JEE Advanced Mix Test 5

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JEE Advanced Mix Test 5

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Question 1
4 / -1

Directions For Questions

Directions: The following question has four choices, out of which ONLY ONE is correct.

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An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in water, such that one half of its volume is submerged. The new fundamental frequency (in Hz) is

A

B

C

D

Solution

The diagrammatic representation of the given problem is shown in figure. The expression of fundamental frequency is
Question 2
4 / -1

What is the minimum possible radius of wire passed over a pulley which supports two blocks of masses m and 2m, if the breaking stress of the wire is σ?

A

B

C

D

Solution

Let tension in the wire is T, then using free body diagrams for both the blocks:

1) For mass m,

ma = T − mg

2) For mass 2m,

2ma = 2mg − T

Thus,
Question 3
4 / -1

The probability of survival of a radioactive nucleus for one mean life is:

A
1/e

B
1 - 1/e

C
ln(2) / e

D
1 - ln(2) / e

Solution
Question 4
4 / -1

The electrochemical cell shown below is a concentration cell.

M | M2+ (saturated solution of a sparingly soluble salt, MX₂) || M²⁺ (0.001 mol dm^-3) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. (The emf of the cell at 298 K is 0.059 V)

The value of ΔG (kJ mol^-1) for the given cell is

(Take 1F = 96500 C mol^-1)

A
-5.7

B
5.7

C
11.4

D
-11.4

Solution

ΔG = -nFE = -(2 x 96500 x 0.059) = -11387 J = -11.387 kJ mol^-1
Question 5
4 / -1

'I' is an aromatic aldehyde, which does not have alpha hydrogens. When reacted with an anhydride of a carboxylic acid in the presence of a sodium salt of the same acid to give an unsaturated acid 'J', which undergoes a series of reactions as depicted in the equation below, to form 'K'.

J gives effervescence on treatment with NaHCO₃ and a positive Baeyer's test.

In the following reaction sequence, the compound K is

A

B

C

D

Solution

Step (i) leads to the selective reduction of the double bond by hydrogenation in the presence of Pd/C.

Step (ii) involves the substitution of the -OH group to form acid chloride.

Step (iii) involves acylation of the benzene ring, in the presence of a Lewis acid as a catalyst, following the Friedel Craft's mechanism.
Question 6
4 / -1

AValue of gas constant R, is

A
0.082 litre atm

B
0.987 cal mol⁻¹K⁻¹

C
8.314 J mol⁻¹K⁻¹

D
83 erg mol⁻¹K⁻¹

Solution

Units of R

(i) In L atm ⇒ 0.082 L atm mol⁻¹K⁻¹

(ii) In C.G.S. system ⇒ 8.314×107erg mol⁻¹K⁻¹

(iii) In M.K.S. system ⇒ 8.314 J mol⁻¹K⁻¹

(iv) In calories ⇒ 1.987 cal mol⁻¹K⁻¹
Question 7
4 / -1

Statement-1: If chlorine is passed into toluene at room temperature in the presence of Anhydrous AlCl₃, electrophilic substitution reaction takes place giving o- and p-chlorotoluenes.

Statement-2: If chlorine is passed through boiling toluene in the presence of UV light, substitution reaction in the aliphatic side chain occurs.

A
Statement-1 is True; Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is True; Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

C
Statement-1 is True; Statement-2 is False.

D
Statement-1 is False; Statement-2 is True.

Solution

In presence of UV light, free radical substitution reaction occurs in aliphatic side chain where as in positive q presence can halogen carrier electrophilic substitution reaction takes place.
Question 8
4 / -1

If the transversal y = m_rx; r = 1, 2, 3 cut off equal intercepts on the transversal x + y = 1, then 1 + m₁, 1 + m₂, 1 + m₃ are in

A
A.P.

B
G.P.

C
H.P.

D
None of these

Solution
Question 9
4 / -1

m & n are integers with 0< n < m. AA is the point (m,n) on the Cartesian plane. B is the reflection of A in the line y = x. C is the reflection of B in the yy-axis, D is the reflection of C in the x-axis and E is the reflection of D in the yy-axis. The area of the pentagon ABCDE is

A
2 m(m + n)

B
m(m + 3n)

C
m(2m + n)

D
2 m(m + 3n)

Solution

A is the point (m,n) on the Cartesian plane.

B is the reflection of A in the line y = x. So, co-ordinates of B ≡ (n, m)

C is the reflection of B in the y-axis. So, co-ordinates of C ≡ (−n, m)

D is the reflection of C in the x-axis. So, co-ordinates of D ≡ (−n, −m)

E is the reflection of D in the y-axis. So, co-ordinates of E ≡ (n, −m)

Sketching the diagram:

From diagram, height (h) of △ABE = (m − n)

So, area of rectangle BCDE is BC × CD = 2m × 2n = 4mn

Also, area of △ABE = 1/2 × BE × h = 1/2 × 2m × (m − n) = m(m − n)

Therefore, the area of pentagon = area of rectangle BCDE + area of △ABE

= 4mn + m(m − n)
Question 10
4 / -1

Consider the polynomial: f(x) = 1 + 2x + 3x² + 4x³

Let s be the sum of all distinct real roots of f(x) and let t = |s|.

The real number 's' lies in the interval

A

B

C

D

Solution

f(x) = 1 + 2x + 3x² + 4x³f’(x) = 2 + 6x + 12x2 > 0 [as a > 0, D < />

f(x) i increasimg function so it can have almost one real root.

Using intermediate value theorem