JEE Main Test 105

Result

JEE Main Test 105

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Weekly Quiz Competition

Question 1
4 / -1

The angle between the lines √3x - y - 2 = 0 and x - √3y + 1 = 0 is

A
90^o

B
60^o

C
45^o

D
30^o

Solution
Question 2
4 / -1

The equation of bisectors of the angles between the lines |x| = |y| are

A
y = ± x and x = 0

B
x = 1/2 and y = 1/2

C
y = 0 and x = 0

D
None of these

Solution

Given equations of lines are
Question 3
4 / -1

The density of O₂ is 16 at NTP. At what temperature its density will be 14 ? Consider that the pressure remains constant, at

A
50°C

B
39°C

C
57°C

D
43°C

Solution
Question 4
4 / -1

Graham's law deals with the relation between

A
pressure and volume

B
density and rate of diffusion

C
rate of diffusion and volume

D
4) rate of diffusion and viscosity

Solution
Question 5
4 / -1

The location for the l₁x + m₁y + n₁ = 0 to be conjugate with respect to the circle x² + y² = r², is

A
r² (ll₁ + mm₁) = nm₁

B
r² (ll₁ - mm₁) = nm₁

C
r² (ll₁ + mm₁) + nm₁ = 0

D
r² (ll₁ + mm₁) = nn₁

Solution

We know,condition for conjugate lines is
Question 6
4 / -1

If I denotes the semi - latusrectum of the parabola y² = 4ax and SP and SQ denote the segments of any focal chord PQ, S being the focus, then SP, I and SQ are in the relation

A
AP

B
GP

C
HP

D
I² = SP² + SQ²

Solution

Hence, SP, I and SQ are in HP
Question 7
4 / -1

A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

A
393.4 nm

B
885.0 nm

C
442.5 nm

D
776.8 nm

Solution
Question 8
4 / -1

In Young's double slit experiment with sodium vapour lamp of wavelength 589 run and the slits 0.589 mm apart, the half angular width of the central maximum is

A
sin^-1(0.01)

B
sin^-1(0.0001)

C
sin^-1(0.001)

D
sin^-1(0.1)

Solution
Question 9
4 / -1

If the bond dissociation energies of XY, X₂ and Y₂ (all diatomic molecules) are in the ratio of 1:1:0.5 and ∆H_f for the formation of XY is -200 kJ mol^-1 . The bond dissociation energy of X₂ will be

A
400 kJ mol^-1

B
300 kJ mol^-1

C
20 kJ mol^-1

D
None of these

Solution
Question 10
4 / -1

Consider the following Ellingham's diagram for carbon.

A
Upto 710°C, the reaction of formation of CO₂ is energetically more favourable but above 710°C, the formation of CO is preferred

B
Carbon can be used to reduce any metal oxide at a sufficiently high temperature

C
Carbon reduces many oxides at elevated temperature because ∆G° vs temperature line has a negative slope

D

Solution