BITSAT Mock Test - 4 - JEE MCQ

If the train is going away from the observer, the apparent frequency is
...(i)
It is observed that v = 1.2 v₁.
In the second case, the apparent frequency is

or ...(ii)
Now, from (i) we have = 1 + or 1.2 = 1 + , or u = 0.2 v or = 0.2.
Using this in (ii), we get = 1.25.
Hence the correct choice is (2).

The equilibrium condition FBD of 2m

When the system is equilibrium, the spring force = 3 mg.
When the string is cut, the net force on block A = 3 mg - 2 mg = mg.
Hence, the acceleration of this block at this instant is

When the string is cut, the block B falls freely with an acceleration equal to g.

According to Lenz's law,


At t = 3,

The horizontally acting tension, T = Mg
The downward vertical force = (M + m)g
So, resultant force, which is choice (4).

If F > μ_s mg, the block m will start moving on block M. It will, therefore, exert a force on block M due to kinetic friction m and M.
This force f_k = μ_k mg.
Thus, the acceleration of block M =
Hence, the correct choice is (1).

The speed of the wave is


v = λf
Frequency of the wave does not change, so

Assuming ideal behaviour,
P_benzene = X_benzene P^o_benzene = 0.5 × 119 torr = 59.5 torr
P_toluene = X_toluene = 0.5 37 torr = 18.5 torr
P_total = 59.5 + 18.5 = 78.0 torr
= P_toluene/P_total = 18.5/78.0 = 0.237
Thus, (1) is the correct option.

From the Gibb's-Helmholtz equation
_{, at equilibrium}
Hence,
T =
_{= 30,560 J mol^-1}

Since, -CH₃ group is ortho, para directing the minor product will be

Hence, the option (1) is correct.

The number of S=O bonds present in an oxoacid of sulphur is inversely proportional to the number of H-atoms present per oxygen atom in the molecule.

The sentence, 'Most hybrid vehicles use a conventional gasoline engine as well as an electric motor to provide power to the vehicle' verifies the correct choice.
Thus, option (3) is the correct answer.

Problem figure 2 is the water image of Problem figure 1. Similarly, Answer figure (b) is the water image of Problem figure (3). Hence, Answer figure (b) is correct.

In the given series, the first four numbers are related to each other, then the next four and so on as shown below:
3, 4, 7, 14:
3 + 4 = 7
7 × 2 = 14
4, 5, 9 18:
4 + 5 = 9
9 × 2 = 18
Similarly, 5 + 6 = 11
11 × 2 = 22
Therefore, the missing term is 22.

According to question,

x = 1
Variance of sample 1 = (Standard deviation)² = (1)²

Variance of sample 1 = (Standard deviation)²
= (1.5)² = 2.25
Difference of variances = 2.25 – 1 = 1.25

Hence, , where .

Putting xe^x = t.
we get (xe^x + e^x) dx = dt
⇒ (x + 1) e^x.dx = dt
⇒
=
= tan t
= tan (xe^x).

The equation of circle is x² + y² + 2gx + 2fy + c = 0 with centre (-g, -f) ......(1)
Comparing the equation x² + y² + 2x + 8y − 25 = 0 with (1), we get,
g = 1 and f = 4
∴ Centre of first circle C₁ (−1, −4).
Also, radius = =
Comparing the equation x² + y² − 4x − 10y + 19 = 0 with (1), we get
g = −2 and f = −5
∴ Centre of second circle C₂ (2, 5).
Also, radius =
Now,
Also,





It means circles intersecting each other.
Hence, number of common tangents = 2