed{1.75 \, \text{hours}} \end{align*}\)

Calculation:

\(g=\frac{GM}{R^2} ​ \)

G is the gravitational constant,

M is the mass of the body,

R  is the radius of the body.

Given that the average density \( 𝜌 \) of the planet is the same as that of Earth, we have:

\(\rho=\frac{M}{\frac{4}{3}\pi R^3}\)

Since the planet has \(\frac{1}{8}\) of the mass of Earth, we express the mass of the planet as \(M_p=\frac{1}{8}M_e\)  where \(M_e\) is the mass of Earth.

Also, since the densities are the same, we can write:

\(\frac{M_p}{R_p^3}=\frac{M_e}{R_e^3}\)

This simplifies to:

\(M_p=M_e(\frac{R_p}{R_e})^3\) then;

\(\frac{1}{8}M_e=M_e(\frac{R_p}{R_e})^3\)

\((\frac{R_p}{R_e})^3=\frac{1}{8}\);

\((\frac{R_p}{R_e})=\frac{1}{2}\)

Now, the gravitational acceleration \(g_p\) for the planet is given by:

\(g_p=\frac{GM}{R_p^2}\)

\(g_p=\frac{G(\frac{1}{8}M_e)}{(\frac{1}{2}R_e)^2}=\frac{G\frac{1}{8}M_e}{\frac{1}{4}}R_e^2\)

\(g_p=\frac{1}{8}\times \frac{4}{1}\times \frac{GM_e}{R_e^2}=\frac{1}{2}g_e\)

\(\frac{g_p}{g_e}=\frac{1}{2}=0.5\)

Thus, the required ratio is 0.5.

Explanation:

The rate at which the raindrop's mass changes is given by:

\(\frac{dm(t)}{dt} = \lambda m(t)\)

This leads to the solution for the mass as a function of time:

\(m(t) = m_0 e^{\lambda t}\)

Now, consider the forces acting on the raindrop:

The gravitational force is \(F_{\text{gravity}} = m(t) g = m_0 e^{\lambda t} g\)

As the raindrop accelerates, its velocity increases, and since its mass is also increasing exponentially, the net acceleration is affected by both gravity and the changing mass. 

From Newton’s second law, the equation for the net force becomes: 

\(F = m(t) a(t) = m_0 e^{\lambda t} g\)

which simplifies to:

\(a(t) = g\)

This shows that the acceleration remains constant and equal to \(g\)  the acceleration due to gravity.

Now, to understand the speed over time:

The speed increases linearly with time initially because the acceleration is constant.

However, as time progresses and \(m(t) \) increases exponentially, the rate of increase in speed becomes less significant as the added mass requires more time for the raindrop to reach higher speeds.

For very large \(\lambda t\) the raindrop’s speed will approach a terminal value because, despite the increasing mass, the acceleration remains constant and doesn't continue to grow indefinitely. This behavior is captured by Option 3: **"The speed of the raindrop approaches a constant value when \(\lambda t \gg 1\)"

Thus, option '3' is correct.

Calculation:

For a planet in a circular orbit, the period of revolution $T$ is related to its orbital radius $r$ by Kepler's third law:

\(T^2\propto r^3\) then;

\(T\propto r^{\frac{3}{2}}\)

According to the question the distances are: a, 4a, and 9a.

\(T_1\propto (a)^{\frac{3}{2}}=T\);

\(T_2\propto (4a)^{\frac{3}{2}}=8T\)

\(T_3\propto (9a)^{\frac{3}{2}}=27T\)

The planets realign after a time equal to the LCM of the periods of the three planets. Since \(T_1=T\), \(T_2=8T\) and \(T_3=27T\) the LCM is:

\(T_1.T_2.T_3=T\times8T\times27T=216T\)

Thus, option '3' is correct.

Concept :
This problem deals with damped harmonic oscillations, specifically with a spring mass system experiencing damping. The motion is described by a second-order differential equation, with the solution depending on the nature of the damping (underdamped, critically damped, or overdamped). The damping coefficient b, spring constant k, and mass m influence the behavior of the system, including its natural frequency \(\omega_0\) and the damping ratio.

The differential equation for damped motion is:

\(\ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0\)

Where:
m is the mass,
b is the damping constant,
k is the spring constant.

Solution :

Given:
\(b = 40 \, \text{N s/m} ,\\ k = 80 \, \text{N/m} ,\\ m = 5 \, \text{kg} . \)1. Damping factor :
  \( \ r = \frac{b}{2m} = \frac{40}{2 \times 5} = 4\)
   

2. Natural frequency :
\( \ \omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{80}{5}}\) = 4
   

3. Equation of motion :
   The second order differential equation becomes:
\( \ \frac{d^2x}{dt^2} + 8 \frac{dx}{dt} + 16x = 0\)
   

4. Characteristic equation :
   The characteristic equation is:
 \( \ D^2 + 8D + 16 = 0\)
   
   Solving the quadratic equation:
\( \ D = -4\)
   
   Therefore, the general solution is:
\( \ x(t) = (C_1 + C_2 t)e^{-4t}\)
   
   This indicates critical damping , as the system returns to equilibrium without oscillations.

5. Velocity equation :
   The velocity is given by:
   \(\ \frac{dx}{dt} = C_1 e^{-4t} + C_2 te^{-4t} - 4(C_1 + C_2 t)e^{-4t}\)
   

6. Initial conditions :
   At t = 0 , x(0) = 0.5 gives:
  \( \ C_1 = 0.5\)
   
   At t = 0 , v(0) = 0 , applying this gives:
  \( \ C_2 = 2\)
 Thus, the final solution is:

\(\ x(t) = (0.5 + 2t) e^{-4t}\)

The system is critically damped because the damping is just enough to prevent oscillations while allowing the system to return to equilibrium as quickly as possible.

Correct options :
The correct answers are (3) and (4), as the system is critically damped and the given initial conditions are satisfied.

Concept:

Simple Harmonic Motion (SHM) is a type of oscillatory motion that is characterized by a restoring force that is directly proportional to the displacement from an equilibrium position and acts in the opposite direction. It’s a fundamental concept in physics and appears in various systems like pendulums, springs, and even molecules.

Here’s a quick breakdown of the key concepts:

Restoring Force : In SHM, the force that brings an object back to its equilibrium position is proportional to its displacement. Mathematically, this is often expressed as ( F = -kx ), where ( k ) is the force constant and ( x ) is the displacement from the equilibrium.

Explanation:

This picture is zoom-out picture. Let's displace the 1.2 m block downward by x distance . 

The equation of motion for 1.2m block is :

\(1.2m\ a=1.2mg-2Tsin\theta \quad \quad \quad (1)\\ \)

where T is the tension in string, \(T=mg\)

Then putting the value of tension in (1) we get,

\(1.2ma=1.2mg-2mgsin\theta \quad \quad \quad (1)\\ \quad \quad \ a=g-1.66\frac{g}{l}x\\ \quad \quad \ \ddot{x}=g-1.66\frac{g}{l}x\\ \)

By changing the variable x to y is :

Taking \(g-1.66\frac{g}{l}x=y\\ \ddot{x}=-\ddot{y} \frac{l}{1.66g}\)

this gives ,

\(\ddot{y}=-\frac{1.66g}{l}y\)

this is the equation of S.H.M.

The correct option is (3).

CONCEPT:

• Rotatory Motion: Rotational motion or circular motion, is a physical motion that happens when an object rotates or spins on an axis. Example: Rotation of a fan.
• Spin Motion: Motion of an object as it rotates around the axis through its center of mass. Example: Rotation of a top.
• Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. Example: a cannonball fired at some angle.
• Translatory motion: Motion with a fixed direction of
  the net force.All the particles of a body executing translatory motion move in the same direction traversing parallel paths. Example: the motion of bus or train.

EXPLANATION:

• The motion of an object is said to be translatory if the position of the object is changing with respect to a fixed point or object. So the motion of train is an example of translatory motion.

So option 4 is correct.