100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Suppose that at time ` t ul( gt)0` the position of a particle moving on the x-axis is `x=(t-1)(t-4)^(4)m`. (a) When is the particle at rest ? (b) During what time interval does the particle move to the left ? (c) Find maximum velocity of particle while moving to the left.
Answer» `(a) (dx)/(dt)=(t-4)^(4)+4(t-4)^(3)(t-1)=0` `:. " " (t-4)^(3)[t-4+4t-4]=0` `rArr " " t=4s or t=(8)/(5)s` (b) `(dx)/(dt) lt 0 rArr (t-4)^(3)(5t-8)lt0` For this , `tlt 4` and `t gt (8)/(5)` `rArr " " t in ((8)/(5),4)` (c) Particle will be faster when `((dx)/(dt))` is maximum `:. " " (d)/(dt)((dx)/(dt))=0` `:. " " (d^(2)x)/(dt^(2))=3(t-4)^(2)(5t-8)+5(t-4)^(3)=0` `rArr " " (1-4)^(2)[15t-24+5t-20]=0` `rArr " " t=4s " " or" " t=(44)/(20)=2.2s` But at `t=4s(dx)/(dt)=0` `:. ((dx)/(dt))` is maximum at t=2.2 s and maximum value of `(dx)/(dt)=(2.2-4)^(3)(5xx2.2-8)` `=-17.49m//s`

Discussion

2.	Electric field at distance L and 2 L from uniformly charged large non conducting sheet of surface charge density \(\sigma\) will be(1) \(\frac{\sigma}{\epsilon_0}, \frac{\sigma}{2\epsilon_0}\)(2) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{2\epsilon_0}\)(3) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{\epsilon_0}\)(4) \(\frac{\sigma}{\epsilon_0}, \frac{\sigma}{\epsilon_0}\)
Answer» Correct option is (2) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{2\epsilon_0}\) \(E = \frac{\sigma}{2\epsilon_0}\)

2.

Electric field at distance L and 2 L from uniformly charged large non conducting sheet of surface charge density \(\sigma\) will be(1) \(\frac{\sigma}{\epsilon_0}, \frac{\sigma}{2\epsilon_0}\)(2) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{2\epsilon_0}\)(3) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{\epsilon_0}\)(4) \(\frac{\sigma}{\epsilon_0}, \frac{\sigma}{\epsilon_0}\)

Answer»

Correct option is (2) \(\frac{\sigma}{2\epsilon_0}, \frac{\sigma}{2\epsilon_0}\)

\(E = \frac{\sigma}{2\epsilon_0}\)

Discussion

3.	what must be added to 0.0099 to get 9.9?1. 9.89012. 9.99003. 7.89004. 9.1901
Answer» Correct Answer - Option 1 : 9.8901 Given: The given two numbers are 0.0099 and 9.9 Concept Used: Basic concept of arithmetic Calculation: Let x must be added to 0.0099 to get 9.9 ∴ x = 9.9000 – 0.0099 = 9.8901 So, 9.8901 must be added to 0.0099 to get 9.9 Hence, option (1) is correct

3.

what must be added to 0.0099 to get 9.9?1. 9.89012. 9.99003. 7.89004. 9.1901

Answer» Correct Answer - Option 1 : 9.8901

Given:

The given two numbers are 0.0099 and 9.9

Concept Used:

Basic concept of arithmetic

Calculation:

Let x must be added to 0.0099 to get 9.9

∴ x = 9.9000 – 0.0099 = 9.8901

So, 9.8901 must be added to 0.0099 to get 9.9

Hence, option (1) is correct

Discussion

4.	In an A.P the sum of the first n terms bears a constant ratio `lamda` with the sum of the next n terms then `lamda` =A. `1/2`B. `1/3`C. `1/4`D. `2/5`
Answer» Correct Answer - 2 a,a + d,a+ 2d……… in A.P `lambda=("the sum of hte first n terms")/("the sum of the next n terms")` `=((n)/(2)[2a + (n -1)d])/((n)/(2) [(2(a +nd)+(n-1)d)]]= (2a-d+nd)/( 2a-d+3nd)` `=(1)/(3)` if `d=2a`

Discussion

5.	what should be subtracted from 39811 to make it divisible by 7?1. 12. 23. 34. 4
Answer» Correct Answer - Option 2 : 2 Given: Here the dividend is 39811 and divisor is 7 Formula used: Dividend = Divisor × Quotient + Remainder Calculation: When 39811 is divided by 7 then quotient will be 5687 and the remainder is 2 ∴ 39811 = 5687 × 7 + 2 So, 2 must be subtracted from 39811 to make it completely divisible by 7 Hence, option (2) is correct

5.

what should be subtracted from 39811 to make it divisible by 7?1. 12. 23. 34. 4

Answer» Correct Answer - Option 2 : 2

Given:

Here the dividend is 39811 and divisor is 7

Formula used:

Dividend = Divisor × Quotient + Remainder

Calculation:

When 39811 is divided by 7 then quotient will be 5687 and the remainder is 2

∴ 39811 = 5687 × 7 + 2

So, 2 must be subtracted from 39811 to make it completely divisible by 7

Hence, option (2) is correct

Discussion

6.	If 10 persons can do a job in 20 days. Then 20 persons with the twice efficiency can do the same job in:1. 152. 103. 54. 20
Answer» Correct Answer - Option 3 : 5 Given: 10 person complete a work in 20 days Formula used: Total work = Time × Efficiency × Men Calculation: Let efficiency of men be 'a'. and the time is taken to complete the work be 'w' Total work = Time × Efficiency × Men ⇒ Total work = 20 × a × 10 ----(i) Now after work efficiency is doubled Total work = Time × Efficiency × Men ⇒ Total work = w × 2a × 20 ----(ii) equating equation 1 and 2 ⇒ 20 × a × 10 = w × 2a × 20 ⇒ w = (20 × a × 10)/(2a × 20) ⇒ w = 5 ∴ 20 persons will complete the same work in 5 days.

6.

If 10 persons can do a job in 20 days. Then 20 persons with the twice efficiency can do the same job in:1. 152. 103. 54. 20

Answer» Correct Answer - Option 3 : 5

Given:

10 person complete a work in 20 days

Formula used:

Total work = Time × Efficiency × Men

Calculation:

Let efficiency of men be 'a'.

and the time is taken to complete the work be 'w'

Total work = Time × Efficiency × Men

⇒ Total work = 20 × a × 10 ----(i)

Now after work efficiency is doubled

Total work = Time × Efficiency × Men

⇒ Total work = w × 2a × 20 ----(ii)

equating equation 1 and 2

⇒ 20 × a × 10 = w × 2a × 20

⇒ w = (20 × a × 10)/(2a × 20)

⇒ w = 5

∴ 20 persons will complete the same work in 5 days.

Discussion

7.	The sum of the digits in the unit place of all numbers formed with the help of 3,4,5,6 taken all at a time isA. 18B. 432C. 108D. 144
Answer» Correct Answer - C Required sum is `3!(3+4+5+6)=6xx16xx18=108` [ if we fix 3 in the unit place, other three digits can be arranged in 3! Ways similarly for 4,5,6]

Discussion

8.	Seema’s age is twice of Radha’s age. Radha’s age is 40 years. Find their average age?1. 402. 453. 604. 30
Answer» Correct Answer - Option 3 : 60 Given: Radha’s age = 40 years Seema’s age = 2 × Radha’s age Concept: Average age = Total age / Total persons Solution: Radha’s age = 40 years Seema’s age = 2 × Radha’s age Seema’s age = 2 × 40 ⇒ 80 years Average = Total age / Total persons = (40 + 80) / 2 = 120/2 = 60 ∴ Their average age is 60.

8.

Seema’s age is twice of Radha’s age. Radha’s age is 40 years. Find their average age?1. 402. 453. 604. 30

Answer» Correct Answer - Option 3 : 60

Given:

Radha’s age = 40 years

Seema’s age = 2 × Radha’s age

Concept:

Average age = Total age / Total persons

Solution:

Radha’s age = 40 years

Seema’s age = 2 × Radha’s age

Seema’s age = 2 × 40 ⇒ 80 years

Average = Total age / Total persons

= (40 + 80) / 2

= 120/2

= 60

∴ Their average age is 60.

Discussion

9.	The line `x+y=6` is normal to the parabola `y^(2)=8x` at the point.A. (4,2)B. (2,4)C. (2,2)D. (3,3)
Answer» Correct Answer - B `y=-x+6` ..(1) is normal to `y^(2)=4ax` the of line (1) Now equation of any normal to the parabola `y^(2)=4ax` is `y=mx-2am-am^(3)` Comparing (1) and (2) we get `mx=-ximpliesm=-1` `6=-2am-am^(3)` `6=-2a(-1)-a(-1)^(3)` `6=2a+a=3aimpliesa=2` `because` point `(am^(2),-2am)=(2(-1)^(2),-2xx2xx-1)=(2,4)`

Discussion

10.	The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number isA. `6^(n)-1`B. `3^(n)-1`C. `6^(n)-3^(n)`D. `6^(n)-5^(n)`
Answer» Correct Answer - C The total number of ways is `6xx6xx…` to n times `=6^(n)` the total number of ways to show only even number is `3xx3`… to n times `=3^(n)` therefore the required number of ways is `6^(n)-3^(n)`.

Discussion

11.	If Ram secures 100 marks n math then he will get a mobile. The logically equivalent statement isA. if ram will get a mobile then he will not secures 100 marks in mathB. if ram will not get a mobile then he will secures 100 marks in mathC. if ram will get a mobile then he sucures 100 marks in math.D. if ram will not get a mobile then he will nor secures 100 marks in math.
Answer» Correct Answer - D Let p: Ram secure 100 marks in mathmatics. q, Ram will get a mobile. Converse of statement p and q if `p` then `q` implies` if `q` then `p` `because` converse of above statement is if ram will get a mobile then he secure 100 marks in mathmatics.

Discussion

12.	The coefficient of `x^(4)` in the expansion of `(1+x+x^(2)+x^(3))^(11)` is (where `({:(n),(r):})`=^(n)C_(r))`A. `({:(11),(4):})`B. `({:(11),(4):})+({:(11),(2):})`C. `({:(11),(4):})+({:(11),(2):})+({:(11),(4):}).({:(11),(2):})`D. `({:(11),(4):})=({:(11),(2):})+({:(11),(1):})({:(11),(2):})`
Answer» Correct Answer - D `1+x+x^(2)+x^(3)=(1+x)+x^(2)(1+x)=(1+x)(1+x^(2))` `(1+x+x^(2)+x^(3))^(11)=(1+x)^(11)(1+x^(2))^(11)` We want the coefficient of `x^(4)` `(1+^(11)C_(1).x+^(11)C_(2).x^(2)+^(11)C_(3).x^(3)+^(11)C_(4).x^(4)+…)` `xx(1xx^(11)C_(1).x^(2)+^(11)C_(2).x^(4)+....)` Collecting the terms which give `x^(4)` `1.^(11)C_(2).x^(4)+^(11)C_(2).x^(2).^(11)C_(1).x^(2)+^(11)C_(4)x^(4).1` `because^(11)C_(2)+^(11)C_(2)` .^(11)C_(1)+^(11)C_(4)=55+605+330=990`

Discussion

13.	A curve is represented parametrically by the equations `x=t+e^(at) and y=-t+e^(at)` when `t in R and a > 0.` If the curve touches the axis of x at the point A, then the coordinates of the point A areA. (1,0)B. `((1)/(e,0))`C. `(e,0)`D. `(2e,0)`
Answer» Correct Answer - D `X=t+e^(at),y=-t+e^(at)` `(dx)/(dt)=1+ae^(at),(dy)/(dx)=(-1+ae^(at))/(1+ae^(at))` At the point `A,y=0` and `(dy)/(dx)=0` for some `t=t_(1)` `becauseae^(at_(1))=1` (i). Also `0=-t+e^(at_(1))` `t_(1)=e^(at_(1))` On putting this value in Eq. (i) we get `at_(1)=1impliest_(1)=(i)/(a)` Now from eq `(i), `ae=1impliesa=(1)/(e)` Hence, `x_(A)=t_(1)+e^(at_(1))=e+e=2e` `impliesA=(2e,0)`

Discussion

14.	If the line `x+y=a`touches the parabola `y=x-x^2,`then find the value of `adot`
Answer» Correct Answer - 2 Eliminating y, we have `a- x= x- x^(2)` or `x^(2) - 2x + a =0`. Since the line touches the parabola, we must have equal roots . Therefore , `4- 4a =0 ` or `a=1`

Discussion

15.	Point A lies on the line `y=2x` and the sum of its abscissa and ordinate is 12. Point `B` lies on the x-axis and the line `AB` is perpendicular to the line `y=2x`. Let `O` be the origin. The area of the Delta AOB isA. 20B. 40C. 60D. 80
Answer» Correct Answer - D Let `A(x_(1),y_(1))` `x_(1)+y_(1)=12` `y_(1)=12-x_(1)` A lies on `y=2x` `y_(1)=2x_(1)` `12-x_(1)=2x_(1)` `x_(1)=4` Hence, `A=(4,8)` equation of AB is `x+2y+lamda=0` Eq. (i) passes through (4,8) `becauselamda=-20` `x+2y=20impliesB` is `(20,0)` Area of `DeltaAOB=(20.8)/(2)=80` sq unit

Discussion

16.	A committee of three persons is to be randomly selected from a group of three men and two women and the chair person will be randomly selected from 2 woman and 1 men for the committee. The probability that the committee will have exactly two woman and one man and that the chair person will be a woman, isA. `(1)/(5)`B. `(8)/(15)`C. `(2)/(3)`D. `(3)/(10)`
Answer» Correct Answer - A `n(S)=^(5)C_(3)=10,n(A)=^(3)C_(1).^(2)C_(2)=3` `becauseP(2W` and `1M)=(3)/(10)` So, `P(2W` and `1M` and chair person is woman) `=(3)/(10)(2)/(3)=(1)/(5)`

Discussion

17.	Two spherical black bodies of radii `R_(1)` and `R_(2)` and with surface temperature `T_(1)` and `T_(2)` respectively radiate the same power. `R_(1)//R_(2)` must be equal toA. `(T_(1))/(T_(2))`B. `(T_(2))/(T_(1))`C. `(T_(1)/T_(2))^(2)`D. `((T_(2))/(T_(1)^(2))`
Answer» Correct Answer - C `(Q)/(t)=sigma A T^(4)` i.e, power `=sigmaAT^(4)` `therefore` for same power `therefore A alpha (1)/(T^(4))` `therefore (A_(2))/(A_(1))=(T_(1)^(4))/(T_(2)^(4))` `therefore (4pi r_(2)^(2))/(4pir_(1)^(2))=(T_(1)^(4))/(T_(2)^(4))` `therefore (r_(2))/(r_(1))=((T_(1))/(T_(2)^(2))`

Discussion

18.	The area bounded by `y=\|log_(e)\|x\|\|` and `y=0` is
Answer» Correct Answer - 2 `y=\|log_(e)\|x\|\|` and `y=0` From the figure. Required area `=` area of shaded region `=1+1=2` sq. units.

Discussion

19.	If the coefficients of the rth, `(r+1)t h ,(r-2)t h`terms is the expansion of `(1+x)^(14)`are in A.P, then the largest value of `r`is.
Answer» Correct Answer - 4 Coefficient of `r^(th)` and `(r+2)^(th)` terms are .^(14)C_(t-1),^(14)C_(r)` and .^(14)C_(r+1)` If these coefficients are in A.P. then `2(` .^(14)C_(r))=^(14)C_(r-1)+^(14)C_(r+1)` `implies(2(14)!)/(r!(14-r)!)=((14)!)/((r-1)!(15-r)!)+((14)!)/((r+1)!(13-r)!)` `implie3s(2(14)!)/(r!(14-r)!)+((14)![(r+1)r+(15-r)(14-r)])/((r+1)!(15-r)!)` `implies2(15-r)(r+1)=2r^(2)-28r+210` `impliesr^(2)-14r+45=0` `implies(r-5)(r-9)=0` `impliesr=5` or `9`

Discussion

20.	Let f(x) = minimum `(x+1, sqrt(1-x))" for all "x le 1.` Then the area bounded by y=f(x) and the x-axis is
Answer» Correct Answer - 7 Required area`=` shaded region `=int_(0)^(1)(x_(2)-x_(1))dy=int_(0)^(1)[(1-y^(2))-(y-1)]dy` `=(7)/(6)` sq. unit

Discussion

21.	The number of times the digit 5 will be written when listing the integers from 1 to 100, isA. 271B. 272C. 300D. None of these
Answer» Correct Answer - 3 `=---:`digits are from 0 to 9 = (only once )+(twice )+(thrice) `=""^(3)C_(1) .9.9(1)+3C_(2).9(2)+3C_(3).1(3)` =3.81 + 3.18 + 1.3 =300

Discussion

22.	If `A(cosalpha,sinalpha),B(sinalpha,-cosalpha),C(1,2)`are thevertices of ` A B C ,`then as `alpha`varies,find the locus of its centroid.A. `x^(2)+y^(2)-2x-4y+1=0`B. `x^(2)+y^(2)-2x-4y+3=0`C. `3(x^(2)+y^(2))-2x-4y+1=0`D. None
Answer» Correct Answer - 3 SO let centroid is (h,k) `h=(cosalpha+sin alpha+1)/(3),K=(sin alpha-cosalpha +2)/(3)` ` implies cos alpha +sin alpha =3h -1 …..(i)` `implies sin alpha -cos alpha = 3K -2 …….(ii)` squaring and adding `2=(3h-1)^(2)+(3K-2)^(2)` locus` (3x-1)^(2)+(3y-2)^(2)=2`

Discussion

23.	`f(x)`is a continuous function for all real values of `x`and satisfies `int_n^(n+1)f(x)dx=(n^2)/2AAn in Idot`Then `int_(-3)^5f(\|x\|)dx`is equal to`(19)/2`(b) `(35)/2`(c) `(17)/2`(d) none of these
Answer» Correct Answer - 7 `int_(-3)^(5)f(\|x\|)dx=int_(-3)^(3)(f\|x\|)dx+int_(3)^(5)f(\|x\|)dx` `=2int_(0)^(3)f(x)dx+int_(3)^(5)f(x)dx` `=2(int_(0)^(1)f(x)dx+int_(1)^(3)f(x)dx+int_(2)^(3)f(x)dx)+(int_(3)^(4)f(x)dx+int_(4)^(5)f(x)dx)` `=(0+(1)/(2)+(2^(2))/(2))+(9)/(2)+(16)/(2))=(35)/(2)`

Discussion

24.	The intercept made by theplane ` vec rdot vec n=q`on the x-axis isa. `q/( hat idot vec n)`b. `( hat idot vec n)/q`c. `( hat idot vec n)/q`d. `q/(\| vec n\|)`A. `q/(hati.overset(rarr)n)`B. `(hati.overset(rarr)n)/q`C. `q/(abs(overset(rarr)n)`D. None of these
Answer» Correct Answer - 1 `vecr.vecn=q` Let , intercept made by the plane (1) on the x-axis is `x_(1)` `implies` plane (1) passes through it `x_(1),hati.vecn =q implies x_(1) =(q)/(hati.vecn)`

Discussion

25.	Use product `[[1,-1, 2],[ 0, 2,-3],[ 3,-2, 4]] [[-2, 0, 1],[ 9, 2,-3],[ 6, 1,-2]]` to solve the system of equation:`x-y+2z=1`; `2y-3z=1`; `3x-2y+4z=2`
Answer» `[[1,-1,2],[0,2,-3],[3,-2,4]][[-2,0,1],[9,2,-3],[6,1,-2]]= [[-2-9+12,0-2+2,1+3-4],[0+18-18,0+4-3,0-6+6],[-6-18+24,0-4+4,3+6-8]]` `=[[1,0,0],[0,1,0],[0,0,1]] = I` So, this a form `A A^-1 = I`, where `A = [[1,-1,2],[0,2,-3],[3,-2,4]], A^-1 = [[-2,0,1],[9,2,-3],[6,1,-2]]` Now, for the given system of linear equation, `AX = B` `=>A A^-1 X = BA^-1` `=>X = BA^-1` Here, `X = [[x],[y],[z]] , B = [[1],[1],[2]], A^-1 = [[-2,0,1],[9,2,-3],[6,1,-2]]` `:.[[x],[y],[z]] = [[1],[1],[2]] [[-2,0,1],[9,2,-3],[6,1,-2]]` `=>[[x],[y],[z]] = [[-2+0+2],[9+2-6],[6+1-4]] ` `=>[[x],[y],[z]] = [[0],[5],[3]] ` `:. x = 0, y = 5 and z = 3` is the solution for the given equations.

Discussion

26.	Find the vector equation of the line passingthrough the point (1,2,3) and parallel to the planes ` vec rdot( hat i- hat j+2 hat k)=5`and ` vec rdot(3 hat i+ hat j+ hat k)=6.`
Answer» `P_1=vecr(hati-hatj+2hatk)=5` `vecn_1=hati-hatj+2hatk` `P_2=vecr(3hati+hatj+hatk)=6` `vecn=3hati+hatj+hatk` `l=vecr=veca+lambdavecb` `vecb=vecn_1*vecn_2` `=i(-1-2)-j(1-6)+k(1-(-3))` `vecb=-3hati+5hatj+4hatk` `vecr=veca+lambdavecb` `vecr=(hati+2hatj+3hatk)+lambda(-3hati+5hatj+4hatk)` `vecr=(1-3lambda)hati+(2+5lambda)hatj+(3+4lambda)hatk`.

Discussion

27.	Writethe direction cosines of a line parallel to z-axis.
Answer» let`(x-9)/C_1=(y-6)/C_2=(z-6)/C_3=t` `x=a+c_1t=a` `y=b+c_2t=b` `c_1=0,c_2=0,c_3=k` `sqrt(c_1^2+c_2^2+c_3^2=1` `sqrt(0+0+k^2)=1` k=1 the direction cosines of a line parallel to z-axis(0,0,1).

Discussion

28.	If `x=a (theta-sintheta), y=a (1+costheta), `find `(d^2 y)/(dx^2)`
Answer» `y = a+acostheta` `=>dy/(d theta) = - asin theta` `x = atheta - asintheta` `=>dy/(d theta) = a - acostheta = a( 1- costheta)` `:. dy/dx = (dy/(d theta))/(dx/(d theta)) = (-asin theta)/(a(1-costheta))` `=>dy/dx = -sintheta/(1-costheta)(1+costheta)/(1+costheta)` `=>dy/dx = -(sintheta (1+costheta))/(1-cos^2theta)` `=>dy/dx = -(sintheta (1+costheta))/(sin^2theta)` `=>dy/dx = -cosec theta - cottheta ` Differentiating both sides w.r.t. `theta`, `=>d/(d theta) (dy/dx) = cosecthetacottheta + cosec^2theta` `=>d/(d theta) (dy/dx) = cosectheta(cottheta + cosectheta)` `=>(d^2y)/dx^2 = cosectheta(cottheta + cosectheta)(d theta)/dx` `=>(d^2y)/dx^2 = (cosectheta(cottheta + cosectheta))/(a(1-costheta))` `=>(d^2y)/dx^2 = (1/sintheta((1+costheta)/sintheta))/(a(1-costheta))` `=>(d^2y)/dx^2 = (1+costheta)/(asin^2theta(1-costheta))` `=>(d^2y)/dx^2 =(1+costheta)/(asin^2theta(1-costheta))**(1-costheta)/(1-costheta)` `=>(d^2y)/dx^2 =sin^2theta/((asin^2theta)(1-costheta)^2)` `=>(d^2y)/dx^2 = 1/(a(1-costheta)^2)` As `y = a(1+costheta) => costheta = y/a-1` Putting this value in our equation, `=>(d^2y)/dx^2 = 1/(a(1-(y/a-1))^2)` `=>(d^2y)/dx^2 = a/(2a-y)^2`

Discussion

29.	Evaluate: `int_(pi/6)^(pi/3)(dx)/(1+sqrt(tanx))`
Answer» `I=int_(pi/6)^(pi/3) (dx/(1+sqrttanx))-(1)` `I=int_(pi/6^(pi/3)(dx/(1+sqrttan(pi/3+pi/6-x)))` `I=int_(pi/6)^(pi/3)(dx/(1+sqrtcotx))` `I=int_(pi/6)^(pi/3) ((dxsqrttanx)/(sqrttanx+1))` Adding 1 and 2 `2I=int_(pi/6)^(pi/3)((1+sqrttanx)/(1+sqrttanx))dx` `2I=int_(pi/6)^(pi/3)dx` `2I=[x]_(pi/6)^(pi/3)` `2I=pi/3-pi/6=pi/6` `I=pi/12`.

Discussion

30.	If the standard deviation of six observations is increased by 5.Find the new standard deviation.
Answer» If a constant, k, is added to each number in a set of data, the mean will be increased by k and the standard deviation will be unaltered (since the spread of the data will be unchanged). Hence, S.D. of the new data = 11

Discussion

31.	The eccentricity of the parabola y^2=16
Answer» As we know, eccentricity is the distance from any point on the parabola to its focus, divided by the perpendicular distance from that point to the directrix. Any point on parabola possesses equal distance to its focus and directrix. Hence, the eccentricity of the parabola comes out to be 1.

Discussion

32.	In which market AR = MR ? (A) Monopoly (B) Monopolistic competition (C) Both (A) and (B) (D) Perfect competition
Answer» (D) Perfect competition

Discussion

33.	Change in quantity of production has effects on - (A) Both Fixed and Variable Cost (B) Only Variable Cost (C) Only Fixed Cost (D) None of above
Answer» (B) Only Variable Cost

Discussion

34.	Write the parametric equation of the line passing through the point (3, 2) and making an angle 120° with the positive direction of the x-axis measured in counter clock sense and also find the coordinates of the points on the line which are at unit distance from the point (3, 2).
Answer» We have (x₁, y₁) = (3, 2), cosθ = cos 120° = −1/2 and sinθ = sin 120° = √3 /2. Therefore, the parametric equations of the line are x = x₁ + γ cosθ = 3 + γ(-1/2) and y = y₁ + γ sinθ = 2 + γ(√3/2) Solving these equations, we get x = 3 - γ/2 and y = 2 + γ (√3/2) Substituting γ = 1 and γ = −1 in the above coordinates, the points on the line which are at a distance of 1 unit from the point (3, 2) are obtained, respectively, (5/2, 4 + √3/2) and (7/2, 4 -√3/2)

34.

Write the parametric equation of the line passing through the point (3, 2) and making an angle 120° with the positive direction of the x-axis measured in counter clock sense and also find the coordinates of the points on the line which are at unit distance from the point (3, 2).

Answer»

We have (x₁, y₁) = (3, 2), cosθ = cos 120° = −1/2 and sinθ = sin 120° = √3 /2. Therefore, the parametric equations of the line are

x = x₁ + γ cosθ = 3 + γ(-1/2)

and y = y₁ + γ sinθ = 2 + γ(√3/2)

Solving these equations, we get

x = 3 - γ/2

and y = 2 + γ (√3/2)

Substituting γ = 1 and γ = −1 in the above coordinates, the points on the line which are at a distance of 1 unit from the point (3, 2) are obtained, respectively,

(5/2, 4 + √3/2) and (7/2, 4 -√3/2)

Discussion

35.	In which type of goods, price fall does not make any increase in demand? (A) Necessity Goods (B) Comfort Goods (C) Luxury Goods (D) None of these
Answer» (A) Necessity Goods

Discussion

36.	Which of the following is included in money cost ? (A) Normal Profit (B) Explicit Cost (C) Implicit Cost (D) All of these
Answer» (D) All of these

Discussion

37.	How many types elasticity of demand has ? (A) Three (B) Five (C) Six (D) Seven
Answer» Five types elasticity of demand has.

Discussion

38.	Reasons of decrease in supply is :- (a) Increase in production cost (b) Increase in price substitution (c) Fall in number of firm in the industry (d) All of the above
Answer» (d) All of the above

Discussion

39.	Law of variable problem is related to - (A) Both short run & long run (B) Lon run (C) Short run (D) Very long run
Answer» (C) Short run

Discussion

40.	Who basically propounded the concept of law of equi-marginal utility ? (a) Marshall (b) Gossen (c) Ricardo (d) Mill
Answer» Gossen propounded the concept of law of equi-marginal utility.

Discussion

41.	Who basically propounded the concept of law of Equimarginal utility? (A) Marshall (B) Gossen (C) Ricardo (D) Mil
Answer» The Correct option is (B) Gossen

Discussion

42.	Who basically propounded the concept of law of equi-marginal utility? (a) Marshall (b) Gossen (c) Ricardo (d) J.S. Mill
Answer» Ricardo propounded the concept of law of equi-marginal utility.

Discussion

43.	Which of the following is true ? (a) TU increases till MU is positive (b) TU is maximum when MU=0 (c) TU declines when MU is negative (d) All of these
Answer» (d) All of these

Discussion

44.	When T.U. becomes maximum, MU is : (a) Positive (b) Negative (c) Zero (d) None of these
Answer» When T.U. becomes maximum, MU is Zero.

Discussion

45.	A train of mass 100 metric tons is ascending uniformly on an incline of 1 in 250, and the resistance due to friction, etc is equal to 6 kg per metric ton. If the engine be of `7.84xx10^(4)` watts and be working at full power, find the speed at which the train is goingA. 2 m/sB. 16 m/sC. 8 m/sD. 10 m/s
Answer» Correct Answer - C

Discussion

46.	What is the coordination number of Cr in K3[Cr(Ox)3]?(a) 6(b) 5(c) 4(d) 3
Answer» Answer is (d) 3

Discussion

47.	The highest magnetic moment shown by the transition metal ion with the outermost electronic configuration is:(a) 3d5(b) 3d2(c) 3d7(d) 3d9
Answer» Answer is (a) 3d⁵

Discussion

48.	The general electronic configuration of transition elements is:(a) (n-1)d5(b) (n-1)d(1-10) ns0.1, or 2(c) (n-1)d(1-10) ns1(d) None of these
Answer» Answer is (b) (n-1)d^(1-10) ns^{0.1, or 2}

Discussion

49.	Increasing order of acid strength among p-methoxyphenol, p-methylphenol and pnitrophenol is: a). p-nitrophenol, p-methoxyphenol, p-methylphenol b). p-methylphenol, p-methoxyphenol, p-nitrophenol c). p-nitrophenol, p-methylphenol, p-methoxyphenol d). p-methoxyphenol, p-methylphenol, p-nitrophenol.
Answer» d). p-methoxyphenol, p-methylphenol, p-nitrophenol.

Discussion

50.	Which of the following is more basic than aniline? a). Benzylamine b). Diphenylamine c). Triphenylamine d). p-Nitroaniline
Answer» a). Benzylamine

Discussion

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