NEET Test 100

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NEET Test 100

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Weekly Quiz Competition

Question 1

4 / -1

Directions For Questions

Match Column-I with Column-II.

Column-I		Column-II
(a)	Nitrococcus	(i)	Denitrification
(b)	Rhizobium	(ii)	Conversion of ammonia to nitrite
(c)	Thiobacillus	(iii)	Conversion of nitrite to nitrate
(d)	Nitrobacter	(iv)	Conversion of atmospheric nitrogen to ammonia

...view full instructions

Choose the correct answer from options given below

Solution

(a) NH₃ is converted to (nitrite) by nitrifying bacteria such as Nitrococcus.

(b) Nitrogen fixation is conversion of atmospheric N₂ to NH₃ (ammonia). It is carried out by N₂ fixers such as Rhizobium.

(d) Then is converted to (nitrate) by nitrfying bacteria called Nitrobacter.

Hence, the correct option is (B).

Question 2
4 / -1

Ploidy level of endosperm in angiosperm is generally:

A
n

B
2n

C
3n

D
6n

Solution

The ploidy level of endosperm in angiosperm is 3n.

Angiosperms endosperms are triploid in nature. In angiosperms, double fertilization occurs and as a result that triple fusion takes place. The female gametophyte or embryo sac has an egg nucleus and two polar nuclei. One sperm fertilizes the egg nucleus and forms the diploid zygote. Another sperm joins the two polar nuclei forming the triploid (3n) nutritive tissue, called the endosperm.

Hence, the correct option is (C).
Question 3
4 / -1

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth's magnetic field H_E at a place. If H_E=0.4G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1G = 10⁻⁴ T.

A
6.28 × 10^-5V

B
4.28 × 10^-5V

C
4.48 × 10^-5V

D
3.28 × 10^-5V

Solution

The number of spokes is immaterial because the emf's across the spokes are in parallel.

Hence, the correct option is (A).
Question 4
4 / -1

Snow-blindness in Antarctic region is due to:

A
Freezing of fluids in the eye by low temperature

B
Inflammation of cornea due to high dose of UV-B radiation

C
High reflection of light from snow

D
Damage to retina caused by infra-red rays
Question 5
4 / -1

Which of the following gates serve as building blocks in digital circuits?

A
OR and AND gates

B
AND and NOT gates

C
OR and NOT gates

D
NAND and NOR gates

Solution

NAND and NOR gates are known as universal gates. Any one of these gates can be used to implement any kind of logic gate. This kind of feasibility does not exist with other gates i.e. any other gate cannot solely implement all logic gates. For example, AND gate cannot be implemented using an OR gate and vice-versa. The implementation of NAND and NOR gates to generate other logic gates is shown above.

Hence, the correct option is (D).
Question 6
4 / -1

Assume that light of wavelength 600nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is

A
3.66 × 10^-7 rad

B
1.83 × 10^-7 rad

C
7.32 × 10^-7 rad

D
6.00 × 10^-7 rad
Question 7
4 / -1

Which compound will liberate oxygen when reacts with ice-cold water?

A
Na₂O₂

B
KO₂

C
Na₂O

D
Cs₂O₂

Solution

KO₂will liberate oxygen when reacts with ice cold water as follows:

Hence, the correct option is (B).
Question 8
4 / -1

Which of the following reagent pairs can be used for distinguishing ethanol from phenol?

A

B
Na, Fe, Mg

C

D
None of these

Solution

Reagent pairs that can be used for distinguishing ethanol from phenol are Neutral FeCl₃, NaOH/I₂, Br₂/H2_O.

Phenol being acidic dissolves in NaOH forming sodium phenoxide whereas C₂H₅OH being netural will not react with NaOH.

C₂H₅OH will give iodoform test and phenol will not give this test due to absence of group. With neutral FeCl₃ phenol forms violet complex whereas C₂H₅OH does not react with neutral FeCl₃.

Hence, the correct option is (A).
Question 9
4 / -1

A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is :

A

B

C

D

Solution

As given, A particle executing a simple harmonic motion has a period of 6 s.
Question 10
4 / -1

At t₁=2.00 s, the acceleration of a particle in counter clockwise circular motion is It moves at constant speed. At time t₂= 5.00 s, the particle's acceleration is What is the radius of the path taken by the particle if t₂− t₁ is less than one period:

A
2.92 m

B
292 m

C
2.92 cm

D
292 cm

Solution

Since, the radius vector in circular motion is anti-parallel to the centripetal acceleration,

Radius vectors at each instant (t₁=2 s and t₂=5 s) are also perpendicular to each other.

points to the south. 'Since, the movement is counterclockwise, therefore, between the two instants is a three-quarter cycle. Therefore, the period is given by the form: