JEE Advanced Mix Test 65

Result

JEE Advanced Mix Test 65

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Weekly Quiz Competition

Question 1
4 / -1

When liquid medicine of density ρ is to be put in the eye it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r << R)

A
2πrT

B
2πRT

C
2πr²T/R

D
2πr²T/r

Solution

The vertical force due to the surface tension on the drop is,
Question 2
4 / -1

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

For a = 0, the value of d (maximum value of ρ as shown in the figure) is

A

B

C

D

Solution
Question 3
4 / -1

Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x << l. Find the rate dq/dt with which the charge leaks off each sphere if their approach velocity varies as v = a/√x , where a is a constant.

A

B

C

D

Solution

Electrostatic force:

Fₑ = (q²) / (4πε₀x²)

Tension:

T sinθ = Fₑ,

T cosθ = mg,

θ ≈ x / 2l

v = dx/dt = a / √x, solve for dq/dt.

Result: dq/dt = (3/2) a √(2πε₀mg / l)
Question 4
4 / -1

(1R, 3S)-Cis-1-Bromo-3-methyl cyclohexane. The product formed in the reaction is

A
(1R, 3S)-Cis-3-methyl cyclohexanol

B
(1S, 3S)-Cis-3-methyl cyclohexanol

C
(1S, 3S)-Trans-3-methyl cyclohexanol

D
(1R, 3R)-Trans-3-methyl cyclohexanol

Solution

In the presence of a polar aprotic solvent, like acetone, the mechanism followed will be SN².

Walden inversion takes place at C₁, where -Br is substituted by -OH.

Hence, the product formed will be (1S, 3S)-Trans-3-methyl cyclohexanol
Question 5
4 / -1

What is the magnetic moment of coordination compound formed during brown ring test?

A
3.87 BM

B
4.92 BM

C
5.92 BM

D
2.83 BM

Solution

Brown ring test When a freshly prepared FeSO₄ solution is added to aqueous solution of NO⁻₃ ion followed by addition of concentrated H₂SO₄ the brown ring is observed at junction between two liquids. This colour is due to charge transfer oxidation state of iron in this complex is + l. In this state Fe⁺, has 3 unpaired electrons, and hence, the magnetic moment will be: √3(3+2) = 3.87 BM
Question 6
4 / -1

The equilibrium constant K for the reaction 2HI(g) ⇌ H₂ (g) + I_2(g) at room temperature is 2.85 and that at 698 K is 1.4 × 10^–2. This implies that -

A
HI is exothermic compound

B
HI is very stable at room temperature

C
HI is relatively less stable than H₂ and I₂

D
HI is resonance stabilised

Solution

K_c drops as temperature rises, so the forward reaction is exothermic. This means HI is less stable than H₂ and I₂, favoring C. A is misleadingly worded, B contradicts K_c = 2.85, and D is irrelevant (no resonance in HI).
Question 7
4 / -1

A

B

C

D
None of these

Solution
Question 8
4 / -1

Out of 3n consecutive integers, three are selected at random. Find the probability that their sum is divisible by 3.

A

B

C

D

Solution

Suppose the sequence of 3n consecutive integers begin with integer m.

Then the 3n consecutive integers are:

m, m + 1, m + 2,…………, m + (3n - 1)

Out of these integers, 3 integers can be chosen in ³ⁿC₃ ways.

Let us divide these 3n consecutive integers in the groups G1, G2 and G3 as follows:

G1: m, m + 3. m + 6, …….., m + (3n - 3)

G2: m + 1, m + 4, m + 7, ……., m + (3n - 2)

G3: m + 2, m + 5, m + 8, …….., m + (3n - 1)
Question 9
4 / -1

If SK be the perpendicular from the focus S on the tangent at any point P on the ellipse then locus of the foot of the perpendicular K is equal to

A
a²x²+ b²y²= (ax − by)²

B
x² + y² = a²

C
x²+ y²= b²

D
x²+ y.= a²+ b²

Solution

By eliminating θ from both the eqs. (i) and (ii) [by squaring and adding (i) & (ii)], we get the locus of K as x²+ y²= a² i.e., the auxiliary circle of the ellipse.
Question 10
4 / -1

A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m which is stretched under a load by 0.7 mm. Calculate the load, given that the Young's modulus of elasticity for copper is 1.1 × 10¹¹ N m^–2 and that for steel is 2 × 10¹¹ N m^–2.

A
1.8 × 10² N

B
1.5 × 10^–2 N

C
2.8 × 10² N

D
1.0 × 10² N

Solution