JEE Mains 2 April 2025 Shift 1 (Previous Year Paper)

Calculation:

Given,

Number, n = 50

Prime number, p = 3

The exponent of 3 in 50! is given by,

(B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{3^2} \right\rfloor + \left\lfloor \frac{50}{3^3} \right\rfloor + \left\lfloor \frac{50}{3^4} \right\rfloor)

(B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor)

(B = 16 + 5 + 1 + 0 = 22)

∴ The maximum value of n is 22.

Concept:

Hyperbola parameters:

• Given one focus of hyperbola ( H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ) is at ( (\sqrt{10}, 0) ).
• The corresponding directrix is ( x = \frac{9}{\sqrt{10}} ).
• Recall the relationships: ( ae = c ) (distance of focus from origin), ( \frac{a}{e} ) is directrix, and ( c^2 = a^2 + b^2 ).
• The length of the latus rectum is ( l = \frac{2b^2}{a} ).

Calculation:

Given:

( ae = \sqrt{10} ) and ( \frac{a}{e} = \frac{9}{\sqrt{10}} )

Then,

( a^2 = 9 ) and ( e = \frac{\sqrt{10}}{3} )

Now,

( (ae)^2 = a^2 + b^2 \implies 10 = 9 + b^2 \implies b^2 = 1 )

( l = \frac{2b^2}{a} = \frac{2 \times 1}{3} = \frac{2}{3} )

Therefore,

( 9(e^2 + l) = 9 \left( \left( \frac{\sqrt{10}}{3} \right)^2 + \frac{2}{3} \right) = 9 \left( \frac{10}{9} + \frac{2}{3} \right) = 10 + 6 = 16 )

Hence, the correct answer is Option 3.

Calculation: 

11111 2 2 2 0 0 

No. of sequences (= \frac{10!}{5!3!2!} = 2520)

Note: The Sequence can start with 0.

Hence, the Correct answer is Option 3.

Calculation:

Given,

We are given the equation:

( (\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y)) )

Taking terms together:

( f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\cos x \sin y + \sin x \cos y) )

Using identities:

( f(2x+2y)\sin(x - y) = f(2x-2y)\sin(x + y) )

Rewriting,

( \frac{f(2x+2y)}{\sin(x + y)} = \frac{f(2x-2y)}{\sin(x - y)} )

Let,

( 2x + 2y = m ) and ( 2x - 2y = n )

Then,

( \frac{f(m)}{\sin\left(\frac{m}{2}\right)} = \frac{f(n)}{\sin\left(\frac{n}{2}\right)} = K ) (constant)

( \Rightarrow f(m) = K \sin\left(\frac{m}{2}\right) )

( \therefore f(x) = K \sin\left(\frac{x}{2}\right) )

Now,

( f'(x) = \frac{K}{2} \cos\left(\frac{x}{2}\right) )

Putting ( x = 0 ):

( \frac{K}{2} = \frac{1}{2} \Rightarrow K = 1 )

So,

( f'(x) = \frac{1}{2} \cos\left(\frac{x}{2}\right) )

( f''(x) = -\frac{1}{4} \sin\left(\frac{x}{2}\right) )

Now,

( 24 f''\left(\frac{5\pi}{3}\right) = 24 \left( -\frac{1}{4} \sin\left(\frac{5\pi}{6}\right) \right) )

( = -6 \times \frac{1}{2} = -3 )

∴ The correct answer is Option 2.

Solution:

(|A| = 0)

(\alpha \beta - (-1)(6) = 0)

(\alpha \beta + 6 = 0)

(\alpha \beta = -6)

(\alpha + \beta = 1)

(\implies \alpha = 3, \beta = -2)

(A = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix})

(A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix})

(\therefore A^2 = A)

(A = A^2 = A^3 = A^4 = A^5 = \cdots )

((I + A)^8)

(= I + {}^8C_1 A + {}^8C_2 A^2 + \cdots + {}^8C_8 A^8)

(= I + A ({}^8C_1 + {}^8C_2 + \cdots + {}^8C_8))

(= I + A (2^8 - 1))

(= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} (256 - 1))

(= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} 255)

(= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \ 1530 & -510 \end{bmatrix})

(= \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix})

Hence, the Correct answer is Option 4.

Calculation:

( \left(\frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x-1}{x-x^{1/2}}\right)^{10}, ; x>1 )

Simplify the terms inside the bracket:

( \frac{x+1}{x^{2/3}+1-x^{1/3}} = x^{1/3}+1 )

( \frac{x-1}{x-x^{1/2}} = 1 + x^{-1/2} )

Subtracting:

( (x^{1/3}+1)-(1+x^{-1/2}) = x^{1/3}-x^{-1/2} )

So the expression reduces to

( (x^{1/3}-x^{-1/2})^{10} )

General term:

( T_{r+1} = \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^{r} )

( = \binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac{r}{2}} )

For the term independent of x:

( \frac{10-r}{3}-\frac{r}{2}=0 )

( 2(10-r)-3r=0 \Rightarrow 20-5r=0 \Rightarrow r=4 )

Required term:

( T_5=\binom{10}{4}(-1)^4=\binom{10}{4}=210 )

The correct answer is 210.

Solution:

(2\sqrt{2} \cos^2 \theta + 2 \cos \theta - \sqrt{6} \cos \theta - \sqrt{3} = 0 )

( \ (2 \cos \theta - \sqrt{3})(\sqrt{2} \cos \theta + 1) = 0)

(\ \cos \theta = \frac{\sqrt{3}}{2}, -\frac{1}{\sqrt{2}} )

Number of solutions = 8.

Hence, the correct answer is Option 3.

Concept:

Arithmetic Progression (A.P.):

• An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant, called the common difference (d).
• The sum of first n terms of an A.P. is given by (S_n = \frac{n}{2} [2a + (n-1)d]), where a is the first term.

 

Calculation:

Sum of first 12 odd terms in A.P. is

 

( S_{12} = \frac{12}{2} [2a + (12-1) \times 2d] = 6[2a + 22d] )

Equate to the given sum,

( 6[2a + 22d] = -\frac{72}{5} a \Rightarrow 12a + 132d = -\frac{72}{5} a )

Multiply both sides by 5,

( 60a + 660d = -72a \Rightarrow 132a + 660d = 0 \Rightarrow 132a + 132 \times 5 d = 0 \Rightarrow a = -5d )

Now, sum of first n terms (S_n) = 0

( \frac{n}{2} (2a + (n-1)d) = 0 \Rightarrow \frac{n}{2} (2(-5d) + (n-1)d) = 0 )

( \Rightarrow \frac{n}{2} (-10d + nd - d) = 0 \Rightarrow \frac{n}{2} (nd - 11d) = 0 \Rightarrow \frac{nd(n-11)}{2} = 0 )

Since d ≠ 0 and n ≠ 0  ,

( n - 11 = 0 \Rightarrow n = 11 )

Hence, the correct answer is Option 1.

Concept:

Finding local maxima and minima of a cubic function:

• To find local maxima and minima of a cubic function, first differentiate the function and find the critical points by setting the derivative equal to zero.
• The critical points correspond to values of xx

where the function attains local maxima or minima.

• Use given conditions on the critical points to find the parameters and evaluate the function at required points.

 

Calculation:

Given,

( f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1 )

First derivative,

( f'(x) = 6x^2 - 18ax + 12a^2 )

Set ( f'(x) = 0 ) to find critical points:

( 6x^2 - 18ax + 12a^2 = 0 )

Dividing by 6:

( x^2 - 3ax + 2a^2 = 0 )

Let the roots be ( p ) and ( q ). Then, by Vieta's formulas,

• ( p + q = 3a )
• ( pq = 2a^2 )

Given condition,

( p^2 = q )

Substitute ( q = p^2 ):

( p + p^2 = 3a )

( p \cdot p^2 = p^3 = 2a^2 )

From sum,

( a = \frac{p + p^2}{3} )

From product,

( a^2 = \frac{p^3}{2} )

Squaring the sum and equating to product:

( \left( \frac{p + p^2}{3} \right)^2 = \frac{p^3}{2} )

Solving for ( p ) gives ( p = 2 ).

Then,

( a = \frac{2 + 4}{3} = 2 )

Finally, evaluate ( f(3) ):

( f(3) = 2(3)^3 - 9(2)(3)^2 + 12(2)^2 (3) + 1 )

( = 54 - 162 + 144 + 1 = 37 )

∴ The correct answer is Option 4.

Calculation:

( \frac{2 + k^2 z}{k + \bar{z}} = kz )

( |z|^2 = 1 )

( k = 2 )

Point P = (2, 4), center of circle C = (1, 2) 

Equation of a circle: ( (x - 1)^2 + (y - 2)^2 = 1 )

Maximum distance from point Pto the circle is given by:

( OP + r = \sqrt{(2 - 1)^2 + (4 - 2)^2} + 1 = \sqrt{1 + 4} + 1 = \sqrt{5} + 1 )

Hence, the correct answer is Option 1.

Solution:

( \text{Let } \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} )

( a_1^2 + a_2^2 + a_3^2 = 1 )

( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}, \quad \vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}, \quad \vec{d} = \hat{k} )

( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{\vec{a} \cdot \vec{d}}{|\vec{d}|} )

Calculate magnitudes:

( |\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 )

( |\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3 )

( |\vec{d}| = \sqrt{0^2 + 0^2 + 1^2} = 1 )

Set equal projections:

( \frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = \frac{a_3}{1} )

From the first equality:

( 2a_1 - a_2 + 2a_3 = 3a_3 \implies 2a_1 - a_2 = a_3 )

From the second equality:

( a_1 + 2a_2 - 2a_3 = 3a_3 \implies a_1 + 2a_2 = 5a_3 )

Solving 

( a_1 = \frac{7}{\sqrt{155}}, \quad a_2 = \frac{9}{\sqrt{155}}, \quad a_3 = \frac{5}{\sqrt{155}} )

∴ a unit vector along  (\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k}) \qquad)

Hence, the correct answer is Option 3.

Solution:

R= {(f,g): f(0) = g(1) and f(1) = g(0)}

Reflexive: (f, f) € R

⇒ f(0) = f(1) and f(1) = f(0) ⇒ must hold

⇒ But this is not true for all functions

So not reflexive

Symmetric: If (f,g) € R ⇒ (g, f) € R

Now, g(0) = f(1) and g(1) = f(0) ⇒ true

∴ symmetric

Transitive: If (f,g) € R and (g,h) € R

⇒ (f,h) € R

Now (f,g) € R= f(0)= g(1) and f(1)= g(0)

⇒ (g,h) € R ⇒ g(0)= h(1) and g(1) = h(0)

⇒ For (f,h) € R we need f(0) = h(1) and f(1) = h(0)

⇒ Now f(0)= g(1) = h(0) nd f(1) = g(0) = h(1)

Hence transitive

Hence, the correct answer is Option 2

Concept:

Limits and Expansion of Elementary Functions:

• To solve limits involving indeterminate forms like 0/0, use series expansion (Maclaurin expansion) of functions such as sinx and e^x.
• Standard expansions:
• sin x = x − x³/3! + x⁵/5! − ...
• e^x = 1 + x + x²/2! + x³/3! + ...
• Compare powers of x from numerator and denominator and simplify accordingly.

 

Calculation:

Given,

lim_x→0 [ x²·sin(αx) + (γ−1)·e^x2 ] / [ sin(2x) − βx ] = 3

Numerator:

x²·sin(αx) = x²·(αx − (αx)³/6 + ... ) = αx³ − α³x⁵/6 + ...

(γ − 1)·e^x2 = (γ − 1)·(1 + x² + x⁴/2! + ...) = (γ − 1) + (γ − 1)x² + ...

So, numerator ≈ (γ − 1) + (γ − 1)x² + αx³ + ...

Denominator:

sin(2x) = 2x − (2x)³/6 + ... = 2x − 8x³/6 + ...

So, denominator = 2x − βx − (8x³/6) + ... = (2 − β)x − (4/3)x³ + ...

Now rewrite limit as:

lim_x→0 [ (γ−1) + (γ−1)x² + αx³ + ... ] / [ (2−β)x − (4/3)x³ + ... ] = 3

Now, use expansion up to x⁰ term after dividing:

⇒ lim_x→0 (γ−1) / (2−β)x = 0 unless (γ−1) = 0

So, γ − 1 = 0 ⇒ γ = 1

Then numerator becomes αx³ + ...

Denominator: (2 − β)x − (4/3)x³ + ...

⇒ lim = αx³ / [ (2 − β)x − (4/3)x³ ]

Now multiply numerator and denominator by x⁻³:

⇒ α / [ (2 − β)x⁻² − (4/3) ]

Now take limit x → 0, x⁻² → ∞

For limit to be finite, coefficient of x⁻² in denominator must be 0

⇒ (2 − β) = 0

⇒ β = 2

Now denominator = −(4/3),

numerator = α ⇒ α / (−4/3) = 3

⇒ α = −4

Given γ = 1, β = 2, α = −4

β + γ − α = 2 + 1 − (−4) = 7

∴ β + γ − α = 7

Calculation:

Given,

( \Delta = \begin{vmatrix} 3 & 1 & \beta \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 0 )

( 3(\alpha + 2) - 1(2 + 1) + \beta (4 - \alpha) = 0 )

( 3\alpha + 6 - 3 + 4\beta - \alpha \beta = 0 )

( 3\alpha + 4\beta - \alpha \beta + 3 = 0 )

Also,

( \Delta_3 = \begin{vmatrix} 3 & 1 & 3 \ 2 & \alpha & -3 \ 1 & 2 & 4 \end{vmatrix} = 0 )

( 3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 0 )

( 12\alpha + 18 - 11 + 12 - 3\alpha = 0 )

( 9\alpha + 19 = 0 \Rightarrow \alpha = -\frac{19}{9} )

Putting in previous equation:

( 3\alpha + 4\beta - \alpha \beta + 3 = 0 )

( \Rightarrow \beta = \frac{6}{11} )

Now,

( 22\beta - 9\alpha = 22 \left(\frac{6}{11}\right) - 9 \left(-\frac{19}{9}\right) )

( = 12 + 19 = 31 )

∴ The correct answer is Option 2.

Concept:

Sequence and Series – Recurrence Relations:

• For a sequence defined by (P_n = \alpha^n + \beta^n), the terms satisfy the linear recurrence (P_n = (\alpha+\beta),P_{n-1} - (\alpha\beta),P_{n-2}.)
• Here, (\alpha+\beta) and (\alpha \beta ) are the sum and product of the roots of the characteristic quadratic.
• Given specific values of (P_n), one can solve for these two parameters and hence recover the quadratic.

Calculation:

Given:

• (P_{10}=123,;P_{9}=76,;P_{8}=47,;P_{1}=1.)

Let (s = \alpha+\beta) and (p = \alpha\beta).

Since (P_1 = \alpha+\beta), we have:

(s = P_1 = 1.)

Next, use the recurrence at (n=10):

(P_{10} = s,P_9 - p,P_8)

(\Rightarrow; 123 = 1\cdot76 - 47p)

(\Rightarrow; 47p = 76 - 123 = -47 ;\Rightarrow; p = -1.)

Therefore, the characteristic equation is:

(x^2 - s,x + p = 0 \Rightarrow x^2 - x - 1 = 0.)

We want the quadratic whose roots are (\tfrac{1}{\alpha}) and (\tfrac{1}{\beta}).

If (\alpha,\beta) satisfy (x^2 - x - 1=0), then:

(\tfrac{1}{\alpha} + \tfrac{1}{\beta} = \tfrac{\alpha+\beta}{\alpha\beta} = \tfrac{s}{p} = \tfrac{1}{-1} = -1)

(\tfrac{1}{\alpha\beta} = \tfrac{1}{p} = -1)

Hence, the required equation is:

(x^2 - \left(\tfrac{1}{\alpha} + \tfrac{1}{\beta}\right)x + \tfrac{1}{\alpha\beta} = 0 \Rightarrow x^2 + x - 1 = 0.)

∴ The correct answer is Option 2.

Calculation:

<br>

 

PS + PS' = 2 × 3 √2 

⇒ b² = a² (1 - e²) ⇒ 9 = 18( 1 - e²)

⇒ e = (\frac{1}{\sqrt2})

Diretrix x = (\frac{a}{e} = \frac{3\sqrt2}{\frac{1}{\sqrt2}{}} = 6)

⇒ PS.PS' = (PS.PS' = | \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta - 6)|)

 = (\frac{1}{2}| 18 cos^2\theta - 36|)

⇒ (PS. PS')_max = 18 and (PS .PS')_min = 9

Sum = 27

Hence, the correct answer is Option 4.

Concept:

Coordinate Geometry - Area of Triangle using Vector Approach:

• To find the area of triangle PQR, we use the coordinates of points P, Q, and R.
• Points Q and R lie on the line given by the parametric equations: (\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}).
• Distance QR is given as 5 units.
• Point P is fixed at (0, 2, 3).
• Area of triangle PQR is given by (\frac{1}{2} \times | \overrightarrow{PQ} \times \overrightarrow{PR}|)

• We find the point M on the line such that vector (\overrightarrow{PM} )is perpendicular to the line.
• Using the condition of perpendicularity, we solve for the parameter(\lambda)
• Calculate the length PM which is the height of the triangle from P to line QR.
• Use QR = 5 and height PM to calculate the area.

 

Calculation:

Given,

(M(5\lambda - 3, 2\lambda + 1, 3\lambda - 4))

Directions of PM:

(\implies 5\lambda - 3, 2\lambda + 1 - 2, 3\lambda - 4 - (-3) \implies 5\lambda - 3, 2\lambda - 1, 3\lambda - 1)

Directions of line L:

(5, 2, 3)

Since (\overrightarrow{PM}) is perpendicular  ( \overrightarrow{L}),

((5\lambda - 3) \times 5 + (2\lambda - 1) \times 2 + (3\lambda - 1) \times 3 = 0)

Simplifying,

(25\lambda - 15 + 4\lambda - 2 + 9\lambda - 3 = 0)

(38\lambda - 20 = 0)

(\lambda = \frac{20}{38} = \frac{10}{19})

Coordinates of M:

(M\left(5 \times \frac{10}{19} - 3, 2 \times \frac{10}{19} + 1, 3 \times \frac{10}{19} - 4 \right) = M\left(\frac{50 - 57}{19}, \frac{20 + 19}{19}, \frac{30 - 76}{19}\right) = M\left(-\frac{7}{19}, \frac{39}{19}, -\frac{46}{19}\right))

Calculate length PM:

(PM = \sqrt{\left(-\frac{7}{19} - (-3)\right)^2 + \left(\frac{39}{19} - 2\right)^2 + \left(-\frac{46}{19} - (-3)\right)^2})

(PM = \sqrt{\left(\frac{50}{19}\right)^2 + \left(\frac{1}{19}\right)^2 + \left(\frac{11}{19}\right)^2} = \sqrt{\frac{2500 + 1 + 121}{361}} = \frac{\sqrt{2622}}{19})

Calculate area of triangle:

(\text{Area} = \frac{1}{2} \times QR \times PM = \frac{1}{2} \times 5 \times \frac{\sqrt{2622}}{19} = \frac{5 \sqrt{2622}}{38} = \frac{m}{n})

Given relation involving m) and n:

(2m - 5 \sqrt{21} n = 0)

Hence, the correct answer is Option 2.

Concept:

Area of Tetrahedron from Areas of Perpendicular Triangles:

• In a tetrahedron where edges AB, AC, and AD are mutually perpendicular, the area of the face opposite to vertex A (i.e., triangle BCD) is related to the areas of the triangles ABC, ACD, and ADB.
• These three triangles form right-angled faces and their areas can be used to find the area of the face BCD using Pythagorean relation in three dimensions.

 

The relation for the area of triangle BCD is,

( \text{Ar}(\triangle BCD) = \sqrt{ \text{Ar}(\triangle ABC)^2 + \text{Ar}(\triangle ACD)^2 + \text{Ar}(\triangle ADB)^2 } )

 

Calculation:

Given,

( \text{Ar}(\triangle ABC) = 5, \quad \text{Ar}(\triangle ACD) = 6, \quad \text{Ar}(\triangle ADB) = 7 )

⇒ ( \text{Ar}(\triangle BCD) = \sqrt{5^2 + 6^2 + 7^2} = \sqrt{25 + 36 + 49} = \sqrt{110} )

∴ The area of triangle BCD is (\sqrt{110}), hence the correct answer is Option 3.

A +  I = (\begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix} )

⇒A = (\begin{bmatrix} 0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1 \end{bmatrix} )

Given det(A)= −4.

det(A)= 0⋅(0. 1 − 0. 1)−a. (2. 1−0. a)+1. (2. 1−0. a)

= - a. (2)+1. (2) =−2a + 2

Set equal to −4−4-4 -4

⇒ 2a + 2 =  -4 ⟹ −2a = −6 ⟹a = 3:

Also

det((a+1) adj ((a−1)A)) =2^m3ⁿ

• For a 3 × 3 Matrix, det(adj)(B) = (det(B))²
• For a scalar k det(kB) = k³det(B)

Let’s plug in values:

⇒ det((a−1) A) = (a−1)³.det(A) = (3−1)³.(−4)=2³.(−4) = 8.(−4) = −32

⇒ det(adj(a - 1)A)) = (-32)² = 1024

⇒ (a +1)³ = (3 + 1)³ = 64

⇒ det((a+1)adj((a−1)A)) = 64⋅1024 = 65536

Thus, 65536 = 2¹⁶3⁰

⇒ Then m = 16,n = 0

⇒ m + n  = 16 + 0 = 16

Concept:

Focal Chord of Parabola and Circle Properties:

• The focal chord of the parabola ( y^2 = 4ax )passes through points on the parabola making a certain angle with the x-axis.
• The circle with one diameter as the focal chord touches the y-axis at a certain point.

• The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.
• The equation of the circle with diameter PS and its intersection with the y-axis is used to find (\alpha)

 

Calculation:

Given,

The parabola: ( y^2 = 4x )

Focal chord PQ makes an angle of (60^\circ) with positive x-axis.

Using slope( m = \tan 60^\circ = \sqrt{3})

( \tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies t = \sqrt{3} )

Coordinates of P: ( (3, 2\sqrt{3}) )

Circle with diameter PS and focus S:

( (x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0 )

At x = 0,

( (-1)(-3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0 \implies (y - \sqrt{3})^2 = 0 \implies y = \sqrt{3} = \alpha )

Calculate ( 5\alpha^2:)

( 5 \alpha^2 = 5 (\sqrt{3})^2 = 5 \times 3 = 15 )

∴ The correct answer is Option 1.

CONCEPT:

Aromaticity and Hückel's Rule

• Aromatic compounds are cyclic, planar molecules with a conjugated π-electron system.
• They must follow Hückel’s rule: a compound is aromatic if it contains (4n + 2) π-electrons, where n is a non-negative integer (n = 0, 1, 2, ...).
• The compound must be fully conjugated (i.e., every atom in the ring has a p-orbital) and must be planar to allow continuous overlap of p-orbitals.

EXPLANATION:

• (a) Cyclopentadienyl cation: Has 6 π-electrons → follows Hückel’s rule → Aromatic
• (b) Cyclopentadienyl anion: Has 4 π-electrons → does not satisfy (4n + 2) rule → Not Aromatic
• (c) 4-membered ring with 6 π-electrons → satisfies Hückel’s rule → Aromatic
• (d) 4-membered ring with 6 π-electrons (similar to c) → Aromatic
• (e) Cycloheptatrienyl cation (tropylium ion) → has 6 π-electrons → Aromatic
• (f) Cycloheptatrienyl anion: Has 8 π-electrons → does not follow Hückel’s rule → Not Aromatic
• (g) 4n π-electron system (non-planar rectangle) → Not Aromatic
• (h) Cyclopropyl cation: Has 2 π-electrons → satisfies Hückel’s rule for n = 0 → Aromatic

Therefore, the correct answer is: a, c, d, e, h are aromatic and b, f, g are not aromatic.

CONCEPT:

Reaction Pathway Involving Alkyl Halide → Alkyne → Ketone

• Alkyl halides can undergo elimination reactions with alcoholic KOH to form alkenes.
• Alkenes react with bromine to give vicinal dibromides (bromination across the double bond).
• Vicinal dibromides treated with strong bases like NaNH₂ in alcohol can undergo double elimination to yield alkynes.
• Alkynes undergo hydration in the presence of H₂SO₄ and HgSO₄ to yield ketones (Markovnikov addition of water).

EXPLANATION:

• Given compound [A] is an optically active C₄H₉Br, most likely 2-bromobutane.
• On treatment with hot alcoholic KOH, [A] undergoes dehydrohalogenation to give but-2-ene [B].
• [B] reacts with Br₂ to form vicinal dibromide [C], i.e., 2,3-dibromobutane.
• [C] reacts with alcoholic NaNH₂ to undergo two eliminations, forming but-2-yne [D].
• [D] undergoes hydration with dilute H₂SO₄ and HgSO₄ at 333 K to form butan-2-one [E].

Therefore, the IUPAC name of compound [E] is Butan-2-one.

CONCEPT:

Trends in Group-17 Elements (Halogens)

• The halogens include Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).
• Covalent Radius: Increases down the group due to the addition of new shells (F < Cl < Br < I).
• Ionic Radius: Also increases as you go down the group (F⁻ < Cl⁻ < Br⁻ < I⁻).
• First Ionization Energy: Decreases down the group due to increasing atomic size (F > Cl > Br > I).
• Electron Affinity: Typically increases down the group, but fluorine shows an irregularity—it has lower electron affinity than chlorine despite being smaller.

EXPLANATION:

• Electron Affinity: The irregularity arises because fluorine is very small, and adding an extra electron leads to high electron–electron repulsion, reducing the energy released.
• All other properties—covalent radius, ionic radius, and first ionization energy—follow regular trends down the group without deviation in the first four elements.

Therefore, the correct answer is: B only

CONCEPT:

Henry’s Law and Temperature Dependence

• Henry’s Law Constant (K_H): It relates the solubility of a gas in a liquid to the pressure of the gas above the liquid.
• According to Henry's Law: C = K_H × P, where C is the concentration (solubility), P is the partial pressure, and K_H is Henry’s constant.
• Temperature Dependence: Solubility of gases in liquids generally decreases with increase in temperature, hence K_H values typically increase with temperature.
• The increase in K_H with temperature is not linear and varies for different gases. Light gases like He show a steeper increase.

EXPLANATION:

• Among the options, only graph 4 shows the correct qualitative trend: Henry’s constant (K_H) increases with temperature for all gases.
• Also, gases like He (non-polar and light) exhibit lower solubility and hence higher K_H, followed by N₂ and CH₄.
• This trend is consistent with the observation that lighter gases have higher Henry’s constants and are less soluble in water.

Therefore, the correct answer is: Option 4

CONCEPT:

Bohr’s Model and Radius of Electron Orbits

r ∝ n²

rₙ = r₁ × n²

• According to Bohr’s model of the hydrogen atom, the radius of the nth orbit is proportional to the square of the principal quantum number n:
• This means the radius of orbit n can be expressed as:
• where r₁ is the radius of the first orbit (ground state).

EXPLANATION:

• Radius of 3rd orbit relative to 1st orbit:

r₃ / r₁ = (3)² / (1)² = 9 / 1 = 9

Statement 1 is correct.

• Radius of 8th orbit relative to 4th orbit:

r₈ / r₄ = (8)² / (4)² = 64 / 16 = 4

Statement 2 is correct.

• Radius of 6th orbit relative to 4th orbit:

r₆ / r₄ = (6)² / (4)² = 36 / 16 = 2.25

Statement 3 says “three times larger”, but the actual ratio is 2.25.

Therefore, Statement 3 is incorrect.

• Radius of 4th orbit relative to 2nd orbit:

r₄ / r₂ = (4)² / (2)² = 16 / 4 = 4

Statement 4 is correct.

Therefore, the incorrect statement is: Radius of 6th orbit is three times larger than that of 4th orbit (Option 3).

CONCEPT:

Free Expansion of Gas and Thermodynamic Changes

• When a gas expands into a vacuum (free expansion), there is no external pressure opposing the expansion, i.e., P_ext = 0.
• In free expansion:
• No work is done by the gas (dw = 0) because the external pressure is zero.
• No heat is exchanged with the surroundings (dq = 0) if the system is insulated or thermally isolated.
• For an ideal gas undergoing free expansion, the internal energy (ΔU) remains unchanged since internal energy depends only on temperature and temperature is constant.

EXPLANATION:

• In the problem, two vessels A and B are connected with a stopcock.
• Initially, vessel A contains gas at certain pressure, and vessel B is evacuated (pressure in B is zero).
• Upon opening the stopcock, gas expands freely into vessel B (vacuum), so P_ext = 0.
• Since the assembly is immersed in water and allowed to equilibrate, temperature remains constant.
• Therefore, no work is done (dw = 0), no heat is exchanged (dq = 0), and internal energy change is zero (ΔU = 0).

Hence, the pressure in vessel B before opening the stopcock must be zero.

The correct answer is The pressure in the vessel B before opening the stopcock is zero.

CONCEPT:

Raoult's Law and Vapor Pressure of Solutions

• Raoult's law states that the total vapor pressure of an ideal solution is the sum of the partial vapor pressures of each component.
• The partial vapor pressure of each component is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
• Mathematically, P_total = P_Δ⁰ × X_Δ + P_B⁰ × X_B
• The component with the higher vapor pressure is the more volatile component.
• The least volatile component has the lower vapor pressure.

EXPLANATION:

• Given:
• Moles of volatile liquid Δ = 1
• Moles of volatile liquid B = 3
• Vapor pressure of pure Δ (P_Δ⁰) = 200 mm Hg
• Total vapor pressure of solution (P_solution) = 500 mm Hg
• Mole fractions:
• X_Δ = 1 / (1 + 3) = 1/4 = 0.25
• X_B = 3 / (1 + 3) = 3/4 = 0.75
• Therefore, vapor pressure of pure B is 600 mm Hg.
• Comparing vapor pressures:
• P_Δ⁰ = 200 mm Hg
• P_B⁰ = 600 mm Hg

B has the higher vapor pressure, so B is the more volatile component.

Δ, having the lower vapor pressure, is the least volatile component.

Therefore, the vapor pressure of pure B is 600 mm Hg, and the least volatile component of the solution is A.

CONCEPT:

Stoichiometry and Limiting Reactant in Chemical Reactions

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

• The reaction between calcium carbonate and hydrochloric acid is a typical acid-carbonate reaction:
• Mole ratio between CaCO₃ and HCl is 1:2.
• The limiting reactant determines the amount of product formed.
• Molar masses are used to convert between moles and grams.

EXPLANATION:

• Given:
• Mass of CaCO₃ = 1000 g
• Concentration of HCl = 0.76 M (moles per liter)
• Volume of HCl = 250 mL = 0.250 L
• Molar masses (g/mol): Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5
• Calculate moles of CaCO₃:

Molar mass of CaCO₃ = 40 + 12 + (16 × 3) = 100 g/mol

Moles of CaCO₃ = 1000 g / 100 g/mol = 10 mol

• Calculate moles of HCl:

Moles of HCl = Molarity × Volume = 0.76 mol/L × 0.250 L = 0.19 mol

• Determine limiting reactant:
• Reaction requires 2 moles of HCl per mole of CaCO₃.
• Moles of HCl required for 10 moles CaCO₃ = 10 × 2 = 20 moles.
• Available moles of HCl = 0.19 moles < 20 moles required, so HCl is limiting reactant.
• Calculate moles of CaCl₂ formed (equal to moles of CaCO₃ reacted):

Moles of CaCl₂ formed = moles of HCl / 2 = 0.19 / 2 = 0.095 mol

• Calculate molar mass of CaCl₂:

Ca = 40, Cl = 35.5 × 2 = 71, Total = 40 + 71 = 111 g/mol

• Calculate mass of CaCl₂ produced:

Mass = moles × molar mass = 0.095 × 111 = 10.545 g

Therefore, the mass of CaCl₂ formed is 10.545 g.

When equal volumes are mixed molarity reduce to half.

CONCEPT:

• When two salt solutions AB₂ and XY are mixed, precipitation of AY₂ will occur if the ionic product (Q_sp) exceeds the solubility product constant (K_sp).
• Q_sp is calculated as:
  

Q_sp = [A⁺][Y⁻]²

• Since equal volumes of solutions are mixed, the concentration of ions is halved due to dilution.

EXPLANATION:

(\text{For precipitation } Q_{SP} = [A^{+2}] [Y^-]^2 > K_{SP} \ (1) , Q_{SP} = (1.8 \times 10^{-3}) \left( \frac{5}{2} \times 10^{-4} \right)^2 < K_{SP} \ (2) , Q_{SP} = (10^{-4}) (0.4 \times 10^{-3})^2 < K_{SP} \ (3) , Q_{SP} = (10^{-2}) (10^{-2})^2 > K_{SP} \ (4) , Q_{SP} = \left( \frac{1.5}{2} \times 10^{-4} \right) \left( \frac{1.5}{2} \times 10^{-3} \right)^2 < K_{SP})

Conclusion:

Only option 3 gives Q_sp > K_sp, so precipitation of AY₂ occurs.

CONCEPT:

Hybridization, Molecular Geometry, and Dipole Moment

• Hybridization describes the mixing of atomic orbitals on the central atom to form new hybrid orbitals in a molecule.
• The presence of lone pairs affects molecular geometry and the resultant dipole moment.
• Dipole moment depends on the geometry and the difference in electronegativities; molecules with symmetrical geometry or lone pairs arranged symmetrically often have zero dipole moment.
• Hybridizations considered here are sp², sp³, sp³d, and sp³d² corresponding to trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral geometries, respectively.
• Lone pairs on the central atom create asymmetry, often leading to non-zero dipole moments.

EXPLANATION:

• Among molecules with non-zero dipole moment, ClF₃ has the highest number of lone pairs (2) on the central atom.
• Its hybridization is sp³d.

Therefore, the correct answer is: sp³d (Option 4).

CONCEPT:

• Vanillin is an aromatic aldehyde with a phenolic -OH group and a methoxy (-OCH3) group attached to the benzene ring.
• The phenolic group (-OH attached to benzene) is acidic and soluble in NaOH due to phenol's acidic nature.
• Tollen's reagent detects aldehyde groups (-CHO) by oxidizing them to carboxylic acids, giving a silver mirror. Vanillin contains an aldehyde group, so it reacts with Tollen's reagent.
• Self-aldol condensation requires the presence of acidic alpha-hydrogen atoms adjacent to the carbonyl group to form enolate ions under basic conditions.
• Vanillin lacks alpha-hydrogen atoms adjacent to the aldehyde group because the carbonyl carbon is directly bonded to the aromatic ring, thus it does not undergo self-aldol condensation easily.

EXPLANATION:

​

Phenolic group soluble in NaOH   

Benzaldehyde derivative react with Tollen’s reagent.

Vanillin does not give self-aldol reaction due to lack of acidic H for condensation. 

• Statement I: Vanillin will react with NaOH due to the acidic phenolic -OH group which forms phenolate ions soluble in NaOH.
• It also reacts with Tollen's reagent because of the aldehyde (-CHO) group, which is oxidized to carboxylate ions, producing the silver mirror effect.
• Statement II: Vanillin will not undergo self-aldol condensation easily because it lacks acidic alpha-hydrogens required to form the enolate ion necessary for aldol condensation.

Therefore, Statement I is correct but Statement II is incorrect.

CONCEPT:

• Most naturally occurring amino acids contain one chiral center, except glycine, which has two hydrogen atoms attached to the alpha carbon, making it achiral.
• Some amino acids, like isoleucine and threonine, contain two chiral centers.
• Optical activity arises from chirality, so glycine is optically inactive, whereas other amino acids with chiral centers are optically active.
• Glutamic acid and aspartic acid both contain a –COOH group in their side chains (carboxylic acid groups), so glutamic acid is not unique in this feature.
• Cysteine contains a thiol (-SH) group in its side chain, which can form disulfide bonds by dimerization or oxidation, contributing to protein structure stabilization.

EXPLANATION:

• Statement 1: Incorrect. While most amino acids have one chiral center, isoleucine has two chiral centers, and glycine has none.
• Statement 2: Incorrect. Glycine is achiral and thus optically inactive.
• Statement 3: Incorrect. Both glutamic acid and aspartic acid have –COOH groups on their side chains.
• Statement 4: Correct. Cysteine contains a free thiol (-SH) group that readily undergoes dimerization to form disulfide bonds.

Therefore, the correct statement is: Amino acid cysteine easily undergoes dimerization due to the presence of free SH group.

CONCEPT:

• The basicity of amines in aqueous solution depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (H⁺).
• Alkyl groups like ethyl (CH₃CH₂–) are electron-donating via +I (inductive) effect, which increases electron density on nitrogen and hence increases basicity.
• However, steric hindrance and solvation effects influence basicity. Primary and secondary amines are generally more basic than tertiary amines in aqueous solution due to better solvation of the protonated form.
• Hydrazine (H₂N–NH₂) shows slightly different basicity due to electron interaction between the two nitrogen atoms.

EXPLANATION:

• NH₃ (Ammonia): Least basic among the given due to no alkyl groups.
• H₂N–NH₂ (Hydrazine): Slightly more basic than NH₃ because the lone pair on one nitrogen can be delocalized to some extent.
• CH₃CH₂NH₂ (Ethylamine, primary amine): More basic due to +I effect of ethyl group increasing electron density.
• (CH₃CH₂)₂NH (Diethylamine, secondary amine): Generally most basic as two alkyl groups donate electrons, increasing basicity.
• (CH₃CH₂)₃N (Triethylamine, tertiary amine): Less basic than secondary amine because of steric hindrance and poor solvation.

Therefore, the correct order of basicity is:

NH2–NH2,<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH

ONCEPT:

• Metallic radius refers to the size of an atom in a metallic state, influenced by electron shielding and effective nuclear charge.
• Gallium (Ga) has filled 3d¹⁰ subshell electrons that poorly shield the nucleus, leading to a stronger effective nuclear charge on the outer electrons, causing contraction in atomic size.
• Due to this d-electron poor shielding, Ga has a smaller metallic radius than Al, despite being below it in the periodic table.
• Ionic radius depends on the number of electrons and nuclear charge after ionization. Both Al³⁺ and Ga³⁺ lose three electrons.
• Since Ga has more protons and poor shielding, its ionic radius is larger than that of Al³⁺.

EXPLANATION:

(\implies \text{The metallic radius order of Al & Ga is} \ \mathrm{B} < \mathrm{Ga} < \mathrm{Al} < \mathrm{In} < \mathrm{Tl} \ \underbrace{\mathrm{Ga} < \mathrm{Al}}_{\text{(due to poor shielding of d-subshell electrons)}} \ \implies \text{The ionic radius order of } \mathrm{Al}^{+3} & \mathrm{Ga}^{+3} \text{ is} \ \mathrm{B}^{+3} < \mathrm{Al}^{+3} < \mathrm{Ga}^{+3} < \mathrm{In}^{+3} < \mathrm{Tl}^{+3})

• Statement I: Metallic radius of Al is less than that of Ga — This is incorrect. The order is Ga < Al due to d-electron poor shielding in Ga causing contraction.
• Statement II: Ionic radius of Al³⁺ is less than Ga³⁺ — This is correct. Ga³⁺ has larger ionic radius because the poor shielding effect expands the size compared to Al³⁺.

Therefore, the correct answer is: Statement I is incorrect but Statement II is correct.

CONCEPT:

• In coordination chemistry, Δ_o (octahedral crystal field splitting energy) is the energy difference between the higher and lower sets of d-orbitals in an octahedral complex.
• P refers to the pairing energy, which is the energy required to pair two electrons in the same orbital.
• For octahedral complexes (coordination number 6):
• If Δ_o < P, electrons prefer to occupy higher energy orbitals unpaired → high spin complexes.
• If Δ_o > P, electrons pair in lower energy orbitals → low spin complexes.
• In tetrahedral complexes (coordination number 4):
• The crystal field splitting energy Δ_t is smaller than Δ_o and always less than the pairing energy (Δ_t < P).
• Therefore, tetrahedral complexes are almost always high spin because electrons occupy orbitals singly to minimize pairing.
• Low spin tetrahedral complexes are rare.

Explanation:-

• Statement I: Correct. It accurately describes the relationship between Δ_o and P in octahedral complexes and the formation of high spin or low spin complexes.
• Statement II: Correct. It correctly states that because Δ_o < P in tetrahedral complexes, low spin complexes are rarely formed.

Therefore, both Statement I and Statement II are correct.

CONCEPT:

• Potassium ferrocyanide, K₄[Fe(CN)₆], reacts with various metal salts to form characteristic precipitates based on the metal ion involved.
• The nature and color of the precipitate help in identifying metal ions qualitatively.
• Specific reactions:
• CuSO₄ (acidified with acetic acid): Reacts with K₄[Fe(CN)₆] to form a chocolate brown precipitate of Cu₂[Fe(CN)₆].
• FeCl₃: Reacts with K₄[Fe(CN)₆] to form Prussian blue precipitate, Fe₄[Fe(CN)₆]₃.
• ZnCl₂ (neutralized with NH₄OH): Forms white or bluish white precipitate K₂Zn₃[Fe(CN)₆]₂.
• MgCl₂ and BaCl₂: Do not give such characteristic precipitates with K₄[Fe(CN)₆].

EXPLANATION:

(2\mathrm{CuSO}_4 + \mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{CH}_3\mathrm{COOH}} \mathrm{Cu}_2[\mathrm{Fe(CN)}_6] \downarrow + 2\mathrm{K}_2\mathrm{SO}_4 \ \text{(Chocolate brown ppt.)} \ 4\mathrm{FeCl}_3 + 3\mathrm{K}_4[\mathrm{Fe(CN)}_6] \rightarrow \mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3 \downarrow + 12\mathrm{KCl} \ \text{(Prussian Blue ppt.)} \ 3\mathrm{ZnCl}_2 + 2\mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{NH}_4\mathrm{OH}} \mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe(CN)}_6]_2 \downarrow + 6\mathrm{KCl} \ \text{(White or bluish white ppt.)})

• Option A: Correct. CuSO₄ with K₄[Fe(CN)₆] gives chocolate brown precipitate.
• Option B: Correct. FeCl₃ with K₄[Fe(CN)₆] gives Prussian blue precipitate.
• Option C: Correct. ZnCl₂ neutralized with NH₄OH reacts with K₄[Fe(CN)₆] to give white/bluish white precipitate.
• Option D: Incorrect. MgCl₂ does not give blue precipitate with K₄[Fe(CN)₆].
• Option E: Incorrect. BaCl₂ neutralized with NaOH does not give white precipitate with K₄[Fe(CN)₆].

Therefore, the correct tests with respective observations are A, B, and C only.

CONCEPT:

• Complete combustion of an organic compound produces CO₂ and H₂O.
• The masses of CO₂ and H₂O formed help determine the amount of carbon and hydrogen in the compound.
• Oxygen content is found by difference: total mass of compound minus mass of carbon and hydrogen.
• Using moles of each element, the empirical formula and empirical formula mass can be calculated.

EXPLANATION:

• Moles of C atoms = 0.033 mol (since 1 mol CO₂ contains 1 mol C)
• Moles of H atoms = 2 × 0.0315 = 0.063 mol (since 1 mol H₂O contains 2 mol H)
• Mass of C = 0.033 × 12 = 0.396 g
• Mass of H = 0.063 × 1 = 0.063 g
• Mass of O = Total mass - (mass of C + mass of H) = 1.0 - (0.396 + 0.063) = 0.541 g
• Moles of O = 0.541 / 16 = 0.0338 mol

Ratio of moles: C : H : O = 0.033 : 0.063 : 0.0338

Divide by smallest (0.033):

C : H : O ≈ 1 : 1.91 ≈ 2 : 1 : 1 (approximately)

Empirical formula: CH₂O

Empirical formula mass = (12 × 1) + (1 × 2) + (16 × 1) = 30 g/mol

Therefore, the empirical formula mass of compound (X) is 30 g/mol.

CONCEPT:

• Radical stability depends on resonance, hyperconjugation, and the type of carbon (primary, secondary, tertiary, or sp hybridized).
• Allylic and benzylic radicals are more stable due to resonance stabilization.
• sp hybridized carbon radicals (like alkynyl radicals) are less stable due to higher s-character and less electron density to stabilize the radical.
• Tertiary radicals are more stable than secondary, which are more stable than primary due to hyperconjugation and inductive effects.

EXPLANATION:

​

<br>

II most stable carbon radical due to resonance stablise 

I least stable carbon radical due to no stabilising factor. 

• Radical at carbon II (•CH) is allylic to the triple bond (C≡C), allowing resonance stabilization, making it the most stable radical.
• Radical at carbon I (H–C•≡C–) is an alkynyl radical (sp hybridized), which is the least stable due to lack of resonance and high s-character.
• Radicals at carbons III and IV are alkyl radicals with moderate stability, less than the allylic radical at II but more than alkynyl radical at I.

Therefore, the most stable radical is at carbon II and the least stable radical is at carbon I.

CONCEPT:

• The rate of hydrolysis of acyl compounds depends mainly on the nature of the leaving group.
• Better leaving groups increase the rate of hydrolysis because they stabilize the transition state and intermediate more effectively.
• The leaving group ability generally follows the order: Cl^– (acyl chloride) > carboxylate ester (–O–C=O) > alkoxy group (–O–alkyl) > amide (–NH₂).

EXPLANATION:

Rate of hydrolysis(\propto)  Leaving group ability 

(\mathrm{CH}_3 - \mathrm{CH}_2 - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{Cl} \quad \text{(p)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{O} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{CH}_3 \quad \text{(q)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{OEt} \quad \text{(r)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{NH}_2 \quad \text{(s)})

• (p) Acyl chloride (CH₃−CH₂−C=O−Cl): Has the best leaving group (Cl^–), so it hydrolyzes fastest.
• (q) Acid anhydride (CH₃−CH₂−C=O−O−C=O−CH₃): Next best leaving group (carboxylate), hydrolyzes slower than acyl chloride.
• (r) Ester (CH₃−CH₂−C=O−O−CH₂−CH₃): Alkoxy leaving group, less reactive than anhydrides.
• (s) Amide (CH₃−CH₂−C=O−NH₂): Amide group is poor leaving group, so hydrolyzes slowest.

Therefore, the correct order of rate of hydrolysis is p > q > r > s

CONCEPT:

• The molecular formula AX₄Y indicates a central atom A bonded to four atoms X and one atom Y.
• Element A is described as rarest, monoatomic, non-radioactive p-block element with the lowest ionization enthalpy among A, X, and Y — this corresponds to Xenon (Xe), a noble gas.
• Elements X and Y have the first and second highest electronegativities among all elements — these are Fluorine (F) and Oxygen (O), respectively.
• The compound formed is XeOF₄, where Xenon is bonded to four fluorines and one oxygen.
• The shape of XeOF₄ is determined by the VSEPR theory. Xenon has 6 electron pairs (5 bonding pairs and 1 lone pair), resulting in a square pyramidal molecular geometry.

EXPLANATION:

• Xenon (Xe) is rare, monoatomic, non-radioactive noble gas with lower ionization energy than F and O.
• Fluorine (F) has the highest electronegativity; Oxygen (O) has the second highest electronegativity.
• In XeOF₄, Xe is bonded to 4 fluorines and 1 oxygen, with one lone pair on Xe.
• This arrangement leads to square pyramidal molecular geometry according to VSEPR theory.

Therefore, the shape of the molecule AX₄Y (XeOF₄) is square pyramidal.

Calculation:

The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y, and z axis.

Angle made by the light with x, y & z axis is same.

Thus, cosα = cosβ = cosγ (α, β, and γ are the angles made by light with x, y, and z axes respectively)

Also, cos²α + cos²β + cos²γ = 1 [Sum of direction cosines]

⇒ α = cos⁻¹(1/√3)

Calculation:

(P + a/V²)(V - b) = RT

Therefore, [a] = [P][V²] = M L⁻¹ T⁻² L⁶ = M L⁵ T⁻²

[b] = [V] = L³

[ab⁻²] = M L⁵ T⁻² L⁻⁶ = M L⁻¹ T⁻²

⇒ Dimension of energy density

Calculation:

W_F = 20 × 1 = 20 J

Therefore, ΔKE = 20 J = (1/2) I ω²

I = M R² = 10 × 0.1² = 0.1 kg m²

Therefore, 20 = (1/2) × 0.1 × ω²

Implying, ω = 20 rad/sec

Calculation:

<br>

Location of image of A :-

(1/v) - (1/u) = (1/f)

⇒ (1/v) - (1/(v - 30)) = (1/20)

⇒ (1/v) = (1/60)

⇒ v = 60 cm

The magnification m = 2.

Since the size of the object is small with respect to the location, hence

dv = m² du ⇒ dv = 4 × 1 = 4 cm

h_i = m h₀ ⇒ h_i(dy) = 2 × 2 = 4 cm

Therefore, Angle made with principle a×is = -45°

Calculation:

Given:

Surface charge density on Plate 1, σ₁ = +σ

Surface charge density on Plate 2, σ₂ = -2σ

Point charge, q = +q at midpoint between plates

Distance between plates = d (charge at midpoint, so at d/2 from each plate)

Electric field due to Plate 1 (positive charge):

⇒ E₁ = σ / (2ε₀) directed away from Plate 1

Electric field due to Plate 2 (negative charge):

⇒ E₂ = 2σ / (2ε₀) = σ / ε₀, directed toward Plate 2 (since negative charge)

Since charge q is at midpoint, the net electric field E_net is vector sum:

⇒ E_net = E₁ (rightward) + E₂ (rightward, since Plate 2's field points toward it)

⇒ E_net = (σ / 2ε₀) + (σ / ε₀) = (3σ) / (2ε₀)

Force on point charge q:

⇒ F = q × E_net

⇒ F = q × (3σ) / (2ε₀)

∴ Force on charge q = (3σ q) / (2ε₀)

Calculation:

Time taken to cross = 30 seconds = 30 / 3600 hr = 1 / 120 hr

Perpendicular velocity, V_⊥ = 27 / 2 km/hr

⇒ Width of river = V_⊥ × time

⇒ Width = (27 / 2) × (1 / 120) km

⇒ Width = (27 × 1000) / (2 × 120) m

⇒ Width = 112.5 m

∴ Width of the river is 112.5 m

Calculation:

Let the position where electric field is zero be at distance x from charge +q at origin.

Then, distance from +9q is (d − x)

Equating magnitudes of electric fields,

⇒ (kq) / x² = (k × 9q) / (d − x)²

⇒ 1 / x² = 9 / (d − x)²

⇒ (d − x)² = 9x²

⇒ d − x = 3x

⇒ x = d / 4

∴ The coordinates of point P are (d / 4, 0, 0)

Calculation:

Given, V = 4.2 V

Battery capacity = 5800 mAh = 5800 × 3600 × 10⁻³ C

⇒ Q = 5800 × 3.6 = 20880 C

⇒ Energy = V × Q = 4.2 × 20880 = 87700 J

⇒ Energy = 87.7 kJ

∴ Energy stored in the battery when fully charged is 87.7 kJ

Calculation:

Given:

x₁ = √7 sin 5t cm

x₂ = 2√7 sin(5t + π/3) cm

Angular frequency, ω = 5 rad/s

Using phasor diagram:

Angle between x₁ and x₂ = 60°

Resultant amplitude A:

A = √[ (√7)² + (2√7)² + 2 × √7 × 2√7 × cos 60° ]

⇒ A = √[ 7 + 4 × 7 + 2 × √7 × 2√7 × (1/2) ]

⇒ A = √49 = 7 cm

Maximum acceleration:

a_max = A × ω²

⇒ a_max = 7 × 25 cm/s²

⇒ a_max = 175 × 10^-2 m/s²

∴ The value of x is 175

Calculation:

We are given: μ_r = 1 + χ

⇒ χ = μ_r - 1

Also, μ = μ₀ μ_r

⇒ μ_r = μ / μ₀

Substitute into the equation for χ:

⇒ χ = (μ / μ₀) - 1

∴ χ = (μ / μ₀) - 1

Calculation:

From the circuit diagram,

⇒ 5i = 20 / 400

⇒ 5i = 1 / 20

⇒ i = 1 / 100 = 0.01 A = 10 mA (Load current)

Given load voltage, V_L = 5 V

⇒ Load resistance, R_L = V_L / i

⇒ R_L = 5 / (10 × 10^-3) = 500 Ω

∴ Load current i = 10 mA and Load resistance R_L = 500 Ω

Calculation:

For adiabatic process,

⇒ dQ = 0

⇒ Molar heat capacity = 0

⇒ From first law, dQ = dU + dW

⇒ 0 = dU + dW ⇒ dU = -dW

Also,

⇒ dU = (f / 2) × nR × dT

∴ Only option (3) is correct.

Concept

• For rotational equilibrium, the net torque about the pivot must be zero.
• Torque = Force × perpendicular distance from pivot.

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Explanation

The square lamina of side (10,cm) is pivoted at point (O).

Forces of (10,N) act at different points producing clockwise and anticlockwise torques about (O).

Calculations

Taking moments about point (O):

• The forces at A and B produce clockwise moments.
• The forces at C and the unknown force (F) produce anticlockwise moments.

Balancing torques for equilibrium:

(F \times 10 = 10 \times 10 )

(\Rightarrow F = 10,N )

Thus, the magnitude of the force (F) required for equilibrium is (10,N).

​Calculation:

Given:

⇒ B₁ = (μ₀ i) / (2R)

⇒ B₂ = B₁ × sin³ θ

⇒ Therefore, B₂ / B₁ = sin³ θ

⇒ sin θ = 4 / 5

⇒ (sin θ)³ = (4 / 5)³ = 64 / 125

∴ The ratio B₂ / B₁ = 64 / 125.

Concept Used:

Energy levels in hydrogen-like atoms are proportional to the atomic number divided by the square of the principal quantum number

E = k × Z² / n² (where k is a constant)

Excited states correspond to n values greater than 1

Explanation:

From the given data:

⇒ For 1^st excited state, n = 2

⇒ For 2^nd excited state, n = 3

∴ Only statements A and B are correct.

Calculation:

 

Before cutting

(\alpha = \frac{M\ell^2}{12} \quad \cdots (i))

After cutting

For each half:

⇒ Mass = M / 2

⇒ Length = ℓ / 2

Moment of inertia for each half, α^' (single part) = (M/2) × (ℓ/2)² / 12

⇒ α^' = (M/2) × (ℓ² / 4) / 12 = (M × ℓ²) / 96

Total moment of inertia, α' = 2 × (M × ℓ² / 96) = (M × ℓ²) / 48

⇒ α' = α / 4

∴ Correct option is (2).

Calculation:

⇒ (1.5 / v) - (1 / -0.2) = (1.5 - 1) / 0.4

⇒ (1.5 / v) = 1.25 - 5 = -3.75

⇒ v = 1.5 / (-3.75) = -0.4 m

∴ v = -1.2 m.

Explanation:

Coefficient of viscosity (η):

Expressed as force per unit area per velocity gradient.

⇒ Dimensional formula: [M¹ L^-1 T^-1].

Intensity (I):

Power delivered per unit area perpendicular to wave propagation.

SI unit is watt per square meter (W/m²).

⇒ Dimensional formula: [M¹ L⁰ T^-3].

Pressure gradient:

Represents the rate of change of pressure with distance.

Its unit is pascal per meter (Pa/m).

⇒ Dimensional formula: [M L^-2 T^-2].

Compressibility (K):

Defined as the fractional change in volume per unit increase in pressure.

Inverse of bulk modulus (elastic modulus of volume).

⇒ Dimensional formula: [M^-1 L¹ T²].

Calculation:

The FBD is shown below

From equilibrium condition:

⇒ q × (σ / 2 ε₀) = mg

⇒ σ = (2 ε₀ mg) / q

⇒ σ = (2 × 8.85 × 10^-12 × 100 × 10^-6 × 10) / (10 × 10^-6)

⇒ σ = 17.7 × 10^-10 C/m²

⇒ σ = 1.77 nC/m²

∴ Surface charge density σ = 1.77 nC/m².

Calculation:

Using KE_max:

⇒ KE_max = (hc / λ) - φ

Momentum of electron:

⇒ p = √(2m KE_max) = √[2m ((hc / λ) - φ)]

Diameter of path:

⇒ d_A-B = 2r = 2 (p / qB) = 2 [√(2m ((hc / λ) - φ)) / eB]

Thus,

⇒ d_A-B = (2 √(2m ((hc / λ) - φ))) / (eB)

⇒ d_A-B = √[ (8m ((hc / λ) - φ)) / (eB) ]