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Sample Mock Test CET
15 questions · Physics, Chemistry, Maths · tap an option to check your answer
Physics
Q1. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is?
Answer: C
The work done on the charged particle by the electric field is equal to the change in kinetic energy. Thus, K.E. = \( F \cdot d = qE \cdot y \). Therefore, the kinetic energy attained after moving a distance y is \( K.E. = qE y \).
The work done on the charged particle by the electric field is equal to the change in kinetic energy. Thus, K.E. = \( F \cdot d = qE \cdot y \). Therefore, the kinetic energy attained after moving a distance y is \( K.E. = qE y \).
Q2. An oscillator is nothing but an amplifier with
Answer: A
A transistor amplifier with proper positive feedback can act as an oscillator, i.e. it can generate oscillations without any external signal source. A positive feedback amplifier is one that produces a feedback voltage that is in phase with the original input signal. A phase shift of 180° is produced by the amplifier and a further phase shift of 180° is introduced by the feedback network. Consequently, the signal is shifted by 360° and fed to the input, i.e. feedback voltage is in phase with the input signal.
A transistor amplifier with proper positive feedback can act as an oscillator, i.e. it can generate oscillations without any external signal source. A positive feedback amplifier is one that produces a feedback voltage that is in phase with the original input signal. A phase shift of 180° is produced by the amplifier and a further phase shift of 180° is introduced by the feedback network. Consequently, the signal is shifted by 360° and fed to the input, i.e. feedback voltage is in phase with the input signal.
Q3. A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the fall in temperature as registered by the Centigrade thermometer?
Answer: C
Using the relation between Celsius and Fahrenheit, we have \( C = \frac{100}{180}(F - 32) \). Substituting \( F = 140° \), we find that \( C = 60° \). Hence, the fall in temperature from 100°C to 60°C is 40°C.
Using the relation between Celsius and Fahrenheit, we have \( C = \frac{100}{180}(F - 32) \). Substituting \( F = 140° \), we find that \( C = 60° \). Hence, the fall in temperature from 100°C to 60°C is 40°C.
Q4. Two equal electric currents are flowing perpendicular to each other. Where do we expect the resultant magnetic field to be zero?
Answer: A
Applying the right-hand grip rule to the perpendicular currents, the magnetic fields due to each current will cancel out at the line AB, resulting in a zero resultant magnetic field there. Thus, the correct answer is that the resultant magnetic field is zero on AB.
Applying the right-hand grip rule to the perpendicular currents, the magnetic fields due to each current will cancel out at the line AB, resulting in a zero resultant magnetic field there. Thus, the correct answer is that the resultant magnetic field is zero on AB.
Q5. When the light of frequency \( 2 \nu_0 \) (where, \( \nu_0 \) is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is \( v_1 \). When the frequency of the incident radiation is increased to \( 5 \nu_0 \), the maximum velocity of electrons emitted from the same plate is \( v_2 \). The ratio of \( v_1 \) to \( v_2 \) is
Answer: C
According to the Einstein’s photoelectric equation, \( K_{max} = \frac{1}{2} mv_{max}^2 = h
u - \phi \). For \( \nu = 2 \nu_0 \), we find \( K_{max1} = h(2\nu_0) - \phi \) and for \( \nu = 5 \nu_0 \), \( K_{max2} = h(5\nu_0) - \phi \). Dividing these equations gives the ratio \( \frac{v_1}{v_2} = \frac{1}{2} \).
According to the Einstein’s photoelectric equation, \( K_{max} = \frac{1}{2} mv_{max}^2 = h
u - \phi \). For \( \nu = 2 \nu_0 \), we find \( K_{max1} = h(2\nu_0) - \phi \) and for \( \nu = 5 \nu_0 \), \( K_{max2} = h(5\nu_0) - \phi \). Dividing these equations gives the ratio \( \frac{v_1}{v_2} = \frac{1}{2} \).
Chemistry
Q6. Which one of the following statements is not correct?
Answer: B
SnCl4 is obtained by passing chlorine over tin, so the statement that Sn(IV) chloride is made by dissolving tin solution in concentrated hydrochloric acid is incorrect.
SnCl4 is obtained by passing chlorine over tin, so the statement that Sn(IV) chloride is made by dissolving tin solution in concentrated hydrochloric acid is incorrect.
Q7. Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
Answer: B
The correct order of electron gain enthalpy or (electron affinity) is O > S > F > Cl with their respective values in eV being 1.48, 2.07, 3.45 and 3.61.
The correct order of electron gain enthalpy or (electron affinity) is O > S > F > Cl with their respective values in eV being 1.48, 2.07, 3.45 and 3.61.
Q8. For the reaction, \( 2A + B \rightarrow C + D \), the reaction rate is written as follows:
Answer: A
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For reactant A, the rate is given by \( -\frac{1}{2} \frac{d[A]}{dt} \) since two moles of A are consumed. The other expressions are for B (consumed), C, and D (produced). Therefore, option A is not a valid representation for the rate of the reaction.
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For reactant A, the rate is given by \( -\frac{1}{2} \frac{d[A]}{dt} \) since two moles of A are consumed. The other expressions are for B (consumed), C, and D (produced). Therefore, option A is not a valid representation for the rate of the reaction.
Q9. Which one of the following gives positive Fehling’s solution test?
Answer: B
Glucose reduces Fehling's solution to give a red precipitate of Cu₂O.
Glucose reduces Fehling's solution to give a red precipitate of Cu₂O.
Q10. Which of the following statements is correct for a reversible process in a state of equilibrium? [CBSE AIPMT 2015]
Answer: A
Mathematical expression of the thermodynamic equilibrium is \Delta G = \Delta G^
\circ + RT \ln Q. At equilibrium when \Delta G = 0 and Q = K, then \Delta G^
\circ = -RT \ln K.
Mathematical expression of the thermodynamic equilibrium is \Delta G = \Delta G^
\circ + RT \ln Q. At equilibrium when \Delta G = 0 and Q = K, then \Delta G^
\circ = -RT \ln K.
Maths
Q11. If the vectors \( p, q, r \) are coplanar, then the value of \( pqrp + q + r \) is
Answer: C
For vectors to be coplanar, their scalar triple product must equal zero. Thus, we can derive the value of the expression based on the conditions of coplanarity.
For vectors to be coplanar, their scalar triple product must equal zero. Thus, we can derive the value of the expression based on the conditions of coplanarity.
Q12. The sum of the squares of the lengths of the chords intercepted on the circle \( x^2 + y^2 = 16 \), by the lines \( x + y = n \), where \( n \in N \), is
Answer: C
The lengths of the chords can be calculated using the intersection points of the lines with the circle. By applying the chord length formula and summing the squares, we find that the total is 160. Thus, the answer is 160.
The lengths of the chords can be calculated using the intersection points of the lines with the circle. By applying the chord length formula and summing the squares, we find that the total is 160. Thus, the answer is 160.
Q13. Where does the function f(x) = sin(x) have a relative maximum?
Answer: B
To find the relative maximum of f(x) = sin(x), we differentiate: f'(x) = cos(x). Setting f'(x) = 0 gives x = π + 2kπ for k ∈ Z. The maximum occurs at x = π.
To find the relative maximum of f(x) = sin(x), we differentiate: f'(x) = cos(x). Setting f'(x) = 0 gives x = π + 2kπ for k ∈ Z. The maximum occurs at x = π.
Q14. If a and b are real numbers such that (2 + 4a) = a + b, where a = -1 + 3i, then what is the value of a + b?
Answer: C
It is given that (2 + 4a) = a + b, where a = -1 + 3i is a non-real complex root of unity. We know that a^3 = 1 and 10 + 2 + a + a = 3 if r is multiple of 3. Therefore, we can simplify: a + b = 9.
It is given that (2 + 4a) = a + b, where a = -1 + 3i is a non-real complex root of unity. We know that a^3 = 1 and 10 + 2 + a + a = 3 if r is multiple of 3. Therefore, we can simplify: a + b = 9.
Q15. An open box is to be made out of a piece of paper of cardboard measuring 24cm by 24cm by cutting off equal squares from the corners and turning up the sides. The height of the box when it has maximum volume is
Answer: B
Let the length of the side of each square cut off from the corners be x cm. Then, the height of the box is x cm. The volume V of the box is given by \( V = x(24 - 2x)(24 - 2x) \). Differentiating this, we find \( \frac{dV}{dx} = 24x - 4x^2 \). Setting this equal to zero and solving yields x = 4. Therefore, the height of the box is maximum at 4 cm.
Let the length of the side of each square cut off from the corners be x cm. Then, the height of the box is x cm. The volume V of the box is given by \( V = x(24 - 2x)(24 - 2x) \). Differentiating this, we find \( \frac{dV}{dx} = 24x - 4x^2 \). Setting this equal to zero and solving yields x = 4. Therefore, the height of the box is maximum at 4 cm.
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