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Sample Mock Test CET
15 questions · Physics, Chemistry, Maths · tap an option to check your answer
Physics
Q1. Two similar coils of radius \( R \) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \( I \) and \( 2I \), respectively. The resultant magnetic field induction at the centre will be?
Answer: A
The magnetic field at the center of a circular coil is given by \( B = \frac{\mu_0 I}{2R} \). For coil 1, with current \( I \), \( B_1 = \frac{\mu_0 I}{2R} \) (i). For coil 2, with current \( 2I \), \( B_2 = \frac{\mu_0 (2I)}{2R} = \frac{\mu_0 I}{R} \) (ii). The resultant magnetic field is \( B = B_1 + B_2 = \frac{\mu_0 I}{2R} + \frac{\mu_0 I}{R} = \frac{5\mu_0 I}{2R} \).
The magnetic field at the center of a circular coil is given by \( B = \frac{\mu_0 I}{2R} \). For coil 1, with current \( I \), \( B_1 = \frac{\mu_0 I}{2R} \) (i). For coil 2, with current \( 2I \), \( B_2 = \frac{\mu_0 (2I)}{2R} = \frac{\mu_0 I}{R} \) (ii). The resultant magnetic field is \( B = B_1 + B_2 = \frac{\mu_0 I}{2R} + \frac{\mu_0 I}{R} = \frac{5\mu_0 I}{2R} \).
Q2. A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F, the net force on the remaining three arms of the loop is?
Answer: B
When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by \( \tau = NBiA \sin(\theta) \). Torque is maximum when \( \theta = 90^\circ \), i.e., the plane of the coil is parallel to the field. The net force on one arm is \( F \), thus the net force on the remaining three arms is \( -F \).
When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by \( \tau = NBiA \sin(\theta) \). Torque is maximum when \( \theta = 90^\circ \), i.e., the plane of the coil is parallel to the field. The net force on one arm is \( F \), thus the net force on the remaining three arms is \( -F \).
Q3. A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is 10 m/s. What is the maximum height attained by the stone? (g=10 m/s²)
Answer: B
Let u be the initial velocity and H be the maximum height attained. At height \( h = \frac{H}{2} \), we have \( v = 10 \text{ m/s} \). From the third equation of motion: \( v^2 = u^2 - 2g h \). Substituting \( h = \frac{H}{2} \), we get \( 10^2 = u^2 - gH \) (1). At height H, \( v = 0 \) gives us \( 0 = u^2 - gH \) (2). Subtracting (2) from (1): \( 10^2 = gH \implies H = \frac{100}{10} = 10 \text{ m}.\
Let u be the initial velocity and H be the maximum height attained. At height \( h = \frac{H}{2} \), we have \( v = 10 \text{ m/s} \). From the third equation of motion: \( v^2 = u^2 - 2g h \). Substituting \( h = \frac{H}{2} \), we get \( 10^2 = u^2 - gH \) (1). At height H, \( v = 0 \) gives us \( 0 = u^2 - gH \) (2). Subtracting (2) from (1): \( 10^2 = gH \implies H = \frac{100}{10} = 10 \text{ m}.\
Q4. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is?
Answer: B
The acceleration of a particle executing SHM at position x is given by \( a_x = -\omega^2 x \). Here, we have \( x = 5 m \) and \( a = -20 m/s^2 \). Substituting the values, we get \( -20 = -\omega^2 \cdot 5 \) which leads to \( \omega^2 = 4 \) or \( \omega = 2 \text{ rad/s} \). The time period \( T \) is given by \( T = \frac{2\pi}{\omega} \), substituting \( \omega \), we find \( T = \pi s \).
The acceleration of a particle executing SHM at position x is given by \( a_x = -\omega^2 x \). Here, we have \( x = 5 m \) and \( a = -20 m/s^2 \). Substituting the values, we get \( -20 = -\omega^2 \cdot 5 \) which leads to \( \omega^2 = 4 \) or \( \omega = 2 \text{ rad/s} \). The time period \( T \) is given by \( T = \frac{2\pi}{\omega} \), substituting \( \omega \), we find \( T = \pi s \).
Q5. The angle between \( A \) and \( B \) is \( \theta \). The value of the triple product \( A \cdot (B \times A) \) is:
Answer: B
In scalar triple product of vectors, the positions of dot and cross can be interchanged. Hence, \( A \cdot (B \times A) = 0 \) since \( A \times A = 0 \). Therefore, the answer is zero.
In scalar triple product of vectors, the positions of dot and cross can be interchanged. Hence, \( A \cdot (B \times A) = 0 \) since \( A \times A = 0 \). Therefore, the answer is zero.
Chemistry
Q6. Which of the following ions will exhibit colour in aqueous solutions?
Answer: B
Color in aqueous solutions is due to d-d or f-f transitions, which require the presence of unpaired electrons. Ti3+ has one unpaired electron, while La3+, Lu3+, and Sc3+ have no unpaired electrons, making them colorless in solution.
Color in aqueous solutions is due to d-d or f-f transitions, which require the presence of unpaired electrons. Ti3+ has one unpaired electron, while La3+, Lu3+, and Sc3+ have no unpaired electrons, making them colorless in solution.
Q7. Zn gives H₂ gas with H₂SO₄ and HCl but not with HNO₃ because
Answer: D
Zn have lower value of E°cell and easily gives oxidation. Zn is present above H₂ in electrochemical series. So, it liberates hydrogen gas from dilute HCl/H₂SO₄. But HNO₃ is an oxidising agent. The hydrogen obtained in this reaction is converted into H₂O. In HNO₃, NO₃⁻ ion is reduced and gives NH₄NO₃, NO₂, NO and NO₂ based upon the concentration of HNO₃.
Zn have lower value of E°cell and easily gives oxidation. Zn is present above H₂ in electrochemical series. So, it liberates hydrogen gas from dilute HCl/H₂SO₄. But HNO₃ is an oxidising agent. The hydrogen obtained in this reaction is converted into H₂O. In HNO₃, NO₃⁻ ion is reduced and gives NH₄NO₃, NO₂, NO and NO₂ based upon the concentration of HNO₃.
Q8. In a set of the given reactions, acetic acid yielded a product C. What would be Product C?
Answer: B
The reaction is: CH3COOH + PCl3 → CH3COCl → A. Then A reacts with MgBr2 to give product C. The reaction leads to the formation of CH3C(OH)C2H5 as the product through the specified reactions.
The reaction is: CH3COOH + PCl3 → CH3COCl → A. Then A reacts with MgBr2 to give product C. The reaction leads to the formation of CH3C(OH)C2H5 as the product through the specified reactions.
Q9. Deficiency of which vitamin causes osteomalacia?
Answer: B
Deficiency of Vitamin D causes osteomalacia and rickets, while Vitamin A deficiency causes xerophthalmia, Vitamin E deficiency causes fragility of RBCs, and Vitamin K deficiency increases blood clotting time.
Deficiency of Vitamin D causes osteomalacia and rickets, while Vitamin A deficiency causes xerophthalmia, Vitamin E deficiency causes fragility of RBCs, and Vitamin K deficiency increases blood clotting time.
Q10. According to molecular orbital theory which of the following lists rank the nitrogen species in terms of increasing bond order?
Answer: B
According to the molecular orbital theory (MOT), the bond order of N 2 is 3, for N 2 2 it is 2.5, and for N 2 2 − it is 2. Therefore, the increasing order of bond order is N 2 2 − < N 2 2 < N 2.
According to the molecular orbital theory (MOT), the bond order of N 2 is 3, for N 2 2 it is 2.5, and for N 2 2 − it is 2. Therefore, the increasing order of bond order is N 2 2 − < N 2 2 < N 2.
Maths
Q11. The number of functions from {1, 2, 3, ..., 20} onto {1, 2, 3, ..., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is
Answer: D
According to the given information, if k is in {4, 8, 12, 16, 20}, then f(k) must be in {3, 6, 9, 12, 15, 18}. We can assign values for k in 6 ways (choosing from 6 multiples of 3) and the remaining 15 elements can be associated in 15! ways. Therefore, the total number of onto functions is calculated as 15 * 6! * 15!
According to the given information, if k is in {4, 8, 12, 16, 20}, then f(k) must be in {3, 6, 9, 12, 15, 18}. We can assign values for k in 6 ways (choosing from 6 multiples of 3) and the remaining 15 elements can be associated in 15! ways. Therefore, the total number of onto functions is calculated as 15 * 6! * 15!
Q12. If the average of the numbers 148, 146, 144, 142, ... in AP, be 125, then the total numbers in the series will be
Answer: C
In an arithmetic progression, the average can be calculated as the sum of the first and last term divided by 2. Given that the average is 125, we need to find the total numbers. The first term is 148 and the common difference is -2. Setting up the equation gives the total number of terms as 30.
In an arithmetic progression, the average can be calculated as the sum of the first and last term divided by 2. Given that the average is 125, we need to find the total numbers. The first term is 148 and the common difference is -2. Setting up the equation gives the total number of terms as 30.
Q13. What is the sum of the angles in a triangle?
Answer: A
In Euclidean geometry, the sum of the angles in a triangle is always 180 degrees. This can be proven by drawing a line parallel to one side of the triangle through the opposite vertex and using the properties of alternate interior angles.
In Euclidean geometry, the sum of the angles in a triangle is always 180 degrees. This can be proven by drawing a line parallel to one side of the triangle through the opposite vertex and using the properties of alternate interior angles.
Q14. Which of the following is the inverse of the proposition ‘If a number is a prime, then it is odd.’
Answer: B
The inverse of a statement 'If p then q' is 'If not p then not q'. Thus, the inverse of 'If a number is a prime, then it is odd' is 'If a number is not a prime, then it is not odd'.
The inverse of a statement 'If p then q' is 'If not p then not q'. Thus, the inverse of 'If a number is a prime, then it is odd' is 'If a number is not a prime, then it is not odd'.
Q15. The probability of getting the first card an ace and the second a honored one is (before drawing the second card first card is not placed again in the pack)
Answer: C
To find the probability, we calculate: P(Ace first) = \frac{4}{52} and P(Honored second | Ace first) = \frac{12}{51}. Thus, P(Ace first and Honored second) = P(Ace first) * P(Honored second | Ace first) = \frac{4}{52} * \frac{12}{51} = \frac{48}{2652} = \frac{5}{221}. Hence, the correct answer is C: \frac{5}{221}.
To find the probability, we calculate: P(Ace first) = \frac{4}{52} and P(Honored second | Ace first) = \frac{12}{51}. Thus, P(Ace first and Honored second) = P(Ace first) * P(Honored second | Ace first) = \frac{4}{52} * \frac{12}{51} = \frac{48}{2652} = \frac{5}{221}. Hence, the correct answer is C: \frac{5}{221}.
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