100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	While explaining interspecific interaction of population, (+) sign is assigned for beneficial interaction, sign is assigned for detrimental interaction and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and (-) for another species involved in the interaction?(1) Amensalism(2) Commensalism(3) Competition(4) Predation
Answer» Correct option is (4) Predation

Discussion

2.	An observer A is at rest in gournd frame, observer B is moving with constant another observer C is moving with constant velocity of 1m/s. A. work done by all forces (both real and pseudo) in 1 sec in frame of A is 0.5JB. work done by all forces (both real and pseudo) in 1 sec in frame of B is 0JC. work done by all force (both real and pseudo) in 1 sec in frame of C is -0.5JD. change in K.E. in frame of B is zero
Answer» `S=(1)/(2)at^(2)=(1)/(2)xx1xx1^(2)=(1)/(2)m` `omega_(A)=1xx(1)/(2)=0.5J` `omega_(B)=0` because in frame 1 ke is at rest. `omega_(C)=DeltaK=0-(1)/(2)=-(1)/(2)J` `k_(i)=(1)/(2)xx1xx1^(2)` `k_(r)=0`

Discussion

3.	A constant force F acts for 1 sec, on a body of mass 1kg moving with perpendicular to its initial velocity, thenA. distance covered by the body is `(u+(F)/(2))`B. displacement of the body is `sqrt(u^(2)+((F)/(2))^(2))`C. change in kinetic energy of the body is `sqrt((1)/(2)(u^(2)+F^(2)))`D. momentum of the body is increased by F/2.
Answer» m=1kg `s_(y)=0+(1)/(2)at^(2)` `s_(y)=(1)/(2)((F)/(1))(I)^(2)=(F)/(2)` t=1sec, `s_(x)=u` `s=sqrt((s_(x))^(2)+(s_(y))^(2))=sqrt(u^(2)+((F)/(2))^(2))` `W_(F)=DeltaK` `Fxx(F)/(2)=(1)/(2)mV^(2)-(1)/(2)mu^(2)` `(F^(2))/(2)+(1)/(2)u^(2)=(V^(2))/(2)` `V=sqrt(F^(2)+u^(2))` J=`DeltaP` `FxxI=DeltaP=F` As constant force acts on the body and angle between u `&` F is neither `0^(@)` nor `180^(@)`, path is parabolic.

Discussion

4.	A ring of radius r is moving without slipping on a circular track of radius R speed of the centre of the ring is `v = omega r`. Ignoring the fact that the ring axis is inclined A. The radial acceleration of centre of the ring is `(omega^(2)r^(2))/(R )`B. The radial acceleration of centre of the ring is `(R-r)^(2)/(R )omega^(2)`C. The angular velocity of centre of the ring about vertical axis passing through the centre of track is `romega//R`.D. The angular velocity of centre of the ring about vertical axis passing through the centre of track is `((R-r))/(R )omega`
Answer» Correct Answer - A::C velocity of centre of ring is = `romega`

Discussion

5.	Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are `v_(1)` and `v_(2)` at angles `theta_(1)` and `theta_(2)` respectively from the horizontal, then ansewer the following questions If `v_(1)costheta_(1) = v_(2)cos theta_(2)`, then choose the incrorrect statementA. the time of flight of both the particles will be sameB. the maximum height attained by the particles will be sameC. the trajectory of one with respect to another will be a horizontal straight lineD. none of these
Answer» Correct Answer - B `y = xtan theta-(gx^(2))/(2x^(2)cos^(2)theta)` `75 = x tan53^(@) - (gx^(2))/(2xx50^(2)cos^(2)53^(@))` `75 = x (4)/(3) - (x^(2))/(2xx2500xx9)` `75 = (4x)/(3)-(x^(2))/(180)` solving `x_(1) =150 m " "x_(2) = 90m` `Deltax= 60m`

Discussion

6.	A 1000 kg car starts from rest at point 1 and moves without friction down the track as shown in the figure. At point 2, the radius od curvature of the track is 6m. A. the force exerted by the track on the car at point 2 is 4000 kg wtB. the force exerted by the track on the car at point 2 is 5000 kg wtC. the minimum safe value of the radius of curvature at point 3 is 15 mD. the minium safe value of the radius of curvature at point 3 is 10 m
Answer» Correct Answer - B::C By work and energy theorem from point 1 to 2 `0+mg(12)=(1)/(2)mv_(2)^(2)` and by NLM at 2 `N-mg=mv_(2)^(2)//R` N = 5 mg At point 3 `0+ mg(12-4.5)=(1)/(2)mv_(3)^(2)` `v_(3)=sqrt(15g)m//s` NLM at point 3 `mg=m(v_(3)^(2)//r)` `r = 15 m`

Discussion

7.	Two projectiles are projected with velocity `v_(A), v_(B)` at angles `theta_(A)` (from horizontal) and `theta_(B)` (from vertical) as shown in the figure below, such that `v_(A) gt v_(B)` but having same horizontal component of velocity. Which of the following can be correct ? A. `T_(A) gt T_(B)`B. `H_(A) gt H_(B)`C. `R_(A) gt R_(B)`D. `R_(B) gt R_(A)`
Answer» Correct Answer - A::B::C `v_(AX)=v_(BX)` `v_(Ay)^(2)+v_(Bx)^(2)gtv_(By)^(2)+v_(Bx)^(2)` `thereforev_(Ay)gtv_(By)` `R=(2v_(x)v_(y))/(g)" "thereforeR_(A) gtR_(B)` `H=(2v_(y)^(2))/(g)" "thereforeH_(A)gtH_(B)` `T=(2v_(y))/(g)" "thereforeT_(A)gtT_(B)`

Discussion

8.	AB is a light rigid rod, which is rotating about a vertical axis passing through A. A spring of force constant K and natural length l is attached at A and its other end is attached to a small bead of mass m. The bead can slide without friction on the rod. At the initial moment the bead is at rest (w.r.t. the rod) and the spring is unstretched. Select correct options : A. The maximum velocity attained by the bead w.r.t the rod is given by `V_("max")=sqrt((momega^(4)l^(2))/(K-momega^(2)))`B. The maximum velocity attained by the bead w.r.t the rod is given by `V_("max")=sqrt(((momega^(4)+K)/(momega^(2)-K))omega^(2)l^(2))`C. The maximum extension in the spring is given by `X_("max")=(2momega^(2)l)/(K-momega^(2))`D. The maximum value of contact force between the bead and the rod is greater than mg
Answer» Correct Answer - A::C::D At equilibrium forces balance `m(l+x)omega^(2)=Kximpliesx=(mlomega^(2))/(K-momega^(2))` Also till this point by work energy theorem `underset(0)overset(x)intm(l+x)omega^(2)dx-(Kx^(2))/(2)=(1)/(2)mv^(2)` `V=sqrt((ml^(2)omega^(4))/(K-momega^(2)))` and at maximum elongation velocity becomes zero apply above concept of work & energy from start to zero velocity.

Discussion

9.	If `f(x) = log ((1+x)/(1-x))`, where `-1 lt x lt 1` then `f((3x+x^(2))/(1+3x^(2))) - f((2x)/(1+x^(2)))` is equal toA. `[f(x)]^(3)`B. `[f(x)^(2)]`C. `-f(x)`D. `f(x)`
Answer» Correct Answer - B Given `f(x)= log ((1+x)/(1-x))` `therefore f((3x + x^(3))/(1+3x^(2))) - f((2x)/(1+x^(2)))` `=log[(1+((3x +x^(3))/(1+3x^(2))))/(1-((3x +x^(2))/(1+3x^(2))))] - log((1+(2x)/(1+x^(2)))/(1-(2x)/(1+x^(2))))` `= log ((1+x)/(1-x))^(3)- log ((1+x)/(1-x))^(2)` ltbr. ` = log ((1+x)/(1-x)) = f(x) `

Discussion

10.	The equation of second degree `x^2+2sqrt2x+2y^2+4x+4sqrt2y+1=0` represents a pair of straight lines.The distance between them isa. 4b. `4/sqrt3`c. 2d. `2sqrt3`A. 1 unitsB. 2 unitsC. 3 unitsD. 4 units
Answer» Correct Answer - B Here, `h = sqrrt(2) , g =2 , a = 1 , c = 1,b =2 , f = 2sqrt(2)` `therefore ` Distance `= 2 sqrt((g^(2) - ac)/(a(a+b)))` ` = 2sqrt((4-1)/(1(1+2))) = 2` units

Discussion

11.	If `y = 1 +(1)/(x) +(1)/(x^(2)) + (1)/(x^(3)) + ……..+oo` with `\|x\| gt 1` then `(dy)/(dx)` is .A. `(x^(2))/(y^(2))`B. `x^(2)y^(2)`C. `(y^(2))/(x^(2))`D. `-(y^(2))/(x^(2))`
Answer» Correct Answer - D Given ` y = 1 +(1)/(x^(1)) +(1)/(x^(2)) +(1)/(x^(3))..........oo` `rArr " "y = (1)/(1-(1)/(x)) = (x)/(x-1)` On differentianting , we get `(dy)/(dx) = ((x-1)-x)/((x-1)^(2)) = (1)/((x-1)^(2))` `rArr (dy)/(dx) = -((y)/(x))^(2) = -(y^(2))/(x^(2))" "[because (y)/(x) =(1)/(X-1)]`

Discussion

12.	`int_(1)^(x) (log(x^(2)))/(x)` dx is equal toA. `(log x)^(2)`B. `(1)/(2) (log x)^(2)`C. `(log x^(2))/(2) `D. None of these
Answer» Correct Answer - A Let ` I = int_(1)^(x) (log (x^(2)))/(x)dx` ` = 2 int_(1)^(x) log x (1)/(x)dx` Put `log x = t rArr (1)/(x) dx = dt` ` therefore 2 int_(0)^(log x) t dt = 2[(t^(2))/(2)]_(0)^( log x)` `= (log x)^(2)`

Discussion

13.	If in a trial the probability of success is twice the probability of failure. In six trials the probability of at least four successes isA. `(496)/(729)`B. `(400)/(729)`C. `(500)/(729)`D. `(600)/(729)`
Answer» Correct Answer - A Let the probability of success and failur are p and q respectively. `therefore p = 2 q and p + q = 1` `rArr 3q = 1 rArr q = (1)/(3) and p =(2)/(3)` `therefore ` Required probability `=""^(6)C_(4)((2)/(3))^(4) ((1)/(3))^(2) = ""^(6)C_(5)((2)/(3))^(5) ((1)/(3)) + ""^(6)C_(6)((2)/(3))^(6)` `=(240)/(729) + (192)/(729)+ (64)/(729) = (496)/(729)`

Discussion

14.	`int {(1+2 tan x(tan x + sec x))^(1//2)` dx is equal toA. `log (sec x + tan x)+c`B. `log (sec x + tanx)^(1//2) + c`C. `log sec x (sec x + tanx) + c`D. None of these
Answer» Correct Answer - C Let ` I = int{1 +2 tan x (tan x + sec x)}^(1//2) dx` ` = int{(1+2 tan^(2) x + 2 tan x sec x}^(1//2) dx` ` = int{sec^(2) x + tan^(2) x + 2 tan x secx}^(1//2) dx` `= int (sec x tan x) dx ` `log ( sec x + tan x) + log sec x + c` `log sec x (sec x + tan x) + c`

Discussion

15.	If a is perpedicular to b and `c\|a\| = 2, \|b\| = 3 \|c\| = 4` and the angle between b and c is `(2pi)/(3)` , then `[a,b,c]` is equal toA. `4sqrt(3)`B. `6sqrt(3)`C. `12sqrt(3)`D. `18sqrt(3)`
Answer» Correct Answer - C `therefore [a,b,c] = a(\|b\|\|c\| sin .(2pi)/(3)hatn)` `=\|a\|\|b\|\|c\|(sin.(2pi)/(3))` `[because a hatn =\|a\|\|hatn\| cos 0^(@) =\|a\|]` ` = 2xx 3 xx 4 xx (sqrt(3))/(2)` ` = 12 sqrt(2)`

Discussion

16.	The equations `x + 2y + 3z = 1,2x + y + 3x = 2, 5x + 5y + 9 z = 5` haveA. unique solutionB. infinite many solutionC. inconsistentD. None of these
Answer» Correct Answer - A The given system of equatins in the matrix form are written as below. `[{:(1,2,3),(2,1,3),(5,5,9):}] [{:(,x),(,y),(,z):}] = [{:(,1),(,2),(,3):}]` Let `A = [{:(,1,2,3),(,2,1,3),(,5,5,9):}]` Now , `\|A\| = \|{:(,1,2,3),(,2,1,3),(,5,5,9):}\|` ` = 1(9-15)-2(18-15) + 3(10-15)` ` = 3ne 0` When know that if `\|A\| ne 0` , then the system of equation has unique solution.

Discussion

17.	The angle made by the vector `vecA=2hati+3hatj` with Y-axis isA. `tan^(-1)(3//2)`B. `tan^(-1)(2//3)`C. `sin^(-1)(2//3)`D. `cos^(-1)(3//2)`
Answer» Correct Answer - B Angle with y-axis `rArr tantheta=(x-"comp")/(y-comp")=(2)/(3)` `rArr theta=tan^(-1)((2)/(3))`

Discussion

18.	A bag contains a total of 120 coins in the denominations of Rs. 0.50 and Rs. 1. Find the number of Rs. 0.50 coins in the bag if the total value of the coin is Rs. 100.1. 502. 403. 204. 305. None
Answer» Correct Answer - Option 2 : 40 Calculation: Let the number of 50 p coins be x. ⇒ Total value of coins = Rs. [(50x)/100 + 1 × (120 -x)] ⇒ (50x)/100 + 120 – x = 100 ⇒ x = 40 ∴ The Required result will be 40.

18.

A bag contains a total of 120 coins in the denominations of Rs. 0.50 and Rs. 1. Find the number of Rs. 0.50 coins in the bag if the total value of the coin is Rs. 100.1. 502. 403. 204. 305. None

Answer» Correct Answer - Option 2 : 40

Calculation:

Let the number of 50 p coins be x.

⇒ Total value of coins = Rs. [(50x)/100 + 1 × (120 -x)]

⇒ (50x)/100 + 120 – x = 100

⇒ x = 40

∴ The Required result will be 40.

Discussion

19.	If `x=(1-t^(2))/(1+t^(2))` and `y=(2t)/(1+t^(2))`, then `(dy)/(dx)` is equal toA. `-(y)/(x)`B. `(y)/(x)`C. `-(x)/(y)`D. `(x)/(y)`
Answer» Correct Answer - C Given ` x = (1-t^(2))/(1+t^(2))` and ` y = (2t)/(1+t^(2))` Put `t = tan theta` in both the equations, we get `x = (1- tan^(2) theta)/(1+ tan^(2) theta) = cos 2 theta " ".....(i)` and ` y = (2 tan theta)/(1+tan^(2) theta) = sin 2 theta " ".....(ii)` On differenting both the Eqs. (i) and (ii) , we get `(dx)/(d theta) = - 2 sin 2 theta` and `(dy)/(d theta) = 2 cos 2 theta` `therefore " "(dy)/(dx) = ((dy)/(d theta))/((dx)/(d theta))` `= - (cos 2 theta)/(sin 2 theta) = - (x)/(y)`

Discussion

20.	The side of a square is increasing at ther rate of `0.2 cm//s`. The rate of increase of perimeter w.r.t. time is :A. 0.2 cm/sB. 0.4 cm/sC. 0.6 cm/sD. 0.8 cm/s
Answer» Correct Answer - D If side=a then rate of increase of perimeter w.r.t. time `-4((da)/(dt))-4(0.2)=0.8cm//s`

Discussion

21.	Let a,b,c be three vectors such that `a ne 0` and `a xx b = 2a xx c,\|a\| = \|c\| = 1, \|b\| = 4 and \|b xx c\| = sqrt(15)`. If `b - 2 c = lambda a,` then `lambda` is equal otA. 1B. `+4`C. 3D. `-2`
Answer» Correct Answer - B Given that `\|a\| = \|c\| = 1` and `\|b\| = 4` Let angle between b and c is `alpha` then `\|b xx c\| = sqrt(15)` `rArr " "\|b\|\|c\| sin alpha = sqrt(15)` `rArr sin alpha (sqrt(15))/(4 xx1) = (sqrt(15))/(4)` `therefore cos alpha = sqrt(1-sin^(2) alpha) =(1)/(4)` We have , `b - 2c = lambda a` On squaring both sides, we get `(b-2c)^(2) = lambda^(2)(a)^(2)` `rArr " " b^(2) + 4c^(2) - 4bc = lambda^(2)a^(2)` `rArr 16+4 - 4 \|b\|\|c\| cos alpha = lambda^(2)` `rArr 16 + 4-4 xx 4 xx 1 xx (1)/(4) = lambda^(2)` `rArr" "lambda^(2) = 16` `rArr" "lambda = pm 4`

Discussion

22.	`lim_(x to (pi)/(2))(a^(cot x) -a^(cosx))/(cot x- cot x )` is equalt oA. `log a`B. `log 2`C. aD. log x
Answer» Correct Answer - A `underset(x to (x)/(2))(lim)((a^(cotx)-a^(cos x))/(cot x - cot x))` `=underset(x to (pi)/(2))(lim) a^(cosx)((a^(cot x - cos x) -1)/(cotx - cosx))` `=a^(cos.(pi)/(2))underset(x to (pi)/(2))(lim)(because underset(x to0)(lim) (a^(x)-1)/(x) = log a)` `= 1 log a = loga( because underset(x to0)(lim) (a^(x)-1)/(x) = log a)`

Discussion

23.	If velocity of a particle is given by `v=(2t+3)m//s`. Then average velocity in interval `0letle1` s is :A. `(7)/(2)m//s`B. `(9)/(2) m//s`C. `4 m//s`D. `5 m//s`
Answer» Correct Answer - C `v_(av)=(intvdt)/(intdt)=(int_(0)^(1)(2t+3)dt)/(int_(0)^(2)dt)=((t^(2)+3t)_(0)^(1))/((t)_(0)^(1))=4m//s`

Discussion

24.	If velocity of a particle is given by `v=(2t+3)m//s`. Then average velocity in interval `0letle1` s is :A. `(7)/(2) m//s `B. `(9)/(2) m//s `C. `4 m//s`D. `5 m//s`
Answer» Correct Answer - 3 `v_("av")= (intvdt)/(intdt) = (overset(1)underset(0) int (2t+ 3) dt)/( overset1 underset0 int dt) = ((t^(2) + 3t)_(0)^(1))/((t)_(0)^(1))= 4m//s `

Discussion

25.	He says he will go elsewhere if we don’t lower our price but I don’t think he will. I think we should ______. A) call his bluff B) call it quits C) called it a day D) calls the shots E) chicken
Answer» Correct option is A) call his bluff

Discussion

26.	Consider the inequalities `5x_(1) + 4x_(2) ge 9,x_(1) x_(2) ge 3`, `x_(1) gt 0 , x_(2) gt 0`. Which of the following point does not lie inside the solution set ?A. `(1,3)`B. `(1,2)`C. `(1,4)`D. `(1,1)`
Answer» Correct Answer - D Since (1,1) is not satisfyng the given lie inside the solution set.

Discussion

27.	The value of `[100(x-1)]` is where [x] is the greatest integer less than or equal to x and `x=(sum_(n=1)^44 cos n^@)/(sum_(n=1)^44 sin n^@)`
Answer» Correct Answer - 2 `x=(underset(n=1)overset(44)(.sum)cosn^(0))/(underset(n=1)overset(44)(.sum)sinn^(0))=(cos1^(0)+cos2^(0)+"...."cos44^(0))/(sin1^(0)+sin2^(0)+"...."sin44^(0))` `((sin((44)/(2))^(0))/(sin((1)/(2))^(0))cos((1^(0)+44^(0))/(2)))/((sin((44)/(2))^(0))/(sin((1)/(2))^(0))sin((1^(0)+44^(0))/(2)))` `rArr cot((45^(@))/(2))rArr cot(22(1^(@))/(2))` `rArr x=(sqrt2+1)=2.414`

Discussion

28.	The total capacity of a tank is 200 litres. At present it is 60% full with water. Some water is taken from the tank and it is found the tank is now 35% full. Find the quantity of water taken
Answer» 78 liters of water

Discussion

29.	Find the loss or gain as percent,if the c.p of 8 articles,all of the same kind,is equal to s.p of 10 articles
Answer» C.P. of 8 articles = S.P. of 10 articles = ₹ 80 (suppose) ∴ C.P. of 1 article = 80/8 = ₹10 And S.P. of 1 article = 80/10 = ₹8 ∴ Loss = C.P. – S.P. = ₹10 - ₹8 = ₹2 Loss% = (Loss × 100)/C.P. = (2 × 100)/10 = 20%

29.

Find the loss or gain as percent,if the c.p of 8 articles,all of the same kind,is equal to s.p of 10 articles

Answer»

C.P. of 8 articles = S.P. of 10 articles = ₹ 80 (suppose)

∴ C.P. of 1 article = 80/8

= ₹10

And S.P. of 1 article = 80/10 = ₹8

∴ Loss = C.P. – S.P. = ₹10 - ₹8

= ₹2

Loss% = (Loss × 100)/C.P.

= (2 × 100)/10

= 20%

Discussion

30.	A lens of power `16D` is used as a simple microscope. In order to obtain maximum magnification, at what distance from the lensA. `5 cm`B. `10 cm`C. `16 cm`D. `25 cm`
Answer» Correct Answer - A

Discussion

31.	Solve (D3 − 3D2 + 3D − 1)y = 0
Answer» Given: (D³ − 3D² + 3D − 1)y = 0 The auxiliary equation is m³ - 3m² + 3m - 1 = 0. (m - 1)³ = 0 m = 1, 1, 1. The general solution is given by y = C.F y = e^x[A + Bx + Cx²]

31.

Solve (D3 − 3D2 + 3D − 1)y = 0

Answer»

Given: (D³ − 3D² + 3D − 1)y = 0

The auxiliary equation is m³ - 3m² + 3m - 1 = 0.

(m - 1)³ = 0

m = 1, 1, 1.

The general solution is given by y = C.F

y = e^x[A + Bx + Cx²]

Discussion

32.	why NaHCO3 & NaOH can't exist together in a solution:
Answer» NaHCO₃ and NaOH both undergo chemical reaction and therefore cannot exist together in solution. NaHCO₃ is an acidic salt while NaOH is a base, therefore the two reacts. The reaction is shown below : NaHCO3 + NaOH →Na2CO3 + H2O

Discussion

33.	Zn amalgam is prepared by elctrtolysis of aqueous `ZnCI_(2)` using 9 gram Hg cathode how much current is to be pased through `ZnCI_(2)` solution for 1000 seconds to prepare a Zn amalagam with 25% by weight ? (atomic masss Zn =65.4 g)A. Current off `8.85` amp is passed in the process.B. Current of `5.65` amp is passed in the processC. Mass of `Zn` in Amalgam is `3gm`D. Mass of `Zn` in Amalgam is `6gm`.
Answer» Correct Answer - A::C::D Let `x` gm of `Zn` deposit of `9gm` of `Hg` `%` of `Zn` in Amalgam`=(m_(Zn)xx100)/(m_(Zn)xxm_(Hg))=x/(9+x)xx100=25` `:.x=3gm` Eq. of `Zn=(3xx2)/(65.4)` `(n_(Zn))/1=(n_(e^(-)))/2=Q/(2f)=("it")/(2f)` Current `i=6/(65.4)xx96500/1000=8.85amp.`

Discussion

34.	What is the volume of water consumed during acid hydrolysis of 1.368 kg of surose ?A. 0.072 `dm^(3)`B. 0.0720 `dm^(3)`C. 0.18 `dm^(3)`D. 0.018 `dm^(3)`
Answer» Correct Answer - a `"Sucrose" +H_(2)OrarrC_(6)H_(12)O_(6)+c_(6)H_(12)O_(6)` `because` 342 g sucrose consume =18 g `H_(2)O` `therefore` 1 g sucrose consume `=(18)/(342) g H_(2)O` `therefore 1.368 xx1000g` sucrose consume `=18/342xx1.368xx1000=72 g H_(2)O` Now given density of `H_(2)O=1 g/cm^(3)` and we know `d=(m)/(v)` `v=(m)/(d)=(72g)/(1g//cm^(3))=72 cm^(3)=0.072 dm^(3)`

Discussion

35.	Which of the following compounds will not yield iodoform on heating with iodine and dilute NaOH?A. B. C. D.
Answer» Correct Answer - 1 2-alkanal/ methyl ketone gives +ve test with `I_(2)/NaOH` (iodoform test)

Discussion

36.	A. ferrous sulphateB. Copper sulphateC.Magnesium sulphateD.Sodium sulphate
Answer» Copper Sulfate does not exist as hydrate

Discussion

37.	Select a ferromagnetic material from the followingA. dioxygenB. chromium (IV) oxideC. benzeneD. dihydrogen monoxide
Answer» Correct Answer - b Substacnes which show permanent magnetism even in the absence of the magenetic field are called ferromagenitic substances e.g Fe ,Ni Co Gd and `CrO_(2)`

Discussion

38.	The correct order of the acidic nature of oxides is in the orderA. `N_(2)O_(5) lt N_(2)O_(3)lt NO_(2) lt N_(2)O`B. `N_(2)O lt NO lt N_(2)O_(3) lt NO_(2) lt N_(2)O_(5)`C. `N_(2)O_(5) lt N_(2) OltN_(2)O_(3) lt NO lt NO_(2)`D. `NO lt N_(2)O lt N_(2) O_(3) lt NO_(2) lt N_(2)O_(5)`
Answer» Correct Answer - 2 As oxidation state of central atom increases acidic strenght increases.

Discussion

39.	Which of the following carbides gives propyne on hydrolysis?A. `CaC_(2)`B. `Be_(2)C`C. `MgC_(2)`D. `Mg_(2)C_(3)`
Answer» Correct Answer - 3 `Mg_(2)C_(3) + 4H_(2)O to 2 Mg(OH)_(2) + CH_(3) - C-= CH`

Discussion

40.	A.CaCl2B.CaBr2C.CaI2D.CaF2
Answer» Correct Answer is CaI₂ As covalent character increases, melting point decreases, thus order of melting point is CaF2 > CaCl2 > CaBr2 > CaI2

40.

A.CaCl2B.CaBr2C.CaI2D.CaF2

Answer»

Correct Answer is CaI₂

As covalent character increases, melting point decreases, thus order of melting point is

CaF2 > CaCl2 > CaBr2 > CaI2

Discussion

41.	For the reaction `O_(3)(g)+O(g)rarr2O_(2)(g)` if the rate law expression is rate =`K[O_(3)][O]` eh molecularity and order of the reaction respectively areA. 2 and 2B. 2 and 133C. 2 and 1D. 1 and 2
Answer» Correct Answer - a Given reaction is `O_(3)(g)+O(g) rarr2O_()(g)` also rate law expression is given as `rate = k[O_(3))] [O]` now since molecularity is simply the sum of reactant molecules in a single step of a chemical reaction so molecularity is two for this reaction on the other hand order is the sum of the pwoers raised to the concentration of eractant molecules from eq (i) order is 1+1 =2 `therefore` the molecularity is 2 and order is also 2

Discussion

42.	How many isomeric monochloro derivatives are possible for n-butane
Answer» In case of n-butane (CH3−CH2−CH2−CH3)CH3CH2CH2CH3 are two isomers will be obtained depending on whether clcl atom adds on to carbon-2 or carbon -1

Discussion

43.	Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.
Answer» The heat of neutralization of strong acid and strong base is 57.0 KJ/mol. The rest released when 0.5 moles at the nitric acid solution mixed with 0.2 moles of potassium hydroxide solution. HNO₃ and KOH are strong acid and strong base respectively. 0.2 mole KOH neutralized only 0.2 moles of HNO₃ 0.3 mole HNO₃remain unreacted. \(\because\) Heat of neutralization of strong acid and strong base is = 57.0 KJ/mol. Heat released on neutralization of 0.2 mol HNO₃ and 0.2 mol KOH = 57.0 KJ/mol x 0.2 mol = 11.4 KJ Hence, heat released = 11.4 KJ.

43.

Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.

Answer»

The heat of neutralization of strong acid and strong base is 57.0 KJ/mol. The rest released when 0.5 moles at the nitric acid solution mixed with 0.2 moles of potassium hydroxide solution.

HNO₃ and KOH are strong acid and strong base respectively.

0.2 mole KOH neutralized only 0.2 moles of HNO₃

0.3 mole HNO₃remain unreacted.

\(\because\) Heat of neutralization of strong acid and strong base is = 57.0 KJ/mol.

Heat released on neutralization of 0.2 mol HNO₃ and 0.2 mol KOH = 57.0 KJ/mol x 0.2 mol

= 11.4 KJ

Hence, heat released = 11.4 KJ.

Discussion

44.	1. Which one has greater size, Na or K? 2. Justify your answer.
Answer» 1. K 2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

44.

1. Which one has greater size, Na or K? 2. Justify your answer.

Answer»

1. K

2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

Discussion

45.	A. 15B. 16C. 17D. All
Answer» Answer is group 15

Discussion

46.	(& how is H2O2 related to it?)
Answer» Reaction mechanism is a sequence of single reaction steps that sum to the overall equation. For example, a possible mechanism for the overall reaction 2A + B → E + F might involve these three simpler steps: (1) A + B → C (2) C + A → D (3) D → E + F Adding the steps and canceling common substances gives the overall equation: A + B + C + A + D → C + D + E + F or 2A + B → E + F A mechanism is a hypothesis about how a reaction occurs; chemists propose a mechanism and then test to see that it fits with the observed rate law. Key points The reaction mechanism describes the sequence of elementary reactions that must occur to go from reactants to products. Reaction intermediates are formed in one step and then consumed in a later step of the reaction mechanism. The slowest step in the mechanism is called the rate determining or rate-limiting step. The overall reaction rate is determined by the rates of the steps up to (and including) the rate determining step. The rapid evolution of oxygen gas is produced by the following reaction: 2 H₂O₂ (aq) = 2 H₂O (l) + O₂ (g) + heat The decomposition of hydrogen peroxide in the presence of iodide ion occurs in two steps: H₂O₂ (aq) + I- (aq) = H₂O (l) + OI^- (aq) H₂O₂ (aq) + OI- (aq) = H₂O (l) + O₂ (g) + I^- (aq)

46.

(& how is H2O2 related to it?)

Answer»

Reaction mechanism is a sequence of single reaction steps that sum to the overall equation. For example, a possible mechanism for the overall reaction
2A + B → E + F
might involve these three simpler steps:
(1) A + B → C
(2) C + A → D
(3) D → E + F
Adding the steps and canceling common substances gives the overall equation:
A + B + C + A + D → C + D + E + F or 2A + B → E + F
A mechanism is a hypothesis about how a reaction occurs; chemists propose a mechanism and then test to see that it fits with the observed rate law.

Key points

The reaction mechanism describes the sequence of elementary reactions that must occur to go from reactants to products.
Reaction intermediates are formed in one step and then consumed in a later step of the reaction mechanism.
The slowest step in the mechanism is called the rate determining or rate-limiting step.
The overall reaction rate is determined by the rates of the steps up to (and including) the rate determining step.

The rapid evolution of oxygen gas is produced by the following reaction:

2 H₂O₂ (aq) = 2 H₂O (l) + O₂ (g) + heat

The decomposition of hydrogen peroxide in the presence of iodide ion occurs in two steps:

H₂O₂ (aq) + I- (aq) = H₂O (l) + OI^- (aq)
H₂O₂ (aq) + OI- (aq) = H₂O (l) + O₂ (g) + I^- (aq)

Discussion

47.	Match the following: Sodium f-block Oxygen s-block Uranium d-block Silver p-block
Answer» Sodium – s-block Oxygen – p-block Uranium – f-block Silver – d-block

47.

Match the following: Sodium f-block Oxygen s-block Uranium d-block Silver p-block

Answer»

Sodium – s-block

Oxygen – p-block

Uranium – f-block

Silver – d-block

Discussion

48.	Calculate the pH at \(\cfrac{M}{1000}\) sodium hydroxide solution assuming complete ionisation (Kw = 1 x 10-14).
Answer» NaOH monobasic so that, normality = molarity \(\therefore\) concentration of NaOH = concentration of \(\bar O\)H [\(\bar O\)H] = [NaOH] = \(\cfrac{M}{1000}\) = 0.001 M = 0.001 M P^OH = - log[\(\bar O\)H] = - log 0.001 = - log 10^-4 = 4 log 10 P^OH= 4 We know that P^H+ P^OH = 14 P^H = 14 - P^OH = 14 - 4 P^H = 10 Hence, P^H at \(\cfrac{M}{1000}\) sodium hydroxide solution will be 10

48.

Calculate the pH at \(\cfrac{M}{1000}\) sodium hydroxide solution assuming complete ionisation (Kw = 1 x 10-14).

Answer»

NaOH monobasic so that, normality = molarity

\(\therefore\) concentration of NaOH = concentration of \(\bar O\)H

[\(\bar O\)H] = [NaOH]

= \(\cfrac{M}{1000}\) = 0.001 M = 0.001 M

P^OH = - log[\(\bar O\)H]

= - log 0.001

= - log 10^-4

= 4 log 10

P^OH= 4

We know that

P^H+ P^OH = 14

P^H = 14 - P^OH

= 14 - 4

P^H = 10

Hence, P^H at \(\cfrac{M}{1000}\) sodium hydroxide solution will be 10

Discussion

49.	Write short notes- a)Banana bond b)Boric acid
Answer» (a) A banana bond is also known as a bent bond. It is a type of chemical bonding where the ordinary hybridization state of two atoms making up a chemical bond are modified with increased or decreased s-orbital character in order to accommodate a particular molecular geometry (that of 3 bananas in a ring shape). (b) Boric acid is low in toxicity if eaten or if it contacts skin. However, in the form of borax, it can be corrosive to the eye. Borax can also be irritating to the skin. People who have eaten boric acid have had nausea, vomiting, stomach aches, and diarrhea.

49.

Write short notes- a)Banana bond b)Boric acid

Answer»

(a) A banana bond is also known as a bent bond. It is a type of chemical bonding where the ordinary hybridization state of two atoms making up a chemical bond are modified with increased or decreased s-orbital character in order to accommodate a particular molecular geometry (that of 3 bananas in a ring shape).

(b) Boric acid is low in toxicity if eaten or if it contacts skin. However, in the form of borax, it can be corrosive to the eye. Borax can also be irritating to the skin. People who have eaten boric acid have had nausea, vomiting, stomach aches, and diarrhea.

Discussion

50.	2. Write IUPAC names of the following compounds:(a) \( CH _{3} CH = C \left( CH _{3}\right)_{2} \)(b) \( CH _{2}= CH - C \equiv C - CH _{3} \)(c)(d)(e)\( (f) \)\( (g) \)
Answer» (a) IUPAC name: 2-Methylbut-2-ene (b) IUPAC name: Pen-1-ene-3-yne

50.

2. Write IUPAC names of the following compounds:(a) \( CH _{3} CH = C \left( CH _{3}\right)_{2} \)(b) \( CH _{2}= CH - C \equiv C - CH _{3} \)(c)(d)(e)\( (f) \)\( (g) \)

Answer»

(a) IUPAC name: 2-Methylbut-2-ene

(b) IUPAC name: Pen-1-ene-3-yne

Discussion

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why NaHCO3 &amp; NaOH can't exist together in a solution:

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How many isomeric monochloro derivatives are possible for n-butane

Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.

1. Which one has greater size, Na or K? 2. Justify your answer.

A. 15B. 16C. 17D. All

(&amp; how is H2O2 related to it?)

Key points

Match the following: Sodium f-block Oxygen s-block Uranium d-block Silver p-block

Calculate the pH at \(\cfrac{M}{1000}\) sodium hydroxide solution assuming complete ionisation (Kw = 1 x 10-14).

Write short notes- a)Banana bond b)Boric acid

2. Write IUPAC names of the following compounds:(a) \( CH _{3} CH = C \left( CH _{3}\right)_{2} \)(b) \( CH _{2}= CH - C \equiv C - CH _{3} \)(c)(d)(e)\( (f) \)\( (g) \)

why NaHCO3 & NaOH can't exist together in a solution:

(& how is H2O2 related to it?)