BITSAT Mock Test - 9 - JEE MCQ

Capacity of the spherical capacitor having inner radius 'a' and outer radius 'b'
b = 5 cm, a = 3 cm


C = 667 × 10^-12 F or C = 667 pF

BC, CD and BA are known resistances. The unknown resistance is connected between A₁ and D.

As the size of alkyl group or halogen atom increases, boiling point increases because the magnitude of Van der Waals forces of attraction increases.
Moreover, the boiling point decreases with branching due to decrease in surface area. Thus, the correct order of boiling points is as follows.

Hybridisations of a molecule can be determined as follows.
X = SA + (G - V + a) for polyatomic anions
X = SA + (G - V - c) for polyatomic cation
SA = Number of atoms or groups around central atom
G = Valence electrons of central atom
V = Valency of central atom (number of mono valent atoms or 2 Number of divalent atoms etc.)
a and c = Charge on anion and cation, respectively
In 4


Therefore, have the same hybridisation, i.e. sp³.

Option (4) is correct.
If -COOH group is directly attached to the ring carbon, suffix used is -carboxylic acid. Carboxylic acid group has higher priority than ketone group.

Option (1) is incorrect.
Ketone group has higher priority than hydroxyl group.

Option (2) is incorrect.
Correct IUPAC name of given compound is 6-ethylcyclohex-2-enol.

Option (3) is incorrect.
If -CHO group is directly attached to the ring carbon, suffix used is -carbaldehyde.

The reaction is called Stephen reaction and is used for the preparation of aldehydes from alkylcyanides.

The number of partitions in the figure is decreasing by one with each step and the number of sides of the figure is increasing by one with every two steps.

'has been living….''
We use the present perfect continuous tense to show that something had started in the past and has continued up until now and its format is 'has/have + been + present participle form of the verb'. The keywords are 'for twenty years'.

The pattern of given series is:
⇒ 18 × 0.5 - 1 = 8
⇒ 8 × 1.5 - 2 = 10
⇒ 10 × 2.5 - 3 = 22
⇒ 22 × 3.5 - 4 = 73
⇒ 73 × 4.5 - 5 = 323.5
⇒ ? = 323.5

Step 1 :

Step 2 :

Step 3 :

After folding along the horizontal axis and then to its vertical axis, figure (X) would look like figure (D).
Hence, option 4 is correct.

Zoology is the branch of science which deals with the study of animals. Physiology is the branch of science which deals with the study of bodies of living organisms.

From (B) and (C), we get the arrangement as:

Combining the above arrangement with the information in (D), the arrangement becomes:

U, R and P get south-facing flats.

At each step, every element moves one space clockwise and one of the elements is replaced by a new element.

Let S_i denote the event of getting an ace in the i^th draw.
∴ Probability of getting aces in both the draws = P(S₁ ∩ S₂) = P(S₁) P(S₂) [Multiplication theorem]
=
=

Let p = q + h, where h is so small that its square and higher powers may be neglected.
Then, =
= = =
= [Expanding the second term and neglecting h², h³, etc.]
= 1 + h - h [Neglecting h²]
= 1 + = =
∴ k =

According to the question we have,
Distance between focus and one end of minor axis is equal to its minor axis
[because triangle formed by the lines are equilateral.]
Then a²e² + b² = (2b)²
a²e² = 3b² = 3a²(1 - e²) [Since b² = a²(1 - e²)]
e² =
Hence, e =