Physics Archives

211. 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"]

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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 $P_A > P_B > P_C$.

212. The motion of a particle of mass m is described by the relation, $y =

The motion of a particle of mass m is described by the relation, $y = ut – \frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is

[amp_mcq option1=”$F = m\left(\frac{du}{dt}\right)$” option2=”$F = mg$” option3=”$F = m\left(\frac{dy}{dt}\right)$” option4=”$F = -mg$” correct=”option4″]

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The force acting on the particle is $F = -mg$.

The given equation of motion, $y = ut – \frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\frac{dy}{dt} = u – gt$. The second derivative gives acceleration: $\frac{d^2y}{dt^2} = -g$. According to Newton’s second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.

The equation $y = ut – \frac{1}{2}gt^2$ is the standard kinematic equation for vertical displacement under constant gravitational acceleration ($g$), with initial velocity $u$. The force responsible for this motion is the gravitational force, which is $mg$ acting downwards. Assuming the upward direction as positive y, the downward force is represented as $-mg$.

213. What will happen if a collection of positive and negative charges are

What will happen if a collection of positive and negative charges are passed at a high speed through a magnetic field which is perpendicular to the direction of motion of the charges ? (Assume that both kind of charges are NOT going to recombine)

[amp_mcq option1=”Both kind of charges will stop moving” option2=”Positive charges and negative charges will separate out” option3=”Positive charges will stop but negative charges will continue moving uninterrupted” option4=”Both kind of charges will keep moving uninterrupted” correct=”option2″]

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When a collection of positive and negative charges are passed at high speed through a magnetic field perpendicular to their motion, the positive and negative charges will separate out.

The Lorentz force ($\vec{F} = q(\vec{v} \times \vec{B})$) acts on a moving charge in a magnetic field. The force is perpendicular to both the velocity ($\vec{v}$) and the magnetic field ($\vec{B}$). For positive charges, the force direction is given by the right-hand rule applied to $\vec{v} \times \vec{B}$. For negative charges (where $q$ is negative), the force is in the opposite direction. This difference in the direction of the force causes the positive and negative charges to be deflected in opposite directions, leading to separation.

This principle is utilized in devices like mass spectrometers or velocity selectors to separate charged particles based on their mass-to-charge ratio or velocity. The charges will move in curved paths, not stop or continue uninterrupted, as long as they are within the magnetic field.

214. 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 :

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

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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.

215. A liquid is heated up to a certain temperature. Which one of the follo

A liquid is heated up to a certain temperature. Which one of the following situation would correspond to the boiling of the liquid?

[amp_mcq option1=”When atmospheric pressure becomes equal to the vapour pressure” option2=”When atmospheric pressure becomes less than vapour pressure” option3=”When atmospheric pressure becomes higher than the vapour pressure” option4=”When vapour pressure becomes equal to the air pressure” correct=”option1″]

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A) When atmospheric pressure becomes equal to the vapour pressure corresponds to the boiling of the liquid.

– Boiling is the phase transition where a liquid turns into a gas when heated to its boiling point.
– The boiling point of a liquid is defined as the temperature at which its vapor pressure equals the surrounding environmental pressure (usually atmospheric pressure).
– At this point, the vapor pressure is sufficient to overcome the external pressure, allowing bubbles of vapor to form within the bulk of the liquid and rise to the surface.

– If atmospheric pressure is less than the vapor pressure at a given temperature, boiling occurs below the standard boiling point (e.g., at high altitudes).
– If atmospheric pressure is higher than the vapor pressure, boiling will not occur at that temperature; the liquid must be heated further to increase its vapor pressure to match the higher external pressure.
– Option D essentially states the same condition as option A, assuming “air pressure” refers to the external pressure, which is typically atmospheric pressure in an open system. Option A uses the more standard terminology.

216. Which of the following statements are correct about the electrical res

Which of the following statements are correct about the electrical resistance and resistivity of a wire?

1. Both quantities depend on the area of cross-section of the wire
2. Both depend on the temperature
3. Resistance of the wire is directly proportional to the resistivity of the wire
4. Resistivity of the wire is directly proportional to the length of the wire

Select the correct answer using the code given below :

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

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C) 2 and 3 are the correct statements.

– The resistance (R) of a wire is given by the formula R = ρ * (L/A), where ρ is the resistivity of the material, L is the length of the wire, and A is the area of its cross-section.
– Resistivity (ρ) is an intrinsic property of the material itself, depending on the material type and temperature, but not on the dimensions (length or area) of the specific wire.

– Statement 1: Resistance depends on the area of cross-section (inversely proportional, R ∝ 1/A), but resistivity does NOT depend on the area of cross-section. So, statement 1 is incorrect.
– Statement 2: Both resistance and resistivity of most metallic conductors are dependent on temperature. For metals, both generally increase with increasing temperature. So, statement 2 is correct.
– Statement 3: From the formula R = ρ * (L/A), Resistance (R) is directly proportional to resistivity (ρ) for a wire of fixed length and area. So, statement 3 is correct.
– Statement 4: Resistivity (ρ) is a material property and does not depend on the length of the wire. So, statement 4 is incorrect.
– Thus, only statements 2 and 3 are correct.

217. Ramesh cannot see distinctly objects kept beyond 2 m. This defect can

Ramesh cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power

[amp_mcq option1=”+ 0.5 D” option2=”- 0.5 D” option3=”+ 0.2 D” option4=”- 0.2 D” correct=”option2″]

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B) – 0.5 D is the correct power of the lens required to correct the defect.

– The person cannot see objects distinctly beyond 2 m, meaning their far point is 2 m. This is a condition of myopia (nearsightedness).
– In myopia, distant objects (effectively at infinity) are focused in front of the retina.
– A diverging lens (concave lens) is used to correct myopia, as it diverges incoming light rays slightly before they reach the eye lens, causing them to focus further back on the retina.
– Concave lenses have negative power.
– The corrective lens should form a virtual image of an object at infinity (u = ∞) at the person’s far point (v = -2 m, as the image is virtual and on the same side as the object).

– Using the lens formula, 1/f = 1/v – 1/u:
– 1/f = 1/(-2) – 1/∞
– 1/f = -1/2 – 0
– 1/f = -1/2
– The power of the lens P is given by P = 1/f (in meters).
– P = 1/(-2 m) = -0.5 Diopters (D).
– Therefore, a concave lens of power -0.5 D is needed.

218. Ms. Rani decides to convert her AC generator into a DC generator. Whic

Ms. Rani decides to convert her AC generator into a DC generator. Which one of the following she would need to use?

[amp_mcq option1=”A split-ring type commutator” option2=”Slip rings and brushes” option3=”A stronger magnetic field” option4=”A rectangular wire loop” correct=”option1″]

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A) A split-ring type commutator is the correct component needed to convert an AC generator into a DC generator.

– An AC generator produces alternating current due to the use of slip rings, which maintain continuous contact with the rotating coil.
– To convert this to direct current (DC), the direction of the current in the external circuit needs to be maintained constant.
– A split-ring commutator achieves this by reversing the connection of the coil to the external circuit every half rotation, ensuring that the current flows in the same direction in the external load.

– Slip rings are used in AC generators to connect the armature coil to the external circuit.
– A stronger magnetic field would increase the magnitude of the induced electromotive force (EMF) and current, but would not change the nature of the output from AC to DC.
– A rectangular wire loop is the armature coil, which is a component of both AC and DC generators. The type of output depends on how the coil is connected to the external circuit (via slip rings for AC, or a commutator for DC).

219. 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?

[amp_mcq option1=”50 km/h” option2=”60 km/h” option3=”70 km/h” option4=”80 km/h” correct=”option1″]

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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.

220. Which one of the following graphs correctly represents the current (I)

Which one of the following graphs correctly represents the current (I) – voltage (V) variation for a rectangular piece of a semiconductor wafer ?

[amp_mcq option1=”Graph (a)” option2=”Graph (b)” option3=”Graph (c)” option4=”Graph (d)” correct=”option1″]

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Graph (a) correctly represents the current (I) – voltage (V) variation for a rectangular piece of a semiconductor wafer under conditions where it behaves ohmically.

While semiconductors are known for non-linear behavior in devices like diodes and transistors, a simple rectangular piece of bulk semiconductor material with ohmic contacts behaves like a resistor. According to Ohm’s Law, the current (I) flowing through a resistor is directly proportional to the applied voltage (V) (V = IR or I = V/R), provided the temperature is constant. This relationship is represented by a straight line passing through the origin on an I-V graph. Graph (a) shows this linear relationship, indicating constant resistance.

Graphs (b), (c), and (d) represent non-linear current-voltage characteristics typical of semiconductor devices (like diodes or transistors) or bulk materials under specific conditions (e.g., high electric fields causing velocity saturation). However, for a basic piece of material operating within its ohmic region, the linear relationship is the expected behavior.