NEET Test 195

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NEET Test 195

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Weekly Quiz Competition

Question 1
4 / -1

The IUPAC name of the given compound is:

A
4-hydroxy-1-methylpentanal

B
3, 4-dihydroxy-2-methylpentanal

C
2-hydroxy-4-methylpentanal

D
2, 3-dihydroxy-4-methylpentanal

Solution

The IUPAC name of the given compound is 3, 4-dihydroxy-2-methylpentanal.

Aldehyde has a higher priority than alcohol, so the numbering is done as shown in the figure. The longest continuous chain contains 5 carbon atoms and at 3rd and 4th position hydroxy group is present. At 2nd carbon atom, one methyl group is present. Alphabetically 'h' comes first then 'm'.
Question 2
4 / -1

A particle is projected vertically with speed V from the surface of the earth. Maximum height attained by the particle, in terms of the radius of earth R,V and g is (V escape velocity, g is the acceleration due to gravity on the surface of the earth).

A

B

C

D

Solution
Question 3
4 / -1

What do hummingbird and hawk moth, Darwin's feathers represent?

A
Convergent evolution

B
Homology

C
Adaptive radiation

D
Parallel evolution

Solution

Humming birds and hawk illustrate adaptive radiation.

Development of different functional structures from a common ancestor is called adaptive radiation. Eg: Hummingbird and hawk moth, Darwin's finches, etc. Whereas development of similar functional structures from different ancestors are known as convergent evolution. Eg: wings of insect and bat.
Question 4
4 / -1

Two cars A and B travel along the same road in the same direction from the same starting place. Car A maintains a speed of 50 km/h and car B of 60 km/h, but B starts one hour later. How many hours will it take for B to overtake A?

A
2

B
3

C
4

D
5

Solution

Car A is ahead of car B by 50 Km before car B leaves the starting point. In the first hour after car B leaves, car A will have traveled 120 Km, 2 Hrs × 50 km/h= 100 Km.

Remember car A left one hour before car B. Car B in the first hour from departure will have traveled 60 Km at 75 Km/h.

The third hour after departure car A will have traveled 180 km. (3 Hrs × 60 Km/h =180 Km)

Car B will be in the second hour since departure traveling at 60 Km/h. (2 Hrs× 60 Km/h= 120 Km).

Car A in its 3rd hr.has now traveled 180 km and car B in its 2nd Hr has now traveled 150 Km. Car B has now closed the distance from 60 Km to 30 Km from Car A.

In the 4th hour after departure Car, A will have traveled 240 Km. (4Hrs.× 60 Km/h= 240 Km.

Car B after leaving the starting point 1 Hr after Car A will be in its 3rd Hr of travel. ( 3 Hrs × 60 Km/h = 180 Km.

Car A Has now completed 5 Hrs of travel and has gone 300 km. at 60 Km/h. ( 5 Hrs × 50 Km/h= 250 km).

Car B after 4 Hrs. of travel at 60 Km/h has now travelled the same distance as car A. ( 4 Hrs ×60 Km/h =240 Km).

Car B would need to complete 4 Hrs of traveling to catch up and overtake Car A who had a 50 Km head start over car B.
Question 5
4 / -1

The movement of water and minerals in xylem and the movement of phloem sap in phloem is ________ and ________ respectively.

A
Unidirectional, bidirectional

B
Bidirectional, bidirectional

C
Bidirectional, unidirectional

D
Unidirectional, unidirectional

Solution

The movement of water and minerals in xylem and the movement of phloem sap in phloem is unidirectional and bidirectional respectively. Plants need water for their growth, and it uptakes water through the roots. Water that is taken by the roots has to reach all the leaves or branches, here the xylem function as transportation of water. Therefore, the movement of water and minerals in xylem is always an unidirectional. Whereas, the movement of phloem is bidirectional this is because, once the food is synthesized it is the phloem that transfers the food to all the parts of plants wherever it is necessary.
Question 6
4 / -1

The oxidation number of sulphur in H₂S₂O₈ is:

A
+ 6

B
+ 8

C
+ 2

D
+ 7

Solution

In H₂S₂O₈, two oxygen are involved in peroxide linkage as shown in the above structure. So, the oxidation number of two oxygen atoms are -1 .
Question 7
4 / -1

For the resistance network shown in the figure, choose the correct option.

A
The current through PQ is 6A

B
I₁= 4A

C
The potential at S is more than that at Q

D
I₂ = 2A

Solution

Due to input and output symmetry P and Q and S and T have the same potential.
Question 8
4 / -1

The first law of motion is also known as the law of:

A
Inertia

B
Momentum

C
Force

D
Impulse

Solution

Newton's first law of motion is also known as the law of inertia. Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. It may be seen as a statement about inertia, that objects will remain in their state of motion unless a force acts to change the motion.
Question 9
4 / -1

In the circuit shown in the figure potential difference between A and B is:

A
30 V

B
90 V

C
10 V

D
60 V

Solution
Question 10
4 / -1

Some magnetic flux is changed from a coil of resistance 10Ω. As a result, an induced current is developed in it, which varies with time as shown in the figure. The magnitude of change in flux through the coil in Wb is:

A
2

B
4

C
6

D
8

Solution

We have a coil in which the current changes with time. Let us say for that small change in time the flux across the cross sectional area of the coil changes by dφ. If at a particular instance of time with respect to the initial the induced emf is ‘E’, we can write the rate of change of flux with time as induced emf. Mathematically this can be represented as,

If the emf E across a particular coil of resistance ‘R’ is applied. Then by ohms law we can say that current ‘I’ in the circuit is proportional to the ratio of the two. Mathematically this can be represented as,

The integration of the factor (-idt) represents the area under the curve of variation of current with time. This from the above graph we get as 0.2 which is area of triangle, hence from the above equation we get change in flux i.e., φ as,