100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	What is the basic difference between the electronic configuration of transition and inner transition elements?
Answer» General electronic configuration of transition elements `=(n-1)d^(1-10)ns^(1-2)` and for inner transition elements `=(n-2)f^(1-14)(n-1)d^(0-1)ns^(0-2)` Thus in transition elements, last electron enters d-orbitals of penultimate shell while in inner transition elements, it enters f-orbital of antepenultimate shell.

Discussion

2.	An exothermic chemical reaction proceeds by two stages. Reactants `overset("stage 1")rarr` Intermediate `overset("state 2")rarr` products. The activation energy of stage 1 is `50k Jmol^(-1)`. The overall enthalpy change of the reaction is `-100kJ mol^(-1)`. Which diagram could represent the energy level diagram for reaction ?A. B. C. D.
Answer» Correct Answer - 3 Two step process with negative `DeltaH` , hence option `(3)` is obvious.

Discussion

3.	In the transition series with an increase in atomic number, the atomic radius does not change very much. Why is it so?
Answer» As we move from left to right along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the d-subshell increases the screening effect which conterbalances the effect of increased nuclear charge.

Discussion

4.	Why do transition metal (elements) show variable oxidation states ?
Answer» Ans. Due to presence of vacant d-orbitals.

Discussion

5.	What is the maximum oxidation state shown by actinides?
Answer» Correct Answer - `+7`

Discussion

6.	Number of isomers represented by molecular formula C4H10O is: (1) 3 (2) 4 (3) 7 (4) 10
Answer» Number of isomers represented by molecular formula C₄H₁₀O is 3

Discussion

7.	Sum of antibonding `pi` electrons (`pi` electrons ) in species `O_2, O_2^(-) and O_2^(2-)` are.
Answer» Correct Answer - 9 `({:("Species"," " "Number of"pi^**"electrons"),(O_2," "2),(O_2^(-)," "3),(O_2^(2-)," "4):})/("Total"," "9"electrons") `

Discussion

8.	In case of concave mirror, if the object is placed at C then image is formedA. At FB. Beyond CC. Between C and FD. At C
Answer» Correct Answer - D

Discussion

9.	On what factors does the resistance of a conductor depends? Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer» The resistance of the conductor depends on the following factors: I. The temperature of the conductor, II. The cross- sectional area of the conductor, III. Length of the conductor IV. Nature of the material of the conductor The current will flow more easily through a thick wire than through a thin wire of the same material when connected to the same source. This is due to the fact that the resistance of a wire is inversely propositional to the square of its diameter. A thick wire has a greater diameter and hence lesser resistance making the current to flow through it more easily.

9.

On what factors does the resistance of a conductor depends? Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer»

The resistance of the conductor depends on the following factors:

I. The temperature of the conductor,

II. The cross- sectional area of the conductor,

III. Length of the conductor

IV. Nature of the material of the conductor

The current will flow more easily through a thick wire than through a thin wire of the same material when connected to the same source. This is due to the fact that the resistance of a wire is inversely propositional to the square of its diameter. A thick wire has a greater diameter and hence lesser resistance making the current to flow through it more easily.

Discussion

10.	The given graph shows the variation of charge q versus potential difference of two capacitors C1 and C2. The two capacitors have same plate separation, but the plate area of C2 is double that of C1.Which of the lines in the graph correspond to C1 and C2 and why?
Answer» As, C = \(\frac{∈_0A}{d}\) and plate area of C₂ is double than that of C₁, Therefore, C₂ >C₁ Now, C = \(\frac{q}{V}\), which is greater for line A. ∴ The line A of the graph corresponds to C₂ and line B corresponds to C₁.

10.

The given graph shows the variation of charge q versus potential difference of two capacitors C1 and C2. The two capacitors have same plate separation, but the plate area of C2 is double that of C1.Which of the lines in the graph correspond to C1 and C2 and why?

Answer»

As,

C = \(\frac{∈_0A}{d}\) and

plate area of C₂ is double than that of C₁,

Therefore,

C₂ >C₁

Now,

C = \(\frac{q}{V}\), which is greater for line A.

∴ The line A of the graph corresponds to C₂ and line B corresponds to C₁.

Discussion

11.	Equipotential surfaces are perpendicular to field lines. Why?
Answer» No work is done in moving a charge from one point on equipotential surface to the other. Therefore, component of electric field intensity along the equipotential surface is zero. It means, the electric field intensity is perpendicular to equipotential surface.

Discussion

12.	The excess pressure inside a soap bubble is twice the excess pressurre inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n isA. 2B. 1C. 0.25D. 4
Answer» Correct Answer - C `Pex_(1) = 2Pex_(2) or (Pex_(1))/(Pex_(2)) = (2)/(1)` `(V_(1))/(V_(2)) = ((r_(1))/(r_(2))^(3) = ((Pex_(2))/(Pex_(1))^(3) = ((1)/(2))^(3) = (1)/(8)` ` n = (1)/(8) = 0.125` (`because` given `v_(1) = nV_(2)`)

Discussion

13.	If 200 MeV energy is released per fission of `._(92)U^(235)` How many fissions must occur per second to produce a power of 1m W?A. `3.5xx10^(10)`B. `4.2xx10^(11)`C. `3.125xx10^(12)`D. `3.125xx10^(13)`
Answer» Correct Answer - D `n(200"Mev")=1"kW"` `impliesn=(1000)/(200xx10^(6)xx1.6xx10^(-19))=3.125xx10^(13)`

Discussion

14.	When a `2 mu C` charge is carried from point A to point B, the amount of work done by the electric field is `50 mu J`. What is the potential difference and which point is at a higher potential ?A. 25V BB. 25V ,AC. 20 V BD. Both are at same potential
Answer» Correct Answer - B `Delta W = qDetla V` `Delta V = (Delta W)/(q)= (50muJ)/( 2 muC) = 25V`

Discussion

15.	Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field B directed downwards perpendicular to the plane of the loop. The arm RS is moved with a uniform speed v. Deduce an expression for :(i) The emf induced across the arm RS. (ii) The external force required to move the arm, and (iii) The power dissipated as heat.
Answer» (i). The mobile free electrons in arm RS are driven from S to R, according to Fleming’s left hand rule. Each charge ‘q’ within the arm RS moves with velocity \(\vec{V}\) and experiences a magnetic force, F = \|q\| vB. Work done in moving the charge +q from R to S, W = F × RS = (q v B)× l ∴ emf induced = work done per unit charge i.e., e = \(\frac{W}{q}\) = \(\frac{qvBl}{q}\) = Blv. (ii). If R is the resistance of the loop PQRS at a given instant, the induced current is : I = \(\frac{e}{R}\) = \(\frac{Blv}{R}\) ∴The magnitude of force on the conductor RS moving in the magnetic field is : F = BIl = B(\(\frac{Blv}{R}\)).l = \(\frac{B^2l^2v}{R}\) The direction of this force is opposite to the velocity of the conductor. ∴ \|F_ext\|= \|F\| (According to Newton’s 3rd law) F_ext= \(\frac{B^2l^2v}{R}\) (iii) Power required to push the conductor with velocity 'v': P =F_ext× v = \(\frac{B^2l^2v}{R}\) \(\times\) v = \(\frac{B^2l^2v^2}{R}\) As the conductor is pushed mechanically, Power dissipated as heat, P = I²R = \((\frac{Blv}{R})^2.R\) = \(\frac{B^2l^2v^2}{R}\)

15.

Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field B directed downwards perpendicular to the plane of the loop. The arm RS is moved with a uniform speed v. Deduce an expression for :(i) The emf induced across the arm RS. (ii) The external force required to move the arm, and (iii) The power dissipated as heat.

Answer»

(i). The mobile free electrons in arm RS are driven from S to R, according to Fleming’s left hand rule.

Each charge ‘q’ within the arm RS moves with velocity \(\vec{V}\) and experiences a magnetic force,

F = |q| vB.

Work done in moving the charge +q from R to S,

W = F × RS

= (q v B)× l

∴ emf induced = work done per unit charge

i.e.,

e = \(\frac{W}{q}\)

= \(\frac{qvBl}{q}\)

= Blv.

(ii). If R is the resistance of the loop PQRS at a given instant, the induced current is :

I = \(\frac{e}{R}\)

= \(\frac{Blv}{R}\)

∴The magnitude of force on the conductor RS moving in the magnetic field is :

F = BIl

= B(\(\frac{Blv}{R}\)).l

= \(\frac{B^2l^2v}{R}\)

The direction of this force is opposite to the velocity of the conductor.

∴ |F_ext|= |F|

(According to Newton’s 3rd law)

F_ext= \(\frac{B^2l^2v}{R}\)

(iii) Power required to push the conductor with velocity 'v':

P =F_ext× v

= \(\frac{B^2l^2v}{R}\) \(\times\) v

= \(\frac{B^2l^2v^2}{R}\)

As the conductor is pushed mechanically,

Power dissipated as heat,

P = I²R

= \((\frac{Blv}{R})^2.R\)

= \(\frac{B^2l^2v^2}{R}\)

Discussion

16.	Mention two applications of polaroids.
Answer» 1. It is used in sunglasses, goggles etc. 2. It is used in windows, head lamps of buses, cars, aeroplane etc.

16.

Mention two applications of polaroids.

Answer»

1. It is used in sunglasses, goggles etc.

2. It is used in windows, head lamps of buses, cars, aeroplane etc.

Discussion

17.	A charge q in placed at the centre of a cube of side ‘l’. What is the electric flux passing through two opposite faces of the cube?
Answer» Flux through each face = \(\frac{q}{6∈_0}\) ∴ Flux through two opposite faces, \(\frac{q}{6∈_0}\) + \(\frac{q}{6∈_0}\) = \(\frac{q}{3∈_0}\)

17.

A charge q in placed at the centre of a cube of side ‘l’. What is the electric flux passing through two opposite faces of the cube?

Answer»

Flux through each face = \(\frac{q}{6∈_0}\)

∴ Flux through two opposite faces,

\(\frac{q}{6∈_0}\) + \(\frac{q}{6∈_0}\) = \(\frac{q}{3∈_0}\)

Discussion

18.	What is the work done in moving a test charge ‘q’ through a distance of 1cm along the equatorial axis of an electric dipole?
Answer» On equatorial axis of a dipole, V = 0 ∴ W = q × v = 0.

18.

What is the work done in moving a test charge ‘q’ through a distance of 1cm along the equatorial axis of an electric dipole?

Answer»

On equatorial axis of a dipole,

V = 0

∴ W = q × v = 0.

Discussion

19.	Define resistivity. State its SI unit. On what factors resistivity depends? Write resistivity range for metals and insulators.
Answer» Resistivity (ρ): It can be defined as the resistance of a conducting material per unit length with unit area of cross- section. Its SI unit is ohm-metre (Ωm). The resistivity of a material depends on its nature and the temperature of the conductor, but not on its shape and size. Range of resistivity for metals = 10^-8 – 10^-6 Ωm For insulators = 10¹² - 10²⁰Ωm

19.

Define resistivity. State its SI unit. On what factors resistivity depends? Write resistivity range for metals and insulators.

Answer»

Resistivity (ρ): It can be defined as the resistance of a conducting material per unit length with unit area of cross- section.

Its SI unit is ohm-metre (Ωm).

The resistivity of a material depends on its nature and the temperature of the conductor, but not on its shape and size.

Range of resistivity for metals = 10^-8 – 10^-6 Ωm

For insulators = 10¹² - 10²⁰Ωm

Discussion

20.	Unit of G isA. `(Nm^2)/(Kg^2)`B. `(Nm)/(Kg^2)`C. `(Nm^2)/(Kg)`D. `(N^2m)/(Kg^2)`
Answer» Correct Answer - A

Discussion

21.	Explain briefly model theory of light.
Answer» Most of the time, light behaves as a wave and it is categorized as one of the electromagnetic waves because it is made of both electric and magnetic fields. Electromagnetic fields perpendicularly oscillate to the direction of wave travel and are perpendicular to each other. As a result of which, they are known as transverse waves. A few characteristics of light are as follows: While dealing with light waves, we deal with the sine waveform. The period of the waveform is one full 0 to 360-degree sweep. Light waves have two important characteristics known as wavelength and frequency. The distance between the peaks of the wave is known as the wavelength. In the case of a light wave, the wavelengths are in the order of nanometers. Frequency is the number of waves that will cross past a point in a second. The relationship between wavelength and frequency is given by the equation: \(f = \frac{1}{T}\) The speed of light in a vacuum is a universal constant which is 3 x 10⁸m/s. As proposed by Einstein, light is made of tiny packets of energy known as photons. The formula devised by Planck determines the energy of a photon and it also shows that the energy is directly proportional to the frequency of the light. E = hf where h is the Planck’s constant 6.63 x 10^-34 Joule-Second.

21.

Explain briefly model theory of light.

Answer»

Most of the time, light behaves as a wave and it is categorized as one of the electromagnetic waves because it is made of both electric and magnetic fields. Electromagnetic fields perpendicularly oscillate to the direction of wave travel and are perpendicular to each other. As a result of which, they are known as transverse waves. A few characteristics of light are as follows:

While dealing with light waves, we deal with the sine waveform. The period of the waveform is one full 0 to 360-degree sweep.
Light waves have two important characteristics known as wavelength and frequency.
The distance between the peaks of the wave is known as the wavelength. In the case of a light wave, the wavelengths are in the order of nanometers.
Frequency is the number of waves that will cross past a point in a second.
The relationship between wavelength and frequency is given by the equation:

\(f = \frac{1}{T}\)

The speed of light in a vacuum is a universal constant which is 3 x 10⁸m/s.
As proposed by Einstein, light is made of tiny packets of energy known as photons. The formula devised by Planck determines the energy of a photon and it also shows that the energy is directly proportional to the frequency of the light.

E = hf

where h is the Planck’s constant 6.63 x 10^-34 Joule-Second.

Discussion

22.	\[\int \frac{d x}{\sqrt{2 a x-x^{2}}}=a^{n} \sin ^{-1}\left[\frac{x}{a}-1\right]\]The value of \( n \) is(a) 0(b) \( -1 \)(c) 1(d) none of these.You may use dimensional analysis to solve the problem.
Answer» \(\int\frac{dx}{\sqrt{2ax-x^2}}\) \(= \int \frac{dx}{\sqrt{a^2 - (x -a)^2}}\) \(= sin^{-1}\left(\frac{x-a}a\right)\) \(= sin^{-1}\left(\frac{x}a -1\right)\) \(\therefore n = 0\)

22.

\[\int \frac{d x}{\sqrt{2 a x-x^{2}}}=a^{n} \sin ^{-1}\left[\frac{x}{a}-1\right]\]The value of \( n \) is(a) 0(b) \( -1 \)(c) 1(d) none of these.You may use dimensional analysis to solve the problem.

Answer»

\(\int\frac{dx}{\sqrt{2ax-x^2}}\)

\(= \int \frac{dx}{\sqrt{a^2 - (x -a)^2}}\)

\(= sin^{-1}\left(\frac{x-a}a\right)\)

\(= sin^{-1}\left(\frac{x}a -1\right)\)

\(\therefore n = 0\)

Discussion

23.	A real gas behaves like an ideal gas if itsA. pressure and temperature are both highB. pressure and temperature are both lawC. pressure is high and temperature is lawD. pressure is high and temperature is high
Answer» Correct Answer - D

Discussion

24.	The atomic number of Nitrogen (N), Oxygen (O) and flourine (F) are 7, 8 & 9 respectively.1. What is the number of valence electrons in N & F?2. Name the element having smallest and largest atomic radii of any of the above three elements. Give reason for your answer.
Answer» (i) Valence electron of N = 5 Valence electron of F = 7 (ii) Smallest radius = F Largest radius = N Reason: Atomic radius decreases from left to right across a period due to increase in the force of attraction between nucleus and valence electrons.

24.

The atomic number of Nitrogen (N), Oxygen (O) and flourine (F) are 7, 8 & 9 respectively.1. What is the number of valence electrons in N & F?2. Name the element having smallest and largest atomic radii of any of the above three elements. Give reason for your answer.

Answer»

(i) Valence electron of N = 5

Valence electron of F = 7

(ii) Smallest radius = F

Largest radius = N

Reason: Atomic radius decreases from left to right across a period due to increase in the force of attraction between nucleus and valence electrons.

Discussion

25.	What is menstruation? What happens when the egg is not fertilised?
Answer» Menstruation is the process of breakdown and removal of the inner lining of the uterus along with the blood vessels in the form of vaginal bleeding. When the egg is not fertilized, its starts dividing and gets implanted in the lining of the uterus.

25.

What is menstruation? What happens when the egg is not fertilised?

Answer»

Menstruation is the process of breakdown and removal of the inner lining of the uterus along with the blood vessels in the form of vaginal bleeding.

When the egg is not fertilized, its starts dividing and gets implanted in the lining of the uterus.

Discussion

26.

Differentiate between metal and non-metal on the basis of their physical properties.

Answer»

Metals	Non-Metals
a. They are electropositive elements.	i. They are electronegative elements.
b. They show the property of malleability and ductility	ii. They do not show the property of malleability and ductility.
c. Metals are usually hard in nature.	iii. They are soft in nature.
d. They are good conductors of heat & electricity	iv. They are bad conductors of heat and electricity
e. They are lustrous	v. They do not have lustre.

Discussion

27.	Write the balanced chemical equations for Reaction when glucose is oxidised/respiration.
Answer» (i) C₆H₁₂O_6(aq)+ 6CO_2(aq) → 6CO_2(aq) + 6H₂O_(l) + energy (ii) 2H_2(g) + O_2(g)→ 2H₂O_(l)

27.

Write the balanced chemical equations for Reaction when glucose is oxidised/respiration.

Answer»

(i) C₆H₁₂O_6(aq)+ 6CO_2(aq) → 6CO_2(aq) + 6H₂O_(l) + energy

(ii) 2H_2(g) + O_2(g)→ 2H₂O_(l)

Discussion

28.	A shiny brown coloured element ‘x’ on heating in air becomes black in colour. Name the element 'x' and the black coloured compound formed. Write the equation.
Answer» Element ‘x’ is copper and the black coloured compound is copper (11) oxide Equation: 2Cu_(s) + O_2(g)→ 2CuO_(s)

28.

A shiny brown coloured element ‘x’ on heating in air becomes black in colour. Name the element 'x' and the black coloured compound formed. Write the equation.

Answer»

Element ‘x’ is copper and the black coloured compound is copper (11) oxide

Equation: 2Cu_(s) + O_2(g)→ 2CuO_(s)

Discussion

29.	A body of mass m has an initial speed v is acted by two forces `vec(F)_(1)` and `vec(F)_(2)`. After something work done by `vec(F)_(1)` is `-1/2 mv^(2)` and speed of the body is 2v. Then, the work done by `vec(F)_(2)` isA. `3/2 mv^(2)`B. `mv^(2)`C. zeroD. `2mv^(2)`
Answer» Correct Answer - D `W_(1)+W_(2)=1/2 m(2v)^(2)-(v)^(2)` `-1/2 mv_(2)+W_(2)=3/2 mv^(2)` `W_(2)=2mv^(2)`

Discussion

30.	An empty box of mass 5kg is found to accelerate up at the rate of `g/6` when a constant force `vec(F)` is applied on it. Howm much mass (in kg ) of sand be filled in it so that box may accelerate down at the same rate ? (g beight the acceleration due to gravity).
Answer» Correct Answer - 5 `I=(ml^(2))/3xx3=ml^(2)=10xx(1.25)^(2) =250/16` `omega=omega_(0)+alphar` `(1900-1400)xx(2pi)/60=(25pi)/250 xx16 t` `t=240/48 = 5 sec.`

Discussion

31.	Two objects of masses 6 and 8kg are hung from the ends of a stick that is 70 cm long and has marked every 10 cm as shown below, if the mass of the stick is negligible, at which of the points should a cord be attached if the stick is to remain horizontal when suspended from the cord?
Answer» To balance the torques on each side, we obviously need to be closer to the heavier mass. Trying point D as a pivot point we have: (m₁g) • r₁ = (m₂g) • r₂ (6kg) (40cm) = (8kg) (30 cm) and we see it works.

31.

Two objects of masses 6 and 8kg are hung from the ends of a stick that is 70 cm long and has marked every 10 cm as shown below, if the mass of the stick is negligible, at which of the points should a cord be attached if the stick is to remain horizontal when suspended from the cord?

Answer»

To balance the torques on each side, we obviously need to be closer to the heavier mass.

Trying point D as a pivot point we have:

(m₁g) • r₁ = (m₂g) • r₂

(6kg) (40cm) = (8kg) (30 cm) and we see it works.

Discussion

32.	A Shopkeeper marks up his goods by 20% and gives a discount of 5%. Also, he uses a false balance, which reads 1000 gins for 750 gins. What is his total profit percent?
Answer» Let the CP per gm be 0.10 Accordingly CP of 1kg i.e. 1000 gms = Rs.100. Selling price of 750 gms = [100 × 120% – 5% of 120] = 120 – 6 = 114. Cost Price of 750 gms = 75. Profit = 114 – 75 = 39 Profit percent = 39/75x100=52%

Discussion

33.	What will come in place of the question mark in the following series?9, 22, 48, 100, 204, ?1. 4102. 4113. 4124. 413
Answer» Correct Answer - Option 3 : 412 EXPLANATION: Series pattern given series 9 9 9 × 2 + 4 = 22 22 22 × 2 + 4 = 48 48 48 × 2 + 4 = 100 100 100 × 2 + 4 = 204 204 204 × 2 + 4 = 412 ? ∴ the answer will be 412

33.

What will come in place of the question mark in the following series?9, 22, 48, 100, 204, ?1. 4102. 4113. 4124. 413

Answer» Correct Answer - Option 3 : 412

EXPLANATION:

Series pattern given series

9 9

9 × 2 + 4 = 22 22

22 × 2 + 4 = 48 48

48 × 2 + 4 = 100 100

100 × 2 + 4 = 204 204

204 × 2 + 4 = 412 ?

∴ the answer will be 412

Discussion

34.	What will come in place of the question mark in the following series?11, ?, 13, 22, 15, 241. 212. 203. 224. 19
Answer» Correct Answer - Option 2 : 20 EXPLANATION: Series pattern given series 11 11 11 + 9 = 20 ? 20 – 7 = 13 13 13 + 9 = 22 22 22 – 7 = 15 15 15 + 9 = 24 24 ∴ the answer will be 20

34.

What will come in place of the question mark in the following series?11, ?, 13, 22, 15, 241. 212. 203. 224. 19

Answer» Correct Answer - Option 2 : 20

EXPLANATION:

Series pattern given series

11 11

11 + 9 = 20 ?

20 – 7 = 13 13

13 + 9 = 22 22

22 – 7 = 15 15

15 + 9 = 24 24

∴ the answer will be 20

Discussion

35.	If x - y = 8, then which of the following must be true? 1. Both x and y must be positive for any value of x and y.2. If x is positive, y must be negative for any value of x and y.3. If x is negative, y must be positive for any value of x and y. Select the correct answer using the code given below. 1. 1 only2. 2 only3. Both 1 and 24. Neither 1 nor 2 nor 3
Answer» Correct Answer - Option 4 : Neither 1 nor 2 nor 3 Given : x - y = 8 Calculations : For this question, we must check each option For option (1) ⇒ x - y = 8 at x = 4 and b = -4, so option 1 does not satisfy. For option (2) ⇒ x - y = 8 at x = 12 and y = 4, so option 2 also does not satisfy. For option (3) ⇒ x - y = 8 at x = -1 and y = -9, so option 3 also does not satisfy. So no option follows. ∴ Option 4 will be the right choice.

35.

If x - y = 8, then which of the following must be true? 1. Both x and y must be positive for any value of x and y.2. If x is positive, y must be negative for any value of x and y.3. If x is negative, y must be positive for any value of x and y. Select the correct answer using the code given below. 1. 1 only2. 2 only3. Both 1 and 24. Neither 1 nor 2 nor 3

Answer» Correct Answer - Option 4 : Neither 1 nor 2 nor 3

Given :

x - y = 8

Calculations :

For this question, we must check each option

For option (1) ⇒ x - y = 8 at x = 4 and b = -4, so option 1 does not satisfy.

For option (2) ⇒ x - y = 8 at x = 12 and y = 4, so option 2 also does not satisfy.

For option (3) ⇒ x - y = 8 at x = -1 and y = -9, so option 3 also does not satisfy.

So no option follows.

∴ Option 4 will be the right choice.

Discussion

36.	In a class of 50 students, 30 take Mathematics, 25 take Biology and 15 take both Mathematics and Biology. How many students take neither Mathematics nor Biology?1. 52. 103. 154. 20
Answer» Correct Answer - Option 2 : 10 Given: The total number of students in a class = 50 The number of students takes Mathematics = 30 The number of students takes Biology = 25 The number of students take both Mathematics and Biology = 15 Formula used: n(A ∪ B) = n(A) + n(B) - n(A ∩ B) n(A ∪ B)' = n(U) - n(A ∪ B) where, n(A) = number of elements in A, n(B) = number of elements in B n(A ∩ B) = number of elements in both A and B, n(A ∪ B) = number of elements in either A or B n(U) = number of total students Calculation: Let the set of total students be U, the set of students takes Mathematics be A And the set of students takes Biology be C. According to the question, n(U) = 50, n(A) = 30, n(B) = 25 and n(A ∩ B) = 15 Now, students take either Mathematics or Biology = n(A ∪ B) ⇒ n(A) + n(B) - n(A ∩ B) ⇒ 30 + 25 - 15 ⇒ 40 So, students take neither Mathematics nor Biology = n(A ∪ B)' ⇒ n(U) - n(A ∪ B) ⇒ 50 - 40 ⇒ 10 ∴ The students take neither Mathematics nor Biology is 10.

36.

In a class of 50 students, 30 take Mathematics, 25 take Biology and 15 take both Mathematics and Biology. How many students take neither Mathematics nor Biology?1. 52. 103. 154. 20

Answer» Correct Answer - Option 2 : 10

Given:

The total number of students in a class = 50

The number of students takes Mathematics = 30

The number of students takes Biology = 25

The number of students take both Mathematics and Biology = 15

Formula used:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

n(A ∪ B)' = n(U) - n(A ∪ B)

where, n(A) = number of elements in A, n(B) = number of elements in B

n(A ∩ B) = number of elements in both A and B, n(A ∪ B) = number of elements in either A or B

n(U) = number of total students

Calculation:

Let the set of total students be U, the set of students takes Mathematics be A

And the set of students takes Biology be C.

According to the question,

n(U) = 50, n(A) = 30, n(B) = 25 and n(A ∩ B) = 15

Now, students take either Mathematics or Biology = n(A ∪ B)

⇒ n(A) + n(B) - n(A ∩ B)

⇒ 30 + 25 - 15

⇒ 40

So, students take neither Mathematics nor Biology = n(A ∪ B)'

⇒ n(U) - n(A ∪ B)

⇒ 50 - 40

⇒ 10

∴ The students take neither Mathematics nor Biology is 10.

Discussion

37.	A lift has the capacity of 18 adults or 30 children. How many children can board the lift with 12 adults?1. 62. 103. 124. 15
Answer» Correct Answer - Option 2 : 10 Given: The capacity of the lift = 30 children or 18 adults Calculation: To find the capacity of the lift, find LCM of 18 and 30 = 90x 1 adult = 90x/18 = 5x 1 child = 90x /30 = 3x If, the lift is occupied by 12 adults, 12 × 5x = 60x Remaining capacity will be = 90x - 60x = 30x So, the number of children that be accommodated with 12 adults are: 30x/3x = 10 ∴ 10 children can board the lift with 12 adults.

37.

A lift has the capacity of 18 adults or 30 children. How many children can board the lift with 12 adults?1. 62. 103. 124. 15

Answer» Correct Answer - Option 2 : 10

Given:

The capacity of the lift = 30 children or 18 adults

Calculation:

To find the capacity of the lift, find LCM of 18 and 30 = 90x

1 adult = 90x/18 = 5x

1 child = 90x /30 = 3x

If, the lift is occupied by 12 adults,

12 × 5x = 60x

Remaining capacity will be = 90x - 60x

= 30x

So, the number of children that be accommodated with 12 adults are:

30x/3x = 10

∴ 10 children can board the lift with 12 adults.

Discussion

38.	The binary equivalent of decimal number 4.625 is1. 100-0012. 100-1103. 100-1014. 100-011
Answer» Correct Answer - Option 3 : 100-101 The correct answer is 100-101. Binary is the simplest kind of number system that uses only two digits of 0 and 1 (i.e. the value of base 2). To convert decimal number 4.625, we convert its integer and fraction parts individually and then add them to get the equivalent binary number, as below: Divide 4 by 2 keeping notice of the quotient and the remainder. Continue dividing the quotient by 2 until you get a quotient of zero. Then just write out the remainders in the reverse order to get the equivalent binary number. 4 / 2 = 2 with remainder 0 2 / 2 = 1 with remainder 0 1 / 2 = 0 with remainder 1 Here is the answer to 4 decimal to binary numbers: 100 For converting decimal fraction 0.625 to a binary number, follow these steps: Multiply 0.625 by 2 keeping notice of the resulting integer and fractional part. Continue multiplying by 2 until you get a resulting fractional part equal to zero (we calculate up to ten digits). Then just write out the integer parts from the results of each multiplication to get the equivalent binary number. 0.625 × 2 = 1 + 0.25 0.25 × 2 = 0 + 0.5 0.5 × 2 = 1 + 0 Here is the answer to 0.625 decimal to binary number: 0.101 Therefore, decimal number 4.625 converted to binary is equal: 100.101

38.

The binary equivalent of decimal number 4.625 is1. 100-0012. 100-1103. 100-1014. 100-011

Answer» Correct Answer - Option 3 : 100-101

The correct answer is 100-101.

Binary is the simplest kind of number system that uses only two digits of 0 and 1 (i.e. the value of base 2).
To convert decimal number 4.625, we convert its integer and fraction parts individually and then add them to get the equivalent binary number, as below:
- Divide 4 by 2 keeping notice of the quotient and the remainder. Continue dividing the quotient by 2 until you get a quotient of zero.
- Then just write out the remainders in the reverse order to get the equivalent binary number.
- 4 / 2 = 2 with remainder 0
- 2 / 2 = 1 with remainder 0
- 1 / 2 = 0 with remainder 1
- Here is the answer to 4 decimal to binary numbers: 100
- For converting decimal fraction 0.625 to a binary number, follow these steps:
  - Multiply 0.625 by 2 keeping notice of the resulting integer and fractional part. Continue multiplying by 2 until you get a resulting fractional part equal to zero (we calculate up to ten digits).
  - Then just write out the integer parts from the results of each multiplication to get the equivalent binary number.
  - 0.625 × 2 = 1 + 0.25
  - 0.25 × 2 = 0 + 0.5
  - 0.5 × 2 = 1 + 0
  - Here is the answer to 0.625 decimal to binary number: 0.101
Therefore, decimal number 4.625 converted to binary is equal: 100.101

Discussion

39.	If tanα = P, then cos2α is equal to(A) \(\frac{1-P^2}{1+P^2}\)(B) \(\frac{2P}{1-P^2}\)(C) \(\frac{2P}{1+P^2}\)(D) \(\frac{1+P^2}{2P}\)
Answer» Correct option is: (A) \(\frac{1-P^2}{1+P^2}\)

Discussion

40.	In ∆ABC if ∠A + ∠C =120° then ∠B =(A) 50° (B) 60° (C) 70° (D) 90°
Answer» Correct answer is (B) 60°

Discussion

41.	What is 234 times 167 ?(a) 42768 (b) 41184(c) 40581 (d) 39078(e) None of these
Answer» (d) 234 × 167 = 39078

Discussion

42.	The y-coordinates of a point which divides the line segment joining A(4, -3) and B(9, 7) in the ratio 3:2 internally is –(A) \(\frac{3X9 + 2X4}{(3 + 2)}\)(B) \(\frac{[3X7 + 2X(-3)]}{(3 + 2)}\)(C) \(\frac{[3X9 + 2X(-3)]}{(3 + 2)}\)(D) \(\frac{[3X7 - 2X(-3)]}{(3 + 2)}\)
Answer» Correct answer is (B) \(\frac{[3X7 + 2X(-3)]}{(3 + 2)}\)

Discussion

43.	Rohan and Amit participate in 300 m race. Amit won the race and takes 1.5 seconds less than Rohan to finish the race and the speed of Rohan is 10 m/sec less than Amit’s speed. Find the speed of Amit.1. 25 m/s2. 50 m/s 3. 30 m/s4. 60 m/s 5. 90 m/s
Answer» Correct Answer - Option 2 : 50 m/s Given : Distance = 300 m Time taken by Rohan is 1.5s more than Amit. The speed of Rohan is 10 m/s less than the speed of Amit. Formula used : Speed = Distance/Time Calculations: Let the speed of Amit is x m/s Then the speed of Rohan is (x - 10)m/s Time taken by Amit to complete the race = 300/x sec Time taken by Rohan to complete the race = 300/(x -10) sec According to question, 300/(x - 10) - 300/(x) = 1.5 ----(1) ⇒ x²- 10x - 2000 = 0 ⇒ x = 50 and -40 ⇒ Eliminating x = - 40 , we get x = 50 ⇒ Speed of Amit = 50 m/s ∴ Speed of Amit is 50 m/s

43.

Rohan and Amit participate in 300 m race. Amit won the race and takes 1.5 seconds less than Rohan to finish the race and the speed of Rohan is 10 m/sec less than Amit’s speed. Find the speed of Amit.1. 25 m/s2. 50 m/s 3. 30 m/s4. 60 m/s 5. 90 m/s

Answer» Correct Answer - Option 2 : 50 m/s

Given :

Distance = 300 m

Time taken by Rohan is 1.5s more than Amit.

The speed of Rohan is 10 m/s less than the speed of Amit.

Formula used :

Speed = Distance/Time

Calculations:

Let the speed of Amit is x m/s

Then the speed of Rohan is (x - 10)m/s

Time taken by Amit to complete the race = 300/x sec

Time taken by Rohan to complete the race = 300/(x -10) sec

According to question,

300/(x - 10) - 300/(x) = 1.5 ----(1)

⇒ x²- 10x - 2000 = 0

⇒ x = 50 and -40

⇒ Eliminating x = - 40 , we get x = 50

⇒ Speed of Amit = 50 m/s

∴ Speed of Amit is 50 m/s

Discussion

44.	P, Q and R started a business. P invested some amount at the beginning. Q invested thrice the amount invested by P and R invested five times the amount invested by P, at the end of a year, there was a loss in the business. Find the ratio of loss to be distributed among them.1. 5 : 3 : 12. 1 : 3 : 53. 5 : 2 : 14. 5 : 3 : 25. 1 : 5 : 3
Answer» Correct Answer - Option 2 : 1 : 3 : 5 Given: Investment of Q = 3 × Investment of P Investment of R = 5 × Investment of P Concept used: Loss = Investment × time Calculation: Q = 3 × P ⇒ Q : P = 3 : 1 R = 5 × P ⇒ R : P = 5 : 1 P : Q : R = 1 : 3 : 5 Investment ratio = Loss ratio ∴ The loss ratio of P, Q and R are 1 : 3 : 5.

44.

P, Q and R started a business. P invested some amount at the beginning. Q invested thrice the amount invested by P and R invested five times the amount invested by P, at the end of a year, there was a loss in the business. Find the ratio of loss to be distributed among them.1. 5 : 3 : 12. 1 : 3 : 53. 5 : 2 : 14. 5 : 3 : 25. 1 : 5 : 3

Answer» Correct Answer - Option 2 : 1 : 3 : 5

Given:

Investment of Q = 3 × Investment of P

Investment of R = 5 × Investment of P

Concept used:

Loss = Investment × time

Calculation:

Q = 3 × P

⇒ Q : P = 3 : 1

R = 5 × P

⇒ R : P = 5 : 1

P : Q : R = 1 : 3 : 5

Investment ratio = Loss ratio

∴ The loss ratio of P, Q and R are 1 : 3 : 5.

Discussion

45.	Vijay and Ajay started a business initially with ₹ 7600 and ₹ 6800 respectively. If the total loss at the end of a year is ₹ 18000, then what is Ajay’s share in the total loss?1. ₹ 85002. ₹ 84003. ₹ 86004. ₹ 88005. ₹ 8200
Answer» Correct Answer - Option 1 : ₹ 8500 Given: Investment of Vijay = ₹ 7600 Investment of Ajay = ₹ 6800 Time of investment = 1 year Total loss = ₹ 18000 Concept used: Loss = Investment × time Calculation: Loss ratio = (7600 × 1) : (6800 × 1) ⇒ Loss ratio = 19 : 17 Total loss = (19 + 17) = 18000 ⇒ 36 = 18000 Ajay’s loss = (18000/36) × 17 = 8500 ∴ Ajay’s share in total loss is ₹ 8500.

45.

Vijay and Ajay started a business initially with ₹ 7600 and ₹ 6800 respectively. If the total loss at the end of a year is ₹ 18000, then what is Ajay’s share in the total loss?1. ₹ 85002. ₹ 84003. ₹ 86004. ₹ 88005. ₹ 8200

Answer» Correct Answer - Option 1 : ₹ 8500

Given:

Investment of Vijay = ₹ 7600

Investment of Ajay = ₹ 6800

Time of investment = 1 year

Total loss = ₹ 18000

Concept used:

Loss = Investment × time

Calculation:

Loss ratio = (7600 × 1) : (6800 × 1)

⇒ Loss ratio = 19 : 17

Total loss = (19 + 17) = 18000

⇒ 36 = 18000

Ajay’s loss = (18000/36) × 17 = 8500

∴ Ajay’s share in total loss is ₹ 8500.

Discussion

46.	A started a business by investing ₹ 25000 and after 7 months, B joined the business. After a year, the loss between A and B is in a ratio of 5 : 6. What is the amount invested by B?1. ₹ 720002. ₹ 660003. ₹ 620004. ₹ 640005. ₹ 68000
Answer» Correct Answer - Option 1 : ₹ 72000 Given: Investment of A = ₹ 25000 The ratio of loss = 5 : 6 Concept used: Loss = Investment × Time period Calculation: Let investment of B be x Time period of investment of B = 12 – 7 = 5 months (Investment of A)/(Investment of B) = 5/6 ⇒ (25000 × 12)/(x × 5) = 5/6 ⇒ 300000/5x =5/6 ⇒ (300000 × 6)/(5 × 5) = x ⇒ x = 72000 ∴ The investment of B is 72000.

46.

A started a business by investing ₹ 25000 and after 7 months, B joined the business. After a year, the loss between A and B is in a ratio of 5 : 6. What is the amount invested by B?1. ₹ 720002. ₹ 660003. ₹ 620004. ₹ 640005. ₹ 68000

Answer» Correct Answer - Option 1 : ₹ 72000

Given:

Investment of A = ₹ 25000

The ratio of loss = 5 : 6

Concept used:

Loss = Investment × Time period

Calculation:

Let investment of B be x

Time period of investment of B = 12 – 7 = 5 months

(Investment of A)/(Investment of B) = 5/6

⇒ (25000 × 12)/(x × 5) = 5/6

⇒ 300000/5x =5/6

⇒ (300000 × 6)/(5 × 5) = x

⇒ x = 72000

∴ The investment of B is 72000.

Discussion

47.	Hello Guys ,can u help me I don't know how prove this problemeShow that:::::1+2lnx
Answer» \(\because\) x² & exponential functions both are increasing functions Also ex² \(\leq\) ex² \(\forall\) x \(\in\) (0, \(\infty\)) ⇒ ln(ex²) \(\leq\) ln ex² (\(\because \) ln x is strictly increasing function) ⇒ ln e + ln x² \(\leq\) x² ln e (\(\because\) ln (AB) = ln A + ln B) ⇒ 1 + 2 ln x \(\leq\) x² (\(\because \) ln e = 1)

47.

Hello Guys ,can u help me I don't know how prove this problemeShow that:::::1+2lnx

Answer»

\(\because\) x² & exponential functions both are increasing functions

Also

ex² \(\leq\) ex² \(\forall\) x \(\in\) (0, \(\infty\))

⇒ ln(ex²) \(\leq\) ln ex² (\(\because \) ln x is strictly increasing function)

⇒ ln e + ln x² \(\leq\) x² ln e (\(\because\) ln (AB) = ln A + ln B)

⇒ 1 + 2 ln x \(\leq\) x² (\(\because \) ln e = 1)

Discussion

48.	Find the square root of 108900.1. 4702. 4303. 3704. 330
Answer» Correct Answer - Option 4 : 330 Given: Given number = 108900 Calculation Factors of 108900 = 2 × 2 × 3 × 3 × 5 × 5 × 11 × 11 = (2 × 3 × 5 × 11)² = (330)² Hence, √108900 = √(330)² = 330 ∴ Square root of 108900 is 330.

48.

Find the square root of 108900.1. 4702. 4303. 3704. 330

Answer» Correct Answer - Option 4 : 330

Given:

Given number = 108900

Calculation

Factors of 108900 = 2 × 2 × 3 × 3 × 5 × 5 × 11 × 11

= (2 × 3 × 5 × 11)²

= (330)²

Hence,

√108900 = √(330)²

= 330

∴ Square root of 108900 is 330.

Discussion

49.	Find the HCF of 1.43, 1.87 and 20.9.A. 11B. 1C.0.11D. 1101. A2. B3. C4. D
Answer» Correct Answer - Option 3 : C Calculation HCF of fraction = HCF of numerator/LCM of denomenator 1.43 = 143/100,, 1.87 = 187/100 and 20.9 = 209/10 HCF of 143, 187, 209 = 11 LCM of 100,, 100, and 10 = 100 HCF of 1.43, 1.87 and 20.9 = 11/100 or 0.11

49.

Find the HCF of 1.43, 1.87 and 20.9.A. 11B. 1C.0.11D. 1101. A2. B3. C4. D

Answer» Correct Answer - Option 3 : C

Calculation

HCF of fraction = HCF of numerator/LCM of denomenator

1.43 = 143/100,, 1.87 = 187/100 and 20.9 = 209/10

HCF of 143, 187, 209 = 11

LCM of 100,, 100, and 10 = 100

HCF of 1.43, 1.87 and 20.9 = 11/100 or 0.11

Discussion

50.	If `(a,b)` and `(c,d)` are two points on the whose equation is `y=mx+k`, then the distance between `(a,b)` and `(c,d)` in terms of `a,c` and `m` isA. `\|a+c\|sqrt(1+m^(2))`B. `\|a-c\|sqrt(1+m^(2))`C. `(\|a-c\|)/(sqrt(1+m^(2))`D. `\|a-c\|(a+m^(2))`
Answer» Correct Answer - B Let `m=tantheta` then equation of line through `(a,b)` is `(x-a)/(costheta)=(y-b)/(sintheta)=r` if `(c,d)` lies on it at `r` distance then `impliesc-a=rcosthetaimpliesr-(c-a)sectheta` `impliesr=\|c-a\|sqrt(1+m^(2))` Alter: Since `(a,b)` and `(c,d)` are on the same line `y=mx+k`, they satisfy the same equation. Therefore `b=ma+kd=mc+k`. Now the distance between `(a,b)` and `(c,d)` is `sqrt((a-c)^(2)+(b-d)^(2))` From the first to equations we obtain `(b-d)=m(a-c)`, so that `sqrt((a-c)^(2)+(b-d)^(2))=sqrt((a-c)^(2)+m^(2)(a-c)^(2))` `=\|a-c\|sqrt(1+m^(2))` Note we are using the fact that `=sqrt(x^(2))=\|x\|` for all real `x`.

Discussion

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