All Exams >
JEE >
National Level Test Series for JEE Advanced 2025 >
All Questions
All questions of JEE Advanced 2016 for JEE Exam
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 100C. Now the end P is maintained at 100C, while the end S is heated and maintained at 4000C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 x 10-5 K-1, the change in length of the wire PQ is- a)0.78 mm
- b)0.90 mm
- c)1.56 mm
- d)2.34 mm
Correct answer is option 'A'. Can you explain this answer?
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 100C. Now the end P is maintained at 100C, while the end S is heated and maintained at 4000C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 x 10-5 K-1, the change in length of the wire PQ is
a)
0.78 mm
b)
0.90 mm
c)
1.56 mm
d)
2.34 mm
|
Krishna Iyer answered |
Extension in a small element of length dx is
An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
a)
b)
c)
d)
|
Yash Patel answered |
For infinite line,
Current through an elemental shell;
This current is radially outwards so;
Current through an elemental shell;
This current is radially outwards so;
A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/e = 1.237×10–6 eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is
Correct answer is '6'. Can you explain this answer?
A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/e = 1.237×10–6 eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is
|
|
Krishna Iyer answered |
Note : Unit of hc/e in original paper is incorrect Energy available 
(4th energy level of hydrogen atom). Hence, the number of lines present in the emission spectrum = 4C2 = 6
(4th energy level of hydrogen atom). Hence, the number of lines present in the emission spectrum = 4C2 = 6Among [Ni(CO)4], [NiCl4]2–, [Co(NH3)4Cl2]Cl, Na3[CoF6], Na2O2 and CsO2, the total number of
paramagnetic compounds is- a)2
- b)3
- c)4
- d)5
Correct answer is option 'B'. Can you explain this answer?
Among [Ni(CO)4], [NiCl4]2–, [Co(NH
paramagnetic compounds is
3
)4Cl2]Cl, Na3[CoF6], Na2
O2
and CsO2
, the total number ofparamagnetic compounds is
a)
2
b)
3
c)
4
d)
5
|
Bhavana Dey answered |
Number of paramagnetic compounds are 3.
Following compounds are paramagnetic.
Following compounds are paramagnetic.
Q. 14 to 18 carry 3 marks eachThe answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.Q.A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2 (P/P0), where P0 is a constant. When the metal surface is at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 °C?- a)9
- b)4
- c)6
- d)7
Correct answer is option 'A'. Can you explain this answer?
Q. 14 to 18 carry 3 marks each
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.
Q.
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2 (P/P0), where P0 is a constant. When the metal surface is at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 °C?
a)
9
b)
4
c)
6
d)
7
|
Anaya Patel answered |
A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratio less than 1, the possible mode(s) of decay is(are)- a)β- -decay ( β emission)
- b)orbital or K-electron capture
- c)neutron emission
- d)β+ -decay (positron emission)
Correct answer is option 'B,D'. Can you explain this answer?
A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratio less than 1, the possible mode(s) of decay is(are)
a)
β- -decay ( β emission)
b)
orbital or K-electron capture
c)
neutron emission
d)
β+ -decay (positron emission)
|
Priya Bose answered |
This upward deviation from linearity in the plot of the number of neutrons (N) against the number of protons (P) of stable nuclei for atomic number, Z, is known as the "neutron excess" or "neutron drip line."
As the atomic number increases, the number of protons in the nucleus also increases. In stable nuclei, the number of protons and neutrons is balanced, with the number of neutrons roughly equal to or slightly greater than the number of protons. However, as the atomic number increases further, the number of neutrons needed to maintain stability increases at a faster rate than the number of protons.
This is due to the strong repulsive forces between the positively charged protons in the nucleus. Neutrons, being electrically neutral, help to counterbalance these repulsive forces, stabilizing the nucleus. As more protons are added, more neutrons are needed to maintain stability.
However, there is a limit to the number of neutrons that can be added before the nucleus becomes unstable. The neutron excess plot shows that there is a point where adding more neutrons no longer stabilizes the nucleus, leading to an upward deviation from linearity.
Beyond this point, the nucleus becomes unstable and can undergo radioactive decay, such as beta decay or neutron emission, to achieve a more stable configuration. This deviation from linearity indicates the limit of stability for nuclei and is known as the "neutron drip line."
The neutron drip line varies depending on the atomic number, Z. For lighter elements, the drip line occurs at higher neutron-to-proton ratios, while for heavier elements, the drip line occurs at lower neutron-to-proton ratios. This is because the repulsive forces between protons become stronger with increasing atomic number, making more neutrons necessary to counterbalance these forces.
In summary, the upward deviation from linearity in the plot of the number of neutrons against the number of protons for stable nuclei indicates the neutron excess or neutron drip line, where adding more neutrons no longer stabilizes the nucleus, leading to radioactive decay.
As the atomic number increases, the number of protons in the nucleus also increases. In stable nuclei, the number of protons and neutrons is balanced, with the number of neutrons roughly equal to or slightly greater than the number of protons. However, as the atomic number increases further, the number of neutrons needed to maintain stability increases at a faster rate than the number of protons.
This is due to the strong repulsive forces between the positively charged protons in the nucleus. Neutrons, being electrically neutral, help to counterbalance these repulsive forces, stabilizing the nucleus. As more protons are added, more neutrons are needed to maintain stability.
However, there is a limit to the number of neutrons that can be added before the nucleus becomes unstable. The neutron excess plot shows that there is a point where adding more neutrons no longer stabilizes the nucleus, leading to an upward deviation from linearity.
Beyond this point, the nucleus becomes unstable and can undergo radioactive decay, such as beta decay or neutron emission, to achieve a more stable configuration. This deviation from linearity indicates the limit of stability for nuclei and is known as the "neutron drip line."
The neutron drip line varies depending on the atomic number, Z. For lighter elements, the drip line occurs at higher neutron-to-proton ratios, while for heavier elements, the drip line occurs at lower neutron-to-proton ratios. This is because the repulsive forces between protons become stronger with increasing atomic number, making more neutrons necessary to counterbalance these forces.
In summary, the upward deviation from linearity in the plot of the number of neutrons against the number of protons for stable nuclei indicates the neutron excess or neutron drip line, where adding more neutrons no longer stabilizes the nucleus, leading to radioactive decay.
The compound(s) with TWO lone pairs of electrons on the central atom is(are)- a)BrF5
- b)ClF3
- c)XeF4
- d)SF4
Correct answer is option 'B,C'. Can you explain this answer?
The compound(s) with TWO lone pairs of electrons on the central atom is(are)
a)
BrF5
b)
ClF3
c)
XeF4
d)
SF4
|
Ankit Chatterjee answered |
ClF3 and XeF4: Compounds with Two Lone Pairs of Electrons
ClF3 and XeF4 are the compounds that have two lone pairs of electrons on the central atom. Let's discuss why these two compounds fit this description.
ClF3 (Chlorine Trifluoride)
- In ClF3, the central atom is chlorine (Cl).
- Chlorine has 7 valence electrons, and in ClF3, it forms three single bonds with fluorine atoms.
- The three bond pairs and two lone pairs on the chlorine atom make a total of five electron pairs around the central atom.
- This arrangement results in two lone pairs of electrons on the central chlorine atom, making ClF3 a compound with two lone pairs.
XeF4 (Xenon Tetrafluoride)
- In XeF4, the central atom is xenon (Xe).
- Xenon has 8 valence electrons, and in XeF4, it forms four single bonds with fluorine atoms.
- The four bond pairs and two lone pairs on the xenon atom make a total of six electron pairs around the central atom.
- This arrangement results in two lone pairs of electrons on the central xenon atom, making XeF4 a compound with two lone pairs.
Conclusion
- ClF3 and XeF4 are the compounds that have two lone pairs of electrons on the central atom.
- The presence of lone pairs affects the geometry and properties of these compounds.
- Understanding the electron pair geometry is essential in predicting the molecular shape and reactivity of these compounds.
ClF3 and XeF4 are the compounds that have two lone pairs of electrons on the central atom. Let's discuss why these two compounds fit this description.
ClF3 (Chlorine Trifluoride)
- In ClF3, the central atom is chlorine (Cl).
- Chlorine has 7 valence electrons, and in ClF3, it forms three single bonds with fluorine atoms.
- The three bond pairs and two lone pairs on the chlorine atom make a total of five electron pairs around the central atom.
- This arrangement results in two lone pairs of electrons on the central chlorine atom, making ClF3 a compound with two lone pairs.
XeF4 (Xenon Tetrafluoride)
- In XeF4, the central atom is xenon (Xe).
- Xenon has 8 valence electrons, and in XeF4, it forms four single bonds with fluorine atoms.
- The four bond pairs and two lone pairs on the xenon atom make a total of six electron pairs around the central atom.
- This arrangement results in two lone pairs of electrons on the central xenon atom, making XeF4 a compound with two lone pairs.
Conclusion
- ClF3 and XeF4 are the compounds that have two lone pairs of electrons on the central atom.
- The presence of lone pairs affects the geometry and properties of these compounds.
- Understanding the electron pair geometry is essential in predicting the molecular shape and reactivity of these compounds.
On complete hydrogenation, natural rubber produces- a)ethylene-propylene copolymer
- b)vulcanised rubber
- c)polypropylene
- d)polybutylene
Correct answer is option 'A'. Can you explain this answer?
On complete hydrogenation, natural rubber produces
a)
ethylene-propylene copolymer
b)
vulcanised rubber
c)
polypropylene
d)
polybutylene
|
Aryan Sharma answered |
Complete Hydrogenation of Natural Rubber
Natural rubber is a polymer composed primarily of polyisoprene units. When natural rubber undergoes complete hydrogenation, it results in the formation of a new polymer known as ethylene-propylene copolymer.
Explanation:
- Natural Rubber: Natural rubber is a polymer derived from the latex of rubber trees. It is composed of repeating isoprene units.
- Hydrogenation Process: Complete hydrogenation involves the addition of hydrogen gas under suitable conditions to break the double bonds present in the polymer chain.
- Ethylene-Propylene Copolymer: The complete hydrogenation of natural rubber leads to the formation of an ethylene-propylene copolymer. This copolymer is made up of repeating units of ethylene and propylene.
- Chemical Transformation: During the hydrogenation process, the double bonds in the isoprene units of natural rubber are saturated with hydrogen atoms, resulting in the conversion of polyisoprene into an ethylene-propylene copolymer.
In conclusion, the complete hydrogenation of natural rubber produces an ethylene-propylene copolymer by saturating the double bonds in the polymer chain.
Natural rubber is a polymer composed primarily of polyisoprene units. When natural rubber undergoes complete hydrogenation, it results in the formation of a new polymer known as ethylene-propylene copolymer.
Explanation:
- Natural Rubber: Natural rubber is a polymer derived from the latex of rubber trees. It is composed of repeating isoprene units.
- Hydrogenation Process: Complete hydrogenation involves the addition of hydrogen gas under suitable conditions to break the double bonds present in the polymer chain.
- Ethylene-Propylene Copolymer: The complete hydrogenation of natural rubber leads to the formation of an ethylene-propylene copolymer. This copolymer is made up of repeating units of ethylene and propylene.
- Chemical Transformation: During the hydrogenation process, the double bonds in the isoprene units of natural rubber are saturated with hydrogen atoms, resulting in the conversion of polyisoprene into an ethylene-propylene copolymer.
In conclusion, the complete hydrogenation of natural rubber produces an ethylene-propylene copolymer by saturating the double bonds in the polymer chain.
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?- a)64
- b)90
- c)108
- d)120
Correct answer is option 'C'. Can you explain this answer?
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
a)
64
b)
90
c)
108
d)
120
|
Malavika Menon answered |
Given Data:
- Half-life of the radioactive material = 18 days
- Radiation level in the laboratory = 64 times more than the permissible level
Calculating the Minimum Number of Days:
- Since the radiation level is 64 times more than the permissible level, we need to wait for the radiation to decrease to a safe level.
- The rule of thumb is that it takes about 10 half-lives for a radioactive material to decay to a safe level.
- Therefore, the minimum number of days required for the laboratory to be considered safe for use can be calculated as 10 times the half-life of the radioactive material.
- 10 x 18 days = 180 days
Final Calculation:
- However, since the radiation level is 64 times more than the permissible level, we need to consider this factor as well.
- Therefore, the minimum number of days required for the laboratory to be considered safe for use is 180 days / 64 = 2.8125 days.
- Since we cannot wait for a fraction of a day, we need to round up to the nearest whole number.
- Therefore, the laboratory can be considered safe for use after approximately 3 half-lives, which is 3 x 18 days = 54 days.
Conclusion:
- The minimum number of days after which the laboratory can be considered safe for use is 54 days.
- Therefore, the correct option is (c) 108 days.
- Half-life of the radioactive material = 18 days
- Radiation level in the laboratory = 64 times more than the permissible level
Calculating the Minimum Number of Days:
- Since the radiation level is 64 times more than the permissible level, we need to wait for the radiation to decrease to a safe level.
- The rule of thumb is that it takes about 10 half-lives for a radioactive material to decay to a safe level.
- Therefore, the minimum number of days required for the laboratory to be considered safe for use can be calculated as 10 times the half-life of the radioactive material.
- 10 x 18 days = 180 days
Final Calculation:
- However, since the radiation level is 64 times more than the permissible level, we need to consider this factor as well.
- Therefore, the minimum number of days required for the laboratory to be considered safe for use is 180 days / 64 = 2.8125 days.
- Since we cannot wait for a fraction of a day, we need to round up to the nearest whole number.
- Therefore, the laboratory can be considered safe for use after approximately 3 half-lives, which is 3 x 18 days = 54 days.
Conclusion:
- The minimum number of days after which the laboratory can be considered safe for use is 54 days.
- Therefore, the correct option is (c) 108 days.
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding (ΔSsurr)in JK–1 is (1L atm = 101.3 J)- a)5.763
- b)1.013
- c)-1.013
- d)-5.763
Correct answer is option 'C'. Can you explain this answer?
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding (ΔSsurr)in JK–1 is (1L atm = 101.3 J)
a)
5.763
b)
1.013
c)
-1.013
d)
-5.763
|
Rajdeep Dasgupta answered |
Given data:
- Temperature (T) = 300 K
- Initial volume (Vi) = 1.0 L
- Final volume (Vf) = 2.0 L
- Pressure (P) = 3.0 atm
- Change in volume (ΔV) = Vf - Vi = 2.0 L - 1.0 L = 1.0 L
- Gas constant (R) = 8.314 J/mol·K
- Number of moles (n) = 1 mol
Calculating work done:
Since the process is isothermal, the work done is given by W = -nRTln(Vf/Vi).
W = -1 mol * 8.314 J/mol·K * 300 K * ln(2.0/1.0) = -1 mol * 8.314 J/mol·K * 300 K * 0.693 = -1727.88 J
Calculating change in entropy of the system (ΔSsys):
ΔSsys = nRln(Vf/Vi) = 1 mol * 8.314 J/mol·K * ln(2.0/1.0) = 1 mol * 8.314 J/mol·K * 0.693 = 5.763 J/K
Calculating change in entropy of the surroundings (ΔSsurr):
Since the process is isothermal, ΔSsurr = -ΔSsys = -5.763 J/K
Converting to JK^-1: ΔSsurr = -5.763 JK^-1
Therefore, the change in entropy of the surroundings (ΔSsurr) is -5.763 JK^-1, which is option 'C'.
- Temperature (T) = 300 K
- Initial volume (Vi) = 1.0 L
- Final volume (Vf) = 2.0 L
- Pressure (P) = 3.0 atm
- Change in volume (ΔV) = Vf - Vi = 2.0 L - 1.0 L = 1.0 L
- Gas constant (R) = 8.314 J/mol·K
- Number of moles (n) = 1 mol
Calculating work done:
Since the process is isothermal, the work done is given by W = -nRTln(Vf/Vi).
W = -1 mol * 8.314 J/mol·K * 300 K * ln(2.0/1.0) = -1 mol * 8.314 J/mol·K * 300 K * 0.693 = -1727.88 J
Calculating change in entropy of the system (ΔSsys):
ΔSsys = nRln(Vf/Vi) = 1 mol * 8.314 J/mol·K * ln(2.0/1.0) = 1 mol * 8.314 J/mol·K * 0.693 = 5.763 J/K
Calculating change in entropy of the surroundings (ΔSsurr):
Since the process is isothermal, ΔSsurr = -ΔSsys = -5.763 J/K
Converting to JK^-1: ΔSsurr = -5.763 JK^-1
Therefore, the change in entropy of the surroundings (ΔSsurr) is -5.763 JK^-1, which is option 'C'.
PARAGRAPH 1Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.Q.P (X = Y) is- a)11/36
- b)1/3
- c)13/36
- d)1/2
Correct answer is option 'C'. Can you explain this answer?
PARAGRAPH 1
Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.
Q.
P (X = Y) is
a)
11/36
b)
1/3
c)
13/36
d)
1/2
|
|
Ananya Das answered |
P(X = Y) = P(T1 and T2 win alternately) + P(Both matches are draws)
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is
Correct answer is '4'. Can you explain this answer?
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is
|
Tejas Verma answered |
Diffusion coefficient (D) is proportional to both mean free path 
and mean speed
and mean speed
If T1 is increased 4 times and P1 is increased 2 times.
Consider two solid spheres P and Q each of density 8 gm cm–3 and diameters 1cm and 0.5cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm–3 and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm–3 and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is
Correct answer is '3'. Can you explain this answer?
Consider two solid spheres P and Q each of density 8 gm cm–3 and diameters 1cm and 0.5cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm–3 and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm–3 and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is
|
Maulik Nair answered |
To compare the masses of two objects, we need to know their volumes and densities. In this case, we are given that both spheres have a density of 8 gm/cm. However, we do not have any information about their volumes, so we cannot determine their masses.
If we had the volume of the spheres, we could calculate their masses using the formula:
Mass = Density x Volume
Without the volume information, we cannot compare the masses of the two spheres.
If we had the volume of the spheres, we could calculate their masses using the formula:
Mass = Density x Volume
Without the volume information, we cannot compare the masses of the two spheres.
Let RS be the diameter of the circle x2 + y2 = 1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s)- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
Let RS be the diameter of the circle x2 + y2 = 1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s)
a)
b)
c)
d)
|
|
Ananya Das answered |
Pragraph 2Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)
Q. The average current in the steady state registered by the ammeter in the circuit will be- a)
- b)
- c)proportional to the potential V0
- d)zero
Correct answer is option 'B'. Can you explain this answer?
Pragraph 2
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)
Q. The average current in the steady state registered by the ammeter in the circuit will be
a)
b)
c)
proportional to the potential V0
d)
zero
|
Rajiv Reddy answered |
is charge on ball then Q ∝ V0 .....(i)
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?- a)The temperature distribution over the filament is uniform
- b)The resistance over small sections of the filament decreases with time
- c)The filament emits more light at higher band of frequencies before it breaks up
- d)The filament consumes less electrical power towards the end of the life of the bulb
Correct answer is option 'C,D'. Can you explain this answer?
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
a)
The temperature distribution over the filament is uniform
b)
The resistance over small sections of the filament decreases with time
c)
The filament emits more light at higher band of frequencies before it breaks up
d)
The filament consumes less electrical power towards the end of the life of the bulb
|
|
Ananya Das answered |
When filament breaks up, the temperature of filament will be higher so according to wein’s law
, the filament emits more light at higher band of frequencies. As voltage is constant, so consumed electrical power is P=V2/R As R increases with increase in temperature so the filament consumes less electrical power towards the end of the life of the bulb.
There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2 respectively, are
- a)2.87 and 2.86
- b)2.87 and 2.87
- c)2.87 and 2.83
- d)2.85 and 2.82
Correct answer is option 'C'. Can you explain this answer?
There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2 respectively, are
a)
2.87 and 2.86
b)
2.87 and 2.87
c)
2.87 and 2.83
d)
2.85 and 2.82
|
Pioneer Academy answered |
In first; main scale reading = 2.8 cm.
Vernier scale reading = 7 x 1/10 = 0.07 cm
So reading = 2.87 cm ;
In second; main scale reading = 2.8 cm
Vernier scale reading = 7 x - 0.1/10= -0.7/10
Vernier scale reading = 7 x 1/10 = 0.07 cm
So reading = 2.87 cm ;
In second; main scale reading = 2.8 cm
Vernier scale reading = 7 x - 0.1/10= -0.7/10
=- 0.07 cm
so reading = (2.80 + 0.10 - 0.07) cm = 2.83 cm
so reading = (2.80 + 0.10 - 0.07) cm = 2.83 cm
A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away
in front of the curved surface of the lens, an image of double the size of the object is produced. Due to
reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away
from the lens. Which of the following statement(s) is(are) true?- a)The refractive index of the lens is 2.5
- b)The faint image is erect and real
- c)The focal length of the lens is 20 cm
- d)none of these
Correct answer is option 'A,C'. Can you explain this answer?
A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away
in front of the curved surface of the lens, an image of double the size of the object is produced. Due to
reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away
from the lens. Which of the following statement(s) is(are) true?
in front of the curved surface of the lens, an image of double the size of the object is produced. Due to
reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away
from the lens. Which of the following statement(s) is(are) true?
a)
The refractive index of the lens is 2.5
b)
The faint image is erect and real
c)
The focal length of the lens is 20 cm
d)
none of these
|
Neha Joshi answered |
For refraction through lens,
The faint image is erect and virtual.Q.No. - 7 - 14 carry 4 marks eachEach questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.Q.While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1S2. which of the following is(are) true of the intensity pattern on the screen?
- a)Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
- b)Semi circular bright and dark bands centred at point O
- c)The region very close to the point O will be dark
- d)Straight bright and dark bands parallel to the x-axis
Correct answer is option 'B,C'. Can you explain this answer?
Q.No. - 7 - 14 carry 4 marks each
Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.
Q.
While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1S2. which of the following is(are) true of the intensity pattern on the screen?
a)
Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
b)
Semi circular bright and dark bands centred at point O
c)
The region very close to the point O will be dark
d)
Straight bright and dark bands parallel to the x-axis
|
Nandini Iyer answered |
Since S1S2 line is perpendicular to screen shape of pattern is concentric semicircle
∴ darkness close to O.
PARAGRAPH 2Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse 
. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. Q.The orthocentre of the triangle F1MN is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
PARAGRAPH 2
Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse . Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.
. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. Q.
The orthocentre of the triangle F1MN is
a)
b)
c)
d)
|
|
Ananya Das answered |
For orthocentre : one altitude is y = 0 (MN is
perpendicular)
perpendicular)
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is- a)380
- b)320
- c)260
- d)95
Correct answer is option 'A'. Can you explain this answer?
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is
a)
380
b)
320
c)
260
d)
95
|
|
Nandini Iyer answered |
If a boy is selected then number of ways = 4C1.6C3
If a boy is not selcected then number of ways = 6C4
Captain can be selected in 4C1 ways
Required number of ways =
If a boy is not selcected then number of ways = 6C4
Captain can be selected in 4C1 ways
Required number of ways =
where 
Suppose Q = [qij] is a matrix such that PQ = kI, where 
and I is the identity matrix of order 3. If 
The increasing order of atomic radii of the following Group 13 elements is- a)Al < Ga < In < Tl
- b)Ga < Al < In < Tl
- c)Al < In < Ga < Tl
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The increasing order of atomic radii of the following Group 13 elements is
a)
Al < Ga < In < Tl
b)
Ga < Al < In < Tl
c)
Al < In < Ga < Tl
d)
none of these
|
|
Ananya Das answered |
Increasing order of atomic radius of group 13 elements Ga < Al < In < Tl.
Due to poor shielding of d-electrons in Ga, its radius decreases below Al.
Due to poor shielding of d-electrons in Ga, its radius decreases below Al.
The crystalline form of borax has- a)tetranuclear [B4O5(OH)4]2– unit
- b)all boron atoms in the same plane
- c)equal number of sp2 and sp3 hybridized boron atoms
- d)one terminal hydroxide per boron atom
Correct answer is option 'A,C,D'. Can you explain this answer?
The crystalline form of borax has
a)
tetranuclear [B4O5(OH)4]2– unit
b)
all boron atoms in the same plane
c)
equal number of sp2 and sp3 hybridized boron atoms
d)
one terminal hydroxide per boron atom
|
Mahi Bajaj answered |
Understanding Borax Structure
Borax, or sodium tetraborate, is a naturally occurring mineral with a complex crystalline structure. The correct answers regarding its crystalline form are options A, C, and D.
Tetranuclear Unit
- The crystalline structure of borax features a tetranuclear [B4O5(OH)4]2– unit.
- This unit consists of four boron atoms interconnected through oxygen atoms and hydroxyl groups, showcasing the interconnected nature of boron in borax.
Boron Hybridization
- In borax, there is an equal number of sp2 and sp3 hybridized boron atoms.
- The sp2 hybridization occurs in the planar arrangement of the boron atoms within the borate groups, while sp3 hybridization occurs in the tetrahedral coordination of some boron atoms with hydroxyl groups.
Terminal Hydroxide Groups
- Each boron atom in borax indeed has one terminal hydroxide (–OH) group.
- This structural feature contributes to the compound's overall stability and reactivity, making it useful in various applications.
Summary
In summary, the crystalline structure of borax is characterized by:
- Tetranuclear [B4O5(OH)4]2– unit: Central to its structure.
- Equal boron hybridization: Both sp2 and sp3 present.
- Terminal hydroxide per boron atom: Enhances stability and reactivity.
These features collectively explain why options A, C, and D are correct regarding the crystalline form of borax.
Borax, or sodium tetraborate, is a naturally occurring mineral with a complex crystalline structure. The correct answers regarding its crystalline form are options A, C, and D.
Tetranuclear Unit
- The crystalline structure of borax features a tetranuclear [B4O5(OH)4]2– unit.
- This unit consists of four boron atoms interconnected through oxygen atoms and hydroxyl groups, showcasing the interconnected nature of boron in borax.
Boron Hybridization
- In borax, there is an equal number of sp2 and sp3 hybridized boron atoms.
- The sp2 hybridization occurs in the planar arrangement of the boron atoms within the borate groups, while sp3 hybridization occurs in the tetrahedral coordination of some boron atoms with hydroxyl groups.
Terminal Hydroxide Groups
- Each boron atom in borax indeed has one terminal hydroxide (–OH) group.
- This structural feature contributes to the compound's overall stability and reactivity, making it useful in various applications.
Summary
In summary, the crystalline structure of borax is characterized by:
- Tetranuclear [B4O5(OH)4]2– unit: Central to its structure.
- Equal boron hybridization: Both sp2 and sp3 present.
- Terminal hydroxide per boron atom: Enhances stability and reactivity.
These features collectively explain why options A, C, and D are correct regarding the crystalline form of borax.
Let P be the point on the parabola y2 = 4x which is at the shortest distance from the center S of the circle x2 + y2 – 4x – 16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then- a)SP = 2√5
- b)SQ:QP = ((√5+1):2)
- c)the x-intercept of the normal to the parabola at P is 6
- d)the slope of the tangent to the circle at Q is 1/2
Correct answer is option 'A,C,D'. Can you explain this answer?
Let P be the point on the parabola y2 = 4x which is at the shortest distance from the center S of the circle x2 + y2 – 4x – 16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then
a)
SP = 2√5
b)
SQ:QP = ((√5+1):2)
c)
the x-intercept of the normal to the parabola at P is 6
d)
the slope of the tangent to the circle at Q is 1/2
|
Avi Chawla answered |
Equation of normal of parabola is
y + tx = 2t + t3
Normal passes through S(2, 8)
8 + 2t = 2t + t3
t = 2
y + tx = 2t + t3
Normal passes through S(2, 8)
8 + 2t = 2t + t3
t = 2Hence P ≡ (4,4) and SQ = eadius = 2
Let bi > 1 for i = 1, 2, …, 101. Suppose loge b1, loge b2, …, loge b101 are in Arithmetic Progression (A. P.) with the common difference loge 2. Suppose a1, a2, …, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + … + b51 and s = a1 + a2 + … + a51, then- a)s > t and a101 > b101
- b)s > t and a101 < b101
- c)s < t and a101 > b101
- d)s < t and a101 < b101
Correct answer is option 'B'. Can you explain this answer?
Let bi > 1 for i = 1, 2, …, 101. Suppose loge b1, loge b2, …, loge b101 are in Arithmetic Progression (A. P.) with the common difference loge 2. Suppose a1, a2, …, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + … + b51 and s = a1 + a2 + … + a51, then
a)
s > t and a101 > b101
b)
s > t and a101 < b101
c)
s < t and a101 > b101
d)
s < t and a101 < b101
|
|
Mansi Sarkar answered |
The binary representation of the number i.
Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length 
a through their centers. This a
ssembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is(see the figure). Which of the following statement(s) is(are) true?
- a)The magnitude of angular momentum of the assembly about its center of mass is 17 ma2 ω/2
- b)The magnitude of the z-component of
is 55 ma2ω - c)The magnitude of angular momentum of center of mass of the assembly about the point O is 81 ma2ω
- d)The center of mass of the assembly rotates about the z-axis with an angular speed of ω/5
Correct answer is option 'A,D'. Can you explain this answer?
Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is(see the figure). Which of the following statement(s) is(are) true?
a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is(see the figure). Which of the following statement(s) is(are) true?a)
The magnitude of angular momentum of the assembly about its center of mass is 17 ma2 ω/2
b)
The magnitude of the z-component of
is 55 ma2ω
is 55 ma2ω
c)
The magnitude of angular momentum of center of mass of the assembly about the point O is 81 ma2ω
d)
The center of mass of the assembly rotates about the z-axis with an angular speed of ω/5
|
Chirag Verma answered |
The circle C1 : x2 + y2 = 3, with centre at O, intersects the parabola x2 = 2y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2√3 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y-axis, then- a)Q2Q3 = 12
- b)R2R3 = 4√6
- c)area of the triangle OR2R3 is 6√2
- d)area of the triangle PQ2Q3 is 4√2
Correct answer is option 'B,C'. Can you explain this answer?
The circle C1 : x2 + y2 = 3, with centre at O, intersects the parabola x2 = 2y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2√3 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y-axis, then
a)
Q2Q3 = 12
b)
R2R3 = 4√6
c)
area of the triangle OR2R3 is 6√2
d)
area of the triangle PQ2Q3 is 4√2
|
Milan Roy answered |
Equation of tangent at P(√2,1) is √2 x+y-3=0
If centre of C2 at (0, α) and radius equal to 2√3
The geometries of the ammonia complexes of Ni2+, Pt2+ and Zn2+, respectively, are- a)octahedral, square planar and tetrahedral
- b)square planar, octahedral and tetrahedral
- c)tetrahedral, square planar and octahedral
- d)octahedral, tetrahedral and square planar
Correct answer is option 'A'. Can you explain this answer?
The geometries of the ammonia complexes of Ni2+, Pt2+ and Zn2+, respectively, are
a)
octahedral, square planar and tetrahedral
b)
square planar, octahedral and tetrahedral
c)
tetrahedral, square planar and octahedral
d)
octahedral, tetrahedral and square planar
|
Atharva Pillai answered |
PARAGRAPH 2Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse . Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at
Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is- a)3:4
- b)4:5
- c)5:8
- d)2:3
Correct answer is option 'C'. Can you explain this answer?
PARAGRAPH 2
Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse . Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.
. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at
Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is
Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is
a)
3:4
b)
4:5
c)
5:8
d)
2:3
|
|
Sarita Yadav answered |
Equation of tangent at M and N are 
R(6, 0)
Equation of normal
Equation of normal
Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let vP, vQ and vR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 m s-1. Which of the following statement(s) is(are) true regarding the sound
heard by the person?- a)vP + vR = 2vQ
- b)The rate of change in beat frequency is maximum when the car passes through Q
- c)The plot below represents schematically the variation of beat frequency with time
- d)The plot below represents schematically the variation of beat frequency with time
Correct answer is option 'A,B,C'. Can you explain this answer?
Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let vP, vQ and vR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 m s-1. Which of the following statement(s) is(are) true regarding the sound
heard by the person?
heard by the person?
a)
vP + vR = 2vQ
b)
The rate of change in beat frequency is maximum when the car passes through Q
c)
The plot below represents schematically the variation of beat frequency with time
d)
The plot below represents schematically the variation of beat frequency with time
|
|
Nandini Iyer answered |
Now, v (between P and Q)
Now, sin θ = 1 at Q (maximum)
∴ rate of change in beat frequency is maximum at Q
The position vector
of a particle of mass m is given by the following equation 
, where α = 10/3 m s-3, β= 5 m s-2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?- a)The velocity
is given by 
- b)The angular momentum
with respect to the origin is given by 
- c)The force
is given by 
- d)The torque
with respect to the origin is given by 
Correct answer is option 'A,B,D'. Can you explain this answer?
The position vector of a particle of mass m is given by the following equation , where α = 10/3 m s-3, β= 5 m s-2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?
of a particle of mass m is given by the following equation , where α = 10/3 m s-3, β= 5 m s-2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?a)
The velocity is given by
is given by b)
The angular momentum
with respect to the origin is given by
with respect to the origin is given by
c)
The force is given by
is given by d)
The torque with respect to the origin is given by
with respect to the origin is given by |
|
Nandini Iyer answered |
A transparent slab of thickness d has a refractive index n (z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2 (> n1), as shown in the figure. A ray of light is incident with angle θi from medium 1 and emerges in medium 2 with refraction angle θf with a lateral displacement l.
Which of the following statement (s) is (are) true?- a)
- b)
- c)
- d)
Correct answer is option 'A,C,D'. Can you explain this answer?
A transparent slab of thickness d has a refractive index n (z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2 (> n1), as shown in the figure. A ray of light is incident with angle θi from medium 1 and emerges in medium 2 with refraction angle θf with a lateral displacement l.
Which of the following statement (s) is (are) true?
a)
b)
c)
d)
|
|
Tejas Verma answered |
From Snell's Law
The deviation of ray in the slab will depend on n (z)
Hence, l will depend on n (z) but not on n2
The deviation of ray in the slab will depend on n (z)
Hence, l will depend on n (z) but not on n2
A solution curve of the differential equation 
passes through the point (1, 3). Then the solution curve- a)intersects y = x + 2 exactly at one point
- b)intersects y = x + 2 exactly at two points
- c)intersects y = (x + 2)2
- d)does NOT intersect y = (x + 3)2
Correct answer is option 'A,D'. Can you explain this answer?
A solution curve of the differential equation passes through the point (1, 3). Then the solution curve
passes through the point (1, 3). Then the solution curvea)
intersects y = x + 2 exactly at one point
b)
intersects y = x + 2 exactly at two points
c)
intersects y = (x + 2)2
d)
does NOT intersect y = (x + 3)2
|
|
Ananya Das answered |
Hence no solution
and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P2 = – I is
Light of wavelength λph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe, which of the following statement(s) is(are) true?
- a)For large potential difference (V >>φ /e), λe is approximately halved if V is made four times
- b)λe increases at the same rate as λph for λph < hc/φ
- c)(C) λe is approximately halved, if d is doubled
- d)λe decreases with increase in φand λph
Correct answer is option 'A'. Can you explain this answer?
Light of wavelength λph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe, which of the following statement(s) is(are) true?
a)
For large potential difference (V >>φ /e), λe is approximately halved if V is made four times
b)
λe increases at the same rate as λph for λph < hc/φ
c)
(C) λe is approximately halved, if d is doubled
d)
λe decreases with increase in φand λph
|
Amrutha Chaudhary answered |
Ans.
Option (a)
A length-scale (
) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for is(are) dimensionally correct? - a)
- b)
- c)
- d)
Correct answer is option 'B,D'. Can you explain this answer?
A length-scale () depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for is(are) dimensionally correct?
) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for is(are) dimensionally correct? a)
b)
c)
d)
|
|
Priya Bose answered |
kBT ≈ FL,ε≈Q2/FL2 and n ≈L-3; where F is force, Q is charge and L is length So (B) and (D) are correct
Let 
be functions defined by f(x) = [x2 - 3] and g(x) = |x| f(x) + |4x-7| f(x) where [y] denotes the greatest integer less than or equal to y for y ∈ R . Then- a)f is discontinuous exactly at three points in
- b)f is discontinuous exactly at four points in
- c)g is NOT differentiable exactly at four points in
- d)g is NOT differentiable exactly at five points in
Correct answer is option 'B,C'. Can you explain this answer?
Let be functions defined by f(x) = [x2 - 3] and g(x) = |x| f(x) + |4x-7| f(x) where [y] denotes the greatest integer less than or equal to y for y ∈ R . Then
be functions defined by f(x) = [x2 - 3] and g(x) = |x| f(x) + |4x-7| f(x) where [y] denotes the greatest integer less than or equal to y for y ∈ R . Thena)
f is discontinuous exactly at three points in
b)
f is discontinuous exactly at four points in
c)
g is NOT differentiable exactly at four points in
d)
g is NOT differentiable exactly at five points in
|
|
Ananya Das answered |
f(x )= [x2 -3]
Which is discontinuous at x = 1,
g(x) = f(x) [|x| + |4x - 7|]
f(x) is non differentiable at
& |x| + |4x - 7| is non differentiable at
Hence g(x) is non differentiable
Which is discontinuous at x = 1,
g(x) = f(x) [|x| + |4x - 7|]
f(x) is non differentiable at
& |x| + |4x - 7| is non differentiable at
Hence g(x) is non differentiable
A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0. Consider two cases: (i) when the block is at x0 ; and (ii) when the block is at x = x0 + A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?- a)The amplitude of oscillation in the first case changes by a factor of
whereas in the second case it remains unchanged - b)The final time period of oscillation in both the cases is same
- c)The total energy decreases in both the cases
- d)The instantaneous speed at x0 of the combined masses decreases in both the cases
Correct answer is option 'A,B,D'. Can you explain this answer?
A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0. Consider two cases: (i) when the block is at x0 ; and (ii) when the block is at x = x0 + A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?
a)
The amplitude of oscillation in the first case changes by a factor of whereas in the second case it remains unchanged
whereas in the second case it remains unchangedb)
The final time period of oscillation in both the cases is same
c)
The total energy decreases in both the cases
d)
The instantaneous speed at x0 of the combined masses decreases in both the cases
|
|
Tejas Verma answered |
A remains same
Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC < R/2, which of the following statement(s) about any one of the galvanometers is(are) true?- a)The maximum voltage range is obtained when all the components are connected in series
- b)The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer
- c)The maximum current range is obtained when all the components are connected in paralle
- d)The maximum current range is obtained when the two galvanometers are connected in series and the
combination is connected in parallel with both the resistors
Correct answer is option 'A,C'. Can you explain this answer?
Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC < R/2, which of the following statement(s) about any one of the galvanometers is(are) true?
a)
The maximum voltage range is obtained when all the components are connected in series
b)
The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer
c)
The maximum current range is obtained when all the components are connected in paralle
d)
The maximum current range is obtained when the two galvanometers are connected in series and the
combination is connected in parallel with both the resistors
combination is connected in parallel with both the resistors
|
|
Devika Iyer answered |
The internal resistance of a galvanometer is usually very high, typically in the range of kilohms to megohms. This is because the galvanometer is designed to have a high sensitivity and measure small currents. So, let's assume that the internal resistance of the galvanometers is RC.
If we connect the galvanometers in parallel to each other and connect them in series to the resistors, the circuit would look like this:
_________
| |
I1 ----| |---R---I2
| |
|_________|
Where I1 and I2 are the currents passing through each galvanometer and R is the resistance of the resistors.
The total resistance of the circuit is given by:
RTotal = RC + R + RC = 2RC + R
The total current passing through the circuit is given by:
ITotal = I1 + I2
Since the galvanometers are identical, the current passing through each galvanometer is the same:
I1 = I2 = I
So, the total current passing through the circuit is:
ITotal = I + I = 2I
By Ohm's Law, the total current passing through the circuit is given by:
ITotal = VTotal / RTotal
Where VTotal is the voltage applied to the circuit.
Therefore, the voltage applied to the circuit is:
VTotal = ITotal * RTotal = 2I * (2RC + R) = 4IRC + 2IR
Now, let's consider the voltage drop across each galvanometer:
V1 = I * RC
V2 = I * RC
Since the galvanometers are identical, the voltage drop across each galvanometer is the same:
V1 = V2 = V
The voltage applied to the circuit is the sum of the voltage drops across the galvanometers and the resistor:
VTotal = V + V + IR = 2V + IR
Comparing this with the expression we obtained for VTotal earlier:
2V + IR = 4IRC + 2IR
Simplifying the equation:
2V = 4IRC
V = 2IRC
So, the voltage drop across each galvanometer is equal to 2 times the internal resistance of the galvanometer multiplied by the current passing through the circuit.
In conclusion, the voltage drop across each galvanometer is equal to 2 times the internal resistance of the galvanometer multiplied by the current passing through the circuit.
If we connect the galvanometers in parallel to each other and connect them in series to the resistors, the circuit would look like this:
_________
| |
I1 ----| |---R---I2
| |
|_________|
Where I1 and I2 are the currents passing through each galvanometer and R is the resistance of the resistors.
The total resistance of the circuit is given by:
RTotal = RC + R + RC = 2RC + R
The total current passing through the circuit is given by:
ITotal = I1 + I2
Since the galvanometers are identical, the current passing through each galvanometer is the same:
I1 = I2 = I
So, the total current passing through the circuit is:
ITotal = I + I = 2I
By Ohm's Law, the total current passing through the circuit is given by:
ITotal = VTotal / RTotal
Where VTotal is the voltage applied to the circuit.
Therefore, the voltage applied to the circuit is:
VTotal = ITotal * RTotal = 2I * (2RC + R) = 4IRC + 2IR
Now, let's consider the voltage drop across each galvanometer:
V1 = I * RC
V2 = I * RC
Since the galvanometers are identical, the voltage drop across each galvanometer is the same:
V1 = V2 = V
The voltage applied to the circuit is the sum of the voltage drops across the galvanometers and the resistor:
VTotal = V + V + IR = 2V + IR
Comparing this with the expression we obtained for VTotal earlier:
2V + IR = 4IRC + 2IR
Simplifying the equation:
2V = 4IRC
V = 2IRC
So, the voltage drop across each galvanometer is equal to 2 times the internal resistance of the galvanometer multiplied by the current passing through the circuit.
In conclusion, the voltage drop across each galvanometer is equal to 2 times the internal resistance of the galvanometer multiplied by the current passing through the circuit.
Pragraph 2Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)Q. Which one of the following statements is correct?- a)The balls will bounce back to the bottom plate carrying the opposite charge they went up with
- b)The balls will execute simple harmonic motion between the two plates
- c)The balls will bounce back to the bottom plate carrying the same charge they went up with
- d)The balls will stick to the top plate and remain there
Correct answer is option 'A'. Can you explain this answer?
Pragraph 2
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)
Q. Which one of the following statements is correct?
a)
The balls will bounce back to the bottom plate carrying the opposite charge they went up with
b)
The balls will execute simple harmonic motion between the two plates
c)
The balls will bounce back to the bottom plate carrying the same charge they went up with
d)
The balls will stick to the top plate and remain there
|
|
Rajiv Reddy answered |
After hitting the top plate, the balls will get negatively charged and will now get attracted to the bottom plate which is positively charged. The motion of the balls will be periodic but not SHM.
Extraction of copper from copper pyrite (CuFeS2) involves- a)crushing followed by concentration of the ore by froth flotation
- b)removal of iron as slag
- c)self-reduction step to produce ‘blister copper’ following evolution of SO2
- d)refining of ‘blister copper’ by carbon reduction
Correct answer is option 'A,B,C'. Can you explain this answer?
Extraction of copper from copper pyrite (CuFeS2) involves
a)
crushing followed by concentration of the ore by froth flotation
b)
removal of iron as slag
c)
self-reduction step to produce ‘blister copper’ following evolution of SO2
d)
refining of ‘blister copper’ by carbon reduction
|
|
Krishna Iyer answered |
Refining of blister copper is done by poling technique
Reagent(s) which can be used to bring about the following transformation is(are)
- a)LiAlH4 in (C2H5)2O
- b)BH3 in THF
- c)NaBH4 in C2H5OH
- d)Raney Ni/H2 in THF
Correct answer is option 'C,D'. Can you explain this answer?
Reagent(s) which can be used to bring about the following transformation is(are)
a)
LiAlH4 in (C2H5)2O
b)
BH3 in THF
c)
NaBH4 in C2H5OH
d)
Raney Ni/H2 in THF
|
|
Neha Joshi answered |
NaBH4 and Raney Ni/H2 do not react with acid, ester or epoxide entities of an organic compound
The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)- a)The number of the nearest neighbours of an atom present in the topmost layer is 12
- b)The efficiency of atom packing is 74%
- c)The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
- d)The unit cell edge length is 2√2 times the radius of the atom
Correct answer is option 'B,C,D'. Can you explain this answer?
The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)
a)
The number of the nearest neighbours of an atom present in the topmost layer is 12
b)
The efficiency of atom packing is 74%
c)
The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
d)
The unit cell edge length is 2√2 times the radius of the atom
|
Roshni Chavan answered |
The middle layers will have 12 nearest neighbours. The top-most layer will have 9 nearest neighbours.
4r = a √2, where ‘a’ is edge length of unit cell and ‘r’ is radius of atom
4r = a √2, where ‘a’ is edge length of unit cell and ‘r’ is radius of atom
This section contains TWO paragraphsBased on each paragraph, there are TWO questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.PARAGRAPH 1Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.Q.P(X > Y) is- a)1/4
- b)5/12
- c)1/2
- d)7/12
Correct answer is option 'B'. Can you explain this answer?
This section contains TWO paragraphs
Based on each paragraph, there are TWO questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.
PARAGRAPH 1
Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.
Q.
P(X > Y) is
a)
1/4
b)
5/12
c)
1/2
d)
7/12
|
Lekshmi Basu answered |
P(X > Y) = P(T1 wins both) + P(T1 wins either of the matches and other is draw)
Q. 6 to 13 carry 4 marks eachEach questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.Q. Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is (are) true?- a)Relative change in the radii of two consecutive orbitals does not depend on Z
- b)Relative change in the radii of two consecutive orbitals varies as 1/n
- c)Relative change in the energy of two consecutive orbitals varies as 1/n3
- d)Relative change in the angular momenta of two consecutive orbitals varies as 1/n
Correct answer is option 'A,B,D'. Can you explain this answer?
Q. 6 to 13 carry 4 marks each
Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.
Q. Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is (are) true?
a)
Relative change in the radii of two consecutive orbitals does not depend on Z
b)
Relative change in the radii of two consecutive orbitals varies as 1/n
c)
Relative change in the energy of two consecutive orbitals varies as 1/n3
d)
Relative change in the angular momenta of two consecutive orbitals varies as 1/n
|
Mohit Patel answered |
Chapter doubts & questions for JEE Advanced 2016 - National Level Test Series for JEE Advanced 2025 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of JEE Advanced 2016 - National Level Test Series for JEE Advanced 2025 in English & Hindi are available as part of JEE exam.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
National Level Test Series for JEE Advanced 2025
3 videos|2 docs|40 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup