BITSAT Mock Test - 1 - JEE MCQ

The magnetic force on a charged particle is:

Compute the cross product:

Let velocity of the river be and velocity of the person w.r.t. river
Velocity of the person w.r.t. ground

or,

Displacement of the person along x-axis in time t w.r.t. earth
------1
Displacement of the person along y-axis in time t
----- 2, where a is the acceleration of the person
From 1 and 2, we have

Hence, trajectory of the person w.r.t. earth is a parabola.

Angle of dip (θ) = tan^-1 (B_V / B_H) where B_V and B_H are the vertical and horizontal components of earth's field respectively. Thus
B_V = B_H tan(θ) = 0.314 × 10^-4 × tan 26.6°
= 0.157 × 10^-4 T
If the plane of the vertical coil A is perpendicular to the magnetic meridain, the field produced by it can neutralize the horizontal component of earth's field if
μ0I A ⁿ / 2 r = B_H
or (4π x 10^-7 x I _A x 10) / (2 x 0.2) = 0.314 × 10^-4
which gives I_A = 1 A.
Similarly, the magnetic field produced by the horizontal coil B will be vertical and will neutralize the vertical component BV of the earth's field, if
μ0I B ⁿ / 2 r = B_V
or (4π x 10^-7 x I B x 10) / (2 x 0.2) = 0.157 × 10^-4
which gives IB = 0.5 A.
Hence the correct choices is (4).

If M is the mass of the earth, the gain in potential energy is given by

Hence the correct choice is (2). Remember, the expression PE = mgh is an approximate one which is valid if h << R.

The electric field E₁ at (a, b) due to q₁ has a magnitude


And is directed along + x-axis. The electric field E₂ at (a, b) due to q₂ has a magnitude
And is directed along + y-axis. The angle θ subtended by the resultant field E with the x-axis is given by
Hence, the correct choice is (B).

Let dw and do be the densities of water and oil, then the pressure at the bottom of the tank = h_wd_wg + h₀d₀g
Let this pressure be equivalent to pressure due to water of height h. Then hd_wg = h_wd_wg + h₀d₀g

= 100 + 360 = 460
According to Toricelli's theorem,

According to Ostwald's dilution law,
K_a = , where K_a = Equilibrium constant for weak acid
C = Concentration
α = Degree of ionisation of weak acid (HA)
For very weak acid, α <<<< 1 or 1 - α = 1
K_a = Cα²
Or,
C = 0.1 M, α = 1.34% = 0.0134
K_a = 0.1 x (1.34 10^-2)²
= 1.79 x 10^-5

The increasing order of the acidic strength is:

-NO₂ group is electron withdrawing at all positions and increases the acidic strength. Hence, benzoic acid has the least acidic strength among the given options.
The effect is more significant at the ortho and para positions due to the -R effect.
However, o-Benzoic is the most acidic due to the ortho effect. Moreover, the carboxylate ion formed in this case is the most stable due to the intramolecular H-bonding.

The sentence is in past tense. The use of 'before' indicates an earlier past event. To show the earlier activity in the past, we use 'past perfect'. So, the correct phrase should be 'as if she had seen me before'.

The series should follow increments of squares: 0²,1²,2²,3²,4²,5². thus:
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.

As,
R + 3 = U
S + 3 = V
Q + 3 = T
Similarly,
X + 3 = A
Y + 3 = B
W + 3 = Z
Hence, ABZ is the correct answer.

Expression: FJUL : BOQQ :: LHRX :
The pattern followed is:

Similarly, LHRX : HMNC

(6)² = 36 and (6)³ = 216
Similarly,
(9)² = 81 and (9)³ = 729

From the given information, two conclusions can be obtained:
1. Anand is not married to Bharti or Chhaya, i.e. Anand can be married to either Divya or Alka.

2. Basant and Diganto are not married to Divya or Bharti, i.e. Basant and Diganto are married to Alka or Chhaya.

From these two conclusions, we can see that since Alka would be married to either Basant or Diganto, Anand would surely be married to Divya, which leaves Chetan and Bharti as a pair as well. This gives two final possible sets of pairs of males and females.
1. When Basant is married to Alka

2. When Basant is married to Chhaya

Following these two possibilities, the partners of Basant and Diganto, i.e. Chhaya and Alka cannot be known for sure without any more information.
Therefore, it cannot be determined whom Chhaya is married to.

Whenever the subject is missing in the passive voice, we provide it while changing into the active voice. Indefinite pronoun 'someone' is the correct subject supplied. 'His pocket' will become the object in passive voice. Hence, option D is the correct answer.

Since the number of terms in (x₁ + x₂ + x₃ + ... + x_m)ⁿ is ^{n + m - 1}C_{m - 1,} the number of terms in (x - 2y + 3z)ⁿ, where m = 3, is ^{n + 3 - 1}C_{3 - 1}.
= (n + 1) (n + 2) / 2
We must have (n + 1) (n + 2) / 2 = 45
⇒ (n + 1)(n + 2) = 90
⇒ n = 8

sin θ/2 = √[(x - 1) / 2x)]
sin θ/2 = √[(1 - (1 / x) / 2)]
⇒ θ is acute
Hence, sin θ/2 = + (Positive)
Therefore, by formula, sin θ/2 = √[(1 - cosθ) / 2]
cosθ = 1 / x
or, sec θ = x
Also, 1 + tan² θ = sec² θ
1 + tan² θ = x²
tan² θ = x² - 1
tan θ = √(x² - 1)
Hence, option (2) is correct.

∵ (dy/dx) + 1 = cosec (x + y)
Let x + y = t
⇒ 1 + (dy/dx) = dt/dx = cosec t
∴ dt / cosec t = dx
On integrating both sides, we get
∫ sin t dt = ∫ dx
⇒ - cos t = x + c'
⇒ cos (x + y) + x = c

Focus of y² = 16x is (4, 0)
∴ Equation of straight line is
y - 0 = m(x - 4)
mx - y - 4m = 0;
it is tangent to (x - 6)² + y² = 2
Distance of tangent from centre = radius
[(|6m - 0 - 4m| / √(m² + 1))] = √2
⇒ Squaring we get m² + 1 = 2m²
m = ±1

For each book, we may take 0, 1, 2, 3, ... m copies.
∴ We may deal with each book in (m + 1) ways and with all the books in (m + 1)ⁿ ways.
But, this includes the case where all books are rejected and no selection is made.
So, the number of ways in which selection can be made = (m + 1)ⁿ − 1

if f coefficient of x2 = 0 and coefficient of x = 0
⇒ (1 - a) = 0 and - (a + b) = 0
⇒ a = 1 and b = -1

x² - y² + 2y = 1
⇒ x² - (y² - 2y + 1) = 0
⇒ x² - (y - 1)² = 0
⇒ (x + y - 1)(x - y + 1) = 0
So the lines are x + y - 1 = 0 and x - y + 1 = 0
Their angle bisectors are
(x + y - 1) / √(1² + 1²) = ±  (x - y + 1) / √(1² + 1²)
(x + y - 1) / √2 = ±  (x - y + 1) / √2
⇒ x + y - 1 = ± (x - y + 1) x + y - 1 = x - y + 1
or
x + y - 1 = -x + y - 1
⇒ 2y = 2 or 2x = 0 y = 1 or x = 0
Given x + y = 3
⇒ Base of triangle = 2 units
⇒ Height of triangle = 3 - 1 = 2 units
⇒ Area = 1/2 x base x height
= 1/2 x 2 x 2 = 2 sq. units.

In the word MATHEMATICS, there are eight distinct letters.
M, A, T, H, E, I, C and S
We have to arrange 4 letters from them.
Therefore, number of ways =  ⁸C₄ × 4! = 1,680

Let
x = r cosθ, y = r sinθ
dx = cosθ dr - r sinθdθ,
dy = sinθ dr + r cosθdθ,
Hence,
xdx + ydy = r cosθ(cosθ dr - r sinθdθ) + r sinθ(sinθdr + r cosθdθ),
 = dθ ⇒ sin-1 r = θ + α
(xdx + ydy)/√(x² + y²) = r²dr
Similarly, (xdy - ydx)√1 - (x² + y²) = r² √1 - r² dθ
dr/√1 - r² = dθ ⇒ sin⁻¹ r = θ + α
r = sin(θ + α)
⇒ x² + y² - x sinα - y cosα = 0
∴ radius = 1/2