Mechanics

141. The motion of a car along a straight path is shown by the following fi

The motion of a car along a straight path is shown by the following figure :
[Figure shows a line starting from O (0km) and points A (25km), B (35km), C (60km) along the path]
The car starts from O and reaches at A, B and C at different instants of time. During its motion from O to C and back to B, the distance covered and the magnitude of the displacement are, respectively

[amp_mcq option1=”25 km and 60 km” option2=”95 km and 35 km” option3=”60 km and 25 km” option4=”85 km and 35 km” correct=”option4″]

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The correct answer is D) 85 km and 35 km.
– The car starts from O (0 km). It goes to C (60 km) and then comes back to B (35 km).
– Distance is the total path length covered. The car travels from O to C (60 km) and then from C back to B (60 km – 35 km = 25 km). Total distance covered = 60 km + 25 km = 85 km.
– Displacement is the straight-line distance and direction from the initial position to the final position. Initial position is O (0 km). Final position is B (35 km). The displacement is B – O = 35 km – 0 km = 35 km in the positive direction. The magnitude of displacement is 35 km.
Distance is a scalar quantity representing the total length of the path traveled. Displacement is a vector quantity representing the change in position from the starting point to the ending point. Its magnitude is the shortest distance between the initial and final points, and its direction is from initial to final.

142. The impulse on a particle due to a force acting on it during a given t

The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its

[amp_mcq option1=”force” option2=”momentum” option3=”work done” option4=”energy” correct=”option2″]

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The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its momentum.
This is a fundamental principle in physics known as the Impulse-Momentum Theorem. Impulse (J) is defined as the product of the force (F) and the time interval (Δt) over which the force acts (J = FΔt). Momentum (p) is the product of mass (m) and velocity (v) (p = mv). The theorem states that the impulse applied to an object is equal to the change in its momentum: J = Δp = mv_f – mv_i, where v_f is the final velocity and v_i is the initial velocity.
Impulse is a vector quantity, having the same direction as the force. The unit of impulse is Newton-second (N·s), which is equivalent to the unit of momentum, kilogram-meter per second (kg·m/s). The Impulse-Momentum Theorem is derived from Newton’s second law of motion (F = ma = m * Δv/Δt) rearranged as FΔt = mΔv.

143. A racing car accelerates on a straight road from rest to a speed of 50

A racing car accelerates on a straight road from rest to a speed of 50 m/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be

[amp_mcq option1=”625 m” option2=”1250 m” option3=”2500 m” option4=”50 m” correct=”option1″]

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The distance covered by the racing car is 625 m.
We are given the initial velocity (u = 0 m/s, since it starts from rest), final velocity (v = 50 m/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50/25 = 2 m/s². Then, find the distance (s) using s = ut + (1/2)at²: s = (0 * 25) + (1/2) * 2 * (25)² = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)/2 = (0+50)/2 = 25 m/s. Distance = Average velocity * time = 25 m/s * 25 s = 625 m.
This problem involves basic kinematics under constant acceleration. The relevant equations of motion are v = u + at, s = ut + (1/2)at², and v² = u² + 2as. Any of these can be used depending on the given and required variables.

144. Two bodies A and B are moving with equal velocities. The mass of B is

Two bodies A and B are moving with equal velocities. The mass of B is double that of A. In this context, which one of the following statements is correct?

[amp_mcq option1=”Momentum of B will be double that of A.” option2=”Momentum of A will be double that of B.” option3=”Momentum of B will be four times that of A.” option4=”Momenta of both A and B will be equal.” correct=”option1″]

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The correct answer is A) Momentum of B will be double that of A.
Momentum (p) of an object is defined as the product of its mass (m) and velocity (v): p = m * v. Given that both bodies A and B are moving with equal velocities (v_A = v_B = v) and the mass of B is double that of A (m_B = 2 * m_A).
Calculating the momentum for each body:
Momentum of A: p_A = m_A * v_A = m_A * v
Momentum of B: p_B = m_B * v_B = (2 * m_A) * v = 2 * (m_A * v)
Comparing p_B with p_A, we find that p_B = 2 * p_A. Therefore, the momentum of B will be double that of A.

145. Mass of a particular amount of substance 1. is the amount of matter

Mass of a particular amount of substance

  • 1. is the amount of matter present in it.
  • 2. does not vary from place to place.
  • 3. changes with change in gravitational force.

Select the correct answer using the code given below :

[amp_mcq option1=”1, 2 and 3″ option2=”1 and 2 only” option3=”2 and 3 only” option4=”1 only” correct=”option2″]

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Let’s analyze each statement about the mass of a particular amount of substance:
1. is the amount of matter present in it: This is the fundamental definition of mass. Mass is a measure of the inertia of an object, which is directly related to the quantity of matter it contains. This statement is true.
2. does not vary from place to place: Mass is an intrinsic property of an object. Unlike weight, which depends on the gravitational acceleration, mass remains constant regardless of the location (e.g., on Earth, the Moon, or in space). This statement is true.
3. changes with change in gravitational force: This statement is false. Weight (W) changes with gravitational force (g) according to the formula W = m * g. As gravity changes, weight changes, but mass (m) itself remains constant. For example, an object weighs less on the Moon than on Earth because the Moon’s gravity is weaker, but its mass is the same in both locations.
Mass is a measure of the amount of matter and is constant for a given object, independent of gravity. Weight is the force of gravity acting on an object’s mass and varies with gravitational acceleration.
Mass is measured in units like kilograms (kg) or grams (g). Weight is a force and is measured in units like Newtons (N) or pounds (lb).

146. Suppose the force of gravitation between two equal masses is F. If eac

Suppose the force of gravitation between two equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become

[amp_mcq option1=”F” option2=”2 F” option3=”4 F” option4=”” correct=”option3″]

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According to Newton’s Law of Universal Gravitation, the force of gravitation (F) between two masses (m₁ and m₂) separated by a distance (r) is given by the formula:
F = G * (m₁ * m₂) / r²
where G is the gravitational constant.
Initially, we have two equal masses, let’s call them m. So, m₁ = m and m₂ = m. The force is F = G * (m * m) / r² = G * m² / r².
Now, each mass is doubled, so the new masses are m₁’ = 2m and m₂’ = 2m. The distance of separation (r) remains unchanged.
The new force (F’) is:
F’ = G * (m₁’ * m₂’) / r²
F’ = G * (2m * 2m) / r²
F’ = G * (4 * m²) / r²
We can rewrite this as:
F’ = 4 * (G * m² / r²)
Since F = G * m² / r², we have F’ = 4 * F.
The force becomes four times the original force.
The gravitational force between two masses is directly proportional to the product of their masses. If both masses are doubled, their product becomes (2m) * (2m) = 4m², quadrupling the force, assuming the distance remains constant.
The gravitational force is also inversely proportional to the square of the distance between the centres of the masses. If the distance were, for example, doubled instead of the masses, the force would become F/4.

147. A container is first filled with water and then the entire water is re

A container is first filled with water and then the entire water is replaced by mercury. Mercury has a density of 13·6 × 10³ kg/m³. If X is the weight of the water and Y is the weight of the mercury, then

[amp_mcq option1=”X = Y” option2=”X = 13·6 Y” option3=”Y = 13·6 X” option4=”None of the above” correct=”option3″]

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The weight of an object is given by W = m * g, where m is the mass and g is the acceleration due to gravity. The mass of a substance filling a container of volume V is given by m = ρ * V, where ρ is the density of the substance.
For water, the weight X is given by X = ρwater * V * g.
For mercury, the weight Y is given by Y = ρmercury * V * g.
The density of water is approximately 1.0 × 10³ kg/m³.
The density of mercury is given as 13.6 × 10³ kg/m³.
Taking the ratio of the weights:
Y / X = (ρmercury * V * g) / (ρwater * V * g)
Y / X = ρmercury / ρwater
Y / X = (13.6 × 10³) / (1.0 × 10³) = 13.6
Therefore, Y = 13.6 X.
Weight depends on mass and gravitational acceleration. Mass is proportional to density for a fixed volume. The ratio of weights for the same volume of different substances is equal to the ratio of their densities.
The density of water varies slightly with temperature and pressure, but 1000 kg/m³ is a standard value at 4°C. The acceleration due to gravity (g) is the same for both substances in the same location, so it cancels out in the ratio.

148. The radius of the Moon is about one-fourth that of the Earth and accel

The radius of the Moon is about one-fourth that of the Earth and acceleration due to gravity on the Moon is about one-sixth that on the Earth. From this, we can conclude that the ratio of the mass of Earth to the mass of the Moon is about

[amp_mcq option1=”10″ option2=”100″ option3=”1,000″ option4=”10,000″ correct=”option2″]

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The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM/R², where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm ≈ Re/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm ≈ ge/6.
From g = GM/R², we can write M = gR²/G.
The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is:
Me/Mm = (ge * Re² / G) / (gm * Rm² / G)
Me/Mm = (ge/gm) * (Re/Rm)²
Substitute the given ratios: ge/gm ≈ 6 and Re/Rm ≈ 4.
Me/Mm ≈ 6 * (4)² = 6 * 16 = 96.
Among the options, 96 is closest to 100.
The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM/R²).
The actual ratio of Earth’s mass to Moon’s mass is approximately 81.3, so the given approximate values lead to a result (96) that is closest to 100, indicating the question uses approximate figures typical for simplified calculations.

149. A spring can be used to determine the mass m of an object in two ways

A spring can be used to determine the mass m of an object in two ways : (i) by measuring the extension in the spring due to the object; and (ii) by measuring the oscillation period for the given mass. Which of these methods can be used in a space-station orbiting Earth ?

[amp_mcq option1=”Both” option2=”Only the extension method” option3=”Only the oscillation method” option4=”Neither” correct=”option3″]

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Method (i) involves measuring the extension of a spring caused by the weight of the object (F = mg, where F is the force causing extension, m is mass, and g is acceleration due to gravity). In a space station orbiting Earth, objects are in a state of apparent weightlessness because they are in free fall. The effect of gravity is effectively absent for this type of measurement, so there would be no significant extension of the spring due to the object’s mass in the absence of gravitational force acting downwards. Method (ii) involves measuring the oscillation period of a mass attached to a spring. The period of oscillation (T) of a simple harmonic oscillator like a mass-spring system is given by T = 2π√(m/k), where m is the mass and k is the spring constant. This formula depends on mass (m) and spring constant (k), but not on gravity (g). Therefore, the oscillation method can be used to determine mass in a weightless environment like an orbiting space station.
Measuring extension relies on weight (force due to gravity), which is effectively zero in orbit. Measuring oscillation period relies on inertial mass and spring constant, which are unaffected by gravity.
The oscillation method is actually used on the International Space Station (ISS) with a device called the Body Mass Measurement Device (BMMD) to measure the mass of astronauts. The BMMD is essentially a chair attached to springs that oscillates, and the oscillation period is used to calculate the mass. This demonstrates the practicality of the oscillation method for mass determination in a microgravity environment.

150. When the sun is 30° above the horizon, shadow of one tree is 17·3 m lo

When the sun is 30° above the horizon, shadow of one tree is 17·3 m long. What is the height of this tree ?

[amp_mcq option1=”20 m” option2=”17·30 m” option3=”10 m” option4=”1·73 m” correct=”option3″]

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This problem can be solved using trigonometry, specifically the tangent function in a right-angled triangle.
The tree stands vertically, forming a right angle with the ground. The sun’s rays create a shadow on the ground. The angle of elevation of the sun (30°) is the angle between the ground (shadow) and the line from the end of the shadow to the top of the tree. Let H be the height of the tree (opposite side) and L be the length of the shadow (adjacent side). We have the relationship: tan(angle) = Opposite / Adjacent. So, tan(30°) = H / L.
Given L = 17.3 m and tan(30°) = 1/$\sqrt{3}$.
H = L * tan(30°) = 17.3 m * (1/$\sqrt{3}$).
The value of $\sqrt{3}$ is approximately 1.732.
H $\approx$ 17.3 / 1.732.
Notice that 17.3 is very close to 10 * 1.73. If we assume the shadow length is precisely $10\sqrt{3}$ meters (which is approximately 17.32 m), then:
H = $(10\sqrt{3}) \times (1/\sqrt{3}) = 10$ meters.
Given the options, 10 m is the most likely intended answer, implying the shadow length 17.3 m was an approximation for $10\sqrt{3}$ m.
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