BITSAT Mock Test - 1 Free Online Test 2026

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).

Using Poiseuille’s law, flow rate

Flow rate in series with total pressure P:

Option C is correct.

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).

Given:

• Cylindrical tank open at the top

• Cross-sectional area of the tank = a₁

• Cross-sectional area of the hole at the bottom = a₂

• a₁ = 3a₂

• Initial height of water = h

• Acceleration due to gravity = g

• Find: Time taken to empty the tank in terms of h and g

Step 1: Use Torricelli’s theorem

According to Torricelli's theorem, the speed v of the water flowing out from the hole is:
v = √(2gh)
(where h is the instantaneous height of the water level above the hole)

Step 2: Volume flow rate

The volumetric flow rate Q through the hole is:
Q = a₂ × v = a₂ × √(2gh)

Step 3: Rate of decrease of water volume in the tank

Volume of water in the tank at height h is:
V = a₁ × h

The rate of change of volume with respect to time is:
dV/dt = a₁ × dh/dt

Since water is flowing out of the tank, this rate is negative and equal to -Q:
a₁ × dh/dt = -a₂ × √(2gh)

Step 4: Set up the differential equation

Divide both sides by a₁:
dh/dt = -(a₂ / a₁) × √(2gh)

Since a₁ = 3a₂,
dh/dt = - (1/3) × √(2gh)

Step 5: Separate variables and integrate

Rewriting:
dh / √h = - (1/3) × √(2g) dt

Now integrate both sides.
Left side: integrate from h to 0 (final height),
Right side: integrate from 0 to T (time to empty)

Integral of h^(-1/2) is 2√h, so:

-2√h = - (1/3) × √(2g) × T

Cancel negatives:

2√h = (1/3) × √(2g) × T

Step 6: Solve for T

Multiply both sides by 3:

6√h = √(2g) × T

Divide both sides by √(2g):

T = 6√h / √(2g)

Now simplify:

T = √(h / g) × 6 / √2 = 3 × √(2h / g)

Final Answer:

This is the time taken to empty the tank.

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.

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,

Hydration energy ^.
With the increase in size of the alkali metal, the charge density decreases and the extent of hydration decreases.
Decreasing order of size of alkali metal is
Rb⁺ > K⁺ > Na⁺ > Li⁺
Therefore, decreasing order of hydration energy is
Li⁺ > Na⁺ > K⁺ > Rb⁺

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 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

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