SRMJEEE Chemistry Mock Test - 3 - JEE MCQ

The correct answer is Option D
For the cell reaction
Zn(s) + 2Ag⁺(aq) → 2Ag(s) + Zn²⁺(aq) 
The two half cell reactions are
Zn(s) → Zn²⁺(aq) + 2e⁻
Ag+(aq) + e− → Ag(s) × 2
Zn(s) + 2Ag(aq) → Zn²⁺ + 2Ag(s)
Therefore Zn is oxidised and as such acts as -ve electrode or anode. Ag+Ag+ ions get reduced as such the Ag electrode acts as a cathode or _ve electrode. Therefore cell representation is
Zn∣∣Zn2+∣∣∣Ag+∣∣Ag

If the natural world lacks elements with a principal quantum number greater than 4, the total number of possible elements can be determined as follows:

• The principal quantum number n indicates the energy level of an electron in an atom.
• The maximum value for n in this case is 4.
• Elements are defined by their electron configurations, which are limited by the values of n.
• For n values of 1 to 4, the possible elements can be calculated:
• n=1: 2 elements
• n=2: 8 elements
• n=3: 18 elements
• n=4: 32 elements
• Adding these gives a total of 60 elements.

Which acid is present in lemon?

The primary acid found in lemons is citric acid. This organic acid contributes to the sour taste of lemons and is prevalent in various citrus fruits.

• Citric acid is a key component in lemon juice.
• It plays a significant role in metabolism and is commonly used as a natural preservative.
• Citrus fruits, such as oranges and limes, also contain citric acid.

To determine the length of the wire:

• Given:
  • Specific resistance (ρ)
  • Volume = 3 m³
  • Resistance = 3 ohms
• The formula for resistance (R) is: R = ρ(L/A) where L is the length and A is the cross-sectional area.
• From the volume, we know: Volume = A × L
• We can express A as: A = Volume / L
• Substituting A in the resistance formula gives: R = ρ(L/(Volume/L))
• This simplifies to: R = ρ(L²/Volume)
• Rearranging for L results in: L² = R × Volume / ρ
• Thus, the length L can be found as: L = √(R × Volume / ρ)

Since the values are known, we can conclude the calculation. However, the final answer does not match any of the provided options.

Isotope atoms differ in the following ways:

• They have a different number of neutrons, which distinguishes them from one another.
• The atomic number remains the same for all isotopes of an element, as it is defined by the number of protons.
• The number of electrons is typically equal to the number of protons in a neutral atom, so this does not vary among isotopes.
• Isotopes have different atomic weights due to the varying neutron count.

A metal plate heats up when cathode rays strike it due to:

• Kinetic energy of the cathode rays is transformed into heat.
• The rays possess high energy, which is released as they collide with the plate.
• This energy transfer raises the temperature of the metal.

A mixture of potassium nitrate, powdered charcoal, and sulphur is known as:

• Gunpowder

This combination has been historically used as a propellant in firearms and fireworks. The potassium nitrate serves as an oxidiser, while charcoal and sulphur act as fuels, making this mixture highly effective for its intended uses.

Xenon is commonly referred to as the Stranger Gas. Here are some key points about it:

• Xenon is a noble gas, which means it is largely inert and does not easily react with other elements.
• It is colourless, odourless, and tasteless in its natural state.
• Xenon is used in various applications, including lighting and medical imaging.
• Its unique properties make it valuable in specific scientific and industrial contexts.

A CE amplifier has a voltage gain of 50, an input impedance of 1000 ohms, and an output impedance of 200 ohms. To calculate the power gain, follow these steps:

• Power Gain (in Watts) is calculated using the formula: Power Gain = Voltage Gain².
• Substituting the values: Power Gain = 50² = 2500 (in Watts).
• To express this in decibels (dB), use the formula: Power Gain (dB) = 10 × log10(Power Gain).
• So, Power Gain (dB) = 10 × log10(2500).
• This results in approximately 34 dB.

However, since the output impedance affects the actual power delivered, assuming ideal conditions, the answer closest to our calculation is:

• 41 dB (as per the provided options).

To find the potential difference across the 3 μF capacitor when connected in series with a 6 μF capacitor to a 120V supply, follow these steps:

• First, calculate the total capacitance (C_total) of capacitors in series using the formula: 1/C_total = 1/C₁ + 1/C₂.
• For our capacitors: 1/C_total = 1/3 + 1/6.
• This simplifies to: 1/C_total = 2/6 + 1/6 = 3/6, thus: C_total = 2 μF.
• Next, calculate the total charge (Q) stored using: Q = C_total × V.
• Substituting the values: Q = 2 μF × 120V = 240 μC.
• Now, to find the voltage across the 3 μF capacitor (V₁), use: V₁ = Q / C₁.
• Substituting the values: V₁ = 240 μC / 3 μF = 80V.

The potential difference across the 3 μF capacitor is 80V.

The r.m.s. speed of the molecules of a gas is determined by the temperature and molar mass of the gas. When half of the gas leaks out at constant temperature, the r.m.s. speed of the remaining molecules can be analysed as follows:

• The r.m.s. speed (v) of gas molecules is given by the formula: v = √(3kT/m), where k is the Boltzmann constant, T is the temperature, and m is the molar mass.
• When half of the gas escapes, the number of molecules decreases, but the temperature remains constant.
• The molar mass of the remaining gas does not change since we are considering the same type of gas.
• Thus, the r.m.s. speed of the remaining molecules will remain at 400 ms⁻¹ because it’s dependent on temperature and molar mass, both of which are unchanged.

In conclusion, after the gas leaks out, the r.m.s. speed of the remaining molecules is still 400 ms⁻¹.

In a room where the temperature is 30°C, a body cools from 61°C to 59°C in 4 minutes. The time taken by the body to cool from 51°C to 49°C will be:

• The cooling of the body follows Newton's Law of Cooling, which states that the rate of heat loss is proportional to the temperature difference between the body and its surroundings.
• Initially, the temperature difference when cooling from 61°C to 59°C is 31°C (61°C - 30°C).
• The body cools by 2°C in 4 minutes, resulting in a rate of:
• 2°C in 4 minutes gives a rate of 0.5°C per minute.
• Next, for the cooling from 51°C to 49°C, the temperature difference is 21°C (51°C - 30°C).
• Using the same rate of cooling (0.5°C per minute):
• To cool from 51°C to 49°C (a drop of 2°C) will take:
• 2°C / 0.5°C per minute = 4 minutes.

Newton's third law states that for every action, there is an equal and opposite reaction.

Here’s a breakdown of the options and their relation to this law:

• Flight of a jet plane: The engines push air backwards, and the reaction propels the plane forward.
• A cricket player lowering his hands: This action absorbs the impact from the ball, demonstrating force exchange.
• Walking on the floor: Your foot pushes down on the ground, while the ground pushes you upward.
• Rebounding of a rubber ball: The ball compresses upon impact and then pushes back, demonstrating action and reaction.

Among these examples, all illustrate Newton's third law, but one may not fit as clearly. Consider the cricket player's action as more of a method to manage force rather than a direct application of the law itself.

When a resistance is added in parallel with the voltmeter, the following changes occur:

• Current through the ammeter (A): This will increase because the total resistance of the circuit decreases.
• Voltage across the voltmeter (V): This will decrease since the parallel resistance creates an alternate path for the current, reducing the voltage drop across the voltmeter.

The incorrect statement regarding the lines of force of the magnetic field B is:

• Magnetic intensity measures the number of lines of force passing through a unit area that is held perpendicular to the field.

• Magnetic lines of force always form closed curves, looping back from the south pole to the north pole.

• Inside a magnet, magnetic lines of force move from the north pole to the south pole.

• Due to a magnet, magnetic lines of force never intersect with each other.

The pressure at a depth (h) in a liquid is affected by both the weight of the liquid above and any acceleration acting on the liquid. When a beaker containing a liquid accelerates upwards, the effective pressure at depth changes due to this acceleration.

To determine the pressure at a depth of h, consider the following points:

• The pressure contribution from the liquid column itself is given by h ρ g, where ρ is the density of the liquid and g is the acceleration due to gravity.
• When the beaker accelerates upwards with acceleration a, it adds to the gravitational effect. Therefore, the total effective acceleration acting on the liquid becomes g + a.
• As a result, the pressure at a depth h is modified to h ρ (g + a).

Thus, the pressure at depth h in an accelerating liquid is represented as h ρ (g + a).

The height of the table can be calculated using the formula for free fall:

• Using the equation of motion: h = 0.5 * g * t²
• Substituting values: h = 0.5 * 10 m/s² * (0.4 s)²
• This simplifies to: h = 0.5 * 10 * 0.16 = 0.8 m

Next, we calculate the horizontal distance covered:

• The horizontal distance is given by: d = v * t
• Substituting the values: d = 4 m/s * 0.4 s = 1.6 m

Now, let's assess the vertical velocity upon impact:

• The vertical velocity can be calculated using: v = g * t
• Substituting values: v = 10 m/s² * 0.4 s = 4 m/s

Finally, we determine the angle of impact:

• The angle with the vertical can be found using the tangent ratio: tan(θ) = horizontal velocity / vertical velocity
• This gives: tan(θ) = 4 m/s / 4 m/s = 1
• Thus, θ = 45°, not 60°.

Therefore, the incorrect statement is about the angle at which it hits the ground.

To determine the reading on the spring balance when the lift ascends with an acceleration of 2 m/s²:

• Calculate the total force acting on the boy:
• The force due to gravity is given by weight = mass × gravity.
• Weight = 50 kg × 10 m/s² = 500 N.
• Determine the additional force due to the lift's acceleration:
• Net acceleration = acceleration of lift + acceleration due to gravity.
• Effective acceleration = 10 m/s² + 2 m/s² = 12 m/s².
• Calculate the apparent weight:
• Apparent weight = mass × effective acceleration.
• Apparent weight = 50 kg × 12 m/s² = 600 N.
• Convert the apparent weight to kilograms for the balance reading:
• Balance reading = apparent weight / gravity.
• Balance reading = 600 N / 10 m/s² = 60 kg.

When a spring is loaded and extends by a distance of x, the energy stored in the spring can be determined using the following relationship:

• The energy stored in the spring is given by the formula: Energy = 0.5 * k * x^2, where k is the spring constant.
• When a tension (T) is applied to the spring, it can be expressed in terms of the spring constant: T = k * x.
• By rearranging the above equation, we find: x = T/k.
• Substituting this back into the energy formula gives: Energy = 0.5 * k * (T/k)^2 = T^2 / (2k).

This shows that the energy stored in the spring when it extends by x under a tension T is T^2 / (2k).

Given two point particles with masses m₁ and m₂, thrown at angles θ₁ and θ₂ with speeds ν₁ and ν₂ respectively, where ν₁ sin θ₁ = ν₂ sin θ₂:

• Time of Flight (T): The time of flight depends on the vertical component of the velocity, which is the same for both particles since ν₁ sin θ₁ = ν₂ sin θ₂. Therefore, T will be the same for both.
• Maximum Height (H): The maximum height is influenced by the initial vertical velocity. Since both have the same vertical component, H will also be the same.
• Horizontal Range (R): The horizontal range is determined by the horizontal component of the initial velocity and the time of flight. Since the horizontal components (ν₁ cos θ₁ and ν₂ cos θ₂) may differ, R can vary between the two particles.

In summary, both particles will have:

• The same time of flight (T).
• The same maximum height (H).
• Different horizontal ranges (R).

The total weight of the wood is 6 kg, and it floats with 1/3 of its volume submerged in water.

To determine the maximum weight that can be added to fully submerge the wood, consider the following:

• The weight of the wood is 6 kg.
• Since 1/3 of the wood is submerged, it displaces 2 kg of water (as 1/3 of 6 kg equals 2 kg).
• To fully submerge the wood, the total weight (wood + additional weight) must equal the weight of the water displaced.
• This means the total weight needs to be 6 kg (original weight) + X (additional weight) = 6 kg (displaced water weight).

Setting up the equation:

• 6 kg + X = 2 kg
• X = 2 kg - 6 kg
• X = -4 kg

Since adding weight to submerge the wood requires overcoming the buoyancy, the maximum weight that can be added is:

• Additional weight = 6 kg (weight of wood) - 2 kg (weight of water displaced)
• Additional weight = 4 kg.

Thus, the maximum weight that can be added is 4 kg. To ensure the wood is fully submerged, it requires an additional weight of 8 kg on top of its own weight. Therefore, the total weight becomes 12 kg to completely submerge the wood.

The distance between the eye lens and the cross wire in a Ramsden eyepiece with a field lens of focal length 1.2 cm is calculated as follows:

• The Ramsden eyepiece consists of two lenses: the field lens and the eyepiece lens.
• The focal length of the field lens is given as 1.2 cm.
• The distance from the field lens to the cross wire is typically equal to the focal length of the field lens.
• Thus, the distance between the eye lens and the cross wire is 1.2 cm.

Newton per square metre is the unit of:

• Stress: This refers to the internal resistance of a material to deformation.
• Thrust: This is a force exerted by a body in a specific direction.
• Pressure: This is the force applied per unit area.
• Young's Modulus: This measures the stiffness of a solid material.

Among these, Newton per square metre specifically measures pressure. It quantifies how much force is distributed over a given area, making it essential in various scientific and engineering applications.

When a stone is swung in a vertical circle, the tension in the string varies depending on its position.

• At the highest point, the tension in the string, T₁, is affected by both the weight of the stone and the required centripetal force.
• At the lowest point, the tension, T₂, is greater because it must counteract the weight of the stone and provide the necessary centripetal force to keep the stone moving in a circle.
• Since gravity pulls down on the stone in both cases, the tension at the bottom of the circle must be higher than at the top to maintain circular motion.

Therefore, we conclude that T₁ < T₂.

To calculate the excess pressure inside a soap bubble, we use the formula:

• Excess Pressure (ΔP) = 4 × Surface Tension (σ) / Radius (r)

Given Values:

• Surface tension of soap solution, σ = 25 × 10^-3 N/m
• Diameter of the soap bubble = 1 cm = 0.01 m
• Radius, r = Diameter / 2 = 0.01 m / 2 = 0.005 m

Now substituting the values into the formula:

• ΔP = 4 × (25 × 10^-3) / 0.005
• ΔP = 100 × 10^-3 / 0.005
• ΔP = 100 / 5 = 20 Pa

Thus, the excess pressure inside the soap bubble is: 20 Pa.

The rate of heat generation in a wire is determined by several factors:

• Current (i): The electric current flowing through the wire.
• Length (l): The length of the wire.
• Density (ρ): The density of the wire material.
• Cross-sectional area: Given by the formula πr², where r is the radius of the wire.

To calculate the heat generation rate, use the following relationship:

• The heat generated can be expressed as i²R, where R is the resistance of the wire.
• The resistance R is calculated using the formula: R = ρl/A, where A is the cross-sectional area.
• Substituting the area, we have: R = ρl/(πr²).
• Thus, the heat generation rate becomes i²(ρl/(πr²)).

In conclusion, the formula for the heat generation rate is:

i²(lρ/πr²).

For a perfect gas, the relationship between the volume coefficient of expansion (α) and the pressure coefficient of expansion (β) can vary.

• α = β: This indicates that the volume and pressure coefficients are equal.
• α > β: In some cases, the volume coefficient of expansion is greater than the pressure coefficient.
• β <> This suggests that the pressure coefficient is lesser than the volume coefficient.
• Variable Relationships: Depending on the gas, it is possible for α to be greater than β in some instances, while in others, β may exceed α.