JEE Advanced Mix Test 61

Result

JEE Advanced Mix Test 61

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

A steady current I goes through a wire loop PQR having shape of a right-angled triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is find the value of k.

A
7

B
8

C
9

D
10

Solution

The magnetic field (B) due to elements PR and PQ is 0 as the point P is on the line of the conductor.

The magnetic induction due to the current element RQ at P:
Question 2
4 / -1

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s^-2.)

The magnitude (in Newton) of the normal reaction that acts on the block at the point Q is N. Find the value of 2N.

A
20

B
25

C
10

D
15

Solution
Question 3
4 / -1

A point object is placed in front of a thin biconvex lens, of focal length 20 cm. When placed in air, the refractive index of material of lens is 1.5 The further surface of the lens is silvered and is having radius of curvature of 25 cm. The position of final image of object is at 25/x cm from lens. Determine the value of x?

A
5

B
3

C
4

D
2

Solution

We know that a thin silvered lens is equivalent to a combination of two lenses and a mirror. This equation system is working as a mirror whose
Question 4
4 / -1

Illuminate the surface of a certain metal alternately with wave length λ₁ = 0.35 μm and λ₂ = 0.54 μm. It is found that the corresponding maximum velocity is of photo electron having a ratio n = 2. Find the work function of that metal (in eV).

(Round off up to 2 decimal places)

A
2.89

B
1.88

C
1.70

D
1.50

Solution
Question 5
4 / -1

EDTA^4– is ethylenediaminetetraacetate ion. What is the total number of N – Co – O bond angles in [Co(EDTA)]1– complex ion?

A
5

B
6

C
10

D
8

Solution

EDTA⁴⁻ forms octahedral complex with Co³⁺, coordinating via 2 N and 4 O atoms.

Each N forms bonds with 4 O atoms around Co, forming 90° angles.

Total N–Co–O angles: 2 N × 4 O = 8.
Question 6
4 / -1

The value of log10 K for a reaction A ⇌ B is (Given Δ_r = -54.07 kJ mol^-1, Δr = 10JK^-1mol^-1 and R = 8.314JK^-1mol^-1; 2.303 x 8.314 x 298 = 5705)

A
10

B
8

C
5

D
20

Solution

Given:
ΔᵣH = –54.07 kJ mol⁻¹ = –54070 J mol⁻¹
ΔᵣS = 10 J K⁻¹ mol⁻¹
R = 8.314 J K⁻¹ mol⁻¹
T = 298 K
2.303 × R × T = 5705

Step 1: Calculate ΔG
ΔG = ΔᵣH – TΔᵣS
ΔG = –54070 – (298 × 10)
ΔG = –54070 – 2980
ΔG = –57050 J mol⁻¹

Step 2: Use the relation
ΔG = –2.303RT log₁₀K
–57050 = –5705 × log₁₀K
log₁₀K = 57050 ÷ 5705 = 10

Final Answer: log₁₀K = 10
Question 7
4 / -1

The number of paired electrons in O₂ molecule is

A
15

B
20

C
25

D
14
Solution
The molecular orbital (MO) configuration of O₂ is:

(σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)¹ (π*2py)¹
- O₂ has a total of 16 electrons.
- Out of these 16, 2 electrons are unpaired (in π2px and π2py orbitals).
- Therefore, the remaining 14 electrons form 7 pairs, i.e., 7 paired orbitals.
- Each pair has 2 electrons, so:
- 7 pairs × 2 electrons = 14 paired electrons
Question 8
4 / -1

A bag contains n white and n black balls (all different). Pairs of balls are drawn one-by-one without replacement until the bag is empty. If the number of ways to draw the balls in which each pair consists of one black and one white ball is 576, then the value of n is equal to__

A
4

B
5

C
10

D
8

Solution
Question 9
4 / -1

A
5

B
4

C
3

D
2

Solution
Question 10
4 / -1

A
3

B
4

C
5

D
6

Solution