Mechanics Archives

151. A deep sea diver may hurt his ear drum during diving because of

A deep sea diver may hurt his ear drum during diving because of

[amp_mcq option1=”lack of oxygen” option2=”high atmospheric pressure” option3=”high water pressure” option4=”All of the above” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

Pressure in water increases significantly with depth due to the weight of the water column above. A deep sea diver experiences very high external pressure from the surrounding water. The middle ear is normally filled with air at approximately atmospheric pressure (adjusted through the Eustachian tube). If the diver does not equalize the pressure in their middle ear with the external water pressure, the large pressure difference across the ear drum can cause it to bulge inward and potentially rupture.

Hydrostatic pressure in a fluid increases linearly with depth. The significant pressure difference between the external water pressure and the internal ear pressure is the cause of potential ear drum damage for divers.

Divers are trained in techniques (like Valsalva maneuver) to equalize the pressure in their middle ears with the external pressure, preventing barotrauma (pressure injury).

152. A person stands on his two feet over a surface and experiences a press

A person stands on his two feet over a surface and experiences a pressure P. Now the person stands on only one foot. He would experience a pressure of magnitude

[amp_mcq option1=”4 P” option2=”P” option3=”$\frac{1}{2}$ P” option4=”2 P” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

Pressure is defined as force per unit area (P = F/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W / A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} \approx A_{two}/2$), the new pressure is $P’ = W / A_{one} \approx W / (A_{two}/2) = 2W/A_{two} = 2P$. Thus, the pressure is doubled.

Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.

This is a simple application of the definition of pressure. By reducing the contact area by half (from two feet to one foot), the pressure exerted on the surface doubles for the same force (weight).

153. Which one among the following happens when a swing rises to a certain

Which one among the following happens when a swing rises to a certain height from its rest position ?

[amp_mcq option1=”Its potential energy decreases while kinetic energy increases” option2=”Its kinetic energy decreases while potential energy increases” option3=”Both potential and kinetic energy decrease” option4=”Both potential and kinetic energy increase” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The correct answer is Its kinetic energy decreases while potential energy increases. This describes the energy transformation as an object moves against gravity and slows down.

At the rest position (lowest point), the swing has its maximum speed and thus maximum kinetic energy (assuming negligible energy loss). As it rises, its height above the lowest point increases, causing its gravitational potential energy (PE = mgh) to increase. Simultaneously, the swing slows down as it moves upwards against gravity, causing its kinetic energy (KE = 1/2 mv^2) to decrease.

According to the principle of conservation of mechanical energy (in the absence of friction and air resistance), the total mechanical energy (sum of potential and kinetic energy) remains constant. Thus, any decrease in kinetic energy is compensated by an equal increase in potential energy, and vice versa.

154. The displacement-time graph of a particle acted upon by a constant for

The displacement-time graph of a particle acted upon by a constant force is

[amp_mcq option1=”a straight line” option2=”a circle” option3=”a parabola” option4=”any curve depending upon initial conditions” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The correct option is C. The displacement-time graph of a particle acted upon by a constant force is a parabola.

According to Newton’s second law, a constant force (F) acting on a particle of constant mass (m) results in constant acceleration (a), where F = ma.
For motion under constant acceleration, the displacement (s) of a particle from its initial position as a function of time (t) is given by the kinematic equation: s = ut + (1/2)at², where u is the initial velocity and a is the constant acceleration.
This equation is a quadratic function of time (t).

The general form of the displacement-time graph for constant acceleration is a parabola. If the initial velocity is zero (u=0), the equation simplifies to s = (1/2)at², which is clearly a parabola passing through the origin. If there is an initial velocity, the parabola is shifted and possibly rotated, but remains a parabola. A straight line graph represents constant velocity (zero acceleration, thus zero net force).

155. If radius of the earth were to shrink by 1%, its mass remaining the sa

If radius of the earth were to shrink by 1%, its mass remaining the same, g would decrease by nearly

[amp_mcq option1=”1%” option2=”2%” option3=”3%” option4=”4%” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The correct option is B. g would increase by nearly 2%. (Note: The question contains a likely typo stating “decrease” instead of “increase” or “change”).

The acceleration due to gravity (g) on the surface of a sphere of mass M and radius R is given by g = GM/R², where G is the gravitational constant.
If the radius R shrinks by 1%, the new radius R’ = R – 0.01R = 0.99R. The mass M remains the same.
The new gravity g’ on the surface would be g’ = GM/(R’)² = GM/(0.99R)² = GM/(0.9801R²) = (1/0.9801) * (GM/R²) ≈ 1.01928 * g.
The change in gravity is g’ – g = 1.01928g – g = 0.01928g.
The percentage change in gravity is ((g’ – g)/g) * 100% = (0.01928g / g) * 100% ≈ 1.93%.
Using the approximation for small changes, if R changes by a small fraction ε (R’ = R(1-ε)), then g changes by approximately 2ε (g’ ≈ g(1+2ε)). Here ε = 0.01 (1% decrease in R). So g changes by approximately 2 * 0.01 = 0.02, which is a 2% increase.

A decrease in Earth’s radius while keeping the mass constant would result in a stronger gravitational pull at the surface, hence an *increase* in ‘g’. The question states “g would decrease by nearly”, which contradicts the physics. Assuming the question intends to ask about the magnitude of the percentage change calculated from the given change in radius, the change is approximately 2% (an increase). Option B (2%) is the closest value to the calculated change. It is highly probable that “decrease” is a typo.

156. A particle moves in a circle of radius 4 m. Its linear speed is given

A particle moves in a circle of radius 4 m. Its linear speed is given by 4√3 t, where t is the time measured in seconds. At t = 1 s, the angle made by the resultant acceleration – r̂ with direction (+ r̂ is the radial direction) is given by φ. Which one among the following is the correct value of φ ?

[amp_mcq option1=”tan⁻¹ (1/√3)” option2=”tan⁻¹ (1/√2)” option3=”tan⁻¹ (2/√3)” option4=”tan⁻¹ (√3)” correct=”option1″]

This question was previously asked in

UPSC Geoscientist – 2024

Download PDF Attempt Online

In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry.

The linear speed is v = 4√3 t. The tangential acceleration is a_t = dv/dt = d(4√3 t)/dt = 4√3 m/s². At t = 1 s, v = 4√3 m/s. The radial acceleration is a_r = v²/r = (4√3)² / 4 = (16*3) / 4 = 48/4 = 12 m/s². The radial direction +r̂ points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle φ between the resultant acceleration and the radial direction (+r̂, pointing inward) satisfies tan(φ) = a_t / a_r.

tan(φ) = (4√3) / 12 = √3 / 3 = 1/√3. Therefore, φ = tan⁻¹(1/√3).

157. A ball of mass 100 g falls freely from a height of h = 20 m and hits t

A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1·4 √gh. Take g = 10 m s⁻². Which one among the following is the correct value of the work done on the ball by air friction ?

[amp_mcq option1=”0·4 J” option2=”0·5 J” option3=”0·3 J” option4=”1 J” correct=”option1″]

This question was previously asked in

UPSC Geoscientist – 2024

Download PDF Attempt Online

The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction.

Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m/s²)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = ½ m v_f², where v_f = 1.4√gh. With g=10 m/s² and h=20m, v_f = 1.4√(10*20) = 1.4√200. So, v_f² = (1.4)² * 200 = 1.96 * 200 = 392 m²/s². KE_f = ½ (0.1 kg)(392 m²/s²) = 0.05 * 392 = 19.6 J.

According to the Work-Energy Theorem, W_total = W_g + W_air = ΔKE = KE_f – KE_i. 20 J + W_air = 19.6 J – 0 J. W_air = 19.6 J – 20 J = -0.4 J. The question asks for the value, and options are positive, implying magnitude. The magnitude of the work done by air friction is |-0.4 J| = 0.4 J.

158. A planar, irregular shaped object has a mass of 2 kg. Its moment of in

A planar, irregular shaped object has a mass of 2 kg. Its moment of inertia along the two orthogonal axes in the plane of the object are 3 kg m² and 4 kg m², respectively. The moment of inertia of the object along an axis perpendicular to the plane and passing through a point 0·5 m away from its centre of mass is represented by I. Which one among the following is the correct value of I ?

[amp_mcq option1=”7·5 kg m²” option2=”7 kg m²” option3=”5 kg m²” option4=”1 kg m²” correct=”option1″]

This question was previously asked in

UPSC Geoscientist – 2024

Download PDF Attempt Online

For a planar object, the moment of inertia about an axis perpendicular to the plane and passing through a point is related to the moments of inertia about two perpendicular axes in the plane by the perpendicular axis theorem and the parallel axis theorem.

By the perpendicular axis theorem, the moment of inertia about the center of mass (assuming the given orthogonal axes intersect at CM) and perpendicular to the plane is I_CM = Ix + Iy = 3 kg m² + 4 kg m² = 7 kg m². By the parallel axis theorem, the moment of inertia I about a parallel axis perpendicular to the plane and at a distance d = 0.5 m from CM is I = I_CM + m d².

Given mass m = 2 kg and distance d = 0.5 m. I = 7 kg m² + (2 kg)(0.5 m)² = 7 + 2(0.25) = 7 + 0.5 = 7.5 kg m².

159. A frictionless pulley has a rope of negligible mass passing over it. B

A frictionless pulley has a rope of negligible mass passing over it. Blocks of mass 15 kg and 5 kg are attached at the two ends of the rope and held stationary. When the masses are released to move freely, their speed in m s^-1, after the rope has slipped by 0.25 m over the pulley, is closest to :
(take g = 10.0 m s^-2)

[amp_mcq option1=”5.0″ option2=”2.2″ option3=”1.6″ option4=”1.2″ correct=”option3″]

This question was previously asked in

UPSC Geoscientist – 2024

Download PDF Attempt Online

Correct Answer: C

– This problem can be solved using the conservation of mechanical energy, as the pulley is frictionless and the rope has negligible mass (ideal Atwood machine).
– Let the initial height of the 15 kg mass be $y_1$ and the 5 kg mass be $y_2$. When the rope slips by 0.25 m, the 15 kg mass moves down by 0.25 m (new height $y_1 – 0.25$) and the 5 kg mass moves up by 0.25 m (new height $y_2 + 0.25$).
– The initial potential energy (PE) can be set relative to the initial positions. $\text{PE}_{initial} = m_1 g y_1 + m_2 g y_2$.
– The final potential energy is $\text{PE}_{final} = m_1 g (y_1 – 0.25) + m_2 g (y_2 + 0.25) = m_1 g y_1 – 0.25 m_1 g + m_2 g y_2 + 0.25 m_2 g$.
– The change in potential energy is $\Delta \text{PE} = \text{PE}_{final} – \text{PE}_{initial} = -0.25 m_1 g + 0.25 m_2 g = 0.25 g (m_2 – m_1)$.
– Given $m_1 = 15$ kg, $m_2 = 5$ kg, and $g = 10.0$ m/s², $\Delta \text{PE} = 0.25 \times 10 \times (5 – 15) = 2.5 \times (-10) = -25$ J. The system loses 25 J of potential energy.
– The initial kinetic energy (KE) is zero as the masses are held stationary. $\text{KE}_{initial} = 0$.
– Let $v$ be the speed of the masses after moving 0.25 m. Both masses move with the same speed. The final kinetic energy is $\text{KE}_{final} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 = \frac{1}{2} (m_1 + m_2) v^2$.
– $\text{KE}_{final} = \frac{1}{2} (15 + 5) v^2 = \frac{1}{2} (20) v^2 = 10 v^2$.
– By the conservation of mechanical energy, $\Delta \text{KE} + \Delta \text{PE} = 0$.
– $(10 v^2 – 0) + (-25) = 0$.
– $10 v^2 = 25$.
– $v^2 = \frac{25}{10} = 2.5$.
– $v = \sqrt{2.5}$ m/s.
– Calculating the value: $\sqrt{2.5} \approx 1.581$ m/s.
– Comparing with the options, 1.6 m/s is the closest value.

The motion of the masses in an Atwood machine is a classic physics problem demonstrating Newton’s laws or conservation of energy. Assuming an ideal setup (frictionless pulley, massless rope), energy is conserved, meaning the decrease in potential energy of the heavier mass going down is converted into the increase in potential energy of the lighter mass going up and the kinetic energy of both masses.

160. In which of the following situations will an applied force do negative

In which of the following situations will an applied force do negative work on a body?

[amp_mcq option1=”The applied force and displacement of the body are at 135° to each other” option2=”The applied force and displacement of the body are parallel to each other” option3=”The applied force and displacement of the body are perpendicular to each other” option4=”The applied force and displacement of the body are at 45° to each other” correct=”option1″]

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The correct answer is A) The applied force and displacement of the body are at 135° to each other.

Work done by a constant force $\vec{F}$ on a body that undergoes a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos \theta$, where $\theta$ is the angle between the force and displacement vectors.
Work is negative when $\cos \theta$ is negative. This occurs when the angle $\theta$ is between 90° and 270° (exclusive of 90° and 270° where work is zero).
Let’s examine the options:
A) Angle is 135°. $\cos 135° = -1/\sqrt{2}$. Since $\cos \theta$ is negative, the work done is negative.
B) Angle is 0° (same direction) or 180° (opposite direction). If $\theta=0°$, $\cos 0° = 1$, work is positive. If $\theta=180°$, $\cos 180° = -1$, work is negative. So, parallel force and displacement *can* result in negative work (if opposite directions), but doesn’t *always*.
C) Angle is 90°. $\cos 90° = 0$. Work is zero.
D) Angle is 45°. $\cos 45° = 1/\sqrt{2}$. Since $\cos \theta$ is positive, work is positive.
Option A is the only situation listed that guarantees the work done by the applied force is negative.

Examples of negative work include the work done by friction when an object slides, the work done by air resistance on a moving object, or the work done by gravity when an object is lifted upwards. In these cases, the force (friction, air resistance, gravity) is generally opposite in direction to the displacement (motion), corresponding to an angle of 180° or a component of the force opposing the displacement direction.