JEE Main Test 131

Result

JEE Main Test 131

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Weekly Quiz Competition

Question 1
4 / -1

A plane electromagnetic wave of frequency 20 MHz travels through a space along x-direction. If the electric field vector at a certain point in space is 6 Vm^-1, what is the magnetic field vector at that point?

A
2 × 10^-8 T

B

C
2 T

D

Solution
Question 2
4 / -1

The equation of tangents drawn from the origin to the circle x² + y² - 2rx - 2hy + h² = 0 are

A
x = 0, y = 0

B
x = 1, y = 0

C
(h² - r²)x - 2rhy = 0, y = 0

D
(h² - r²)x - 2rhy = 0, x = 0

Solution
Question 3
4 / -1

According to Newton's corpuscular theory, the speed of light is

A
same in all the medium

B
lesser in rarer medium

C
lesser in denser medium

D
independent of the medium

Solution

The Newton’s theory of refraction concludes that velocity of light in air medium is less than velocity of light in denser medium.
Question 4
4 / -1

A parabola has the origin as its focus and the line x = 2 as the directrix. Then, the vertex of the parabola is at

A
(2, 0)

B
(0, 2)

C
(1, 0)

D
(0, 1)

Solution

We know vertex is a mid – point of focus and directrix

i.e. (1, 0) is the vertex
Question 5
4 / -1

Carbon cannot reduce Fe₂ O₃ to Fe at a temperature below 983 K because

A
free energy change for the formation of CO is more negative than that of Fe₂ O₃

B
CO is thermodynamically more stable than Fe₂ O₃

C
carbon has higher affinity towards oxygen than iron

D
iron has higher affinity towards oxygen than carbon

Solution

Below 983K, standard change in free energy for the formation of Fe₂O₃ is more than that of formation of CO₂ from CO. Therefore CO is better reducing agent for Fe₂O₃ below 983K.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Above 983K, standard change in free energy value for the formation of Fe₂O₃ is more than that of CO formation from carbon (coke). At this temperature carbon has a higher affinity towards oxygen than iron. Hence above 983K, carbon (coke) is better reducing agent for Fe₂O₃.

Fe₂O₃ + 3C → 2Fe + 3CO
Question 6
4 / -1

If 2y = x and 3y + 4x = 0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is

A

B

C

D

Solution
Question 7
4 / -1

Which of the following is the strongest base?

A

B

C

D

Solution

In piperidine nitrogen atom is SP³ hybridized which has a less S – character while in pyridine Nitrogen atom is SP² hybridized which has more S – character. Therefore held by the nitrogen nucleus in pyridine which has more S – character. In piperidine ione pair of electrons are less tightly held and are easily available for protonation therefore piperidine is a stronger base than pyridine.
Question 8
4 / -1

With respect to graphite and diamond, which of the statement(s) given is (are) correct?

A
Graphite is harder than diamond

B
Graphite has higher electrical conductivity than diamond

C
Graphite has higher C—C bond order than diamond

D
Both (2) and (3)

Solution
Question 9
4 / -1

IUPAC name of (CH₃)₃ CCl is

A
n-butylchloride

B
3-chlorobutane

C
2-chloro- 2-methyl propane

D
t-butyl chloride

Solution
Question 10
4 / -1

The IUPAC name for tertiary butyl iodide is

A
4–iodo butane

B
2–iodo butane

C
1–iodo–3–methyl propane

D
2–iodo–2–methyl propane

Solution