BITSAT Mock Test - 1 - JEE MCQ

Let r be the internal resistance.
Power across R₁ = Power across R₂


Therefore we have

Area under a - t graph gives the change in velocity.


u = 0
v = 15 m/s
Note: Only solutions to be posted here. For any other discussions, please scroll down to the discussion forum.

A (g) + B (g) ⇌ C (g) + D (g)

_{2 - 3x = x}
∴ x =
K_p =
=
=
= 9

Alkynes are oxidatively cleaved with alkaline permaganate solution to give carboxylic acids and aldehydes, depending on the substitution.
Under mild conditions vicinal diols are formed which undergo tautomerism to form aldehydes or ketones.
However, in the presence of cold KMnO₄ corresponding carboxylic acids are formed.
CH₃CH₂C CH CH₃CH₂COOH + HCOOH
If hot KMnO₄ is used the formic acid will further disintegrate to form CO₂ + H₂O
Hence, option (4) is correct.

From the data, it is clear that the difference between the radii of ground state and excited state is
r_n - r₁ = 0.79 x 10^-9 m
r₁ = 0.0529 x 10^-9 m
r_n - 0.0529 x 10^-9 m = 0.79 x 10^-9 m
r_n = 0.84 x 10^-9 m
Also, r_n = 0.0529 x 10^-9 x n²
n² =  = 15.87 16
n = 4
Thus, electrons excited from first orbit to fourth orbit and number of spectral lines
=  = 6

For a first order reaction:

Let [A_o] = a
If 75% of the reaction was completed in 32 minutes, then

Now, the time required to complete 50% (t_1/2) of the reaction would be:

The pattern followed in the series is: + 5, + 10, + 15, + 20.
The missing term is 40 + 20 = 60.

49 + 0 = 49
49 + 1 = 50
50 + 4 = 54
54 + 9 = 63
63 + 16 = 79
79 + 25 = 104
Thus, 60 is the wrong term in the series.

The movement of elements in the first step is as follows:

In the second step, the elements are moving in anti-clockwise direction and the element at the centre is kept unchanged in its place.
In the third step, the same pattern is followed as in the first step.
In the fourth step, the same pattern is followed as in the second step.
In the fifth step, the same pattern is followed as in the first step.
Hence, answer option (1) is correct.

When we unfold the paper once, it will show us a circle made by the punch on half of the other side because it was punched only on half of the final fold. Further on unfolding it, we would see two holes on the extreme left side. Unfolding it once more, we will see two holes at left and at the top too.
On opening the final fold, the holes will be on all the four sides.

, only when x = 4n, where n is any positive integer.

Given curve is
3x² - y² = 8
On differentiating, we get
6x - 2y = 0
=
Slope of normal =
Slope of given line =
A.T.Q.,
=
=> y = x
Putting the value of y in above equation, we get
3x² - y² = 8
We get x = 2, -2 and y = 2, - 2
Equation of normal
y - y₁ = (x - x₁)
On solving, we get
x + 3y + 8 = 0 and x + 3y - 8 = 0

Let A be the event that a student succeeds in the IIT entrance test and B be the event that a student succeeds in the Roorkee entrance test.
Given that, P(A) = 0.2, P(B) = 0.5, P(A∩B) = 0.3
Thus, the probability that a student does not succeed at both places = P(A'∩B')
P(A'∩B') = 1- P(AUB)
P(A'∩B') = 1 - [P(A) + P(B) - P(A∩B)]
P(A'∩B') = 1- [0.2 + 0.5 - 0.3]
P(A'∩B') = 1 - 0.4
P(A'∩B') = 0.6.

From question,

Total number of egg or meat eaters = 1700 + 1500 + 1000 = 4200
Number of pure vegetarians = 5000 - 4200 = 800