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AIIMS Full Mock Test 1 - NEET MCQ


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30 Questions MCQ Test - AIIMS Full Mock Test 1

AIIMS Full Mock Test 1 for NEET 2024 is part of NEET preparation. The AIIMS Full Mock Test 1 questions and answers have been prepared according to the NEET exam syllabus.The AIIMS Full Mock Test 1 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AIIMS Full Mock Test 1 below.
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AIIMS Full Mock Test 1 - Question 1
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): Electrical resistivity of typical insulator (≃ 1014Ω-m) is about 1022 times that of metal (≃ 10-8Ω-m) while in case of heat it is only about 103 (as KAg ≃ 400 W/mk while Kglass ≃ 1 W/mk)
Reason(R): In electrical conduction only free electrons take part while in thermal conduction both lattice and free electrons contribute.

Detailed Solution for AIIMS Full Mock Test 1 - Question 1

A is the correct option.In electrical conduction only free electrons take part while in thermal conduction both lattice and free electrons contribute and this is the reason that leads to the phenomenon of Electrical resistivity of typical insulator (≃ 1014Ω-m) is about 1022 times that of metal (≃ 10-8Ω-m) while in case of heat it is only about 103 (as KAg ≃ 400 W/mk while Kglass ≃ 1 W/mk).

AIIMS Full Mock Test 1 - Question 2
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Voltage in the secondary coil of a transformer does not depend upon.

Detailed Solution for AIIMS Full Mock Test 1 - Question 2
As, V1/V2 = N1/N2;
so. V2 = (V1 x N2)/N1
Hence we can conclude voltage in secondary coil i.e V2 is independent of Frequency of the source but depends on Voltage in the primary coil and Ratio of number of turns in the two coils.
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AIIMS Full Mock Test 1 - Question 3
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The closest distance of approach of an α - particle travelling with a velocity 'V' to a certain nucleus is 'x'. The distance of closest approach of α - particle travelling with a velocity 3V to the same nucleus is

Detailed Solution for AIIMS Full Mock Test 1 - Question 3
Solution:
To find the distance of closest approach of an α-particle travelling with a velocity 3V to the same nucleus, we can use the concept of the impact parameter.
The impact parameter is defined as the perpendicular distance between the path of the particle and the center of the nucleus at the closest point of approach.
Let's assume the distance of closest approach of the α-particle travelling with velocity V to the nucleus is x.
Now, when the velocity is increased to 3V, the impact parameter can be calculated using the formula:
b' = (3V/V) * x
Simplifying the equation, we get:
b' = 3x
Therefore, the distance of closest approach of the α-particle travelling with a velocity 3V to the same nucleus is 3x.
Hence, the correct answer is option A: 3x.
AIIMS Full Mock Test 1 - Question 4
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A 10 μF capacitor and a 20 μF capacitor are connected in series across 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor?
Detailed Solution for AIIMS Full Mock Test 1 - Question 4

C1 = 10μF, C2 = 20μF
Fquivalent capacitance in series

Charge on each capacitor 
q = CV

In seies charge in both capacitor is same.
∴ toal charge 
Now the capacitors are connected in parallel.
Equivalent capacitance in parallel. 

Potential 

AIIMS Full Mock Test 1 - Question 5
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If the de Broglie wavelengths for a proton and for a α-particle are equal, then the ratio of their velocities will be
Detailed Solution for AIIMS Full Mock Test 1 - Question 5
H/ 4mv2=h/ mv1; V1:V2= 4:1
AIIMS Full Mock Test 1 - Question 6
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Two solid rubber balls A and B having masses 200 and 400 gm respectively are moving in opposite directions with velocity A equal to 0.3 m/s. After collision the two balls come to rest, then the velocity of B is
Detailed Solution for AIIMS Full Mock Test 1 - Question 6
M1= 200gm or 0.2kg, v1= 0.3m/s m2=400gm or 0.4 kg v2=? m1v1=m2v2 because after collision two ball are stop so collision is elastic and it's not depends on mass(action reactions rule) and it's then possible when two ball moving in opposite direction 0.2×0.3=0.4×(-v2) 0.06/0.4=- v2 v2=-0.15 ans.
AIIMS Full Mock Test 1 - Question 7
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If intensity of incident light is increased in PEE then which of the following is true
Detailed Solution for AIIMS Full Mock Test 1 - Question 7
Solution:
To understand the effect of increasing the intensity of incident light in the photoelectric effect (PEE), let's consider the following points:
1. Photoelectric Effect:
The photoelectric effect is the phenomenon where electrons are ejected from a material when it is exposed to light of sufficient energy.
2. Work Function:
The work function (represented by the symbol Φ) is the minimum amount of energy required to remove an electron from the surface of a material. It is a characteristic property of the material.
3. Stopping Potential:
The stopping potential is the minimum potential difference required to stop the flow of photoelectrons from reaching the anode in a photoelectric setup.
Now, let's analyze the given options:
A: Maximum K.E. of ejected electron will increase:
Increasing the intensity of incident light does not change the energy of individual photons. Therefore, the maximum kinetic energy (K.E.) of the ejected electrons, which is determined by the energy of the incident photons and the work function of the material, will remain unchanged.
B: Work function will remain unchanged:
The work function of a material is a characteristic property and does not depend on the intensity of incident light. Therefore, increasing the intensity of incident light does not change the work function.
C: Stopping potential will decrease:
Increasing the intensity of incident light increases the number of photons incident on the material, resulting in more photoelectrons being emitted. As a result, the current in the photoelectric setup increases. To stop this increased current, a higher stopping potential is required. Therefore, increasing the intensity of incident light will increase the stopping potential.
D: Maximum K.E. of ejected electron will decrease:
As mentioned earlier, increasing the intensity of incident light does not change the energy of individual photons or the work function of the material. Therefore, the maximum kinetic energy (K.E.) of the ejected electrons will not decrease.
Hence, the correct answer is B: Work function will remain unchanged.
AIIMS Full Mock Test 1 - Question 8
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An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is
Detailed Solution for AIIMS Full Mock Test 1 - Question 8
Solution:
To find the kinetic energy of the electron, we need to use the de Broglie wavelength equation:
λ = h / p
Where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum.
We know that the de Broglie wavelength is the same for both the electron and the proton. Therefore, we can write:
λe = λp
Now, let's express the momentum in terms of the mass (m) and velocity (v) for both the electron and the proton:
For the electron:
pe = me * ve
For the proton:
pp = mp * vp
Since the de Broglie wavelength is the same for both particles, we have:
me * ve = mp * vp
Now, let's solve for the velocity of the electron in terms of the velocity of the proton:
ve = (mp / me) * vp
The kinetic energy (KE) of a particle can be calculated using the formula:
KE = (1/2) * m * v^2
Substituting the expressions for velocity of the electron and the proton, we have:
KEe = (1/2) * me * ((mp / me) * vp)^2
Simplifying the equation, we get:
KEe = (1/2) * (mp^2 / me) * vp^2
Since mp is greater than me, (mp^2 / me) is greater than 1. Therefore, the kinetic energy of the electron (KEe) is greater than the kinetic energy of the proton (KEp).
Therefore, the correct answer is option D: Greater than the kinetic energy of the proton.
AIIMS Full Mock Test 1 - Question 9
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You are given several identical resistances each of value R = 10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this job
Detailed Solution for AIIMS Full Mock Test 1 - Question 9
Solution:
To find the minimum number of resistances required, we need to consider the formula for combining resistances in parallel and in series.
Formula for resistances in series:
When resistances are connected in series, the total resistance is the sum of individual resistances.
R_total = R1 + R2 + R3 + ... + Rn
Formula for resistances in parallel:
When resistances are connected in parallel, the reciprocal of the total resistance is the sum of reciprocals of individual resistances.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
Given that we have identical resistances of value R = 10 Ω, we can use these formulas to find the minimum number of resistances required.
Step 1: Calculate the total resistance required using the formula for resistances in parallel.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
1/5 Ω = 1/10 Ω + 1/10 Ω + 1/10 Ω + ... + 1/10 Ω
1/5 Ω = n/10 Ω
n = 2
Step 2: Calculate the total current required using the formula for resistances in series.
I_total = I1 + I2 + I3 + ... + In
4 A = 1 A + 1 A + 1 A + 1 A
4 A = n A
n = 4
Therefore, the minimum number of resistances required is 4 (Option A).
AIIMS Full Mock Test 1 - Question 10
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According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus frequency of the incident radiation gives a straight line, whose slope
Detailed Solution for AIIMS Full Mock Test 1 - Question 10
Einstein's Photoelectric Equation and the Plot of Kinetic Energy:
According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line. The slope of this line provides important information about the photoelectric effect.
Explanation:
1. Photoelectric Effect: The photoelectric effect refers to the emission of electrons from a metal surface when it is exposed to light or electromagnetic radiation.
2. Einstein's Photoelectric Equation: Einstein's photoelectric equation explains the relationship between the kinetic energy (KE) of the emitted photoelectrons and the frequency (f) of the incident radiation. The equation is given by KE = hf - φ, where h is Planck's constant and φ is the work function of the metal.
3. Plot of Kinetic Energy versus Frequency: When the kinetic energy of the emitted photoelectrons is plotted against the frequency of the incident radiation, a straight line is obtained.
4. Slope of the Line: The slope of the straight line obtained from the plot is directly related to the value of Planck's constant (h). This means that the slope is the same for all metals and independent of the intensity of the radiation.
5. Implications: The fact that the slope is the same for all metals and independent of the intensity of the radiation indicates that the photoelectric effect is a quantum phenomenon. It suggests that the energy of the incident radiation is transferred to individual electrons, rather than being absorbed collectively by the entire metal.
Conclusion:
According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line. The slope of this line is the same for all metals and independent of the intensity of the radiation. This provides evidence for the quantum nature of the photoelectric effect.
AIIMS Full Mock Test 1 - Question 11
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The working of a dynamo is based on principle of
AIIMS Full Mock Test 1 - Question 12
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A 100 mH coil carries a current of 10 A. The magnetic energy stored in the coil is
Detailed Solution for AIIMS Full Mock Test 1 - Question 12
Solution:
Given:
- Inductance of the coil (L) = 100 mH = 0.1 H
- Current through the coil (I) = 10 A
To find:
- Magnetic energy stored in the coil
Formula:
The magnetic energy stored in an inductor can be calculated using the formula:
Energy (W) = 0.5 * L * I^2
Calculation:
Substituting the given values into the formula, we get:
Energy (W) = 0.5 * 0.1 H * (10 A)^2
= 0.5 * 0.1 H * 100 A^2
= 0.05 * 100 J
= 5 J
Therefore, the magnetic energy stored in the coil is 5 J.
Answer:
The correct option is D:

5 J

AIIMS Full Mock Test 1 - Question 13
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A coil of inductances 300mH and resistance 2 Ω is connected to a source of voltage 2V. The current reaches half of its steady state value in
Detailed Solution for AIIMS Full Mock Test 1 - Question 13


AIIMS Full Mock Test 1 - Question 14
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Which of the following statement is true?
Detailed Solution for AIIMS Full Mock Test 1 - Question 14
Velocity of light in vacuum is maximum is 3×10^8 and it's highest velocity.
AIIMS Full Mock Test 1 - Question 15
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A straight wire of length l and electric dipole moment p is bent to from a semicircle. The new electric dipole moment would be
Detailed Solution for AIIMS Full Mock Test 1 - Question 15
Initial dipole moment, P = q × Lwhen it is bent in the form of a semicircle,L = πRR = L/πNow the charges are present at diametrically opposite points of the semicircleDistance between the charges = 2R = 2L/πNew dipole moment, P' = q × 2R = q × (2L/π) = 2p/π.
AIIMS Full Mock Test 1 - Question 16
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A cylinder of radius r and length a is placed in a uniform in a uniform electric field E parallel to the axis of the cylinder. The total electric flux over the curved surface of cylinder is
Detailed Solution for AIIMS Full Mock Test 1 - Question 16

The electric flux entering and leaving the cylindrical surface is same
Φ = EπR2 - EπR2 = 0
where, E = electric field
R = radius of cylinder
Φ = outward flux

AIIMS Full Mock Test 1 - Question 17
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In an L-R circuit, the value of L is (0.4/π) henry and the value of R is 30 ohm. If in the circuit, an alternating emf of 200 volt at 50 cycles per second is connected, the impedance of the circuit and current will be
Detailed Solution for AIIMS Full Mock Test 1 - Question 17



AIIMS Full Mock Test 1 - Question 18
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A car travelling at a speed of 30 km/hr is brought to a halt in 8 m by applying brakes. If the same car is travelling at 60 km/hr, it can be brought to a halt with the same braking force in
Detailed Solution for AIIMS Full Mock Test 1 - Question 18
Yes option D is correct answer...solution..v^2=u^2-2as0=u^2-2asu^2 is directionally proportional to s(u1÷u2)^2=s1÷s2since;u1=30 and u2=60 and s1=8 and s2=?30×30/60×60=8/s21/4=8/s2so;s2=32m..
AIIMS Full Mock Test 1 - Question 19
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The enegy stored in wound watch spring is
Detailed Solution for AIIMS Full Mock Test 1 - Question 19
Explanation:
The energy stored in a wound watch spring is potential energy (PE). Here's a detailed explanation:
Potential Energy:
- Potential energy is the energy stored in an object due to its position or condition.
- It is the energy that an object possesses because of its potential to do work.
- Potential energy can be converted into other forms of energy, such as kinetic energy (KE), when the object moves or changes its position.
Wound Watch Spring:
- A watch spring is a tightly wound coil that stores potential energy.
- When the watch is wound, the spring is compressed and stores potential energy.
- This potential energy is then gradually released as the watch spring unwinds, providing the necessary energy to power the watch's movement.
Other Options:
- K.E (Kinetic Energy): Kinetic energy is the energy an object possesses due to its motion. In the case of a wound watch spring, the energy is stored and not in motion, so it is not kinetic energy.
- Heat Energy: Heat energy is the energy associated with the random motion of particles within a substance. Although some heat may be generated due to friction within the watch mechanism, the primary energy stored in the wound watch spring is potential energy, not heat energy.
- Chemical Energy: Chemical energy is the energy stored in the bonds of chemical compounds. While chemical energy may be used in batteries or other power sources for watches, it is not directly associated with the energy stored in a wound watch spring.
Conclusion:
The correct answer is P.E (Potential Energy), as the energy stored in a wound watch spring is primarily potential energy.
AIIMS Full Mock Test 1 - Question 20
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The rate of transfer of heat is more in
Detailed Solution for AIIMS Full Mock Test 1 - Question 20
Radiation is electromagnetic wave traveling at highest possible speed. Example is sun ray coming through space nd falling on earth. It takes 8.5min approx to cover the distance nd governed by radiation.
AIIMS Full Mock Test 1 - Question 21
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Time taken by a 846 W heater to heat one litre of water from 10ºC to 40ºC is
Detailed Solution for AIIMS Full Mock Test 1 - Question 21

AIIMS Full Mock Test 1 - Question 22
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Two similar bar magnets P and Q each of magnetic moment M are taken. If P is cut along its axial line and Q is cut along its equatorial line, all the four pieces obtained have each of
Detailed Solution for AIIMS Full Mock Test 1 - Question 22
Explanation:

In order to answer this question, we need to understand the concept of magnetic moment and the effect of cutting a magnet.


Magnetic Moment:


Magnetic moment is a property of a magnet that determines the strength of its magnetic field.


It is given by the formula: Magnetic Moment (M) = pole strength (m) × distance between the poles (d)


Effect of Cutting a Magnet:


When a magnet is cut along its axial line, it forms two smaller magnets, each with a magnetic moment equal to half of the original magnet.


When a magnet is cut along its equatorial line, it also forms two smaller magnets, but the magnetic moment of each smaller magnet remains the same as the original magnet.


Analysis:


In this question, we have two similar bar magnets, P and Q, each with a magnetic moment M.


If P is cut along its axial line, it will form two smaller magnets, each with a magnetic moment of M/2.


If Q is cut along its equatorial line, it will also form two smaller magnets, but the magnetic moment of each smaller magnet will remain M.


Conclusion:


From the analysis above, we can conclude that all the four pieces obtained from cutting magnets P and Q will have a magnetic moment of M/2.


Therefore, the correct answer is option C: Magnetic moment M/2.

AIIMS Full Mock Test 1 - Question 23
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A bar magnet, of magnetic moment M, is placed in a magnetic field of induction B. The torque exerted on it is
Detailed Solution for AIIMS Full Mock Test 1 - Question 23
Solution:
Given:
- Magnetic moment of the bar magnet: M
- Magnetic field induction: B
Formula:
- Torque (τ) exerted on a magnetic dipole in a magnetic field is given by the formula: τ = M x B
Explanation:
- The torque exerted on the bar magnet is given by the cross product of its magnetic moment and the magnetic field induction.
- The cross product of two vectors is a vector perpendicular to both of those vectors.
- The direction of the torque vector is determined by the right-hand rule, where the thumb represents the direction of the cross product.
- In this case, the torque vector is perpendicular to both the magnetic moment vector (M) and the magnetic field induction vector (B).
- So, the torque exerted on the bar magnet is in the direction perpendicular to the plane containing M and B.
Answer:
The torque exerted on the bar magnet is M̅ x B̅ (Option C).
AIIMS Full Mock Test 1 - Question 24
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Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio of
Detailed Solution for AIIMS Full Mock Test 1 - Question 24

AIIMS Full Mock Test 1 - Question 25
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The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration 'a' is 3:2. The value of 'a' is (g-Acceleration due to gravity of the earth)
Detailed Solution for AIIMS Full Mock Test 1 - Question 25
Solution:
Let's assume the weight of the man in the stationary lift is W1 and the weight of the man when the lift is moving downward with uniform acceleration 'a' is W2.
Given that the ratio of W1 to W2 is 3:2, we can write the equation as:
W1/W2 = 3/2
Now, let's calculate the weight of the man in both cases:
When the lift is stationary:
W1 = mg --------(1) (where m is the mass of the man and g is the acceleration due to gravity)
When the lift is moving downward with uniform acceleration 'a':
W2 = m(g - a) ---------(2)
Substitute equation (1) and (2) into the ratio equation:
mg / m(g - a) = 3/2
Cancel out the common terms 'm' and simplify the equation:
g / (g - a) = 3/2
Cross multiply:
2g = 3(g - a)
2g = 3g - 3a
3a = g
a = g/3
Therefore, the value of 'a' is g/3.
So, the correct answer is option B: g/3.
AIIMS Full Mock Test 1 - Question 26
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Half life of a radioactive substance which disintegrates by 75% in 60 minutes will be
Detailed Solution for AIIMS Full Mock Test 1 - Question 26
Yes option B is correct answer....because 75 percentage of the substance is disintegrate by 2 half lives..2half lives=60÷t1t1=60÷2=30therefore: t=30 minutes...
AIIMS Full Mock Test 1 - Question 27
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If a spring is extended to length l, then according to Hooke's law
Detailed Solution for AIIMS Full Mock Test 1 - Question 27
The Correct option is A. Hooke's law states that the force(F) needed to extend or compress a spring by some distance l sacked linearly with respect to that distant. That is F= kl. Here the -ve sign indicates the spring restoring force is opposite to the stretch direction.
AIIMS Full Mock Test 1 - Question 28
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): We use a thick wire in the secondary of a step down transformer to reduce the production of heat.
Reason(R): When the plane of the armature is parallel to the line of force of magnetic field, the magnitude of the induced emf is maximum.
Detailed Solution for AIIMS Full Mock Test 1 - Question 28
Assertion: We use a thick wire in the secondary of a step down transformer to reduce the production of heat.
Reason: When the plane of the armature is parallel to the line of force of magnetic field, the magnitude of the induced emf is maximum.
Solution:
The given statement discusses the use of a thick wire in the secondary of a step down transformer to reduce the production of heat. Let's analyze the assertion and reason provided:
1. Assertion: We use a thick wire in the secondary of a step down transformer to reduce the production of heat.
- This statement implies that using a thick wire in the secondary winding of a step down transformer helps in reducing heat production.
- A thick wire has lower resistance compared to a thin wire, which leads to less power loss in the form of heat due to lower Joule heating.
- Therefore, the assertion is true.
2. Reason: When the plane of the armature is parallel to the line of force of the magnetic field, the magnitude of the induced emf is maximum.
- This statement explains the relationship between the orientation of the armature and the induced electromotive force (emf).
- When the plane of the armature (coil) is parallel to the magnetic field lines, the flux linkage is maximum, resulting in the maximum induced emf.
- Therefore, the reason is true.
Based on the above analysis, we can conclude that:
- Both the assertion and reason are true.
- However, the reason provided is not the correct explanation of the assertion.
- The thickness of the wire in the secondary winding of a step down transformer is primarily related to reducing heat production, while the reason discusses the maximum induced emf in a different context.
Hence, the correct answer is Option B: Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
AIIMS Full Mock Test 1 - Question 29
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): Efficiency of an engine with sink temperature of zero Kelvin is 100%.
Reason(R): Keeping the sink at ice point with the source at 100oC will bring 100% efficiency.
Detailed Solution for AIIMS Full Mock Test 1 - Question 29
Explanation:
The correct answer is option C: Assertion is true but Reason is false.
Assertion (A): Efficiency of an engine with sink temperature of zero Kelvin is 100%.
Reason (R): Keeping the sink at ice point with the source at 100°C will bring 100% efficiency.
To understand the correctness of the assertion and reason, let's analyze each statement separately:
Assertion (A): Efficiency of an engine with sink temperature of zero Kelvin is 100%.
- This statement is incorrect because according to the Carnot efficiency formula, the efficiency of a heat engine is given by (1 - Tc/Th), where Tc is the temperature of the sink and Th is the temperature of the source. If the sink temperature is zero Kelvin, the efficiency would be (1 - 0/Th), which is 100%. However, this is not practically achievable as reaching absolute zero temperature is impossible.
Reason (R): Keeping the sink at ice point with the source at 100°C will bring 100% efficiency.
- This statement is incorrect because the efficiency of a heat engine cannot be 100% when the sink temperature is at the ice point (0°C). In order to achieve maximum efficiency, the sink temperature should be as low as possible (approaching absolute zero) and the source temperature should be as high as possible. The closer the sink temperature is to absolute zero, the higher the efficiency.
Therefore, the correct answer is option C: Assertion is true but Reason is false.
AIIMS Full Mock Test 1 - Question 30
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): Potential energy is defined for only conservative forces.
Reason(R):

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