NEET question bank, practice MCQ , Notes, Lectures

1 Electrochemistry

The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous H₂SO₄ and fused MgSO₄ are:

1) 1:8
2) 1:12
3) 1:16
4) 1:1

Option 1

Option 2 Solution:

SOLUTION
Acc. to Faraday 2nd Law of electrolysis.

w1 / w2=E1 /E2 w1= At deposited of element H.
w1 / w2=1 / 12 w2 = At deposited of element Mg.

Option 3

Option 4

2 Ray Optics

A plane mirror is moving with velocity $$\displaystyle{4}{\left(\hat{{i}}\right)}+{4}{\left(\hat{{j}}\right)}+{8}{\left(\hat{{k}}\right)}$$
A point object in front of the mirror moves with a velocity $$\displaystyle{3}{\left(\hat{{i}}\right)}+{4}{\left(\hat{{j}}\right)}+{5}{\left(\hat{{k}}\right)}$$
. Here, kˆ is along the normal to the plane mirror and facing towards the object. The velocity of the image is

1) $$\displaystyle-{3}{\left(\hat{{i}}\right)}-{4}{\left(\hat{{j}}\right)}+{5}{\left(\hat{{k}}\right)}$$
2) $$\displaystyle{3}{\left(\hat{{i}}\right)}+{4}{\left(\hat{{j}}\right)}+{11}{\left(\hat{{k}}\right)}$$
3) $$\displaystyle-{4}{\left(\hat{{i}}\right)}+{5}{\left(\hat{{j}}\right)}+{11}{\left(\hat{{k}}\right)}$$
4) $$\displaystyle{7}{\left(\hat{{i}}\right)}+{9}{\left(\hat{{j}}\right)}+{3}{\left(\hat{{k}}\right)}$$

Option 1

Option 2 Solution:

Solution

$$\displaystyle{V}_{{{O}{M}}}={V}_{{{O}}}-{V}_{{{M}}}={\left(-\hat{{{i}}}-{3}\hat{{{k}}}\right)}$$

$$\displaystyle\therefore{V}_{{{I}{M}}}={\left(-\hat{{{i}}}+{3}\hat{{{k}}}\right)}={V}_{{{I}}}-{V}_{{{M}}}$$

$$\displaystyle{V}_{{{I}}}={\left(-\hat{{{i}}}+{3}\hat{{{k}}}\right)}+{v}_{{{m}}}$$

$$\displaystyle={\left({3}\hat{{{i}}}+{4}\hat{{{j}}}+{11}\hat{{{k}}}\right)}$$

Option 3

Option 4

3 Basic Principles of Organic Chemistry

How many stereoisomers are therefor tartaric acid?

1) 2
2) 3
3) 4
4) 5

Option 1

Option 2 Solution:

SOLUTION

d-tartaric acid, l-tartaric acid,meso-tartaric acid

Option 3

Option 4

Type A hepatitis can be prevented by
1) Passive immunization
2) Sanitation
3) Control of reservoirs
4) Isolation

Option 1 Solution:

Option 2

Option 3

Option 4

5 The Living World

Which of the following pair is wrongly matched about the taxonomic arrangement of cat?

1)   Scientific name – Felis domestica.
2)   Family – Felidae.
3)   Order – Mammalia.
4)   Phylum – Chordata.

Option 1

Option 2

Option 3 Solution:

Answer

Mammalia is a class

Option 4

Gibberellins have a role in seed germination because they
1. break the seed coat
2. weaken the seed coat
3. stimulate the synthesis
4. stimulate the synthesis of vernalin

Option 1

Option 2

Option 3

Option 4 Solution:

7 Atomic Structure

Which of the following has maximum number of zero values appear in the curve plotted between ψ² (probability of finding electron) versus radius?

1) 3p
2) 4d
3) 5p
4) 4f

Option 1

Option 2

Option 3 Solution:

Solution

Max^m no. of zero values is given by total no. of nodes which is given by ⇒n−1.

n−1 principle quantum number

Option 4

8 Kinematics

A particle is projected from the ground level. It just passes through upper ends of vertical poles A, B, C of height 20 m, 30 m and 20 m respectively. The time taken by the particle to travel from B to C is double of the time taken from A to B. Find the maximum height attained by the particle from the ground level.
1) 15 m
2) 125/4 m
3) 45/4 m
4) none

Option 1

Option 2 Solution:

Solution
t_AB = t

t_BC = 2t

So, for ABC part,

Option 3

Option 4

9 Gravitation

Consider a drop of rain water having mass 1 g fallingfrom a height of 1 km. It its the ground with a speedof 50 m/s. Take g constant with a value10 m/s². The work done by the (i) gravitational force and the (ii) resistive force of air is

1) (i) – 10 J (ii) –8.25 J
2) (i) 1.25 J (ii) –8.25 J
3) (i) 100 J (ii) 8.75 J
4) (i) 10 J (ii) –8.75 J

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Work done by gravitational force = potential energy of drop
W =mgh = 10^-3×10×10³ = 10 J
Work done by resistive force W_R= P.E. – K.E of drop when it reach surface

W_R = 10 – 1.250 = 8.75. This work is done by resistive force so Work is negative = -8.75 J

10 Chemical Kinetics

The table below shows recorded concentration data for the following chemical reaction:
$$CaCl_{2}(aq) + 2AgNO_{3}(aq) \rightarrow Ca(NO_{3})_{2} (aq) + 2AgCl(s)$$
Use the data to determine the exponents $$x$$ and $$y$$ in the rate law:
$$rate = k[CaCl_{2}]^{x} [AgNO_{3}]^{y}$$

$$[CaCl_{2}]M$$	$$[AgNO_{3}]M$$	Rate M/s
$$0.050$$	$$0.050$$	$$3.33\times 10^{-3}$$
$$0.100$$	$$0.050$$	$$6.66\times 10^{-3}$$
$$0.100$$	$$0.100$$	$$13.32\times 10^{-3}$$

1) x = 2, y = 3
2) x = 1, y = 2
3) x = 1, y = 1
4) x = 2, y = 2

Option 1

Option 2

Option 3 Solution:

Solution

On making the concentration of CaCl₂ double with no change in concentration of AgNO₃ rate of reaction also increases 2 times which implies that x=1

And on making concentration of both the reactants double rate increases by 4 times and since x=1 it can be concluded that y=1

Option 4

NEET

EXAM PREP

Question Bank

Solution

Solution

Solution

Solution

MOCK TEST