BITSAT Mock Test - 10 Free Online Test 2026

If mass m falls through a distance h, mass m rises up through the same distance h.
Let v be the common velocity of the masses when this happens.
Now, loss in PE = gain in KE, i.e.
Mgh - mgh = (M + m) v²
which gives v =

According to Bohr's postulate,
Putting
Moment of inertia, I =
_{KE rotational =
=
⁼}

The electric flux is given by the surface integral. Here, the electric field E is due to charge inside the Gaussian surface only. Hence, the correct option is (4).

Refer to the figure.

As 5/10 = 15/30
Since the bridge is balanced, so no current flows through the galvanometer as potential at point B is equal to potential at point D.

Let the distance between the two cliffs be d. Since, the man is standing midway between the two cliffs, then the distance of man from either end is d/2.
The distance travelled by sound (in producing an echo)

⇒ d = 340 × 1
⇒ 340 m

Mohr's salt is (NH₄)₂SO₄.FeSO₄.6H₂O.
The equation is 5Fe²⁺ + MnO^-₄ + 8H⁺ 5Fe³⁺ + Mn²⁺ + 4H₂O
Total change in oxidation number of iron = (+3) - (+2)
So, equivalent weight of Mohr's salt
=
=
= 392

Carbylamine test is given by aliphatic as well as aromatic primary amines, i.e. amines having --NH₂ group.

Hence, dimethyl amine, being a secondary amine, does not undergo carbylamine reaction.

V.P does not depend on the amount of water.
The volume occupied by water molecules in vapour phase is 1 × 10^-4 dm³ which is approximately equal to 1 × 10^-3 m³(volume of the flask)
Ideal Gas Law -
PV = nRT
where
P - Pressure
V - Volume
n - No. of Moles
R - Gas Constant
T - Temperature
PV = nRT

The mixture consists of weak acid and its salt with strong base, so it acts as an acidic buffer solution.
According to Henderson Hasselbalch equation,
pH = pK_a + log
= 4.57 + log
= 4.57 + (log 10 - log 3) [log 3 = 0.4771]
pH = 4.57 + 0.52
= 5.09

Diethyl ether, when heated with CO at 150^oC and 500 atm pressure in the presence of BF₃, forms ethyl propionate.
C₂H₅OC₂H₅ + CO

Hydride ion, being a stronger base than OH^- ion, will accept a proton from water to form hydrogen gas.
H^- (aq) + H₂O (l) → OH^- (aq) + H₂ (g)

If ∆H/∆T is same for in case of SO₂ (g) and SO₂Cl₂ (g), then change in entropy is related to molar mass of substance. Entropies generally increase with molecular weight. Thus, SO₂Cl₂(g) has greater entropy.

Acidic buffer solution is produced on mixing the aqueous solutions of acid and acidic salt.

The sequence consists of two series: 3, 6, 9, 12 and 5, 10, 15, 20. So, the next term will be 20.

(Degree of intensity) Both 'fury' and 'ire' refer to intense anger or wrath; the first is a more intense form of the second. In option 3, "spasm" refers to a sudden, violent, and involuntary contraction or series of contractions of muscles. A 'convulsion' is a more severe form of spasm.

In terms of height, we have:T, who is taller than P and S, plays Tennis.
T > P, T > S
Only T is between Q, who plays Football, and P in order of height.
Q > T > P
The shortest among them plays Volleyball.
R plays Volleyball, so R is the shortest.
S plays neither Volleyball nor Basketball.
So, Q is not the tallest.
Thus, U is the tallest.
So, the sequence becomes:U > Q > T > P > S > R
S is taller than R but shorter than P.

Given: ∑x = 18 and ∑x² = 30
n = 12
Now,
variance = ∑– (∑)²
=– ()²
=-
=
= = 0.25

Sum of the roots = 3/2
Sum of product (two at a time) of roots = 6/2 = 3
Required Answer = (3/2)² - 2 x 3
= -15/4
(Note: Answer is negative, It means there must be 2 roots which are imaginary. This is because the sum of square of real number can never be negative.
Imaginary roots are always in pair. )
Correct answer is Option (1)

I₁ =
Put log x = z => x = e^z
=> dx = e^z dz
When x = e, z = log e = 1
x = e², z = log e² => z = 2 log e = 2
I₁ = =
I₁ = I₂

Let (r + 1)^th term be the constant term in the expansion of .
Then, T_{r + 1} = ¹⁵C_r (x³)^{15 - r} is independent of x.
Or T_{r + 1} = ¹⁵C_r x^{45 - 5r} (-1)^r is independent of x.
45 - 5r = 0 r = 9
Thus, the tenth term is independent of x and is given by:
T₁₀ = ¹⁵C₉ (-1)⁹ = – ¹⁵C₉

R₃  R₃ - R₂; R₂  R₂ - R₁

D = (2008!)³ (2009)² (2010) (2)

= 2[[(2008 + 1]² (2010) - 2]
= 2[(2008)² (2010) + 2(2008) (2010) + 2010 - 2]
= 2[(2008)² (2010) + 2(2008) (2010) + 2008]
Which is divisible by 2008.

We know that ⁿC_r + ⁿC_{r - 1} = ^{n + 1}C_r
¹³C₉ - ¹²C₈ = (¹²C₉ + ¹²C₈) - ¹²C₈ = ¹²C₉

2x³y dy + (1 - y²)(x²y² + y² - 1) dx = 0

Solving, we get:

tx² = cx - 1
y²x² = (1 - y²) (cx - 1)