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         Horizontal displacement =v₀t [as the acceleration is zero]

          In a horizontal projectile, horizontal velocity remains constant.

          it is given that at a certain time horizontal velocity is equal to vertical velocity

                          v_y = v₀

                      Generally,    v_y = u + gt [u = 0]

                                           v₀ = gt

                                           t =  v₀ /g ----------(1)

     As given above horizontal displacement at time t = v₀t ----------(2)

      substitute the above equation (1) in (2)

                                              s_x = v₀ (v₀ /g)

                                             s_x = v₀ ² /g 

• A body is projected with a velocity v₀ at an angle θ then its horizontal velocity becomes v₀ cos θ and vertical velocity becomes v₀ sin θ
• At certain point, it is given that the velocity makes angle β with vertical   

                 let the velocity at that point be 'v' then the horizontal velocity becomes vsinβ

                            vertical velocity becomes vcosβ

            In a projectile, horizontal velocity is always constant as the acceleration in that direction is 0

                so,    v₀ cos θ = vsinβ 

                              v = v₀ cos θ/sinβ 

As a, b, c are in G.P. 

Therefore b² = ac … (1) 

We have to prove a², b², c² are in GP or 

we need to prove: (b²)² = (ac)² {using idea of GM} 

On squaring equation 1 we get, 

⇒ b⁴ = a²c² ⇒ (b²)² = (ac)² 

Hence a²,b²,c² are in GP

(3) Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2– < N^3– 

Correct order of size for isoelectronic species.

Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2– < N^3– 

Correct answer is (c) tan 90°

Correct option is: (a) -2

Correct answer is (b) 3 : 4

Correct answer is (a) 5.213213213...

Answer is (a) Nitrogen

(High IE of N is because of smallest size in the group and completely half - filled p subshell)

Answer is (b) partial double bond character of C-OH bond

Answer is (3) Au

Atomic radius of Ag and Au is nearly same due to lanthanide contraction.

Answer is (3) Au

Atomic radius of Ag is closest to Au.

The answer is (3) Au

Answer is (4) Atomic radius

In general across a period atomic radius decreases while ionisation enthalpy, electron gain enthalpy and electronegativity increases because effective nuclear charge (Z_eff) increases.

(2) eutrophication

In Eutrophication nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity. If indicates polluted environment.

ICl is more reactive than I₂ because I−Cl bond in ICl is weaker than I−I bond in I₂.

[Hint : I₂O₅]

HOCl, HOClO, HOClO₂, HOClO₃
[Hint : HOCl < HOClO < HOClO₂ < HOClO₃]

[Hint : Due to smallest size of F.]

SI unit would be Nm^-2(m³ )²(K)² / mol 

Nm⁴K²mol^-1

(4) 

Correct order of increasing basic strength is

R–COO^(–) < CH≡C^(–) < 2 NH₂^- < R^(–) 

We have given :

gas mixture made up of 

A_r - 10g

CO₂ - 14.2 g

K_r - 32.1 g

Volume of mixture of gases = 24.5 L

Temperature = 260°C or 533 K

Number of moles of A_r = \(\cfrac{10.0}{39.948}\) = 0.25 mole

Number of moles of K_r = \(\cfrac{32.1}{83.79}\) = 0.38 mole

Number of moles of CO₂ \(\cfrac{14.2}{44}\) = 0.32 mole

Total number of mole of gases = 0.32 + 0.38 + 0.25

= 0.95 mole

assume, the mixture of gases follow ideal gases equation 

PV = nRT

= P = \(\cfrac{0.95}{24.5L}\) x 0.082 x 533

= 1.7 atm

mole fraction of A_r = \(\cfrac{0.25}{0.95}\) = 0.26

mole fraction of K_r = \(\cfrac{0.38}{0.95}\) = 0.40

mole fraction of CO₂ = \(\cfrac{0.32}{0.95}\) = 0.34 

partial pressure of A_r = (P_Ar) =  mole fraction x total pressure at A_r

= 0.26 x 1.7 atm

= 0.442 atm

partial pressure of K_r (P_kr) = 0.40 x 1.7 atm 

= 0.68 atm

partial pressure of CO₂  = (CO₂) = 0.34 x 1.7 atm

= 0.578 atm

Hence, The total pressure of mixture of gases will be 1.7 atm and partial pressure of A_r, K_r and CO₂ will be 0.442 atm, 0.680 atm and 0.578 atm respectively

Answer is : d) 6.5

Correct option is (D) 6.5

using molarity equation-

M₁V₂ = M₂V₂

10 ml x 2.2 x 10^-5 M = 1000 mL x M₂

M₂ = 2.2 x 10^-7M of HCl

\(\therefore\) final concentration of HCl = 2.2 x 10^-7 M.

Here, we also consider the H⁺ions come from water dessociation.

We know that

[H⁺] = [\(\bar O\)H] = 1 x 10^-7

\(\therefore\) Total H⁺ concentration in solution

 = (2.2 x 10^-7) + (1 x 10^-7) M

 = 3.2 x 10^-7 M

\(\therefore\) P^H of solution = - log[H⁺]

 = -[log(3.2) + log 10^-7]

 = -0.5 + 7

 = 6.5

Hence, the P^H of final solution will be 6.5

\(\frac d{dx} x(x^2 + 1)^{\frac{-1}2} = x\frac d{dx} (x^2 + 1)^{\frac{-1}2} + \frac{dx}{dx} \times (x^2 + 1)^{\frac{-1}2}\)

\( = x\times \frac{-1}2 (x^2 + 1)^{\frac{-1}2-1} \frac{d}{dx} (x^2 + 1) + (x^2 + 1)^{\frac{-1}2}\)

\(= -\frac x2 (x^2 + 1)^{\frac{-3}2}.2x + (x^2 + 1)^\frac{-1}2\)

\( = (x^2 + 1)^\frac{-1}2 \left(\frac{-x^2}{x^2 + 1}+ 1\right)\)

\( = (x^2 + 1)^\frac{-1}2 \left(\frac1{x^2 + 1}\right)\)

\( = \frac1{(x^2 + 1)^\frac32}\)

\(F= At + \frac B{t - C}\)

\(F = \frac B{t - C}\)

\([MLT^{-2}] [T] = B\)

\([MLT^{-1}] = B\)

\(x^2 - 3x + 2 > 0\;\text{and}\; x^2 - 3x + 2 \ne 1\)

⇒ \( (x - 1) (x - 2) > 0\;\text{and}\; x^2 - 3x + 1 \ne 0\)

⇒ \(x \in (-\infty , 1) \cup (2, \infty)\;\text{and}\; x \ne \frac{3\pm \sqrt5}2\)

Also,

\(-1 \le \frac{x^2 - 5x + 6}{x^2 - 9} \le 1 \; \text{and}\; x^2 - 9 \ne 0\)

⇒ \(- x^2 + 9 \le x^2 - 5x + 6 \;\text{and}\;x^2 - 5x + 6 \le x^2 - 9\; \text{and}\; x \ne 3 \) & \(-3\)

⇒ \(2x^2 - 5x - 3 \ge 0 \;\text{and}\; -5x + 6 \le -9\)

⇒ \((x - 3) (2x + 1) \ge 0 \; \text{and}\; 5x \ge 15\)

⇒ \(x\in (-\infty, \frac{-1}2]\cup [3, \infty)\;\text{and}\;x\in(3, \infty).\)

∴ Domain of function is 

\(x\in (-\infty, \frac{-1}2]\cup [3, \infty) - \{-3\}.\)

Correct option is (1)

We know that v = w/k

\(\frac{1}{v}=\frac{k}{w}\)

\(\frac{k}{w}=\frac{1}{[LT^1]}\)

\(\frac{k}{w}=[L^{-1}T]\) 

कैलीपर्स द्वारा न्यूनतम मापन = मुख्य पैमाने के एक भाग का माप / वर्नियर पैमाने में कूल भाग

\(= \frac {0.05}{50}\)

\(= 0.01\, mm\)

Correct option is (2) \( 21 \,l/s\)

Given Volume

\(V = t^3 - t^2 + 2\)

\(\left(\frac{dV}{dt}\right) = 3t^2 - 2t\)

at \(t = 3\)

\(\left(\frac{dV}{dt}\right) = 27 - 6\)

\(\frac{dv}{dt} = 21 \,l/s\)

From mass-energy equivalence,

E=mc²

=0.5×10⁻³×9×10¹⁶

=4.5×10¹³J