100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Write about three menstrual disorders
Answer» Menstrual disorders include: Dysmenorrhea refers to painful cramps during menstruation. Premenstrual syndrome refers to physical and psychological symptoms occurring prior to menstruation. Menorrhagia is heavy bleeding, including prolonged menstrual periods or excessive bleeding during a normal-length period.

Discussion

2.	Nature Definition and meaning
Answer» Nature is all the animals, plants, and other things in the world that are not made by people, and all the events and processes that are not caused by people.

Discussion

3.	A wooden block slides more easily on smooth cemented floor then on rollers true or false
Answer» (c) False Rolling friction is less than the sliding friction.

3.

A wooden block slides more easily on smooth cemented floor then on rollers true or false

Answer»

(c) False

Rolling friction is less than the sliding friction.

Discussion

4.	Two strings `A` and `B` are connected together end to end as shown in the figure. The ratio of mass per unit length `(mu_(B))/(mu_(A))=4`. The tension in the string is same. A travelling wave is coming from the string A towards string B. if the fraction of the power of the incident wave that goes in sting B is `(n)/(9)` the value of n is:
Answer» Correct Answer - 8 `A_(T)=(2V_(B))/((V_(A)+V_(B)))A_(1)=(2sqrt((T)/(m_(B))))/(sqrt((T)/(mu_(A)))+sqrt((T)/(mu_(B))))A_(1)=(2)/(3)A_(1)` power `P=(1)/(2)muvomega^(2)A^(2)=(1)/(2)sqrt(T_(epsilon)omega^(2)A^(2)` `(P_(t))/(P_(i))=(At^(2))/(Ai^(2))sqrt((mu_(B))/(mu_(A)))=(8)/(9)`

Discussion

5.	Two strings `A` and `B` are connected together end to end as shown in the figure. The ratio of mass per unit length `(mu_(B))/(mu_(A))=4`. The tension in the string is same. A travelling wave is coming from the string A towards string B. if the fraction of the power of the incident wave that goes in sting B is `(n)/(9)` the value of n is:
Answer» Correct Answer - 8 `A_(T)=(2V_(B))/((V_(A)+V_(B))) A_(1)=(2sqrt(T/(m_(B))))/(sqrt(T/(mu_(A)))+sqrt(T/(mu_(B))))A_(1)=2/3A_(1)` `(P_(t))/(P_(i))=(At^(2))/(Ai^(2)) sqrt((mu_(B))/(mu_(A)))=8/9`

Discussion

6.	When the voltage applied to an X-ray tube increased from `V_(1)=15.5kV` to `V_(2)=31kV` the wavelength interval between the `K_(alpha)` line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of `1.3`. If te atomic number of the element of the target is z. Then the value of `(z)/(13)` will be: (take `hc=1240eVnm` and `R=1xx(10^(7))/(m))`
Answer» Correct Answer - 2 `lamda_(th)=(hc)/(eV_(a))` `1/(lamda_(Kalpha))=R(z-1)^(2)(1/(1^(2))-1/(2)^(2))` `13/10 lamda_(k_(alpha))=(13/10-1/2)_(lamda_(th)` `3/10((4xx10^(-7))/(3(z_(7))^(2)))=(8/10)(12.4xx10^(-7))/(15.5xx10^(3))implies5000/8` `(z-1)^(2)` `625=(z-1)^(2)impliesz=26`

Discussion

7.	The `SI` unit of inductance the Henry can not be written as :A. weber/ampereB. volt-second/ampereC. `"joule"//("ampere")^(2)`D. ohm/second
Answer» Correct Answer - D (A) `L = (phi)/(i)` or Henry `= ("Weber")/("Ampere")` (B) `e = -L ((di)/(dt)) :. L = (e)/(di//dt)` or Henry `=("Volt - second")/("Ampere")` (C) `U = (1)/(2) Li^(2) :. L = (e)/(di//dt)` or Henry `= ("Joule")/(("Ampere")^(2))` (D) `U = (1)/(2) Li^(2) = i^(2) Rt :. [L] = [Rt]` or Henry = ohm-second.

Discussion

8.	Two thin rods of same length `l` but of different uniform mass per unit length `mu_(1)` and `mu_(2)` respectively are joined together. The system is ortated on smooth horizontal plane as shown in figure. The tension at the joint will be A. `3/2 mu_(2)l^(2)omega^(2)`B. `3/2(mu_(1)+mu_(2))l^(2)omega^(2)`C. `3/2 mu_(1)l^(2)omega^(2)`D. `1/2mu_(1)l^(2)omega^(2)`
Answer» Correct Answer - A The tension at joint is due to force exerted by the root of linear density `mu_(2)`. So `F=int_(l)^(2l) mu_(2) dxomega^(2)x=(3mu_(2) omega^(2)l^(2))/2`

Discussion

9.	In the given circuit, the valuue of `R` so that thermal power generated in `R` will be maximum is: A. `10Omega`B. `12Omega`C. `8Omega`D. `6Omega`
Answer» Correct Answer - B For maximum power, `R=r_(eq)=(20xx30)/(20+30)=12Omega`

Discussion

10.	In the given circuit, the valuue of `R` so that thermal power generated in `R` will be maximum is: A. `10Omega`B. `12Omega`C. `8Omega`D. `6Omega`
Answer» Correct Answer - B For maximum power, `R=r_(eq)=(20xx30)/(20+30)=12Omega`

Discussion

11.	A closed organ pipe of length L is vibrating in its first overtone there is a point Q inside the pipe at a distance `7L//9` form the open end the ratio of pressure amplitude at Q to the maximum pressure amplitude in the pipe isA. `1:2`B. `2:1`C. `1:1`D. `2:3`
Answer» Correct Answer - A `DeltaPm=2DeltaP_(0) coskx` (assuming closed end as origin) At point `Q,x=L-(7L)/9=(2L)/9` `DeltaPm=2DeltaP_(0)cos((2pi)/(lamda)xx(2L)/9)=DeltaP_(0)` `:.` Required ratio `=1:2`

Discussion

12.	The shape of a wave propagating in the positive x or negative x-direction is given `y=1/sqrt(1+x^(2))` at t=0 and `y=1/sqrt(2-2x+x^(2))` at t=1s where x and y are in meters the shape the wave disturbance does not change during propagation find the velocity of the waveA. `1m//s` in positive x directionB. `1m//s` in negative x directionC. `1/2 m//s` in positive x directionD. `1/2 m//s` in negative x direction
Answer» Correct Answer - A `y=f(xpmc.t)` is the general wave equation `at t=0y=f(x) Rightarrowy=1/sqrt(1+x^(2))` `y=1/sqrt(2-2x+x^(2)) =1/sqrt(1+(x-1)^(2)) = f(x-1) Rightarrow f(x-ct)=f(x-1)at t=1 Rightarrowc=1m//s]`

Discussion

13.	A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is_____Hz.
Answer» f₁ = f f₂ = f + 4 f₃ = f + 2 × 4 f₄ = f + 3 × 4 f₂₀ = f + 19 × 4 f + (19 × 4) = 2 × f f = 76 Hz. Frequency of last tuning forks = 2f = 152 Hz

13.

A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is_____Hz.

Answer»

f₁ = f

f₂ = f + 4

f₃ = f + 2 × 4

f₄ = f + 3 × 4

f₂₀ = f + 19 × 4

f + (19 × 4) = 2 × f

f = 76 Hz.

Frequency of last tuning forks = 2f

= 152 Hz

Discussion

14.	Write precautions to be taken while a clinical thermometer and laboratory thermometer
Answer» The precautions needed while reading a laboratory thermometer are: 1. It should be kept upright not tilted. 2. Bulb should be surrounded from all sides by the substance of which the temperature is to be measured. 3. The bulb should not touch the surface of the container. 1. Precautions to be taken while using a clinical thermometer 2. Thermometer should be washed before and after each use, preferably with an antiseptic solution 3. do not wash the thermometer with hot water 4. ensure that before use the mercury level is below 35^oC 5. Read the thermometer keeping the level of mercury along the line of sight. 6. Handle the thermometer with care, as you handle any glass articles 7. Don't hold the thermometer by the bulb while reading it.

14.

Write precautions to be taken while a clinical thermometer and laboratory thermometer

Answer»

The precautions needed while reading a laboratory thermometer are:

1. It should be kept upright not tilted.

2. Bulb should be surrounded from all sides by the substance of which the temperature is to be measured.

3. The bulb should not touch the surface of the container.

1. Precautions to be taken while using a clinical thermometer
2. Thermometer should be washed before and after each use, preferably with an antiseptic solution
3. do not wash the thermometer with hot water
4. ensure that before use the mercury level is below 35^oC
5. Read the thermometer keeping the level of mercury along the line of sight.
6. Handle the thermometer with care, as you handle any glass articles
7. Don't hold the thermometer by the bulb while reading it.

Discussion

15.	A dipole is placed in xy plane parallel to the line `y = 2x`. There exists a uniform electric field along z-axis. Net force acting on the dipole will be zero. But it can experience some torque. We can show that the direction of this torque will be parallel to the line.A. `y = 2x + 1`B. `y = -2x `C. `y = -1/2x`D. `y = -1/2 x + 2`
Answer» Correct Answer - C::D c.,d. Torque will be perpendicualr to the line `y=2x` and it should be in `xy` plance, because electric field is in `z-`directio. The lines in options (c) and (d) are perpendicular to `y=2x.

Discussion

16.	Two plates of a parallel plate capacitor carry charges q and -q and are separated by a distance a from each other. The capacitor is connected to a constant voltage source `V_0`. The distance between the plates is changed to `x+dx`. Then in steady state. A. change in electrostatic energy stored in the capacitor is `-Udx//x` .B. change in electrostatic energy in the capacitor is `-Udx//dx`C. attraction force between the plates is `1//2 qE`.D. attraction force between the plates is `qE (where E is electic field between the plates)
Answer» Correct Answer - A::C a.,c. Initial stored energy `U_(i)=(1)/(2)(epsilon_(0)AV_(1)^(2))/(x)` Final stored energy `U_(f)=(1)/(2)(epsilon_(0)AV_(0)^(2))/(2(x+dx))` So `DeltaU=U_(f)-U_(i)=(1)/(2)epsilon_(0)AV_(0)^(2)[(1)/(x+dx)(1)/(x)]` `=(1)/(2)epsilon_(0)AV_(0)^(2)[(x-x-dx)/(x(x+dx))]` `=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x^(2))dx=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x)(edx)/(x)=-(Udx)/(x)` So, option (a) is correct and option (b) is incorrect. `F=(dU)/(dx)=(-(V)/(x))` `=(1)/(2)(epsilon_(0)A)/(x)(V_(0)^(2))/(x)=(-(1)/(2))((epsilon_(0)A)/(x)V_(0)) (V_(0))/(x)=-(1)/(2)qE` So magnitude of attractive force is `1//2qE`. So, option (c) is correct and option (d) is incorrect

Discussion

17.	A charge Q is imparted to two identical capacitors in paralle. Separation of the plates in each capacitor is `d_0`. Suddenly, the first plate of the first capacitor and the second plate of the second capacitor start moving to the left with speed v, then A. charges on the two capacitors as a function of time are `(Q(d_0-vt))/(2d_0), (Q(d_0+vt))/(2d_0)`.B. charges on the two capacitors as a function of time are `(Qd_0)/(2(d_0-vt)), (Qd_0)/(2(d_0+vt))`.C. current in the circuit will increase as time passes onD. current in the circuit will be constant.
Answer» Correct Answer - A::D a.,d. Leq `q_(1)` and `q_(2)` be the instantaneous charges on capacitors. Since they are in parallel, then `(q_(1))/(C_(1))=(q_(2))/(C_(2))` and `q_(1)+q_(2)=Q` `C_(1)=(epsilon_(0)A)/(d_(0)+vt),C_(2)=(epsilon_(0)A)/(d_(0)-vt)` So `(q_(1))/(q_(2))=(C_(1))/(C_(2))=(d_(0)-vt)/(d_(0)+vt)` or `q_(2)((d_(0)-vt)/(d_(0)+vt))+q_(2)=0` So `q_(2)=(Q(d_(0)+vt))/(2d_(0))` and `q_(1)=(Q(d_(0)-vt))/(2d_(0))` Hence, option (a) is correct and option (b) is incorrect. `i=(-dq_(1))/(dt)` or `(dq_(2))/(dt)` or `i=(Q_(v))/(2d_(0))` Which does not depend on time. So option (d) is correct and option (c) is incorrect.

Discussion

18.	A hollow conducting sphere of inner radius r and outer radius 2R is given charge Q as shown in figure, then the A. potential at A and B is same.B. potential at O and B is same.C. potential at O and C is same.D. potential at A,B,C and O is same.
Answer» Correct Answer - A::B::C::D a.,b.,c.,d. points `A` and `B` lie within the same metal hence `V_(A)=V_(B)`. The potential inside a hollow sphere is same as potential at the surface, hence `V_(A)=V_(B)=V_(C)=V_(0)`.

Discussion

19.	If `(1 +x+x^2)^n=sum_(r=0)^(2n) a_r x^r` , then prove that `a_r=a_(2n-r)`A. `n+1`B. `r+1`C. `n+r+1`D. `n+r`
Answer» Correct Answer - C Coeff. `X^(3r)` in `[(1+x+x^(2))^(n)(1-x)^(n)]` `implies` coeff. of `x^(3r)` in `[(a_(0)+a_(1)x+…a_(2)x^(2n))(n_(c_(0))-n_(c_(1)) xx……)]` `.^(n)C_(r)`

Discussion

20.	If `a,b,c` are in AP and `A,B,C` are in G.P. (Common ratio `!=1`). Then which of the following is/are correctA. `A/a,B/b,C/c` are in HP if common of GP is `c//a`B. `a/A,b/B,c/C` are in HP if common ratio of GP is equal to common difference of APC. `(A^(2))/a,(B^(2))/b,(C^(2))/c` are in HP if common ratio of GP is `sqrt(c/a)`D. `a/(A^(2)),b/(B^(2)),c/(C^(2))` are in HP if common ratio of GP is equal to square root of common difference of AP.
Answer» Correct Answer - A::B::D `A/a,B/b,C/c HP` `=(2b)/B=a/A+c/C` `implies2bB=aC+cA` `impliesaB+cB=aC+cA` `impliesa[B-C]=c[A-B]` so `r=c/a` `(A^(2))/a,(B^(2))/b,(C^(2))/c` are in `HPimpliesr^(2)=c/a`

Discussion

21.	Lines `L_(1):(x-6)/3=(y-4)/2=z-2` and `L_(2):(x-8)/4=y-2=(z-4)/2` meets the plane `pi:vecr(2hati+hatj-hatk)=7` at points A and B (2)Area of the triangle formed by the lines `L_(1),L_(2)` & `AB` isA. `7/2 sqrt(19/2)`B. `sqrt(19/2)`C. `(sqrt(11))/4`D. `6sqrt(5)`
Answer» Correct Answer - B `A-=(3,2,1), B-=(4,1,2)` point of intersection `-=(0,0,0)` Area of triangle `=1/2\|(3hati+2hatj+hatk)xx(4hati+hatj+2hatk)\|=sqrt(19/2)` Also image of `(0,0,0)` w.r.t plane `pi` is `(14/3, 7/3, (-7)/3)` Image of `(0,0,0)` w.r.t `AB` is `(14/3, 16/3, 2/3)` Equation of line joining these two points `=(3x-14)/0=(3y-7)/3=(3z+7)/3`

Discussion

22.	Lines `L_(1):(x-6)/3=(y-4)/2=z-2` and `L_(2):(x-8)/4=y-2=(z-4)/2` meets the plane `pi:vecr(2hati+hatj-hatk)=7` at points A and B (1) Equation of line joining the images of the point of intersection of `L_(1)` & `L_(2)` with respect to the plane `pi` and the line `AB` isA. `(3x-14)/4=(3y-7)/5=(3z+7)/1`B. `(3x-14)/3=(3y-7)/3=(3z+7)/6`C. `(3x-14)/0=(3y-7)/3=(3z+7)/4`D. `(3x-14)/0=(3y-7)/3=(3z+7)/3`
Answer» Correct Answer - D `A-=(3,2,1), B-=(4,1,2)` point of intersection `-=(0,0,0)` Area of triangle `=1/2\|(3hati+2hatj+hatk)xx(4hati+hatj+2hatk)\|=sqrt(19/2)` Also image of `(0,0,0)` w.r.t plane `pi` is `(14/3, 7/3, (-7)/3)` Image of `(0,0,0)` w.r.t `AB` is `(14/3, 16/3, 2/3)` Equation of line joining these two points `=(3x-14)/0=(3y-7)/3=(3z+7)/3`

Discussion

23.	Let `x_(i) epsilonR,i=1,2,3……….n` are numbers such that `sum_(i=1)^(n)isqrt(x_(i)-i^(2))=(sum_(i=1)^(n)x_(i))/2` and `x_(1)+x_(2)+……….+x_(n)=280` No. of ways of distributioin of `n` identical objects among 3 persons such that each get at least 1 object isA. 4B. 10C. 20D. 140
Answer» Correct Answer - C `sum_(i=1)^(n-1)(x_(i)-2isqrt(x_(i)-i^(2)))=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2)))^(2)-2isqrt(x_(i)-i^(2))+i^(2)=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2))-i)^(2)=0` so, `x_(i)=2i^(2)` Now, `x_(1)^(2)+….+x_(n)^(2)=280` `2[1^(2)+2^(2)+........n^(2)]=280` `n=7` `y_(1)+y_(2)+y_(3)=7` `y_(1)^(1)+y_(2)^(1)+y_(3)^(1)=4` `.^(4+3-1)C_(3)=.^(6)C_(3)=20` Total triangles formed `=.^(15)C_(3)=(15xx14xx13)/6` `N` of isosceles triangles formed `=15xx7` probability `=(15xx7)/(15xx14xx13)xx6` `3/13`

Discussion

24.	The voltage applied to an X-ray tube is 20 kV. The minimum wavelength of X-ray produced, is given by `(31xxn)/50 Å` then n will be (`h=6.62xx10^(-34) Jx,c=3xx10^(8) m//s, e=1.6xx10^(-19)` coulomb):
Answer» `lambda_("min")=(hc)/(eV)=(6.62xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx20xx10^(3))=12375/(20xx10^(-3)) Å =0.62 Å`

Discussion

25.	`N (lt 100)` molecules of a gas have velocities 1,2,3….N km/s respectively. ThenA. rms speed and average speed of molecules is sameB. ratio of rms speed to average speed is `sqrt((2N+1))(N+1)//6N`C. ratio of rms speed to average speed is `sqrt((2N + 1))(N+1)//6`D. ratio of rms speed to average speed of molecules is `sqrt(((2n+1))/(6(N+1)))`
Answer» Correct Answer - D `V_("rms")=sqrt((V_(1)^(2)+V_(2)^(2)+.......V_(N)^(2))/(N))=sqrt((1^(2)+2^(2)+.......+ N^(2))/(N))` `=sqrt((N(N+1)(2N+1))/(6N))rArr V_("rms")=sqrt(((N+1)(2N+1))/(6))` `V_("avg")=(V_(1)+V_(2)+......+V_(N))/(N)` `=(1+2+......+N)/(N)=(N(N+1))/(2N)=(N+1)/(2)` `(V_("rms"))/(V_("avg"))=2 sqrt(((2N+1))/(6(N+1)))`.

Discussion

26.	A horizontal cylinder is fixed, its inner surface is smooth and its radius is R. A small block is initially at the lowest point. The minimum velocity that should by given to the block at the lowest point, so that it can just cross the point P is u then A. If the block moves anti clockwise then `u = sqrt(3.5 gR)`B. If the block moves anti clockwise then `u = sqrt(3 gR)`C. If the block moves clockwise then `u = sqrt(3.5 gR)`D. If the block moves clockwise then `u = sqrt(5 gR)`
Answer» Correct Answer - A::D `N = (mu^(2))/(R) + mg (3 cos theta - 2)`, at `theta = 120^(@) N` `= 0 rArr N = (m u^(2))/(R) + mg (3 cos 120^(@) - 2) =0` `rArr u = sqrt(3.5 gR)` If the block is moving clockwise, then to cross the point P, the block has to cross the highest point, so to cross the highest point `= sqrt(5gR)`.

Discussion

27.	Let `x_(i) epsilonR,i=1,2,3……….n` are numbers such that `sum_(i=1)^(n)isqrt(x_(i)-i^(2))=(sum_(i=1)^(n)x_(i))/2` and `x_(1)+x_(2)+……….+x_(n)=280` Probability that a randomly selected triangle formed by vertices of a `2n+1` sided regular polygon is isosceles isA. `3/13`B. `5/13`C. `7/13`D. `9/13`
Answer» Correct Answer - A `sum_(i=1)^(n-1)(x_(i)-2isqrt(x_(i)-i^(2)))=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2)))^(2)-2isqrt(x_(i)-i^(2))+i^(2)=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2))-i)^(2)=0` so, `x_(i)=2i^(2)` Now, `x_(1)^(2)+….+x_(n)^(2)=280` `2[1^(2)+2^(2)+........n^(2)]=280` `n=7` `y_(1)+y_(2)+y_(3)=7` `y_(1)^(1)+y_(2)^(1)+y_(3)^(1)=4` `.^(4+3-1)C_(3)=.^(6)C_(3)=20` Total triangles formed `=.^(15)C_(3)=(15xx14xx13)/6` `N` of isosceles triangles formed `=15xx7` probability `=(15xx7)/(15xx14xx13)xx6` `3/13`

Discussion

28.	If `A=[(5, -6),(1,-1)]` then the value of `("det"(A^(m)-5A^(m-1)))/("det"(A^(n)-5A^(n-1)))(m, "n" in N)` is
Answer» Correct Answer - 1 `"det"(A^(mu-1)(A-5I))="det"(A^(mu-1))."de"(A-51)`

Discussion

29.	The value of `[int_(-pi)^(pi) sqrt((\|sinx\|)/(1+tan^(2)x))dx]` is (where [.] is greates integer).
Answer» Correct Answer - 2 `I=int_(-pi)^(pi) sqrt((\|sinx\|)/(1+tan^(2)x))dx` `I=2int_(-(pi)/2)^((pi)/2)sqrt((sinx)/(1+cot^(2)x))dx` `4 int_(0)^((pi)/2) sqrt(cosx) sin x dx` `=8/3` `[I]=2`

Discussion

30.	A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 1/2 years. Find the sum and rate of interests.
Answer» S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156. S.I. for 2 years = Rs.(156(2/3)2)=Rs.208 Principal = Rs. (1008 - 208) = Rs. 800. Now, P = 800, T = 2 and S.I. = 208. Rate =(100* 208)/(800*2)% = 13%

30.

A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 1/2 years. Find the sum and rate of interests.

Answer»

S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.

S.I. for 2 years = Rs.(156*(2/3)*2)=Rs.208

Principal = Rs. (1008 - 208) = Rs. 800.

Now, P = 800, T = 2 and S.I. = 208.

Rate =(100* 208)/(800*2)% = 13%

Discussion

31.	Find the equation of the locus of the point which is at a constant distance of 5 units from the fixed point (-2,3).
Answer» Let A = (-2,3) and let P = (x,y). Now AP = 5 (AP)² = 25 (x + 2)² + (y - 3)² = 25 x² + y² + 4x - 6y - 12 = 0 The equation of the locus is x² + y² + 4x – 6y – 12 = 0.

31.

Find the equation of the locus of the point which is at a constant distance of 5 units from the fixed point (-2,3).

Answer»

Let A = (-2,3) and let P = (x,y). Now

AP = 5 (AP)² = 25

(x + 2)² + (y - 3)² = 25

x² + y² + 4x - 6y - 12 = 0

The equation of the locus is x² + y² + 4x – 6y – 12 = 0.

Discussion

32.	which is/are the correct set of electrophile?A. `BF_(3),H_(2)O`B. `Br^(+),overset(+)(N)H_(4)`C. `overset(+)(C)H_(3),SO_(3)`D. `overset(+)(N)O_(2):CH_(2)`
Answer» Correct Answer - C::D electrophiles `overset(+)(C)H_(3),Br^(+),overset(+)(N)O_(2)` (positively charged species) `BF_(3)SO_(3)` (speccies with vacant orbital)

Discussion

33.	A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.
Answer» Let sum be Rs. x then , S.I.=Rs.(x(27/2) 4(1/100) ) = Rs.27x/50 amount = (Rs. x+(27x/50)) = Rs.77x/50 77x/50 = 2502.50 x = 2502.50 50/77 = 1625 Hence , sum = Rs.1625.

33.

A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Answer»

Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50

amount = (Rs. x+(27x/50)) = Rs.77x/50

77x/50 = 2502.50

x = 2502.50 * 50/77 = 1625

Hence , sum = Rs.1625.

Discussion

34.	Which statement is/are correct:A. Basicity order `I^(-)geBr^(-)geCl^(-)F^(-)`B. nucleophilicity order `CH_(3)-O^(ɵ)gePh-O^(ɵ)geCH_(3)-COO^(ɵ)geCH_(3)-SO_(3)^(ɵ)`C. nucleophilicity order in polar protic solvent is `I^(-)geCl^(-)Br^(-)geF^(-)`D. leaving group ability order `I^(-)geBr^(-)geCl^(-)geF^(-)`
Answer» Correct Answer - B::D `overset(" "F^(-)" "Cl^(-)" "Br^(-)" "I^(-))to` basicity `darr` nucleophilicity `uarr` leaving group ability `uarr`

Discussion

35.	A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest rate is increased by 8%, it would amount to bow much?
Answer» S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs. R = ((100 x 120)/(8003) ) % = 5%. New rate = (5 + 3)% = 8%. New S.l. = Rs. (8008*3)/100 = Rs. 192. New amount = Rs.(800+192) = Rs. 992.

35.

A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest rate is increased by 8%, it would amount to bow much?

Answer»

S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs.

R = ((100 x 120)/(800*3) ) % = 5%.

New rate = (5 + 3)% = 8%.

New S.l. = Rs. (800*8*3)/100 = Rs. 192.

New amount = Rs.(800+192) = Rs. 992.

Discussion

36.	An ideal monoatomic gas initially in state 1 with pressure `P_(1)=20` atm and volume `V_(1)1500cm^(3)` it is then taken to state 2 with pressure `P_(2)=1.5P_(1)` and volume `V_(2)=2V_(1)` find the change in internal energy in this process in KJ. (take `1atm` lit `=100J`)
Answer» Correct Answer - 9 `DeltaE=nCv(T_(2)-T_(1))` `DeltaE=nxx(3)/(2)R((P_(2)V_(2))/(nR)-(P_(1)V_(1))/(nR))` `=(3)/(2)(1.5xx20xx2xx1.5-20xx1.5)` `=90lit-atm` `=9000J` `=9KJ`

Discussion

37.	Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond four years. he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?
Answer» Let the sum borrowed be x. Then, (x26)/100 + (x93)/100 + (x144)/100 = 11400 (3x/25 + 27x/100 + 14x / 25) = 11400 95x/100 = 11400 x = (11400*100)/95 = 12000. Hence , sum borrowed = Rs.12,000.

37.

Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond four years. he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Answer»

Let the sum borrowed be x. Then,

(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400

(3x/25 + 27x/100 + 14x / 25) = 11400

95x/100 = 11400

x = (11400*100)/95 = 12000.

Hence , sum borrowed = Rs.12,000.

Discussion

38.	Electric field given by the vector `vecE=(E_(0))/(l)(xhati+yhatj)` N/C is present in the `x-y` plane. A small ring of mass `M` carrying charge `+Q`, which can slide freely on a smooth non conducting rod, is projected along the rod from the point `(0,l)` such that it can reach the other end of the rod. Assuming there is no gravity in the region. What minimum velocity should be given to the ring? if in `S.I.` unit `(QE_(0)l)/(M)=8`
Answer» For the electric field, `V=-(E_(0)x^(2))/(2l)-(E_(0)y^(2))/(2l)+K` from energy conservation `-(QE_(0))/(l)(l^(2))/(2)+k+(1)/(2)Mv^(2)=-(QE_(0))/(l)*(l^(2))/(8)xx2+K` `v=2m//s`

Discussion

39.	Two identical parallel plate capacitors are connected in one case in parallel and in the other in series. In each case the plates of one capacitors are brought closer by a distance a and the plates of the outer are moved apart by the same distance a. ThenA. total capacitance of first system increasesB. total capacitance of first system decreasesC. total capacitance of second system decreasesD. total capacitance of second system remains costant.
Answer» Correct Answer - A::D a.,d. When capacitors are connected in parallel, initial capacitance is `C=2epsilon_(0)A//d`. After distance between the plates is changed, the capacitance becomes `C=(epsilon_(0)A)/(d+a)+(epsilon_(0)A)/(d-a)` which is greater than initial one. hence option (a) is correct and option (b) is incorrect. When capacitors are connected in series, then `(1)/(C)=(2d)/(epsilon_(0)A)` After the distance between the plates is changed, `(1)/(C)=(d+a)/(epsilon_(0)A)+(d-a)/(epsilon_(0)A)=(2d)/(epsilon_(0)A)`

Discussion

40.	An insulating spherical shell of uniform surface charge density is cut into two parts and place at a distance d apart as shown in figure. `vecE_p and vecE_Q` denote the electric fields at P and Q, respectively. As `d (i.e. PQ) rarr oo` A. `\|vecE_(P)\|gt\|vecE_(Q)\|`B. `\|vecE_(P)\|=\|vecE_(Q)\|`C. `\|vecE_(P)\|lt\|vecE_(Q)\|`D. `vecE_(P) + vecE_(Q)=0`
Answer» Correct Answer - B::D b.,d. The electric field inside any point of the sphere is zero.

Discussion

41.	A plane electromagnetic wave in a non-magnetic dielectric medium is given by `vecE = vecE_(0)(4 xx 10^(-7) x - 50 t)` with distance being in meter and time in seconds. The dielectric constant of the medium is :A. 2.4B. 8.2C. 5.76D. 4.8
Answer» Correct Answer - A `mu = (C)/(V) = (CK)/(omega) = (3 xx 10^(8) xx 4 xx 10^(-7))/(50) = (120)/(50) = 2.4`

Discussion

42.	A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell in not to exceed T, the thickness of the shell should not be less than .......
Answer» At equilibrium energy radiated by point source `=`heat conducted through the thickness of the shell

Discussion

43.	A point charge `mu` is placed at origin. Let `vecE_(A), vec E_(B), and vecE_(C)` be the electirc field at three points A (1,2,3), B (1,1,1), and C(2,2,2) due to charge `mu`. ThenA. `vecE_(A)_\|_vecE_(B)`B. `vecE_(A)\|\|vecE_(B)`C. `\|vecE_(B)\| = 4\|vecE_(C)\|`D. `vecE_(B) = 16 \| vecE_(C)\|`
Answer» Correct Answer - A::C a.,c. `vecE_(A)` is along `OA` and `OA=hati+2hatj+3hatK` `vecE_(B)` is along `OB=hati+hatj-hatK` Since `oAxxOB=(hati+2hatj+3hatK)xx(hati+hatj-hatk)=0` So `OAbotOBrArrvecE_(A)botvecE_(0)` So, option (a) is correct. Since `E_(B)=(kq)/([OB]^(2))=(kq)/(3)` `E_(C)=(kq)/([OC]^(2))=(kq)/(12)` So `(E_(B))/(E_(C))=4` or `\|vecE_(B)\|=4\|vecE_(C)\|` So, option (c) is correct. Option (b) and (d) are wrong from the explanation.

Discussion

44.	At what rate percent per annum will a sum of money double in 16 years.
Answer» Let principal = P. Then, S.I. = P and T = 16 yrs. Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

44.

At what rate percent per annum will a sum of money double in 16 years.

Answer»

Let principal = P. Then, S.I. = P and T = 16 yrs.

Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

Discussion

45.	In the labortary, the high emf of a battery is measured by using potentiometer and two resistance `R_(1)` and `R_(2)`, where `R_(1) lt lt R_(2)` as shown in the figure `R_(1)=(100+-0.10)Omega` and `R_(2)=(9900+-9.90)Omega`, `AB=1m` The voltage across `R_(1)`, when switches `S_(1)` & `S_(2)` are connected to point `1` and point `2` is balanced against `l_(1)=(60+-0.06)cm`. when the switches `S_(1)` and `S_(2)` are shifted to point `3` and point `4` as shown in the figure, the potential difference of standard `E_(0)=2` volt is balanced against length `l_(2)=(75+-0.075)cm`. The maximum error in emf of battery `E` is __________ volts (upto two decimal places)
Answer» `E=2xx(60)/(75)xx((10000)/(100))=160` volt , `E=E_(0)(l_(1))/(l_(2))((R_(1)+R_(2))/(R_(1)))` `(DeltaE)/(E)=\|(Deltal_(1))/(l_(1))\|+\|(Deltal_(2))/(l_(2))\|+\|(Delta(R_(1)+R_(2)))/(R_(1)+R_(2))\|+\|(DeltaR_(1))/(R_(1))\|` `rArr DeltaE=0.64` volts

Discussion

46.	Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region(A) AB (B) BC(C) CD (D) DE
Answer» (A) AB (C) CD

Discussion

47.	The displacement of a particle is `s = (a+bt)^(6)`, where a and b are constants. Find acceleration of the particle as a function of time.A. `30b^(2)(a+bt)^(4)`B. `30b^(2)(a+bt)^(3)`C. `6b^(2)(a+bt)^(5)`D. `30b^(2)(a+bt)^(5)`
Answer» Correct Answer - C Let F be the force of air resistance. For the upward motion Facts downward. Let `a_(1)` = retardation (downward acceleration). `mg+F=ma_(1) " or "a_(1) = g +(F)/(m)` Thus `a_(1) gt a_(2)` Let h = maximum height reached `h=(1)/(2)a_(1)t_(1)^(2)=(1)/(2)a_(2)t_(1)^(2)`

Discussion

48.	Consider the system shown in the figure. The wall is smmoth, but the surfaces of blocks A and B in contact are rough. The friction on B due to A in equilibrium is A. upwardB. downwardC. zeroD. the system cannot remain in equilibrium
Answer» Correct Answer - B `a_(1)=g//1" "a_(2)=g//3" "a_(3)=g//2`

Discussion

49.	What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?
Answer» Let each Installment be Rs. x Then, ( x+ ((x121)/100)) + (x+ ((x122)/100) ) + x = 1092 ((28x/25) + (31x/25) + x) = 1092 (28x+31x+25x)=(109225) x= (109225)/84 = Rs.325. Each installment = Rs. 325.

49.

What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?

Answer»

Let each Installment be Rs. x

Then, ( x+ ((x*12*1)/100)) + (x+ ((x*12*2)/100) ) + x = 1092

((28x/25) + (31x/25) + x) = 1092

(28x+31x+25x)=(1092*25)

x= (1092*25)/84 = Rs.325.

Each installment = Rs. 325.

Discussion

50.	Account for the following/Explain why :Bond enthalpy of F2 is less than that of Cl2.
Answer» This is due to small size of fluorine which results in large electron-electron repulsion among the lone pairs in F₂ molecule.

Discussion

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