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4 g of NaOH is dissolved is 250 ml of a solution, the molarity of the solution will be  0.4

Answer (d) None of these

LiCl is less than ionic character then KCl.

The excess amount of supplied energy to the reactants in which the reactant contain normal energy state is equal called activation energy.

Activation energy = Threshold energy - Normal energy of reactants.

Tetrahedral the structure of CaCl.

Isotonic solutions have same Malar concentration

Factors which affects the conductance of electrolysis are 

(i)  the nature of the electrolyte added

(ii)  size of the ions produced and their sovation 

Pbl₄ halide of group 14 does not exist.

The osmotic pressure of a 1 M solution at 27°C is 24.6 atm

CH₄ cannot be prepared by Wurtz reaction.

1, 3, 5-trimethylbenzene

(b) Dehydration

False Calcium carbide on hydrolysis gives ethylene.

The domain of cos^-1x is [-1, 1]

The Direction cosines are given by

=(cos90°, cos60°, cos30°)

= (0, 1/2, √3/2 )

A vector having Unit magnitude along any vector a is Called a unit vector in the direction of vector a add is denoted vector a|a = 1

=> ∫(2x – 3cosx +e^x).dx 

= ∫ 2 x.dx – 3 ∫ cos x.dx + ∫ e^x .dx 

= 2x x²/2 - 3 x sinx + e + c 

= x² - 3sin x + e^x + c

y = cos(1 - x) 

Diff w.rl.x

dy/dx = d/dx = cox(1 - x)

= -sin(1 - x) x d/dx (1 - x) dx 

= -sin(1 - x)x - 1 = sin(1 - x)

We know that the ratio of area of two similar triangle is the square of the ratio of corresponding sides of these similar triangles. 

Let the area of the larger triangle is A cm² . 

Given that the area of the smaller triangle is 48 cm² and their corresponding sides are in the ratio 2 : 3.

Therefore, \(\frac{48}{A} = (\frac{2}{3})^2\) (The ratio of Area of similar triangles is the square of the ratio of their corresponding sides.) 

⇒ A = 48 × \((\frac{3}{2})^2\) 

⇒ A = 48 × \(\frac{9}{4}\) = 12 × 9 = 108.

Hence, the area of the larger triangle is 108 cm².

Let a, b ∈ Z, where Z is set of integers.

Given binary operation is a ∗ b = ab³ .

Now, b ∗ a = ba³ ≠ ab³ = a ∗ b. ( \(\because\) ba³ ≠ ab³ , where a and b are arbitrary integers )

Therefore, b ∗ a ≠ a ∗ b.

Therefore, the binary operation ‘∗’ is not commutative.

The Relation R is defined by R = {(1,1) (1,2) (3,3)} for Symmetric 

Relation if (1, 2) ∈ R then (2,1) ∈ R but (2,1) ∉ R? 

∴ R is not Symmetric Relation.

f(x) = |x| – |x+1| 

f(c)= |c | – |c+1| 

Let c = 0, f(0) = |0 |-|1| = 0-1= -1 

Lt f(x) = It|x| – |x + 1|= |c| – |c+1| 

x→ c X → C 

let c = 0, Lt f(x)=0 – 1 = -1 

x— +0 

∴ Lt (x→0) f(x) = f(0) = -1 

⇒ f is continuous at x =0 

There is no point of Discontinuity.

If * be a commutative binary operation.

a*b = b*a 

⇒ a*b = ab + 1 = ba + 1= b*a 

∴ * is commutative binary operation on*.

Given lines are 7x + y – 10 = 0 → (1), x – 2y + 5 = 0 → (2), x + y + 2 = 0 → (3) 

Point of intersection of (2) and (3) is B (–3, 1). 

Equation of the altitude through B is (x + 3) – 7(y – 1) = 0 

⇒ x – 7y + 10 = 0 → (4) 

Point of intersection of (3) and (1) is C(2, –4) 

Equation of the altitude through C is 2(x – 2) + (y + 4) = 0 

⇒ 2x + y = 0 → (5)

Solving (4) and (5), orthocenter is (- 2/3, 4/3).

Concept:

We know that the domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. 

Given:

\(f\left( x \right) = \sqrt {{x^2} - 3x + 2} \)

Analysis:

For domain,

f(x) ≥ 0

\(\sqrt {{x^2} - 3x + 2} \ge 0\)

(x - 2) (x - 1) ≥ 0

Case: 1

(x - 2) ≥ 0 and (x – 1) ≥ 0

x ≥ 2 and x ≥ 1

overall: x ≥ 2

{x|2 ≤ x < ∞} = [2, ∞)

Case: 2

(x - 2) ≤ 0 and (x - 1) ≤ 0

x ≤ 2 and x ≤ 1

overall : x ≤ 1

{x|-∞ < x ≤ 1} = (-∞, 1]

∴ The domain of the function is:

(-∞, 1] ∪ [2, ∞)

Concept:

If the root is 2, it means that the polynomial is completely divisible by x - 2.

The given polynomial looking carefully is 3x2 – 13x + 6 = 0 and it does not have any root equal to 2.

The correct answer is not available in the options given in the exam and hence one of the options has been modified.