41. Which one of the following graphs represents the equation of motion v

Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ?

(a)
(b)
(c)
(d)
This question was previously asked in
UPSC NDA-2 – 2023
The equation of motion is given by v = u + at, where:
– v is the final velocity
– u is the initial velocity
– a is the acceleration (constant)
– t is the time
This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), ‘a’ is the slope (m), and ‘u’ is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start.
A graph of v versus t for this equation should be a straight line.
Since ‘u’ is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin.
Since ‘a’ is non-zero, the slope of the line will not be zero (the line is not horizontal).
Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While ‘a’ could be negative (negative slope) or ‘u’ could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero.
– The equation v = u + at represents linear motion with constant acceleration.
– A graph of velocity (v) versus time (t) for this equation is a straight line.
– The y-intercept of the line is the initial velocity (u).
– The slope of the line is the acceleration (a).
If ‘a’ were zero, the graph would be a horizontal line at v = u. If ‘u’ were zero, the line would pass through the origin (v = at). The shape of the line (upward or downward slope) depends on the sign of acceleration ‘a’. Option (c) shows a positive slope (acceleration).

42. A block of wood (dimensions : $40 \text{ cm} \times 20 \text{ cm} \tim

A block of wood (dimensions : $40 \text{ cm} \times 20 \text{ cm} \times 10 \text{ cm}$) is kept on a tabletop in three different positions : (a) with its side of dimensions $20 \text{ cm} \times 10 \text{ cm}$; (b) with its side of dimensions $10 \text{ cm} \times 40 \text{ cm}$; and (c) with its side of dimensions $40 \text{ cm} \times 20 \text{ cm}$. The pressure exerted by the wooden block on the tabletop in these positions is represented by $P_A$, $P_B$ and $P_C$ respectively. The pressure follows the trend

[amp_mcq option1=”$P_A > P_B > P_C$” option2=”$P_A < P_B < P_C$" option3="$P_A = P_B = P_C$" option4="$P_A < P_B = P_C$" correct="option1"]

This question was previously asked in
UPSC NDA-2 – 2023
The pressure follows the trend $P_A > P_B > P_C$.
Pressure is defined as force per unit area ($P = F/A$). The force exerted by the wooden block on the tabletop is its weight, which is constant regardless of its orientation. Therefore, the pressure exerted is inversely proportional to the area of contact. The smallest contact area will exert the highest pressure, and the largest contact area will exert the lowest pressure.
The dimensions are 40 cm x 20 cm x 10 cm. The areas of contact in the three positions are:
(a) $A_A = 20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2$
(b) $A_B = 10 \text{ cm} \times 40 \text{ cm} = 400 \text{ cm}^2$
(c) $A_C = 40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2$
Comparing the areas: $A_A < A_B < A_C$. Since $P \propto 1/A$, the corresponding pressures will be $P_A > P_B > P_C$.

43. An iron nail sinks in water whereas an iron ship floats. Which of the

An iron nail sinks in water whereas an iron ship floats. Which of the following statements is correct in this regard ?

  • 1. Average density of ship is greater than that of the water
  • 2. Average density of iron nail is greater than that of the water
  • 3. Average density of the ship is less than that of the water
  • 4. Average density of the ship is equal to that of the water

Select the correct answer using the code given below :

1 and 2
2 and 3
2 and 4
1 and 4
This question was previously asked in
UPSC NDA-2 – 2023
Statements 2 and 3 are correct.
An object sinks in water if its average density is greater than the density of water. An object floats if its average density is less than the density of water. An iron nail is a solid piece of iron, and the density of iron is greater than the density of water, so it sinks. An iron ship, despite being made of iron, is designed with a large hollow volume filled with air. The total mass of the ship divided by this large volume (its average density) is less than the density of water, allowing it to float.
Archimedes’ principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. An object floats when the buoyant force equals its weight. A ship displaces a volume of water whose weight is equal to the ship’s weight, and this displacement is possible because the ship’s average density is lower than water.

44. Ram records the odometer readings of his car for the distance covered

Ram records the odometer readings of his car for the distance covered from 2000 km at the start of his journey and 2400 km at the end of the journey after 8 hours. What is the average speed of the car?

50 km/h
60 km/h
70 km/h
80 km/h
This question was previously asked in
UPSC NDA-2 – 2023
The distance covered by the car is the difference between the final and initial odometer readings:
Distance = Final reading – Initial reading
Distance = 2400 km – 2000 km = 400 km
The time taken for the journey is given as 8 hours.
Average speed is calculated as Total Distance covered divided by the Total Time taken:
Average speed = Distance / Time
Average speed = 400 km / 8 hours
Average speed = 50 km/h
Average speed is calculated by dividing the total distance covered by the total time taken.
The odometer measures the total distance traveled by a vehicle. Average speed does not reflect the instantaneous speed at any given moment during the journey, which might fluctuate due to varying road conditions or driver actions.

45. A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1

A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1}$ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?

0·3 m s$^{-1}$
3 m s$^{-1}$
−3 m s$^{-1}$
−0·3 m s$^{-1}$
This question was previously asked in
UPSC NDA-2 – 2022
The recoil velocity of the pistol is -3 m/s.
– This problem can be solved using the principle of conservation of linear momentum.
– The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest.
– According to the conservation of momentum, the total momentum of the system after firing must also be zero.
– Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol.
– Momentum before firing = 0
– Momentum after firing = $m_b v_b + m_p v_p$
– By conservation of momentum: $m_b v_b + m_p v_p = 0$
– Given: $m_b = 10 \text{ g} = 0.01 \text{ kg}$, $v_b = 300 \text{ m/s}$, $m_p = 1 \text{ kg}$.
– Substituting the values: $(0.01 \text{ kg})(300 \text{ m/s}) + (1 \text{ kg}) v_p = 0$
– $3 \text{ kg} \cdot \text{m/s} + v_p \text{ kg} = 0$
– $v_p = -3 \text{ m/s}$
The negative sign for the recoil velocity indicates that the pistol moves in the opposite direction to the bullet. The magnitude of the recoil velocity is 3 m/s. This demonstrates Newton’s third law of motion (action-reaction) applied to momentum conservation.

46. A simple pendulum having bob of mass m and length of string l has time

A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be

T
T /√2
2T
√2 T
This question was previously asked in
UPSC NDA-2 – 2022
The correct option is B, stating that the new time period will be T/√2.
The time period (T) of a simple pendulum is given by the formula T = 2π√(l/g), where l is the length of the string and g is the acceleration due to gravity. This formula shows that the time period depends only on the length of the pendulum and the acceleration due to gravity, and it is independent of the mass of the bob.
In the given problem, the initial time period is T for a pendulum with mass m and length l. When the mass of the bob is doubled (to 2m) and the length of the string is halved (to l/2), the new time period T’ will be T’ = 2π√((l/2)/g) = 2π√(l/(2g)). We can rewrite this as T’ = 2π√(l/g) * (1/√2). Since T = 2π√(l/g), the new time period T’ is T/√2.
The independence of the time period from the mass of the bob is a key characteristic of a simple pendulum, provided the amplitude of oscillation is small (usually less than 15 degrees). This property allows pendulums to be used as reliable timekeeping devices. Real-world factors like air resistance and the rigidity of the string can affect the period slightly, but for an ideal simple pendulum, mass is irrelevant.

47. The volume of a sealed packet is 1 litre and its mass is 800 g. The pa

The volume of a sealed packet is 1 litre and its mass is 800 g. The packet is first put inside water with density 1 g cm^-3 and then in another liquid B with density 1.5 g cm^-3. Then which one of the following statements holds true?

The packet will float in both water and liquid B.
The packet will sink in both water and liquid B.
The packet will sink in water but will float in liquid B.
The packet will float in water and sink in liquid B.
This question was previously asked in
UPSC NDA-2 – 2022
The correct option is A, stating that the packet will float in both water and liquid B.
The principle of buoyancy states that an object will float in a fluid if its density is less than the density of the fluid. If its density is greater, it will sink. The packet has a volume of 1 litre (1000 cm³) and a mass of 800 g. Its density is mass/volume = 800 g / 1000 cm³ = 0.8 g/cm³. The density of water is 1 g/cm³. Since 0.8 g/cm³ < 1 g/cm³, the packet will float in water. The density of liquid B is 1.5 g/cm³. Since 0.8 g/cm³ < 1.5 g/cm³, the packet will also float in liquid B.
When an object floats, the buoyant force exerted by the fluid is equal to the weight of the object. The buoyant force is also equal to the weight of the fluid displaced by the object. If the object sinks, the buoyant force is less than the weight of the object. The proportion of the object submerged when floating is equal to the ratio of the object’s density to the fluid’s density. In water, 0.8/1 = 80% of the packet’s volume will be submerged. In liquid B, 0.8/1.5 ≈ 53.3% of the packet’s volume will be submerged.

48. Which one of the following statements about speed and velocity is

Which one of the following statements about speed and velocity is correct?

Speed and velocity both are vector quantities.
Speed and velocity both are scalar quantities.
Speed is vector quantity and velocity is scalar quantity.
Speed is scalar quantity and velocity is vector quantity.
This question was previously asked in
UPSC NDA-2 – 2022
Speed is a scalar quantity and velocity is a vector quantity.
– **Speed** is the magnitude of how fast an object is moving. It only has magnitude and no direction. Quantities with magnitude only are called scalar quantities.
– **Velocity** is the rate of change of displacement. It has both magnitude (speed) and direction. Quantities with both magnitude and direction are called vector quantities.
For example, saying a car is travelling at “50 km/h” describes its speed (a scalar). Saying a car is travelling at “50 km/h North” describes its velocity (a vector).

49. A negative work is done when an applied force F and the corresponding

A negative work is done when an applied force F and the corresponding displacement S are

perpendicular to each other.
parallel to each other.
anti-parallel to each other.
equal in magnitude.
This question was previously asked in
UPSC NDA-2 – 2021
A negative work is done when an applied force F and the corresponding displacement S are anti-parallel to each other.
The work done (W) by a constant force (F) when an object undergoes a displacement (S) is given by the dot product of the force and displacement vectors: $W = \vec{F} \cdot \vec{S} = |\vec{F}| |\vec{S}| \cos\theta$, where $\theta$ is the angle between the force and displacement vectors. Work done is positive when the force and displacement are in the same general direction ($0^\circ \le \theta < 90^\circ$), zero when they are perpendicular ($\theta = 90^\circ$), and negative when they are in opposite directions ($90^\circ < \theta \le 180^\circ$). When the force and displacement are anti-parallel ($\theta = 180^\circ$), $\cos\theta = -1$, and the work done is maximum negative: $W = -FS$.
Examples of negative work include the work done by friction when an object moves (friction acts opposite to motion), or the work done by gravity when an object is lifted upwards (gravity acts downwards while displacement is upwards).

50. Which one of the following is not a conservative force ?

Which one of the following is not a conservative force ?

Frictional force
Electric force
Gravitational force
Spring force
This question was previously asked in
UPSC NDA-2 – 2021
Frictional force is not a conservative force.
A conservative force is one for which the work done in moving an object between two points is independent of the path taken, or equivalently, the work done in moving an object around a closed loop is zero. Gravitational force, electric force, and spring force are examples of conservative forces because the work done by them depends only on the initial and final positions. Frictional force, however, is a dissipative force and the work done by friction depends on the path length; the longer the path, the more work is done against friction. Work done by friction over a closed loop is generally non-zero and negative (energy is dissipated as heat).
Forces like friction and air resistance are examples of non-conservative forces. The work done by non-conservative forces leads to a change in the mechanical energy of the system, often converting mechanical energy into other forms like heat or sound.