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Sample Mock Test JEE
15 questions · Physics, Chemistry, Maths · tap an option to check your answer
Physics
Q1. A lift of mass 1000 kg is moving upwards with an acceleration of 1 m/s². The tension developed in the string, which is connected to the lift is (g = 9.8 m/s²)
Answer: B
Using the equation of motion, T - mg = ma.
Here, m = 1000 kg, a = 1 m/s², and g = 9.8 m/s².
So, T = mg + ma = 1000(9.8) + 1000(1) = 9800 + 1000 = 10800 N.
Using the equation of motion, T - mg = ma.
Here, m = 1000 kg, a = 1 m/s², and g = 9.8 m/s².
So, T = mg + ma = 1000(9.8) + 1000(1) = 9800 + 1000 = 10800 N.
Q2. A 10 eV electron is circulating in a plane at right angle to a uniform field of magnetic induction 10⁻⁴ Wb/m² (1.0 gauss). What is the orbital radius of the electron?
Answer: C
Using the equation \( r = \frac{mv}{qB} \), and substituting the values derived from kinetic energy, we can calculate the radius. The kinetic energy of the electron is \( KE = 10 eV = 1.6 \times 10^{-19} \text{ J} \), leading to the radius calculation.
Using the equation \( r = \frac{mv}{qB} \), and substituting the values derived from kinetic energy, we can calculate the radius. The kinetic energy of the electron is \( KE = 10 eV = 1.6 \times 10^{-19} \text{ J} \), leading to the radius calculation.
Q3. An equi-convex lens has power P. It is cut into two symmetrical halves by a plane containing the principal axis. What will be the power of one part?
Answer: D
When the lens is cut along its principal axis, the focal length of the two halves will remain the same because the radius of curvature of both surfaces are still the same. Therefore, the power also remains the same, which is \( P \).
When the lens is cut along its principal axis, the focal length of the two halves will remain the same because the radius of curvature of both surfaces are still the same. Therefore, the power also remains the same, which is \( P \).
Q4. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient \( \frac{\Delta v}{\Delta z} \) is given by \( F = - \eta A \frac{dv}{dz} \), where \( \eta \) is constant called?
Answer: D
Given the equation \( F = - \eta A \frac{dv}{dz} \), we identify the dimensions of each term. The force \( F \) has dimensions \([MLT] 2\), area \( A \) has dimensions \([L^2]\), and velocity gradient \( \frac{dv}{dz} \) has dimensions \([L][T]^{-1}\). Thus, rearranging gives us \( \eta = \frac{[MLT] 2}{[L^2][L][T]^{-1}} = [ML T] 1 1−\). Therefore, the answer is (d).
Given the equation \( F = - \eta A \frac{dv}{dz} \), we identify the dimensions of each term. The force \( F \) has dimensions \([MLT] 2\), area \( A \) has dimensions \([L^2]\), and velocity gradient \( \frac{dv}{dz} \) has dimensions \([L][T]^{-1}\). Thus, rearranging gives us \( \eta = \frac{[MLT] 2}{[L^2][L][T]^{-1}} = [ML T] 1 1−\). Therefore, the answer is (d).
Q5. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
Answer: D
Given that the half-life T1/2 = 30 minutes, we know for a 40% decay, N/N0 = 0.6, and for an 85% decay, N/N0 = 0.15. Using the decay formula \( N/N_0 = e^{-\lambda t} \), we can relate the decay times. The time taken for the decay from 40% to 85% is calculated by determining the time it takes for two half-lives to elapse, leading to a total time of 60 minutes.
Given that the half-life T1/2 = 30 minutes, we know for a 40% decay, N/N0 = 0.6, and for an 85% decay, N/N0 = 0.15. Using the decay formula \( N/N_0 = e^{-\lambda t} \), we can relate the decay times. The time taken for the decay from 40% to 85% is calculated by determining the time it takes for two half-lives to elapse, leading to a total time of 60 minutes.
Chemistry
Q6. Which of the following statements is false?
Answer: B
Nylon-66 is actually a type of fiber, not an elastomer. This statement is false, while the others are true.
Nylon-66 is actually a type of fiber, not an elastomer. This statement is false, while the others are true.
Q7. Which of the following structures represents neoprene polymer?
Answer: A
Neoprene (synthetic rubber) is a polymer of chloroprene, i.e. 2-chloro-1, 3-butadiene.
Neoprene (synthetic rubber) is a polymer of chloroprene, i.e. 2-chloro-1, 3-butadiene.
Q8. Match the following and identify the correct option.
Answer: D
The correct combinations are: (a)—(iii), (b)—(i), (c)—(ii), (d)—(iv). Therefore, the correct answer is (d).
The correct combinations are: (a)—(iii), (b)—(i), (c)—(ii), (d)—(iv). Therefore, the correct answer is (d).
Q9. Calculate the pOH of a solution at 25°C that contains 1 × 10⁻¹⁰ M of hydronium ion.
Answer: B
Given that [H₃O⁺] = 1 × 10⁻¹⁰ M, we can find the pH using the formula:
\[ pH = -\log[H₃O^+] = -\log(1 \times 10^{-10}) = 10. \]
Using the relation:
\[ pH + pOH = 14, \]
we can find pOH as follows:
\[ pOH = 14 - pH = 14 - 10 = 4. \]
Thus, the answer is (b).
Given that [H₃O⁺] = 1 × 10⁻¹⁰ M, we can find the pH using the formula:
\[ pH = -\log[H₃O^+] = -\log(1 \times 10^{-10}) = 10. \]
Using the relation:
\[ pH + pOH = 14, \]
we can find pOH as follows:
\[ pOH = 14 - pH = 14 - 10 = 4. \]
Thus, the answer is (b).
Q10. For the reaction \( 2N_2O_4 \rightarrow 2NO_2 \), if the rate and rate constant are \( 1.0 \times 10^{-4} \) and \( 3.4 \times 10^{-5} s^{-1} \) respectively, then the concentration of \( NO_2 \) at that time will be:
Answer: B
Using the rate equation and given values, we can calculate the concentration of \( NO_2 \). The rate expression can help us determine the concentration based on the stoichiometry of the reaction, leading us to the conclusion that the concentration of \( NO_2 \) is 3.
Using the rate equation and given values, we can calculate the concentration of \( NO_2 \). The rate expression can help us determine the concentration based on the stoichiometry of the reaction, leading us to the conclusion that the concentration of \( NO_2 \) is 3.
Maths
Q11. Area bounded by the curve \( y = x^3 \) and the X-axis lying between the ordinates \( x = 0 \) and \( x = 3 \) is equal to?
Answer: A
To find the area under the curve from \( x = 0 \) to \( x = 3 \), we need to calculate the definite integral \( \int_0^3 x^3 \, dx \). Evaluating this gives us \( \left[ \frac{x^4}{4} \right]_0^3 = \frac{3^4}{4} - 0 = \frac{81}{4} \). Hence, the area is \( \frac{81}{4} \) sq units.
To find the area under the curve from \( x = 0 \) to \( x = 3 \), we need to calculate the definite integral \( \int_0^3 x^3 \, dx \). Evaluating this gives us \( \left[ \frac{x^4}{4} \right]_0^3 = \frac{3^4}{4} - 0 = \frac{81}{4} \). Hence, the area is \( \frac{81}{4} \) sq units.
Q12. The logical statement [~ (~) ()] (~)pqprqrÚÚÙÙÙ is equivalent to which of the following?
Answer: C
By applying De-Morgan’s law and other logical transformations, we find that the statement is equivalent to () ~p rq.
By applying De-Morgan’s law and other logical transformations, we find that the statement is equivalent to () ~p rq.
Q13. What is the sum of angles in a triangle?
Answer: B
In Euclidean geometry, the sum of the interior angles of 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, creating alternate interior angles. Therefore, the correct answer is B.
In Euclidean geometry, the sum of the interior angles of 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, creating alternate interior angles. Therefore, the correct answer is B.
Q14. The center of a circle is (2, – 3) and the circumference is 10π. Then, the equation of the circle is.
Answer: A
The radius is half of the circumference, so r = 5. The equation of the circle in standard form with center (2, -3) is (x - 2)^2 + (y + 3)^2 = 5^2. Expanding this gives x^2 + y^2 + 4x + 6y - 4 = 0.
The radius is half of the circumference, so r = 5. The equation of the circle in standard form with center (2, -3) is (x - 2)^2 + (y + 3)^2 = 5^2. Expanding this gives x^2 + y^2 + 4x + 6y - 4 = 0.
Q15. The sum of the products of elements of any row (or column) of a determinant with the cofactors of the corresponding elements of any other row (or column) is zero. If D is represented as follows, what can we infer? D = \( \begin{vmatrix} 1 & 2 & 3 \ 2 & 2 & 3 \ 3 & 2 & 3 \ \end{vmatrix} \)
Answer: B
This statement reflects the property of determinants indicating that the sum of the products of the elements of any row (or column) with the cofactors of another row (or column) yields zero. Therefore, the determinant D in this case is zero.
This statement reflects the property of determinants indicating that the sum of the products of the elements of any row (or column) with the cofactors of another row (or column) yields zero. Therefore, the determinant D in this case is zero.
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