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Sample Mock Test NEET
15 questions · Physics, Chemistry, Biology · tap an option to check your answer
Physics
Q1. When a uranium isotope \(\text{U}^{235}_{92} \) is bombarded with a neutron, it generates \(\text{Kr}^{89}_{36} \), three neutrons and?
Answer: D
According to the law of conservation of atomic numbers and masses: \n\(92 + 0 = 36 + Z\) and \(235 + 1 = 89 + A + 3\).\n\nSolving these gives Z = 56 and A = 144.\nSo, the other element generated is \(\text{Ba}^{144}_{56} \). Hence, correct option is (d).
According to the law of conservation of atomic numbers and masses: \n\(92 + 0 = 36 + Z\) and \(235 + 1 = 89 + A + 3\).\n\nSolving these gives Z = 56 and A = 144.\nSo, the other element generated is \(\text{Ba}^{144}_{56} \). Hence, correct option is (d).
Q2. The output of OR gate is 1
Answer: B
The OR gate outputs 1 if at least one of its inputs is 1. From the truth table, we can see that the output Y is 1 for input combinations (0,1), (1,0), and (1,1), confirming that the correct answer is option B.
The OR gate outputs 1 if at least one of its inputs is 1. From the truth table, we can see that the output Y is 1 for input combinations (0,1), (1,0), and (1,1), confirming that the correct answer is option B.
Q3. In the nucleus of 11 23 Na, the number of protons, neutrons and electrons are?
Answer: A
A nucleus of mass number A and atomic number Z contains Z protons and (A-Z) neutrons. For 11 Na 23, Z = 11 (protons) and A = 23. Therefore, the number of neutrons is 23 - 11 = 12, and since there are no electrons in the nucleus, the answer is 11 protons, 12 neutrons, 0 electrons.
A nucleus of mass number A and atomic number Z contains Z protons and (A-Z) neutrons. For 11 Na 23, Z = 11 (protons) and A = 23. Therefore, the number of neutrons is 23 - 11 = 12, and since there are no electrons in the nucleus, the answer is 11 protons, 12 neutrons, 0 electrons.
Q4. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an
Answer: D
Let the radioactive substance be \( Z_A X \). Radioactive transition is given by \( Z_A X \rightarrow Z_A X - 4 \); The atoms of element having the same atomic number but different mass numbers are called isotopes. Therefore, \( Z_A X \) and \( Z_A X - 4 \) are isotopes.
Let the radioactive substance be \( Z_A X \). Radioactive transition is given by \( Z_A X \rightarrow Z_A X - 4 \); The atoms of element having the same atomic number but different mass numbers are called isotopes. Therefore, \( Z_A X \) and \( Z_A X - 4 \) are isotopes.
Q5. Current through 3Ω resistor is 0.8 A, then potential drop through 4Ω resistor is[CBSE AIPMT 1993]
Answer: C
Voltage across the 3Ω resistance is given by Ohm's law as \( V = i \times R = 0.8 A \times 3 \Omega = 2.4 V \). Since 3Ω and 6Ω are in parallel, the voltage across the 6Ω resistance will also be 2.4 V. Therefore, the total current in the circuit is \( i_{total} = 0.8 A + 0.4 A = 1.2 A \). The potential drop across the 4Ω resistance is \( V = i \times R = 1.2 A \times 4 \Omega = 4.8 V \). Thus, the correct answer is (c) 4.8 V.
Voltage across the 3Ω resistance is given by Ohm's law as \( V = i \times R = 0.8 A \times 3 \Omega = 2.4 V \). Since 3Ω and 6Ω are in parallel, the voltage across the 6Ω resistance will also be 2.4 V. Therefore, the total current in the circuit is \( i_{total} = 0.8 A + 0.4 A = 1.2 A \). The potential drop across the 4Ω resistance is \( V = i \times R = 1.2 A \times 4 \Omega = 4.8 V \). Thus, the correct answer is (c) 4.8 V.
Chemistry
Q6. Cane sugar on reaction with nitric acid gives
Answer: C
When nitric acid reacts with cane sugar, it forms oxalic acid. The reaction can be represented as C_6H_{12}O_6 + 18[O] → (COOH)_2 + 5H_2O.
When nitric acid reacts with cane sugar, it forms oxalic acid. The reaction can be represented as C_6H_{12}O_6 + 18[O] → (COOH)_2 + 5H_2O.
Q7. Chloropicrin is obtained by the reaction of
Answer: D
Chloroform reacts with nitric acid to produce chloropicrin (nitro chloroform). The reaction is shown as: CHCl3 + HNO3 → CCl3NO2 + H2O. Hence, the correct answer is nitric acid on chloroform.
Chloroform reacts with nitric acid to produce chloropicrin (nitro chloroform). The reaction is shown as: CHCl3 + HNO3 → CCl3NO2 + H2O. Hence, the correct answer is nitric acid on chloroform.
Q8. Which law states that the volume of a gas is directly proportional to its temperature at constant pressure?
Answer: B
Charles's Law states that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin) when pressure is held constant. This relationship is crucial for understanding gas behavior in thermodynamics.
Charles's Law states that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin) when pressure is held constant. This relationship is crucial for understanding gas behavior in thermodynamics.
Q9. The bond dissociation energies of X, Y, and XY are in the ratio of 1 : 0.5 : 1. ∆H for the formation of XY is −200 kJ mol −1. The bond dissociation energy of X 2 will be
Answer: A
Given the bond dissociation energies of X, Y, and XY are in the ratio 1 : 0.5 : 1, let the bond dissociation energies be a kJ mol −1, 0.5a kJ mol −1 and a kJ mol −1 respectively. The relation \( \Delta r H = - (BE_{Reactants} - BE_{Products}) \) gives us the equation. Solving this leads to the conclusion that the bond dissociation energy of X 2 is 800 kJ mol −1.
Given the bond dissociation energies of X, Y, and XY are in the ratio 1 : 0.5 : 1, let the bond dissociation energies be a kJ mol −1, 0.5a kJ mol −1 and a kJ mol −1 respectively. The relation \( \Delta r H = - (BE_{Reactants} - BE_{Products}) \) gives us the equation. Solving this leads to the conclusion that the bond dissociation energy of X 2 is 800 kJ mol −1.
Q10. Which of the following statement is correct about bakelite?
Answer: A
Bakelite is a cross linked condensation thermosetting copolymer of phenol and formaldehyde, making option (a) the correct answer.
Bakelite is a cross linked condensation thermosetting copolymer of phenol and formaldehyde, making option (a) the correct answer.
Biology
Q11. Which one of the following acids is a derivative of carotenoids?
Answer: D
Abscisic acid is a terpenoid, which is a derivative of steroid (carotenoid).
Abscisic acid is a terpenoid, which is a derivative of steroid (carotenoid).
Q12. Middle piece of mammalian sperm possesses
Answer: B
The middle piece of human sperm contains mitochondria which are coiled around an axial filament called mitochondrial spiral. These provide energy for the movement of sperm.
The middle piece of human sperm contains mitochondria which are coiled around an axial filament called mitochondrial spiral. These provide energy for the movement of sperm.
Q13. Drug called ‘Heroin’ is synthesised by
Answer: A
Drug ‘Heroin’ is synthesised by the acetylation of morphine. Chemically heroin is diacetymorphine and commonly it is called smack. It is an opium derivative which is used as medicine. Excessive use of it causes addiction.
Drug ‘Heroin’ is synthesised by the acetylation of morphine. Chemically heroin is diacetymorphine and commonly it is called smack. It is an opium derivative which is used as medicine. Excessive use of it causes addiction.
Q14. The process of RNA interference has been used in the development of plants resistant to?
Answer: A
RNA interference (RNAi) has been utilized to develop plants resistant to nematodes like Meloidogyne incognitia, which infect tobacco roots and significantly reduce yield.
RNA interference (RNAi) has been utilized to develop plants resistant to nematodes like Meloidogyne incognitia, which infect tobacco roots and significantly reduce yield.
Q15. Which of the following plants produces seeds but not flowers?
Answer: D
Pinus is a gymnospermic plant. Ovules of Pinus are uncovered which lie on the megasporophyll, hence, these plants do not have flowers. However, it produces seeds (from ovule after fertilisation) like the other three plants mentioned, all of which are angiosperms.
Pinus is a gymnospermic plant. Ovules of Pinus are uncovered which lie on the megasporophyll, hence, these plants do not have flowers. However, it produces seeds (from ovule after fertilisation) like the other three plants mentioned, all of which are angiosperms.
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