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Correct option is (D) 1/2 (Y + Z - X)

\(\Lambda^0_{eq} (HCl) = \lambda ^0_{H^+} + \lambda^0_{Cl^-}\)

\(2 \times \Lambda^0_{eq} (HCl) = 2\lambda^0_{H^+} + 2\lambda^0_{Cl^-}\)

\(2 \times \Lambda^0_{eq} (HCl) = 2\lambda^0_{H^+} + \lambda ^0_{SO^{-2}_4}- \lambda ^0_{SO^{-2}_4}+2\lambda^0_{Cl^-} + \lambda^0_{Ba^{+2}} - \lambda^0_{Ba^{+2}}\)

\(2 \times \Lambda^0_{eq} (HCl) = \Lambda^0_{eq}(H_2SO_4) + \Lambda^0_{eq}(BaCl_2) - \Lambda^0_{eq}(BaSO_4) \)

\(2 \times \Lambda^0_{eq} (HCl) = Z + Y - X\)

\( \Lambda^0_{eq} (HCl) =\frac12( Z + Y - X)\)

a. Actual amount of water in the air

In most insects, the sense of smell is localized in the Antennae.

b. Horizontal resistance

Optimum temperature range for majority of the insects is 28-30 °C.

Yellow to blue is in the range from.

Formula : [Co(NH₃)₄Cl_2]Cl
Name : Tetraaminedichloridocobalt(III) chloride Yes, it show geometrical isomerism.

The cell reaction that takes place in a lead storage battery are:

Pb + PbO₂ + 4H⁺ + SO₄^-2 → 2PbSO₄ + 2H₂O

In this reaction Pb shows an oxidation state of +4. It gets reduced to PbSO₄ which shows an oxidation state of +2.  

In this reaction, 2 electrons are necessary for the reduction of PbO₂.

Pb⁴⁺ + 2e^-  -----------> Pb²⁺

Therefore, as 2 electrons were needed for the reduction, the Faraday’s law of electrolysis stated that:

The reduction of 1 mole of PbO₂ required 2F electricity.  

The equation for the reaction will be as follows:

A₂B  ⇌ 2A⁺+B^-

By Stoichiometry,

1 mole of A₂B  gives 2 moles of  and 1 mole of B

Thus if solubility of  A₂B is s moles/liter, solubility of A is 2s moles\liter and solubility of B is s moles/liter

Therefore,

K_sp = [2A⁺]2[B^-]

K_sp = [2s]²[s]

4s³ = 32 x 10^-9 mol³L^-1

s = 2 x 10^-3 moles/liter

Hence, the solubility of A₂B is 2 x 10^-3 moles/liter

\(\underset{38\,gram}{H_3C -\overset{F\\|}{CH} -CH_2 - CH_3 } \xrightarrow{\bar oH,\Delta} \underset{Major product}{CH_3 - \underset{|\\OH}{CH} - CH_2 -CH_3} + \underset{Minor product\\(Y \, gram)}{CH_3 - CH = CH - CH_3}\)

Minor product will be elemination product.

Assuming all \(CH_3 -\overset{F\\|}{CH} -CH_2 - CH_3\) converted into minor product.

\(\therefore\) Molecular weight of \(CH_3 -\overset{F\\|}{CH} -CH_2 - CH_3\) = 76 g/mol

 Molecular weight of CH₃ - CH = CH - CH₃ = 56 g/mol

Number of moles  of \(CH_3 -\overset{F\\|}{CH} -CH_2 - CH_3\) = \(\frac{38}{76}\) = 0.5 mol

\(\therefore\) Number of moles of minor product will  be = 0.5 mol

\(\therefore\) Weight of minor product = 0.5 x 56 = 28 gm

Hence, the value of y = 28 gm.

(e) Let the number of girls = x
 Number of boys = 1.16 x
Required ratio = 1.16 x : x
     = 116 : 100 = 29 : 25

(a) Total number of students = 1224
Total number of girls = 600
Total number of boys = 1224 - 600 = 624
Required ratio = 624 :600 = 26 : 25

395/10500
395/ (2²× 3× 5³× 7)
Here, the factors of the denominator 10500  are 2²× 3× 5³× 7, which is not in the form 2ⁿ 5^m .

So , 395/10500 has non terminating repeating decimal expansion.

Remainder when 5 is divided by 13 is 5,

Remainder when 5² is divided by 13 is 12,

Remainder when 5³ ( or 125) is divided by 13 is 8.

Remainder when 5⁴ ( or 625) is divided by 13 is 1.

∴ Remainder when (5⁴)ⁿ (n∈N) is divided by 13 is 1.

∴  When 5100 = (5⁴)²⁵ is divided by 13 it leaves remainder 1.

\(\frac {3\sqrt2}{\sqrt3+\sqrt6} - \frac {4\sqrt3}{\sqrt6+\sqrt2} + \frac {\sqrt6}{\sqrt2+\sqrt3} \)

= \(\frac {3\sqrt2(\sqrt3 -\sqrt6)}{(\sqrt3+\sqrt6)(\sqrt3-\sqrt6)} -\frac {4\sqrt3(\sqrt6 -\sqrt2)}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)} +\frac {\sqrt6(\sqrt2 -\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)}\) 

= \(\frac {3\sqrt6 - 6\sqrt3}{3-6} -\frac {12\sqrt2 - 4\sqrt6}{6-2} +\frac {2\sqrt3 - 3\sqrt2}{2-3} \) 

= \(\frac {3\sqrt6 + 6\sqrt3}{3} +\frac {4\sqrt6 - 12\sqrt2}{4} +\frac {3\sqrt2 - 2\sqrt3}{1} \) 

=  \(​​\frac {-12\sqrt6 + 24\sqrt3 + 12\sqrt6 - 36\sqrt2 + 36\sqrt2 - 24\sqrt3}{12} = \frac 0{12} =0\)

Correct option is: (C) 1

Shows the resolution of force F = 20 kN into its components in horizontal and vertical components. From the figure it is clear that

F_x = F cos 60° = 20 cos 60° = 10 kN (to the left) 

F_y = F sin 60° = 20 sin 60° = 17.32 kN (downward)

Let ∆ ABC be the triangle of forces drawn to some scale. In this 

∠BAC = α = 20° 

∠ABC = 180 – 50 = 130° 

∴ ∠ACB = 180 – (20 + 130) = 30° 

Applying sine rule to ∆ ABC, we get

\(\cfrac{AB}{sin\,30^\circ}\) = \(\cfrac{BC}{sin\,20^\circ}\) = \(\cfrac{500}{sin\,130^\circ}\)

AB = 326.35 N 

and BC = 223.24 N. 

Thus F₁ = AB = 326.35 N 

and F₂ = BC = 223.24 N

The plane makes an angle of 20° to the horizontal. Hence the normal to the plane makes an angles of 70° to the horizontal i.e., 20° to the vertical. If AB represents the given force W to some scale, AC represents its component normal to the plane and CB represents its component parallel to the plane. 

Thus from ∆ ABC, 

Component normal to the plane = AC

= W cos 20° 

= 10 cos 20° 

= 9.4 kN as shown in 

Component parallel to the plane = W sin 20° = 10 sin 20° 

= 3.42 kN, down the plane

From the above example, the following points may be noted: 

1. Imagine that the arrow drawn represents the given force to some scale. 

2. Travel from the tail to head of arrow in the direction of the coordinates selected. 

3. Then the direction of travel gives the direction of the component of vector. 

4. From the triangle of vector, the magnitudes of components can be calculated.

The force of 3000 N acts along line AB. Let AB make angle α with horizontal. Then, obviously 200 sin α = 80 sin 60° 

∴ α = 20.268° 

Referring to we get Horizontal component = 3000 cos 20.268° = 2814.2 N 

Vertical component = 3000 sin 20.268° = 1039.2 N 

Components along and normal to crank: 

The force makes angle α + 60° = 20.268 + 60 = 80.268° with crank. 

∴ Component along crank = 3000 cos 80.268° = 507.1 N 

Component normal to crank = 3000 sin 80.268° = 2956.8 N

In this problem, note that selecting X and Y axes parallel to the plane and perpendicular to the plane is convenient.

R_x = ΣF_x = T – F – W sin θ 

= 1200 – 100 – 1000 sin 60° = 233.97 N 

R_y = ΣF_y = N – W cos 60° = 500 – 1000 cos 60° = 0. 

∴ Resultant is force of 233.97 N directed up the plane.

5000 N force is shifted to a point B along its line of action (law of transmissibility) and it is resolved into its x and y components (F_x and F_y as shown in)

F_x = 5000 cos θ = 5000 x  4/5 = 4000 N 

and F_y = 5000 sin θ  = 5000 x 3/5 = 3000 N. 

By Varignon's theorem, moment of 5000 N force about A is equal to moment of its component forces about the same point. 

8000 = 4000 × y + 3000 × 0 

∴ y = 2 m.

Let T be the tension in the cable and the reaction at the pair of wheels be R₁ and R₂ as shown in Fig.

Now, ∑ of forces parallel to the track = 0, gives 

T – 60 sin 60° = 0 

T = 51.9615 kN. 

Taking moment equilibrium condition about upper axle point on track, we get 

R₁ × 1200 + T × 600 – 60 sin 60° × 800 – 60 cos 60° × 600 = 0 

R₁ = 23.6603 kN. 

∑ of forces normal to the plane = 0, gives 

R₁ + R₂ – 60 cos 60° = 0 

R₂ = 30 – 23.6603 

R₂ = 6.3397 kN.

M = 20kg

angular speed, co – 100 rad s^-1; R = 0.25m

Moment of inertia of the cylinder about its axis

= \(\frac{1}{2}\)MR² = \(\frac{1}{2}\) × 20 × (0.25)²kgm² = 0.625kgm²

Rotational kinetic energy,

Er = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\) × 0.625 × (100)²J = 3125 J

Angular momentum,

L = Iω = 0.625 × 100 Js = 62.5 Js.

Correct option is (3) 14 x 10² 

Area = Length x Breadth 

= 55.3 x 25 

= 1382.5 

= 14 x 10²

Resultant should have 2 significant figures.

Correct option is (1) B

Initially speed is zero, then increases & after some time it becomes constant. Acceleration (slope of v/t curve) of ball first decreases and after some time it becomes zero.

Correct answer is (3) reverse bias only

In reverse bias external battery attract majority charge carriers. 

So width of the depletion region increase

(1) should be approximately equal and are small

Resistance of P & Q should be approx. equal as it decreases error in experiment.

(b) 2Ag+ 2H₂SO₄→ 2H₂O+ SO₂+ Ag₂SO₄  .

 Au, Pt does not react. Pb forms insoluble PbSO₄

(c) HI gets oxidised to I₂ 

(b) Conc. H₂ SO₄ is a strong dehydrating agent due to which carbohydrates becomes charred on reaction with conc. H₂ SO₄ acid

The main gas responsible for greenhouse effect is CO₂ . Other greenhouse gases are methane, nitrous oxide, water vapours, chlorouorocarbons (CFC’s) and ozone.

Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.

• Gaseous air pollutants. These include mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂) in addition , to hydrogen sulphide (H₂S), Hydrocarbons, ozone and other oxidants. 
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Plants require CO₂ in an optimum amount for the process of photosynthesis. But, high concentration of (more than normal) CO₂ is harmful and considered as a pollutant. Higher concentration of CO₂ is one of the causes of greenhouse effect and global warming as it absorbs the infrared radiations thus increasing the temperature of Earth. This leads to many environmental problems.

(a) BOD is low but DO is high 

(b) the decomposing leaves depleted the levels of oxygen.