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21701.

A `20 cm` long string, having a mass of `1.0g`, is fixed at both the ends. The tension in the string is `0.5 N`. The string is into vibrations using an external vibrator of frequency `100 Hz`. Find the separation (in cm) between the successive nodes on the string.

Answer» Correct Answer - 5
`v = sqrt((T)/(mu)) = 10 m//s`
`lamda = (v)/(f) = (10)/(100) = 10 cm`
Distance between the successive nodes `= lamda//2 = 5 cm`.
21702.

Two highways are perpendicular to each other imagine them to be along the x-axis and the y-axis, respectively. At the instant `t = 0`, a police car P is at a distance `d = 400 m` from the intersection and moving at speed of `8 km//h` towards it at a speed of `60 km//h` along they y-axis. The minimum distance between the cars isA. `300 m`B. `240 m`C. `180 m`D. `120 m`

Answer» Correct Answer - B
If s be their separation at time t then,
`s^(2) = (0.4-80t)^(2) + (0.6-60t)^(2)`
where t is in hour, s in km. Minimise s using calculus.
21703.

thickness of the paper measured by micrometer screw gauge of least count 0.01 mmis 1.03 mm, the percentage error in the measurement of thickness of paper is?

Answer» Percentage error = (delta d /d x100)%

                            =(0.01/1.03 x 100)%

                             = 0.97%

Thickness of the paper measured is by Screw Gauge.

21704.

A block of mass `m` connected to a spring of stiffness `k` is placed on horizontal smooth surface. When block is at `x=0`, spring is in natural length. The block is also subjected to a force `F=(k|x|)/(@)` thenA. Motion of block is SHMB. Time period of motion is `pisqrt((m)/(k))[sqrt(2)+sqrt((2)/(3))]`C. Time period of motion is `pisqrt((m)/(k))[1+sqrt((3)/(2))]`D. Sum of potential energy and kinetic energy of spring block system is maximum at right extreme.

Answer» Correct Answer - B::D
`k_(eq)=(k)/(2)` for displacement to right `impliesT_(1)=pisqrt((2m)/(k))`
`k_(eq)=(3k)/(2)`, for displacement `impliesT_(2)=`
`pisqrt((2m)/(3k))`
`implies` motion is not `SHM`
Time period `T=T_(1)+T_(2)`
Work done by F is maximum when block reaches the right extreme
21705.

Find maximum and minimum values of y: (i) `y=2 sinx` , (ii) `y=4-cosx` , (iii) `y=3sinx+4 conx`

Answer» (i) `y_("max")=2(1)=2` and `y_("min")=2(-1)=-2` , (ii) `y_("max")=4-(-1)=4+1=5` and `y_("min")=4-(1)=3`
(iii) `y_("max")=sqrt(3^(2)+4^(2))=5` and `y_("min")=-sqrt(3^(2)-4^(2))=-5`
21706.

the moon subtends an angle of 57 minute at the base -line equal to the radius of the earth . What is the distance of the moon from the earth ?[ Radius of the earth =6.4 x 106 m]

Answer»

parallactic angle , theta =57 '

=(57/60)o

=(57/60 )x pi/180 rad 

b= Radius of earth =6.4 x 106 m 

Distance of the moon from the earth ,

s= b/theta = 6.4 x106 x 60 x180 / 57x pi 

s= 3.86 x 108 m

21707.

the angular diameter of the sun is 1920'' . If the distance of the sun from the earth is 1.5 x 1011 m, then the liner diameter of the sun is  

Answer»

distance of sun from the earth =1.5 x 1011 m

angular diameter of sun ,

theta =1920''= (1920/60x60)o =1920/3600 xpi/ 180 rad

diameter of sun , D = s x theta 

                             = 1.5 x 1011 x 1920/3600 x pi /180 

                           D= 1.4 x 109 m

21708.

The range for a projectile that lands at the same elevation from which it is fired is given by `R = (u^(2)//g) sin 2theta`. Assume that the angle of projection `= 30^(@)`. If the initial speed of projection is increased by `1%`, while the angle of projection is decreased by `2%` then the range changes byA. `-0.3%`B. `+ 4.3%`C. `+ 0.65%`D. `0.85%`

Answer» Correct Answer - D
`deltaR//R = 2 deltau//u + 2 cot2theta " "delta theta`
21709.

A ball is projected with speed u at an angle `theta` to the horizontal. The range R of the projectile is given by `R=(u^(2) sin 2theta)/(g)` for which value of `theta` will the range be maximum for a given speed of projection? (Here g= constant)A. `pi/2 rad`B. `pi/4 rad`C. `pi/3 rad`D. `pi/6 rad`

Answer» Correct Answer - B
As `sin2theta le 1` so range will be maximum if `sin 2theta=1`. Therefore `2theta=pi/2 rArr theta=pi/4 rad`
21710.

The elongation produced on a steel wire and given stress is 0.01mm what will be the elongation if the radius of wire is half by the same condition?

Answer»

y = Fx/AΔx,

y = Fx/πr2(0.01)  ...(1)

y' \(=\frac{Fx}{\pi(\frac r2)^2\;\Delta x}\)

y' \(=\frac{Fx\times4}{\pi r^2(\Delta x)}\)  ...(2)

y' = y

4/Δx = 1/0.01

Δx = 0.04 mm

21711.

A “coulomb” is: A. one ampere per second B. the quantity of charge that will exert a force of 1 N on a similar charge at a distance of 1 m C. the amount of current in each of two long parallel wires, separated by 1 m, that produces a force of 2 × 10−7 N/m D. the amount of charge that flows past a point in one second when the current is 1 A E. an abbreviation for a certain combination of kilogram, meter and second 

Answer»

D. the amount of charge that flows past a point in one second when the current is 1 A

21712.

Suitable units for µ0 are: A. tesla B. newton/ ampere 2 C. weber /meter D. kilogram·ampere/meter E. tesla· meter/ampere

Answer»

E. tesla ·meter /ampere 

21713.

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed from a point on the positive z axis. Its magnetic dipole moment is: A. 0.25 A · m2, in the positive z direction B. 0.25 A · m2, in the negative z direction C. 2.5 A · m2, in the positive z direction D. 2.5 A · m2, in the negative z direction E. 0.25 A · m2, in the xy plane

Answer»

A. 0.25 A · m2, in the positive z direction

21714.

For a loop of current-carrying wire in a uniform magnetic field the potential energy is a minimum if the magnetic dipole moment of the loop is: A. in the same direction as the field B. in the direction opposite to that of the field C. perpendicular to the field D. at an angle of 45◦ to the field E. none of the above

Answer»

A. in the same direction as the field

21715.

The units of magnetic dipole moment are: A. ampere B. ampere·meter C. ampere· meter 2 D. ampere/meter E. ampere/meter 2

Answer»

C. ampere· meter 2

21716.

The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. If a uniform magnetic field is in the positive x direction the magnetic torque on the loop is: A. 0 B. in the positive y direction C. in the negative y direction D. in the positive z direction E. in the negative z direction

Answer»

B. in the positive y direction

21717.

The direction of the magnetic field in a certain region of space is determined by firing a test charge into the region with its velocity in various directions in different trials. The field direction is: A. one of the directions of the velocity when the magnetic force is zero B. the direction of the velocity when the magnetic force is a maximum C. the direction of the magnetic force D. perpendicular to the velocity when the magnetic force is zero E. none of the above

Answer»

A. one of the directions of the velocity when the magnetic force is zero

21718.

A magnetic field exerts a force on a charged particle: A. always B. never C. if the particle is moving across the field lines D. if the particle is moving along the field lines E. if the particle is at rest

Answer»

C. if the particle is moving across the field lines

21719.

At any point the magnetic field lines are in the direction of: A. the magnetic force on a moving positive charge B. the magnetic force on a moving negative charge C. the velocity of a moving positive charge D. the velocity of a moving negative charge E. none of the above

Answer»

E. none of the above 

21720.

The equilibrium constant for the following two reactions are `K_(1)` and `K_(2)` respectively. `XeF_(6)(g)+H_(2)O(g) hArrXeOF_(4)(g)+2HF(g)` `XeO_(4)(g)+XeF_(6)(g)hArrXeOF_(4)(g)+XeO_(3)F_(2)(g)` The equilibrium constant for the given reactiono is `:`A. `K_(1)K_(2)^(2)`B. `K_(1)-K_(2)`C. `K_(2)//K_(1)`D. `K_(2)//K_(2)`

Answer» Correct Answer - 3
Substracting `1^(st)` reaction from second, we will get desired reaction
`3^(rd)`. So `K=K_(2)//K_(1)`
21721.

A sodium street light gives off yellow light that has a wavelength of `600 n m.` Then .A. frequency of this light is `7xx10^(14)s^(-1)`B. frequency of this light is `5xx10^(14)s^(-1)`C. wavenumber of the light is `3xx10^(8)m^(-1)`D. energy of the photon is approximately `4.14 eV`

Answer» Correct Answer - 2
`V=(C)/(lambda)=(3xx10^(8))/(6000xx10^(-10))=5xx10^(14)S^(-1)`. Energy of photon `=(12400)/(6000)=2.07eV`
21722.

A steel blade floats on the surface of pure water but sinks when detergent added is added to the water. Why?

Answer»

When a steel blade is placed on water horizontally, the surface tension and the buoyancy will be able to balance the wight of the blade and it will oat on the water surface. You know the buoyancy is not sufficient to make the steel blade oat. Now if you add detergent to water, the surface tension of water immediately decreases. The surface tension no longer be able to balance the blade and it will sink.

21723.

If position and momentum of an electron are determined at a time then value of `DeltaP` is obtained `3DeltaX`, now the uncertainity in velocity of electron will be `:-`A. `(1)/(2m)sqrt((h)/(3pi))`B. `(1)/(2m)sqrt((3h)/(pi))`C. `(1)/(4m)sqrt((3h)/(pi))`D. `(1)/(6m)sqrt((h)/(pi))`

Answer» Correct Answer - B
21724.

Verify Rolle's theorem for the function f(x) = x2 + 4 in [–3, 3].

Answer»

Given f(x) = x2 + 4

Since f is a second degree polynomial

∴ f is continuous on [–3, 3] and f is desivable on (–3, 3)

Also f(–3) = 9 + 4 = 13 f(3) = 9 + 4 = 13

∴ f(–3) = f(3)

∴ f satisfies all the conditions of Role's theorum.

∴ There exists c∈(–3, 3) such that f1(c) = 0

But f1(x) = 2x

f1(c) = 2c

0 = 2c

⇒ c = 0 ∈ (–3, 3)

Hence Rolle's theorum is verified.

21725.

Find the equation of the locus of P, if A = (2, 3), B = (2, –3) and PA + PB = 8.

Answer»

Given A = (2, 3) 

B = (2, –3) 

Let P(x, y) be any point on the locus. 

Given geometric condition is 

PA + PB = 8 ⇒ PA = 8 – PB 

⇒ PA2 = (8 – PB)

⇒ PA2 = 64 + PB2 – 16PB

⇒ (x – 2)2+(y – 3)= 64 + (x – 2)2 + (y + 3)2 - 16√((x - 2)2 + (y + 3)2)

⇒ y2 – 6y + 9 – y2 – 6y – 9 – 64 = – 16√(x2 - 4x + 4 + y2 + 6y + 9)

⇒ (–12y – 64) = –16√(x2 + y2 - 4x + 6y + 13)

⇒ (3y + 16) = 4√(x2 + y2 - 4x + 6y + 13) 

S.O.B.S

⇒ (3y + 16)2 = 16 (x2 + y2 – 4x + 6y + 13) 

⇒ 9y2 + 256 + 96y = 16x2 + 16y2 – 64x + 96y + 208 

⇒ 16x2 + 7y2 – 64x – 48 = 0

∴ The locus of P is 16x2 + 7y2 – 64x – 48 = 0

21726.

If y = aenx + be– nx, then prove that yn = n²y.

Answer»

Given y = aenx + be– nx 

y1 = aenx . n + be– nx (–n) 

= anenx – bne– nx 

y11 = anenx . n – bne– nx(–n) 

= an2 enx + bn2 e– nx

= n2 (aenx + be– nx

= n2

∴ y11 = n2y.

21727.

Find dy and ∆y of y = f(x) = x2 + x at x = 10 when ∆x = 0.1

Answer»

Given f(x) = x2 + x, x = 10 and ∆x = 0.1

dy = f'(x) ∆x

= (2x + 1) ∆x

= [2(10) + 1] (0.1)

= (21) (0.1) = 2.1

∆y = f(x + ∆x) – f(x)

= (x + ∆x)2 + (x + ∆x) – (x2 + x)

= x2 + 2x ∆x + (∆x)2 + x + ∆x – x2 – x

= 2x ∆x + (∆x)2 + ∆x

= 2(10) (0.1) + (0.1)2 + 0.1

= 2 + 0.01 + 0.1

= 2.11

21728.

Cosec inverse 5/4  + 2 cot inverse

Answer»

Incomplete question.  What will come after cot-1

21729.

Which of the following is one of the reasons for the celebration of International Women's Day?1. As a member of UNO, India is legally bound to celebrate this day.2. To celebrate acts of courage of ordinary women who played extraordinary roles.3. Women can arrange meetings and spend some time for themselves without disturbance from men.4. Because women want to prove that they are better than men.

Answer» Correct Answer - Option 2 : To celebrate acts of courage of ordinary women who played extraordinary roles.

The correct answer is To celebrate acts of courage of ordinary women who played extraordinary roles.

  • About International Women's Day
    • International Women's Day is a global day celebrating the social, economic, cultural and political achievements of women.
    • The day also marks a call to action for accelerating gender parity. Significant activity is witnessed worldwide as groups come together to celebrate women's achievements or rally for women's equality. 
    • Marked annually on March 8th, International Women's Day (IWD) is one of the most important days of the year to:
      • celebrate women's achievements
      • raise awareness about women's equality
      • lobby for accelerated gender parity
      • fundraise for female-focused charities
      • To celebrate acts of courage of ordinary women who played extraordinary roles.

  • What's the theme for International Women's Day?
    • The campaign theme for International Women's Day 2021 is 'Choose To Challenge'.
    • A challenging world is an alert world. And from challenge comes change.
  • What colours symbolize International Women's Day?
    • Purple, green and white are the colours of International Women's Day.
    • Purple signifies justice and dignity. Green symbolizes hope.
    • White represents purity, albeit a controversial concept.
    • The colours originated from the Women's Social and Political Union (WSPU) in the UK in 1908.
21730.

If `x=a(theta+sintheta)` and `y=a(1-costheta)`, find `dy//dx`.

Answer» `x=a(theta+sintheta)`
`implies (dx)/(d theta)=a(1+costheta)` (1)
`y=a(1-cos theta)implies(dy)/(d theta)=a(0-(-sin theta))=a sin theta` (2)
Dividing (ii) and (i) `(dy//d theta)/(dx//d theta)=(dy)/(dx)=(sin theta)/(1+cos theta)=tan( theta//2)`
21731.

Differentiate the following w.r.t. x. a. `tan^3x` b. `tanx^2` c. `sin^2sqrtx`

Answer» a. Let `y=tan^3x=(tanx)^3`, `(dy)/(dx)=3(tan^2x)(sec^2x)`
b. Let `y=tan(x^2)`, `(dy)/(dx)=sec^2(x^2)2x`
c. Let `y=sin^2(sqrtx)=(sinsqrtx)^2`
`(dy)/(dx)=2(sinsqrtx)cos(sqrtx)1/2x^(-1//2)=(1)/(2sqrtx)sin(2sqrtx)`
21732.

A mixture of 45 litres of spirit and water contains 20% of water. How much water must be added to it to make the 25% in the new mixture?1. 5 litres2. 6 litres3. 4 litres4. 3 litres

Answer» Correct Answer - Option 4 : 3 litres

Given:

Total mixture of spirit and water = 45 litres

Percentage of water in mixture = 20%

Calculation:

Let x litre of water must be added to make it 25% in the new mixture

Percentage of water in mixture = 20% of 45 litres

⇒ (20/100 × 45)

⇒ 9 litres

According to the question

⇒ (9 + x)/(45 + x) = 25/100

⇒ (9 + x)/(45 + x) = 1/4

⇒ 36 + 4x = 45 + x

⇒ 3x = 9

⇒ x = 3 litres

∴ Water should be added is 3 litres

21733.

D invested Rs. 1210 for 3 years and E invested 3450 for 9 years. D again invested Rs. 2410 for 7 years and E invested Rs. 1750 for 1 year. Find the ratio of their profit after 10 years?1. 7 : 52. 7 : 83. 5 : 84. 6 : 7

Answer» Correct Answer - Option 3 : 5 : 8

Given:

D invested Rs. 1210 for 3 years and E invested 3450 for 9 years. D again invested Rs. 2410 for 7 years and E invested Rs. 1750 for 1 year.

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Profit sharing ratio of D and E after 10 years

D                                                        :                         E

1210 × 3 + 2410 × 7                          :                         3450 × 9 +1750 × 1

3630 + 16870                                    :                         31050 + 1750

20500                                                :                         32800

Required ratio = 205 : 328 = 5 : 8

21734.

The ratio of cost price of two articles is 4 ∶ 9. The articles are marked up by 40% and and 15% respectively and the ratio of their Marked price is 112 ∶ 207.If the discount of 12.5% and 11.11% is given respectively the profit earned on first article is Rs.270. What is the profit earned on second article?1. Rs.402. Rs.1503. Rs.604. Rs.805. Rs.200

Answer» Correct Answer - Option 3 : Rs.60

The ratio of cost price of two articles is 4 ∶ 9

⇒ Let the cost price of first article = 4x

The cost price of second article = 9x

The articles are marked up by 40% and 15% respectively

⇒ Marked price of first article = 4x × (140/100) = 5.6x

Marked price of second article = 9x × (115/100) = 10.35x

The ratio of their Marked price is 112 ∶ 207

⇒ Marked price of first article = 112y

Marked price of second article = 207y

∴ 5.6x = 112y

⇒ x = 20y

According to question,

Selling price of first article = 112y × (7/8) = 98y      ----(12.5% = 1/8)

Profit = 98y – 4x = 270

⇒ 98y – 80y = 270

⇒ 18y = 270

⇒ y = 15

∴ x = 20 × 15 = 300

⇒ Cost price of second article = 300 × 9 = 2700

Marked price of second article = 2700 × (115/100) = 3105

Selling price of second article = 3105 × (8/9) = 2760

∴ Profit earned = 2760 – 2700 = 60

∴ The profit earned on second article is Rs.60

21735.

Two brothers K and L invested Rs. 3900 and 4500 respectively. K invested for 8 months and L invested for 4 months. They then interchanged their amount and again invested for 4 months and 8 months respectively. Find the ratio of their investment after 1 year?1. 1 : 12. 1 : 23. 2 : 34. 1 : 3

Answer» Correct Answer - Option 1 : 1 : 1

Given:

Two brothers K and L invested Rs. 3900 and 4500 respectively. K invested for 8 months and L invested for 4 months. They then interchanged their amount and again invested for 4 months and 8 months respectively.

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Given that two brothers invested 3900 and 4500 for 8 and 4 months respectively and they interchanged their amount and again invested for 4 months and 8 months respectively.

K                                           :                L

3900 × 8 + 4500 × 4             :               4500 × 4 + 3900 × 8

49200                                   :                49200

Required ratio = 1 : 1

21736.

The total cost price of mobile and earphone is Rs. 14000. The cost price of an earphone is Rs. 2500 less than the cost price of a mobile. The profit percentage earned by selling mobile and earphone is 10% and 15% respectively. Find the difference between the selling price of a mobile and an earphone.1. Rs. 24632. Rs. 25603. Rs. 22904. Rs. 26705. Rs. 2040

Answer» Correct Answer - Option 1 : Rs. 2463

Given:

Let cost price of a mobile and an earphone be Rs.a and Rs.b respectively.

⇒ a + b = 14000

⇒ b = a - 2500

Solving,

a = 8250 and b = 5750

⇒ Selling price of a mobile = 8250 × 110/100 = Rs.9075

⇒ Selling price of an earphone = 5750 × 115/100 = Rs.6612.5 ≈ 6612

∴ Required difference = 9075 - 6612 = Rs.2463

21737.

A train running at 7/11 of its own speed reached a place in 22 hrs. How much time could be saved if the train runs at its own speed?1. 7 hrs2. 8 hrs3. 14 hrs4. 16 hrs

Answer» Correct Answer - Option 2 : 8 hrs

Given:

The new speed of the train = 7/11 × old speed of the train

With new speed, time taken to reach the place = 22 hours

Concept used:

When traveling the same distance,

Speed ∝ (1/time)

Calculation:

Let the old speed be SO and the new speed be SN

Let the old time taken be TO and new time taken be TN

SN = (7/11) × SO 

⇒ SN ∶ SO = 7 ∶ 11

⇒ TN ∶ TO = 11 ∶ 7

With new speed, time taken to reach, TN = 11x = 22 hrs

⇒ x = 2 hrs

The time that could be saved by him if the train runs on time = 11x – 7x = 4x

⇒ 4x = 4 × 2 = 8 hours.

∴ The time that can be saved by running with old speed is 8 hours.

21738.

Measure of chance of an uncertain event in form of numerical figures is classified as variability, durability, probability, none

Answer»

Answer: probability

21739.

Write the converse and income of the statement “If Maths is easy then child is brave”.

Answer»

Converse = (q → P) = If the child is brave then maths is easy 

Inverse = (~ P → ~q) = If maths is not easy then the child is not brave.

21740.

Selling Price = 1400RsProfit % = 20%Find Cost Price.

Answer» Selling price=cost price+profit

Profit=20percent

20/100×cp+cp=1400

Cp=3500/3=1166 approx.
21741.

If P (E) denotes the probability of an event E, then(a) 0< P(E) ⩽1(b) 0 < P(E) < 1(c) 0 ≤ P(E) ≤1(d) 0 ⩽P(E) <1

Answer»

Correct answer is: (c) 0 ≤ P(E) ≤1

0≤ P( E) ≤1

21742.

For matrix A = \(\begin{bmatrix} 2 & 5 \\[0.3em] -11 & 7 \end{bmatrix}\),(adj A)' is is equal to :A = [(2,5)(-11,7)](a) \(\begin{bmatrix} -2 & -5 \\[0.3em] 11 & -7 \end{bmatrix}\) (b) \(\begin{bmatrix} 7 & 5 \\[0.3em] 11 & 2 \end{bmatrix}\) (c) \(\begin{bmatrix} 7 & 11 \\[0.3em] -5 & 2 \end{bmatrix}\)(d) \(\begin{bmatrix} 7 & -5 \\[0.3em] 11 & 2 \end{bmatrix}\)

Answer»

Option : (c)

adj A = \(\begin{bmatrix} 7& -5 \\[0.3em] -11 & 2 \end{bmatrix}\)

⇒ (adj A)' = \(\begin{bmatrix} 7& 11 \\[0.3em] -5 & 2 \end{bmatrix}\)

21743.

The following distribution gives the daily income of 50 workers of a factory. `{:("Daily income",,400-420,,420-440,,440-460,,460-480,,480-500),("Number of workers",,12,,14,,8,,6,,10):}` Convert this distribution to less than type of cumulative frequency distribution and draw its ogive.

Answer» `{:("Daily Income",,"Number of workers",,"Cumulative Frequency"),(400-420,,12,,12),(420-440,,14,,26),(440-460,,8,,34), (460-480,,6,,40),(480-500,,10,,50):}`
Correct Table
Drawing an ogive with co-ordinates
(420, 12), (440, 26), (460, 34), (480, 40), (500, 50)
21744.

Solve : 2\((\frac{2x-1}{x+3})\) - 3\((\frac{x+3}{2x-1})\)= 5, x ≠ - 3,\(\frac{1}{2}\).

Answer»

2((2x-1)/(x+3)) - 3((x+3)/(2x-1)) = 5, x ≠ - 3,1/2

⇒ \(\frac{2(2x-1)^2-3(x+3)^2}{(x+3)(2x-1)}\) = 5

⇒ 2(4x2 − 4x + 1) − 3(x2 + 6x + 9) = 5(x + 3)(2x − 1) 

⇒ 8x2 − 3x2 − 8x − 18x + 2 − 27 = 10x2 + 25x − 15 

⇒ 5x2 − 26x − 25 = 10x2 + 25x − 15 

⇒ 10x2 − 5x2 + 25x + 26x − 15 + 25 = 0 

⇒ 5x2 + 51x + 10 = 0 

⇒ 5x2 + x + 50x + 10 = 0 

⇒ 5x(5x + 1) + 10(5x + 1) = 0 

⇒ 5(x + 10)(5x + 1) = 0 

⇒ x + 10 = 0 or 5x + 1 = 0

⇒ x = −10 or x = \(\frac{-1}{5}\).

Hence, 

The solutions of given equation are x = \(\frac{-1}{5}\),-10.

21745.

Find the coordinate of the centroid of triangle whose vertices are the following A(4,6), B(6, 3) and C(5,9)1. (5, 6)2. (5, 5)3. (6, 5)4. (6, 6)

Answer» Correct Answer - Option 1 : (5, 6)

Given:

The vertices of the triangle A, B and C = (4, 6), (6, 3) and (5, 9)

Formula used:

[(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3]

where, x1 = 4, x2 = 6, x3 = 5, y1 = 6, y2 = 3 and y3 = 9

Calculation:

According to the question

[(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3]

⇒ [(4 + 6 + 5)/3, (6 + 3 + 9)/3]

⇒ [(15/3), (18/3)]

⇒ (5, 6)

∴ The centroid of the triangle is (5, 6)

21746.

If the nth term of an AP is 4n - 9, then find its common difference1. 42. 53. -54. 6

Answer» Correct Answer - Option 1 : 4

Given:

tn = 4n – 9

Formula used:

Common difference (d) = t2 – t1

Calculation:

According to the question

⇒ t1 = 4 × (1) – 9 = -5

⇒ t1 = 4 – 9 

⇒ t1 = -5

⇒ t2 = 4 × (2) – 9

⇒ t2 = 8 – 9

⇒ t2 = -1

⇒ d = (-1) – (-5)

⇒ -1 + 5

⇒ 4

∴ Common difference is 4

21747.

Is (7×5×3×2+3) a composite number? Justify your answer.

Answer»

We know that,

A number that is divisible by a number other than 1 and itself, is called a composite number. 

Now,

7 × 5 × 3 × 2 + 3 = 3(7 × 5 × 2 + 1) 

= 3(70 + 1) 

= 3 × 71. 

∵ The given number has two factors 3 and 71 other than 1 and itself. 

∴ The given number is a composite number.

21748.

In the given figure, the graph of the polynomial P(x) is shown. What is the number of zeroes of P(x)?

Answer»

Zeros of the polynomial P(x) are the points where the graph of P(x) intersect x-axis. There are two such points. Hence, total number of zeros of P(x) is 2.

21749.

“Today I came bearing on olive branch in one hand and the freedom fighter s gun in the other. Don’t let the olive leaves fall from my hand. ” Whose speech is this? What made him make such a speech?

Answer»

The speech of Yaser Arafat, the leader of PLO in the UN Assembly in 1974.

Israel was formed in 1948. Following this, Israel seized Palestine, expelling the Palestinians from their homeland. The Palestinian refugees migrated to various Arab countries. It was in this context that the Palestine Liberation Organization (PLO) was formed with Yaser Arafat as President to establish a nation for the Palestinians. Yaser Arafat made such a speech in order to create a state for Palestinians through peaceful means.

21750.

1+tan2 θ is equal to (a) sec2 θ (b) cosec2 θ(c) tan2 θ (d) cot2 θ 

Answer»

The correct option is: (a) secθ