BITSAT Mock Test - 3 - JEE MCQ

= v₀ + gt + ft²
x = v₀t + gt² + ft³
At t = 0, x = 0
At t = 1s,

So, displacement is .

Direction of the magnetic moment to current flowing in circular coil is parallel to external field so the torque acting on the loop will be zero.

Since the given mirror is a concave mirror, the image formed will be inverted and double the size of the object.
Height of the object = 2.5 cm
Height of the image = – 5 cm; -ve sign indicates that the image is inverted.

Potassium metabisulphite and potassium sulphite are used as a source of sulphur dioxide, which reacts with H₂O to form sulphurous acid.
Sulphurous acid inhibits the growth of bacteria, fungi, yeast and moulds.
K₂S₂O₅ → K₂SO₃ + SO₂
SO₂ + H₂O → H₂SO₃

8% sulphur by mass means 8 g of sulphur is present in 100 g of compound.
Therefore, 32 g of sulphur (1 mole atom) will be present in 100/8 x 32 = 400 g of compound.
[∵ Compound must have at least one atom of sulphur]
^⇒ Minimum molecular mass = 400 g

Boric acid is a weak monobasic acid. It is not a protonic acid, but acts as a Lewis acid by accepting electrons from a hydroxyl ion:
B(OH)₃ + 2H₂O → [B(OH)₄]^- + H₃O⁺
The number of electrons around the central atom in BF₃ is only six. Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic configuration and thus, behave as Lewis acids.
Hence, both Q and S are incorrect.

NH4Cl comprises NH₄⁺ and Cl^- ions.
NH₄⁺ has 7 electrons of nitrogen and 3 electrons of hydrogen as 1 hydrogen is devoid of electrons.
Cl- has 18 electrons.
So, total = 7 + 3 + 18 = 28 electrons

is used for water softening to remove temporary hardness.

H₂ + O₂ → H₂O(l)
P + O₂ → P₄O₁₀(s)
Ca + O₂ → CaO(s)
S + O₂ → SO₂(g)

Glycine is the only amino acid with no asymmetric (chiral) carbon because it has two hydrogens attached to alpha carbon.

Adenine and thymine form two hydrogen bonds.

'Atrocity' (plural 'atrocities') denotes a behaviour or an action that is wicked, cruel or ruthless.

The first and third letters of the first term are each moved five steps forward, while the second letter is moved three steps forward to obtain the corresponding letters of the second term.
K + 5 = P
N + 3 = Q
O + 5 = T
Hence, option (2) is correct.

The pattern is: - 66, - 55, - 44, - 33, - 22, - 11, ……
So, missing term = 105 - 11 = 94

The required vector c is given by

Let p be any arbitrary element of A - B.
Then p ∈ (A - B)
⇒ p ∈ A and p ∉ B
⇒ p ∈ A and p ∈ B^C
⇒ p ∈ A ∩ B^C
∴ A - B A ∩ B^C ...(i)
Again, let q be any arbitrary element of A ∩ B^C. Then,
q ∈ A ∩ B^C
⇒ q ∈ A and q ∈ B^C
⇒ q ∈ A and q ∉ B
⇒ q ∈ A - B
∴ A ∩ B^C (A - B) ...(ii)
Hence from (i) and (ii), we have
A - B = A ∩ B^C

g(x) = xf(x), where f(x) = x sin1/x
at x = 0
g(0) = 0 f(0)
= 0 (0) = 0
g'(0) =
=
=
also g'(x) = xf'(x) + f(x)
g'(x) = [xf' (x) + f(x)]
= 0 + f(0) = 0 + 0 = g'(0)
∴ g'(x) = g'(0)
Hence g' is continuous and g(x) is differentiable.

Let,
AB be a pole of height h and BC = x be the shadow of pole and θ be the angle of elevation of the source of light. Given:
Lenght of shadow of pole = height of pole
⇒ x = h ........(1)

Now, from figure
tanθ = h/x
⇒ tan θ = 1     [∴ x = h]
⇒ θ = 45º.

Six girls and five boys can sit in a row in 11! ways.
∴ exhaustive number of cases = 11!
Six girls can sit in a row in 6! ways and in each such arrangement there are 7 places between them in whcih boys can be sated in ⁷p₅ ways. 
Therefore, the total number of ways in which two boys sit together = 6! ⁷p₅.
Hence, requires probability
= (6! ⁷p₅) / 11! = (6!7!) / (2!11!)

Required area
=
=
=
= 1 sq. unit

First two lines intersect at (2, - 2)….(A).
The second line and the third line intersect at (1, 1)… (B).
The first line and the last line intersect at (- 2, 2)…(C).
Distance of A and C from B is equal, whereas AC is not equal to them. Hence, it is an isosceles triangle.