JEE Main 2023 12 April Shift 1 Question Paper is here. NTA conducted JEE Main 2023 April 12 Shift 1 from 9 AM to 12 PM. Candidates can download the official JEE Main 2023 Question Paper PDF with Solution and Answer Key for 12 April Shift 1 using the link below.

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JEE Main 2023 12 April Shift 1 Question Paper with Answer Key PDF

JEE Main 2023 12th April Shift 1  Question Paper with Solution PDF download iconDownload Check Solution

Question 1:

The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7, and 9 without repetition, is equal to:

  • (1) 120
  • (2) 132
  • (3) 72
  • (4) 96
Correct Answer: (1) 120
View Solution

Step 1: Determine the possibilities for the unit place.
Since the number must be divisible by 5, the unit digit must be either 0 or 5.

Step 2: Consider the case when the unit digit is 0.
The first digit must be greater than 4 (i.e., 5, 7, or 9). After choosing the first digit, there are 4 remaining digits for the second, third, and fourth places.
Thus, the number of such five-digit numbers is: \[ 3 \times 4 \times 3 \times 2 = 72 \]

Step 3: Consider the case when the unit digit is 5.
The first digit must be greater than 4 (i.e., 7 or 9). After choosing the first digit, there are 4 remaining digits for the second, third, and fourth places.
Thus, the number of such five-digit numbers is: \[ 2 \times 4 \times 3 \times 2 = 48 \]

Step 4: Total number of such five-digit numbers
Adding the results from Step 2 and Step 3: \[ 72 + 48 = 120 \] Quick Tip: To solve this type of problem, break it into cases based on the divisibility condition (here, the number must end in 0 or 5) and calculate the possibilities for each case separately.


Question 2:

Let \( \alpha, \beta \) be the roots of the quadratic equation \( x^2 + \sqrt{6}x + 3 = 0 \). Then, \( \frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}} \) is equal to:

  • (1) 729
  • (2) 72
  • (3) 81
  • (4) 9
Correct Answer: (3) 81
View Solution

Question 3:

Let  < an > be a sequence such that \( a_1 + a_2 + \dots + a_n = \frac{n^2 + 3n}{(n+1)(n+2)} \). If \[ \sum_{k=1}^{10} \frac{1}{a_k} = p_1p_2p_3 \dots p_m \]
where \( p_1, p_2, \dots, p_m \) are the first \( m \) prime numbers, then \( m \) is equal to:

  • (1) 7
  • (2) 6
  • (3) 5
  • (4) 8
Correct Answer: (2) 6
View Solution

Question 4:

Let the lines \( l_1 : \frac{x + 5}{3} = \frac{y + 4}{1} = \frac{z - \alpha}{-2} \) and \( l_2 : 3x + 2y + z - 2 = 0, \, x - 3y + 2z - 13 = 0 \) be coplanar. If the point P(a, b, c) on \( l_1 \) is nearest to the point Q(-4, -3, 2), then \( |a| + |b| + |c| \) is equal to:

  • (1) 12
  • (2) 14
  • (3) 10
  • (4) 8
Correct Answer: (3) 10
View Solution

Question 5:

Let \( P \left( \frac{2\sqrt{3}}{7}, \frac{6}{\sqrt{7}} \right), Q, R, \) and \( S \) be four points on the ellipse \( 9x^2 + 4y^2 = 36 \). Let PQ and RS be mutually perpendicular and pass through the origin. If \[ \frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{p}{q} \]
where \( p \) and \( q \) are coprime, then \( p + q \) is equal to:

  • (1) 143
  • (2) 137
  • (3) 157
  • (4) 147
Correct Answer: (3) 157
View Solution

Question 6:

Let \(a, b, c\) be three distinct real numbers, none equal to one. If the vectors \( a \hat{i} + \hat{j} + \hat{k}, \, \hat{i} + b\hat{j} + \hat{k}, \, a \hat{i} + \hat{j} + c \hat{k} \) are coplanar, then \( \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} \) is equal to:

  • (1) 1
  • (2) -1
  • (3) -2
  • (4) 2
Correct Answer: (1) 1
View Solution

Question 7:

If the local maximum value of the function \[ f(x) = \left( \frac{\sqrt{3e}}{2 \sin x} \right)^{\sin^2 x}, \quad x \in \left( 0, \frac{\pi}{2} \right) \]
is \( \frac{k}{e} \), then \[ \left( \frac{k}{e} \right)^8 + \frac{k^8}{e^5} + k^8 \]
is equal to:

  • (1) \( e^5 + e^6 + e^{11} \)
  • (2) \( e^3 + e^5 + e^{11} \)
  • (3) \( e^3 + e^6 + e^{11} \)
  • (4) \( e^3 + e^6 + e^{10} \)
Correct Answer: (3) \( e^3 + e^6 + e^{11} \)
View Solution

Question 8:

Let \( D \) be the domain of the function \( f(x) = \sin^{-1} \left( \log_{3x} \left( \frac{6 + 2 \log_3 x}{-5x} \right) \right) \). If the range of the function \( g: D \to \mathbb{R} \) defined by \( g(x) = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer function, is \( (\alpha, \beta) \), then \( \alpha^2 + \frac{5}{\beta} \) is equal to:

  • (1) 46
  • (2) 135
  • (3) 136
  • (4) 45
Correct Answer: (2) 135
View Solution

Question 9:

Let \( y = y(x), \, y > 0 \), be a solution curve of the differential equation \( (1 + x^2) \, dy = y(x - y) \, dx \). If \( y(0) = 1 \) and \( y(2\sqrt{2}) = \beta \), then:

  • (1) \( e^{3\beta - 1} = e^{(3 + 2\sqrt{2})} \)
  • (2) \( e^{\beta - 1} = e^{-2(5 + \sqrt{2})} \)
  • (3) \( e^{\beta - 1} = e^{-2(3 + \sqrt{2})} \)
  • (4) \( e^{3\beta - 1} = e^{(5 + \sqrt{2})} \)
Correct Answer: (1) \( e^{3\beta - 1} = e^{(3 + 2\sqrt{2})} \)
View Solution

Step 1: Rearrange the given differential equation.
The given equation is: \[ (1 + x^2) \, \frac{dy}{dx} = y(x - y) \]
We can express it as: \[ \frac{dy}{dx} = \frac{y(x - y)}{1 + x^2} \]

Step 2: Transform into a solvable form.
Rewrite the equation as: \[ \frac{dy}{dx} + y \left( \frac{-x}{1 + x^2} \right) = \left( \frac{-1}{1 + x^2} \right) y^2 \]
Multiply through by \( \frac{1}{y} \): \[ \frac{1}{y} \frac{dy}{dx} + \frac{-x}{(1 + x^2)y} = \frac{-1}{(1 + x^2)} \]

Step 3: Simplify the integrals.
Let \( \frac{1}{y} = t \), then we have the differential equation: \[ \frac{-1}{y^2} \frac{dy}{dx} = \frac{dt}{dx} \]
Integrating both sides: \[ \int \frac{1}{1 + x^2} \, dx = \int \frac{1}{y} \, dt \]
Thus, we obtain the general solution: \[ \sqrt{1 + x^2} = y \ln(e(x + \sqrt{1 + x^2})) \]

Step 4: Apply the boundary conditions.
We know that \( y(0) = 1 \), so we can use this to find the value of the constant.
Substitute \( x = 0 \) and \( y = 1 \) into the equation: \[ 1 = \sqrt{1 + 0^2} \ln(e(0 + \sqrt{1 + 0^2})) \]
This simplifies to \( 1 = \ln(e(1)) = 1 \), confirming the constant is correct.

Step 5: Find the value of \( \beta \).
Now, for \( y(2\sqrt{2}) = \beta \), we substitute \( x = 2\sqrt{2} \) into the solution to find \( \beta \): \[ \beta = \frac{3}{\ln(e(3 + 2\sqrt{2}))} \]
Thus, we obtain \( 3 = \ln(e(3 + 2\sqrt{2})) \), which leads to: \[ e^{3\beta - 1} = e^{(3 + 2\sqrt{2})} \] Quick Tip: When solving first-order differential equations, use the method of integrating factors. This helps in simplifying the equation to an easily solvable form.


Question 10:

Among the two statements

(Statement1): \( (p \Rightarrow q) \land (q \land (\sim q)) \) is a contradiction and

(Statement2): \( (p \land q) \lor (\sim p) \land (\sim q) \) is a tautology,

  • (1) Only (S2) is true
  • (2) Only (S1) is true
  • (3) Both are false
  • (4) Both are true
Correct Answer: (4) Both are true
View Solution

Question 11:

Let \( \lambda \in \mathbb{Z}, \vec{a} = \lambda \hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \). Let \( \vec{c} \) be a vector such that \[ (\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0, \quad \vec{a} \cdot \vec{c} = -17 \quad and \quad \vec{b} \cdot \vec{c} = -20. \]
Then \( \left| \vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k}) \right|^2 \) is equal to:

  • (1) 62
  • (2) 46
  • (3) 53
  • (4) 49
Correct Answer: (2) 46
View Solution

Question 12:

The sum of the coefficients of the first 50 terms in the binomial expansion of \((1-x)^{100}\), is equal to\

  • (1) \( -^{101}C_{50} \)
  • (2) \( ^{99}C_{49} \)
  • (3) \( -^{99}C_{49} \)
  • (4) \( ^{101}C_{50} \)
Correct Answer: (3) \( -^{99}C_{49} \)
View Solution

Question 13:

The area of the region enclosed by the curve \( y = x^3 \) and its tangent at the point \( (-1, -1) \) is:

  • (1) \( \frac{27}{4} \)
  • (2) \( \frac{19}{4} \)
  • (3) \( \frac{23}{4} \)
  • (4) \( \frac{31}{4} \)
Correct Answer: (1) \( \frac{27}{4} \)
View Solution

Question 14:

Let \( A = \begin{bmatrix} 1 & \frac{1}{51}
0 & 1 \end{bmatrix} \). If \( B = \begin{bmatrix} 1 & 2
-1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2
1 & 1 \end{bmatrix} \),
then the sum of all the elements of the matrix \[ \sum_{n=1}^{50} B^n \]
is equal to:

  • (1) 100
  • (2) 50
  • (3) 75
  • (4) 125
Correct Answer: (1) 100
View Solution

Question 15:

Let the plane \( P : 4x - y + z = 10 \) be rotated by an angle \( \frac{\pi}{2} \) about its line of intersection with the plane \( x + y - z = 4 \). If \( \alpha \) is the distance of the point \( (2, 3, -4) \) from the new position of the plane P, then \( 35\alpha \) is:

  • (1) 90
  • (2) 85
  • (3) 105
  • (4) 126
Correct Answer: (4) 126
View Solution

Question 16:

If \(\frac{1}{n+1}\) \({^{n}C_{n}}+\frac{1}{n}\) \({^{n}C_{n-1}}\) + \dots + \(\frac{1}{2}\) \({^{n}C_{1}}\) + \({^{n}C_{0}}\) = \(\frac{1023}{10}\) then n is equal to:

  • (1) 6
  • (2) 9
  • (3) 8
  • (4) 7
Correct Answer: (2) 9
View Solution

Question 17:

Let \( C \) be the circle in the complex plane with centre \( z_0 = \frac{1}{2}(1 + 3i) \) and radius \( r = 1 \). Let \( z_1 = 1 + i \) and the complex number \( z_2 \) be outside the circle \( C \) such that \( |z_1 - z_0| = |z_2 - z_0| = 1 \). If \( z_0, z_1 \) and \( z_2 \) are collinear, then the smaller value of \( |z_2|^2 \) is equal to:

  • (1) \( \frac{13}{2} \)
  • (2) \( \frac{5}{2} \)
  • (3) \( \frac{3}{2} \)
  • (4) \( \frac{7}{2} \)
Correct Answer: (2) \( \frac{5}{2} \)
View Solution

Question 18:

  • (1) 7
  • (2) -7
  • (3) \( -7\sqrt{3} \)
  • (4) \( 7\sqrt{3} \)
Correct Answer: (1) 7
View Solution

Question 19:

Two dice A and B are rolled, Let the numbers obtained on A and B be \( \alpha \) and \( \beta \) respectively. If the variance of \( \alpha - \beta \) is \( \frac{p}{q} \), where \( p \) and \( q \) are coprime, then the sum of the positive divisors of \( p \) is equal to:

  • (1) 36
  • (2) 48
  • (3) 31
  • (4) 72
Correct Answer: (2) 48
View Solution

Question 20:

In a triangle ABC, if \( \cos A + 2 \cos B + \cos C = 2 \) and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then \( \cos A - \cos C \) is equal to:

  • (1) \( \frac{3}{7} \)
  • (2) \( \frac{9}{7} \)
  • (3) \( \frac{10}{7} \)
  • (4) \( \frac{5}{7} \)
Correct Answer: (3) \( \frac{10}{7} \)
View Solution

We are given the equation: \[ \cos A + 2 \cos B + \cos C = 2 \]
Also, the lengths of the sides opposite to angles A and C are \( a = 3 \) and \( c = 7 \), respectively.

We can use the identity: \[ \cos \left( \frac{A + C}{2} \right) = \sin \left( \frac{B}{2} \right) \]
which simplifies to: \[ \cos A - \cos C = 2 \sin \frac{B}{2} \cos \frac{B}{2} \]

Next, we use the given identity \( 2 \cos B / 2 \cos A \) and simplify the calculation to the sum of \( A + C \) and thus \( \cos A - \cos C \).

So, after calculating the entire expression: \[ \boxed{\cos A - \cos C = \frac{10}{7}} \] Quick Tip: For triangle trigonometry problems, use the relationship between the sides and angles, as well as the trigonometric identities for the sum and difference of angles. In this case, use the cosine and sine half-angle identities to simplify.


Question 21:

A fair \( n > 1 \) faces die is rolled repeatedly until a number less than \( n \) appears. If the mean of the number of tosses required is \( \frac{n}{9} \), then \( n \) is equal to:

Correct Answer: (10)
View Solution

Question 22:

Let the digits \( a, b, c \) be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

Correct Answer: (1260)
View Solution

Question 23:

Let \( [x] \) be the greatest integer \( \leq x \). Then the number of points in the interval \( (-2, 1) \), where the function \( f(x) = |[x]| + \sqrt{x - [x]} \) is discontinuous is:

Correct Answer: 3
View Solution

Question 24:

Let the plane \( x + 3y - 2z + 6 = 0 \) meet the co-ordinate axes at the points A, B, C. If the orthocentre of the triangle ABC is \( (\alpha, \beta, \frac{6}{7}) \), then \( 98(\alpha + \beta)^2 \) is equal to:

Correct Answer: 288
View Solution

Question 25:

Let I(x) = \(\int \sqrt{\frac{x+7}{x}}\) \, dx and I(9) = 12 + 7 \(\log_e\) 7.
If I(1) = \(\alpha + 7 \log_e (1+2\sqrt{2})\), then \(\alpha^4\) is equal to:

  • (1) 64
  • (2) 128
  • (3) 32
  • (4) 16
Correct Answer: (1) 64
View Solution

Question 26:

  • (1) 6
  • (2) 8
  • (3) 4
  • (4) 5
Correct Answer: 6
View Solution

Question 27:

Let the positive numbers \( a_1, a_2, a_3, a_4 \), and \( a_5 \) be in a G.P. Let their mean and variance be \( \frac{31}{10} \) and \( \frac{m}{n} \), respectively, where \( m \) and \( n \) are co-prime. If the mean of their reciprocals is \( \frac{31}{40} \) and \( a_3 + a_4 + a_5 = 14 \), then \( m + n \) is equal to:

Correct Answer: 50
View Solution

Question 28:

The number of relations, on the set \( \{1, 2, 3\} \) containing \( (1, 2) \) and \( (2, 3) \), which are reflexive and transitive but not symmetric, is:

Correct Answer: 3
View Solution

Question 29:

If \[ \int_{-0.15}^{0.15} \left| 100x^2 - 1 \right| \, dx = \frac{k}{3000}, then k is equal to: \]

Correct Answer: (575)
View Solution

Question 30:

Two circles in the first quadrant of radii \( r_1 \) and \( r_2 \) touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line \( x + y = 2 \). Then \( r_1^2 + r_2^2 - r_1r_2 \) is equal to:

Correct Answer: 7
View Solution

Question 31:

An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is 12 cm. When viewed from the opposite side, the apparent distance of the bubble is observed as 4 cm. If the side of the ice cube is 24 cm, the refractive index of the ice cube is:

  • (1) \( \frac{4}{3} \)
  • (2) \( \frac{3}{2} \)
  • (3) \( \frac{2}{3} \)
  • (4) \( \frac{6}{5} \)
Correct Answer: (2) \( \frac{3}{2} \)
View Solution

Question 32:

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be:

  • (1) Potential energy
  • (2) Total energy
  • (3) Kinetic energy
  • (4) Speed
Correct Answer: (4) Speed
View Solution

Step 1: Understanding orbital motion of satellites:
The total energy \( E \) and kinetic energy \( K.E. \) of a satellite moving in a circular orbit around the Earth depend only on the radius of the orbit and the gravitational constant, not on the mass of the satellite.

Step 2: Speed of the satellite:
The speed of the satellite \( v \) in orbit is given by: \[ v = \sqrt{\frac{GM_p}{R}} \]
Since the radius of orbit and gravitational constant are the same for both satellites, their speed will be the same. Quick Tip: In orbital motion problems, remember that speed is independent of the mass of the satellite, while potential energy and kinetic energy depend on the mass.


Question 33:

The amplitude of 15 sin (1000 \( \pi t \)) is modulated by 10 sin (4\( \pi t \)) signal. The amplitude modulated signal contains frequencies of:

(1) 500 Hz.

(2) 2 Hz.

(3) 250 Hz.

(4) 498 Hz.

(5) 502 Hz.

  • (1) and (3) only
  • (1) and (4) only
  • (1) and (2) only
  • (1) , (4) and (5) only
Correct Answer: (4) (1), (4) and (5) only
View Solution

Question 34:

In an n-p-n common emitter (CE) transistor, the collector current changes from 5 mA to 16 mA for the change in base current from 100 μA and 200 μA, respectively. The current gain of the transistor is:

  • (1) 110
  • (2) 0.9
  • (3) 210
  • (4) 9
Correct Answer: (1) 110
View Solution

Question 35:

If the r.m.s. speed of chlorine molecules is 490 m/s at 27°C, the r.m.s. speed of argon molecules at the same temperature will be (Atomic mass of argon = 39.9u, molecular mass of chlorine = 70.9u):

  • (1) 751.7 m/s
  • (2) 451.7 m/s
  • (3) 651.7 m/s
  • (4) 551.7 m/s
Correct Answer: (3) 651.7 m/s
View Solution



The r.m.s. speed of gas molecules is related to the temperature and molecular mass by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:

\( R \) is the gas constant.

\( T \) is the temperature.

\( M \) is the molar mass of the gas.


Since both gases are at the same temperature, we can use the ratio of their r.m.s. speeds:

\[ \frac{\nu_{Ar}}{\nu_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} \]

Substitute the given values:

\( M_{Cl} = 70.9 \, u \)

\( M_{Ar} = 39.9 \, u \)

\( \nu_{Cl} = 490 \, m/s \)

\[ \frac{\nu_{Ar}}{490} = \sqrt{\frac{70.9}{39.9}} = 1.33 \]

Thus,
\[ \nu_{Ar} = 1.33 \times 490 = 651.7 \, m/s \]

Thus, the r.m.s. speed of argon molecules is \( \boxed{651.7} \, m/s \). Quick Tip: The r.m.s. speed of gas molecules is inversely proportional to the square root of their molar mass. Use this relation to compare the speeds of different gases at the same temperature.


Question 36:

A proton and an \( \alpha \)-particle are accelerated from rest by 2V and 4V potentials, respectively. The ratio of their de-Broglie wavelength is:

  • (1) 4:1.
  • (2) 2:1.
  • (3) 8:1.
  • (4) 16:1.
Correct Answer: (1) 4:1
View Solution

Question 37:

Given below are two statements:

Statement I: The diamagnetic property depends on temperature.

Statement II: The included magnetic dipole moment in a diamagnetic sample is always opposite to the magnetizing field.

In light of the given statement, choose the correct answer from the options below:

  • (1) Statement I is incorrect but Statement II is true.
  • (2) Both Statement I and Statement II are true.
  • (3) Both Statement I and Statement II are false.
  • (4) Statement I is correct but Statement II is false.
Correct Answer: (1) Statement I is incorrect but Statement II is true.
View Solution

Question 38:

A wire of resistance 160 \( \Omega \) is melted and drawn into a wire of one-fourth of its length. The new resistance of the wire will be:

  • (1) 10 \( \Omega \)
  • (2) 640 \( \Omega \)
  • (3) 40 \( \Omega \)
  • (4) 16 \( \Omega \)
Correct Answer: (1) 10 \( \Omega \)
View Solution

Question 39:

Match List I with List II:





Choose the correct answer from the options given below:

  • (1) A-II, B-I, C-IV, D-III
  • (2) A-IV, B-I, C-III, D-II
  • (3) A-II, B-III, C-I, D-IV
  • (4) A-I, B-III, C-II, D-IV

Question 40:

Three forces \( F_1 = 10 \, N, F_2 = 8 \, N, F_3 = 6 \, N \) are acting on a particle of mass 5 kg. The forces \( F_2 \) and \( F_3 \) are applied perpendicular so that particle remains at rest. If the force \( F_1 \) is removed, then the acceleration of the particle is:

  • (1) 2 \( ms^{-2} \).
  • (2) 0.5 \( ms^{-2} \).
  • (3) 4.8 \( ms^{-2} \).
  • (4) 7 \( ms^{-2} \).
Correct Answer: (1) 2 \( \text{ms}^{-2} \)
View Solution

Question 41:

A body cools from 80°C to 60°C in 5 minutes. The temperature of the surrounding is 20°C. The time it takes to cool from 60°C to 40°C is:

  • (1) 500 s.
  • (2) \( \frac{25}{3} \) s.
  • (3) 450 s.
  • (4) 420 s.
Correct Answer: (1) 500 s
View Solution

Question 42:

An engine operating between the boiling and freezing points of water will have:

(1) efficiency more than 27%.

(2) efficiency less than the efficiency a Carnot engine operating between the same two temperatures.

(3) efficiency equal to 27%.

(4) efficiency less than 27%.

  • (1) \( 2, 3 and 4 only \)
  • (2) \( 2 and 3 only \)
  • (3) \( 2 and 4 only \)
  • (4) \( 1 and 2 only \)
Correct Answer: (3) 2 and 4 only
View Solution

Question 43:

Given below are two statements:

Statement I: A truck and a car moving with the same kinetic energy are brought to rest by applying brakes which provide equal retarding forces. Both come to rest in equal distance.


Statement II: A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.


In light of the given statements, choose the most appropriate answer from the options given below:

  • (1) Statement I is correct but Statement II is incorrect
  • (2) Statement I is incorrect but Statement II is correct
  • (3) Both Statement I is correct but Statement II are incorrect
  • (4) Both Statement I is correct but Statement II are correct
Correct Answer: (1) Statement I is correct but Statement II is incorrect
View Solution

Question 44:

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be:

  • (1) 1 : 1
  • (2) 2 : 1
  • (3) 1 : 4
  • (4) 1 : 3
Correct Answer: (4) 1 : 3
View Solution

Question 45:

A ball is thrown vertically upward with an initial velocity of 150 m/s. The ratio of velocity after 3 s and 5 s is \( \frac{x+1}{x} \). The value of \( x \) is:

Take \( g = 10 \, m/s^2 \).

  • (1) 6
  • (2) 5
  • (3) -5
  • (4) 10
Correct Answer: (2) 5
View Solution

Question 46:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If an electric dipole of dipole moment \( 30 \times 10^{-5} \) Cm is enclosed by a closed surface, the net flux coming out of the surface will be zero.


Reason R: Electric dipole consists of two equal and opposite charges.


In the light of above statements, choose the correct answer from the options given below:

  • (1) Both A and R are true and R is the correct explanation of A
  • (2) A is true but R is false
  • (3) Both A and R true but R is NOT the correct explanation of A
  • (4) A is false but R is true
Correct Answer: (1) Both A and R are true and R is the correct explanation of A
View Solution

Question 47:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: EM waves used for optical communication have longer wavelengths than that of microwave, employed in Radar technology.


Reason R: Infrared EM waves are more energetic than microwaves.


In the light of given statements, choose the correct answer from the options given below:

  • (1) A is false but R is true
  • (2) A is true but R is false
  • (3) Both A and R true but R is NOT the correct explanation of A
  • (4) Both A and R true and R is the correct explanation of A
Correct Answer: (1) A is false but R is true
View Solution

Question 48:

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

  • (1) 2.
  • (2) 1.
  • (3) 3.
  • (4) 4.
Correct Answer: (3) 3
View Solution

Question 49:

The ratio of escape velocity of a planet to the escape velocity of Earth will be:
Given: Mass of the planet is 16 times the mass of Earth and radius of the planet is 4 times the radius of Earth.

  • (1) 4:1.
  • (2) 2:1.
  • (3) 1 : \( \sqrt{2} \).
  • (4) 1 : 4.
Correct Answer: (2) 2:1
View Solution

Question 50:

Given below are two statements:

Statement I: When the frequency of an a.c. source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases.

Statement II: In a series LCR circuit, the value of the power factor at resonance is one.

In light of the given statements, choose the most appropriate answer from the options given below:

  • (1) Statement I is incorrect but Statement II is true.
  • (2) Both Statement I and Statement II are false.
  • (3) Statement I is correct but Statement II is false.
  • (4) Both Statement I and Statement II are true.
Correct Answer: (4) Both Statement I and Statement II are true.
View Solution

Question 51:

For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:5 respectively. If the frequency of the fifth harmonic is 405 Hz and the speed of sound in air is 324 m/s\(^{-1}\), the length of the organ pipe is _______ m.

Correct Answer: 1 m
View Solution

Question 52:

For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is \( \frac{x}{5} \). The value of \( x \) is:

Correct Answer: 2
View Solution

Question 53:

A compass needle oscillates 20 times per minute at a place where the dip is 30° and 30 times per minute where the dip is 60°. The ratio of total magnetic field due to the earth at two places respectively is \( \frac{4}{\sqrt{x}} \). The value of \( x \) is:

Correct Answer: 243
View Solution

Question 54:

A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm/s. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be ___________, \(\mu V\).

Correct Answer: (1) 50 \(\mu V\)
View Solution

Question 55:

To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ________ KJ.
[The coefficient of friction between tyre of bus and road is 0.04].

Correct Answer: 784 KJ
View Solution

Question 56:

A common example of alpha decay is \[ ^{238}_{92} U \rightarrow ^{234}_{90} Th + ^{4}_{2} He + Q. \]
Given: \[ ^{238}_{92} U = 238.05060 \, u, \quad ^{234}_{90} Th = 234.04360 \, u, \quad ^{4}_{2} He = 4.00260 \, u, \quad 1u = 931.5 \, MeV/c^2. \]
The energy released \( Q \) during the alpha decay of \( ^{238}_{92} U \) is ______ MeV.

Correct Answer: 4.0986 MeV
View Solution

Energy released \( Q \) is given by the equation: \[ Q = (\Delta m)_{amu} \times 931.5 \, MeV. \]
Here, the mass defect \( \Delta m \) is: \[ \Delta m = m_u - m_{Th} - m_{He} = 238.05060 \, u - 234.04360 \, u - 4.00260 \, u = 0.0044 \, u. \]

Thus, \[ Q = 0.0044 \times 931.5 \, MeV = 4.0986 \, MeV. \]

Therefore, the energy released during the alpha decay of \( ^{238}_{92} U \) is 4.0986 MeV. Quick Tip: To calculate the energy released during a nuclear decay, find the mass defect and multiply it by \( 931.5 \, MeV \) (the energy equivalent of 1 atomic mass unit).


Question 57:

The current flowing through a conductor connected across a source is 2A and 1.2 A at 0°C and 100°C respectively. The current flowing through the conductor at 50°C will be _______ \( \times 10^2 \) mA.

Correct Answer: 15
View Solution

Question 58:

Two convex lenses of focal length 20 cm each are placed coaxially with a separation of 60 cm between them. The image of the distant object formed by the combination is at _______ cm from the first lens.

Correct Answer: (100)
View Solution

Question 59:

Glycerine of density \( 1.25 \times 10^3 \, kg/m^{-3} \) is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is 10 cm\(^2\) and 5 cm\(^2\) and pressure drop across its length is 3 Nm\(^{-2}\). The rate of flow of glycerine through the pipe is \( x \times 10^{-5} \, m^3 \, s^{-1} \). The value of \( x \) is _______ .

Correct Answer: 4
View Solution

Question 60:

64 identical drops each charged up to a potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be _______ mV.

Correct Answer: (160)
View Solution

Question 61:

The compound shown below undergoes the following reactions:

Correct Answer: (4)
View Solution

The reaction shown involves the conversion of an alkyl bromide to an alcohol in two steps.

1. First, the reaction with magnesium (\( Mg \)) in dry ether will convert the alkyl bromide to a Grignard reagent:
\[ C_6H_5Br + Mg \xrightarrow{ether} C_6H_5MgBr \]

This results in the formation of the Grignard reagent, which is an organomagnesium compound.

2. In the second step, the Grignard reagent reacts with water (\( H_2O \)):
\[ C_6H_5MgBr + H_2O \rightarrow C_6H_5OH + MgBr \]

The final product is phenol (C6H5OH), formed by the addition of a hydroxyl group (\( OH \)) to the carbon atom that was originally bonded to the bromine atom.

Thus, the major product is the phenol, which is option (4).

Therefore, the correct answer is \( \boxed{4} \). Quick Tip: When an alkyl halide reacts with magnesium in ether, it forms a Grignard reagent. This reacts with water to produce an alcohol as the major product.


Question 62:

Four gases A, B, C, and D have critical temperatures 5.3, 33.2, 126.0, and 154.3 K respectively.

For their adsorption on a fixed amount of charcoal, the correct order is:

  • (1) C \(>\) B \(>\) D \(>\) A.
  • (2) C \(>\) D \(>\) B \(>\) A.
  • (3) D \(>\) C \(>\) A \(>\) B.
  • (4) D \(>\) C \(>\) B \(>\) A.
Correct Answer: (4) D \(>\) C \(>\) B\(>\) A.
View Solution

Question 63:

Given below are two statements:

Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons.

Reason R: 5f orbitals are not as buried as 4f orbitals.

In light of the above statements, choose the correct answer from the options given below:

  • (1) Both A and R are true but R is NOT the correct explanation of A.
  • (2) Both A and R are true and R is the correct explanation of A.
  • (3) A is false but R is true.
  • (4) A is true but R is false.
Correct Answer: (2) Both A and R are true and R is the correct explanation of A.
View Solution

Question 64:

The incorrect statement regarding the reaction given below is:

  • (1) The electrophile involved in the reaction is NO\(^+\)
  • (2) 'B' is N-nitroso ammonium compound
  • (3) The reaction occurs at low temperature
  • (4) The product 'B' formed in the above reaction is p-nitroso compound at low temperature
Correct Answer: (2) 'B' is N-nitroso ammonium compound
View Solution

Question 65:

Match List I with List II




Choose the correct answer from the options given below:

  • (1) A-I, B-IV, C-II, D-III.
  • (2) A-II, B-III, C-I, D-IV.
  • (3) A-I, B-II, C-IV, D-III.
  • (4) A-III, B-IV, C-I, D-II.

Question 66:

Match List I with List II




Choose the correct answer from the options given below:

  • (1) A-IV, B-I, C-III, D-II.
  • (2) A-IV, B-III, C-I, D-II.
  • (3) A-II, B-IV, C-I, D-III.
  • (4) A-II, B-I, C-IV, D-III.

Question 67:

The density of alkali metals is in the order:

  • (1) Na \(<\) K \(<\) Cs \(<\) Rb.
  • (2) K \(<\) Na \(<\) Rb \(<\) Cs.
  • (3) K \(<\) Cs \(<\) Na \(<\) Rb.
  • (4) Na \(<\) Rb \(<\) K \(<\) Cs.
Correct Answer: (2) K \(<\) Na \(<\) Rb \(<\) Cs.
View Solution

Question 68:

Given below are two statements:


\text{Statements: \quad \mathrm{SbCl\(_5\) \text{ is more covalent than \mathrm{SbCl\(_3\)

\text{Statements: \quad \text{The higher oxides of halogens also tend to be more stable than the lower ones.


In light of the above statements, choose the most appropriate answer from the options given below:

  • (1) Both statement I and Statement II are correct
  • (2) Both statement I and Statement II are incorrect
  • (3) Statement I is correct but Statement II is incorrect
  • (4) Statement I is incorrect but Statement II is correct
Correct Answer: (1)
View Solution

Question 69:

A metal chloride contains 55.0% of chlorine by weight. 100 mL vapours of the metal chloride at STP weigh 0.57 g. The molecular formula of the metal chloride is: \[ (Given: Atomic mass of chlorine is 35.5 u) \]

  • (1) MCl\(_2\)
  • (2) MCl\(_4\)
  • (3) MCl\(_3\)
  • (4) MCl
Correct Answer: (1)
View Solution

Question 70:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: In the Ellingham diagram, a sharp change in the slope of the line is observed for \( Mg \rightarrow MgO \) at \( \sim 1120^\circ C \).

Reason R: There is a large change of entropy associated with the change of state.

In light of the above statements, choose the correct answer from the options given below:

  • (1) Both A and R are true but R is NOT the correct explanation of A.
  • (2) Both A and R are true and R is the correct explanation of A.
  • (3) A is false but R is true.
  • (4) A is true but R is false.
Correct Answer: (2) Both A and R are true and R is the correct explanation of A.
View Solution

Question 71:

Match List I with List II




Choose the correct answer from the options given below:

  • (1) A-IV, B-III, C-II, D-I.
  • (2) A-II, B-III, C-I, D-IV.
  • (3) A-I, B-IV, C-II, D-III.
  • (4) A-IV, B-II, C-III, D-I.

Question 72:

For lead storage battery pick the correct statements:


\text{A. During charging of battery, PbSO\(_4\) \text{ on anode is converted into PbO\(_2\)


\text{B. During charging of battery, PbSO\(_4\) \text{ on cathode is converted into PbO\(_2\)

\text{C. Lead storage battery, consists of grid of lead packed with PbO\(_2\) \text{ as anode


\text{D. Lead storage battery has \(\sim 38%\) solution of sulphuric acid as an electrolyte


Choose the correct answer from the options given below:

  • (1) B, D only
  • (2) B, C, D only
  • (3) A, B, D only
  • (4) B, C only
Correct Answer: (1)
View Solution

Question 73:

  • (1) Butanoic acid and acetic acid
  • (2) Butanal and acetic acid
  • (3) Butanal and acetaldehyde
  • (4) Butanoic acid and acetaldehyde
Correct Answer: (1) Butanoic acid and acetic acid
View Solution



The reaction involves ozonolysis of 2-hexene, where ozone reacts with the double bond, resulting in the formation of two carbonyl compounds. The products formed from this reaction are:

Acetic acid (CH\(_3\)COOH) from the oxidation of the terminal carbon of the hexene.

Butanoic acid (CH\(_3\)CH\(_2\)COOH) from the oxidation of the other carbon involved in the double bond.


Thus, the correct products are Butanoic acid and acetic acid. Quick Tip: In ozonolysis of alkenes, the double bond undergoes cleavage, leading to the formation of two carbonyl compounds, typically acids or aldehydes depending on the substrate.


Question 74:

Correct statements for the given reaction are:





A. Compound 'B' is aromatic

B. The completion of the above reaction is very slow

C. 'A' shows tautomerism

D. The bond lengths C-C in compound 'B' are found to be same


Choose the correct answer from the options given below:

  • (1) A, B and D only
  • (2) A, B and C only
  • (3) B, C and D only
  • (4) A, C and D only

Question 75:

The bond order and magnetic property of acetylide ion are same as that of:

  • (1) NO\(^+\).
  • (2) O\(_2^+\).
  • (3) O\(_2^-\).
  • (4) N\(_2^+\).
Correct Answer: (1) NO\(^+\).
View Solution

Question 76:

In the given reaction cycle



Question 77:

Given below are two statements:

Statement I: Boron is extremely hard indicating its high lattice energy.

Statement II: Boron has the highest melting and boiling point compared to its other group members.

  • (1) Statement I is incorrect but Statement II is correct
  • (2) Both Statement I and Statement II are correct
  • (3) Statement I is correct but Statement II is incorrect
  • (4) Both Statement I and Statement II are incorrect
Correct Answer: (2) Both Statement I and Statement II are correct
View Solution

Question 78:

Correct Answer: (3)
View Solution



The reaction shown is an example of a Wittig reaction, which forms a cyclic compound through the reaction of the acetylide ion (\(C_2^{2-}\)) with an electrophile (\(OE^{-}\)). The acetylide ion undergoes a [2+2] cycloaddition with the electrophile to form a cyclic intermediate, which upon heating gives a cyclic product containing a carbonyl group.


The acetylide ion (\(C_2^{2-}\)) reacts with the electrophile in a manner that results in the formation of a five-membered cyclic compound, A, containing a carbonyl group. This is consistent with the structure shown in option (3).
Quick Tip: The Wittig reaction typically involves the reaction of an acetylide ion with an electrophile to form a new carbon-carbon bond, which often results in the formation of a cyclic structure. Heating can help drive this reaction to completion.


Question 79:

Match List I with List II

  • (1) A-III, B-II, C-IV, D-I
  • (2) A-II, B-III, C-IV, D-I
  • (3) A-II, B-III, C-I, D-IV
  • (4) A-III, B-II, C-I, D-IV

Question 80:

The major product \(P\) formed in the following sequence of reactions is:


Correct Answer: (4) \(\text{H-N-R} \text{O} \)
View Solution

Question 81:

One mole of an ideal gas at 350K is in a 2.0 L vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0L to 3.0L against a constant pressure of 4 atm. The change in entropy of the surroundings \(\Delta S\) is ______ J K\(^{-1}\) (Nearest integer).

Given \(R = 8.314\) J K\(^{-1}\) Mol\(^{-1}\).

Correct Answer: 3 J K\(^{-1}\)
View Solution

Question 82:

The mass of NH\(_3\) produced when 131.8 kg of cyclohexanecarbaldehyde undergoes Tollens test is ______ kg. (Nearest Integer)
Given: Molar Mass of C = 12 g/mol, N = 14 g/mol, O = 16 g/mol.

Correct Answer: 60 kg
View Solution

The chemical equation for the reaction is:
\[ Cyclohexanecarbaldehyde + 2 [Ag(NH_3)_2] OH \rightarrow Cyclohexanecarboxamide + 3 NH_3 + 2 Ag + H_2O \]

We are given the mass of cyclohexanecarbaldehyde:
\[ Mass of cyclohexanecarbaldehyde = 131.8 \, kg \]

The molecular weight of cyclohexanecarbaldehyde:
\[ Molecular weight = 12 \times 6 + 14 + 16 = 112 \, g/mol \]

From the stoichiometry of the reaction, we know that 1 mole of cyclohexanecarbaldehyde produces 3 moles of NH\(_3\). The number of moles of cyclohexanecarbaldehyde is:
\[ Moles of cyclohexanecarbaldehyde = \frac{131.8 \times 1000}{112} = 1176.8 \, mol \]

Since 1 mole of cyclohexanecarbaldehyde produces 3 moles of NH\(_3\), the moles of NH\(_3\) produced are:
\[ Moles of NH_3 = 3 \times 1176.8 = 3530.4 \, mol \]

The molar mass of NH\(_3\) is:
\[ Molar mass of NH_3 = 17 \, g/mol \]

The mass of NH\(_3\) produced is:
\[ Mass of NH_3 = 3530.4 \times 17 = 60,023.2 \, g = 60 \, kg \]

Thus, the mass of NH\(_3\) produced is:
\[ \boxed{60 \, kg} \] Quick Tip: In reactions like Tollens' test, always pay attention to the stoichiometry between the reactants and products to correctly calculate the mass of the desired product.


Question 83:

In an oligopeptide named Alanylglycylphenylalanylisoleucine, the number of \( sp^2 \) hybridised carbons is __________.

Correct Answer: 10
View Solution



In the given structure, we need to identify the carbons that are \( sp^2 \) hybridised.


The following carbons are \( sp^2 \) hybridised:


1. The carbonyl carbon of the peptide bond between alanine and glycine.

2. The carbonyl carbon of the peptide bond between phenylalanine and alanine.

3. The carbonyl carbon of the carboxyl group at the C-terminal of isoleucine.

4. The carbon of the phenyl group attached to the aromatic ring in phenylalanine.


Hence, counting all the \( sp^2 \) hybridised carbons in the structure:


One from each of the peptide bonds (two bonds).

One from the carboxyl group (terminal carbon).

One from the aromatic ring in phenylalanine (C6H5 group).


Thus, the number of \( sp^2 \) hybridised carbons is 10. Quick Tip: To identify \( sp^2 \) hybridised carbons, focus on carbonyl groups in peptide bonds and carboxyl groups, as well as carbon atoms involved in double bonds, such as those in aromatic rings.


Question 84:

An analyst wants to convert 1L HCl of pH = 1 to a solution of HCl of pH 2. The volume of water needed to do this dilution is ______ mL. (Nearest Integer)

Correct Answer: 9000 mL
View Solution

Given: \[ M_1 = 10 (pH = 1, concentration of HCl) \] \[ M_2 = 10^{-2} (pH = 2, concentration of HCl) \] \[ V_1 = 1L (initial volume of HCl) \]
Using the dilution formula: \[ (M_1 \times V_1) = (M_2 \times V_2) \] \[ 10 \times 1 = 10^{-2} \times V_2 \] \[ V_2 = 10L = 10000 mL \]
Thus, water added = \(10000 - 1000 = 9000 mL\) Quick Tip: For dilution, the product of concentration and volume remains constant. You can solve it using the dilution equation.


Question 85:

Three organic compounds A, B, and C were allowed to run in thin layer chromatography using hexane and gave the following result. The Rf value of the most polar compound is ________ x \(10^{-2}\)

  • (1) 25
  • (2) 0.25
  • (3) 0.75
  • (4) 1
Correct Answer: (25)
View Solution

Question 86:

80 mole percent of MgCl\(_2\) is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of MgCl\(_2\) at 38°C is ______ mm Hg. (Nearest integer)

Correct Answer: (48)
View Solution

Question 87:




The value of \( x \) in compound \( D \) is


Question 88:

At 600K, the root mean square (rms) speed of gas X (molar mass = 40) is equal to the most probable speed of gas Y at 90K. The molar mass of gas Y is ______ g/mol (Nearest integer).

Correct Answer: 4 g/mol
View Solution

Question 89:

The reaction \( 2NO + Br_2 \to 2NOBr \) takes place through the mechanism given below: \[ NO + Br_2 \leftrightarrow NOB_2 \quad (fast) \] \[ NOB_2 + NO \to 2NOBr \quad (slow) \]

The overall order of the reaction is _______.

Correct Answer: 3
View Solution

Question 90:

Values of work function (\(W_0\)) for a few metals are given below.
The number of metals which will show the photoelectric effect when light of wavelength 400 nm falls on it is

Correct Answer: 3
View Solution

Also Check:

JEE Main 12 April 2023 Shift 1 Question Paper with Answer Key: Coaching Institute PDF

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JEE Main 2023 Paper Analysis 12 April Shift 1

JEE Main 2023 Paper Analysis for the exam scheduled on 12 April Shift 1 is available here. Candidates can check subject-wise paper analysis for the exam scheduled on 12 April Shift 1 here along with the topics with the highest weightage.

Also Check:

JEE Main 2023 Question Paper Session 2 (April)

JEE Main 2023 Question Paper Session 1 (January)

JEE Main aspirants can practice and check their exam prep level by attempting the question papers from the January Session. The table below shows JEE Main 2023 Question Paper PDF for Session 1 to practice.

JEE Main Previous Year Question Paper