This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 5951. |
Who can be registered under SANKALP? |
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Answer» Pensioners, Pensioners’ Associations and Non-Government Organisations can be registered under SANKALP. |
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| 5952. |
Nomination means nominating a person to receive the benefits of Life Insurance Policy. The person nominated is called __________. |
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Answer» Nomination means nominating a person to receive the benefits of Life Insurance Policy. The person nominated is called Nominee. |
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| 5953. |
The issue of having regular income during old age is taken care off by ___________Policies. a) Endowment b) Annuityc) Unit linked d) Money back |
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Answer» The issue of having regular income during old age is taken care off by Annuity Policies. |
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| 5954. |
The most appropriate method to determine the financial loss due to loss of human life would be to access the same on the basis of loss of income in the future years. This is also known as __________. |
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Answer» The most appropriate method to determine the financial loss due to loss of human life would be to access the same on the basis of loss of income in the future years. This is also known as human life value. |
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| 5955. |
Figs, (a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why? |
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Answer» Fig (a) is incorrect. This is because at a constriction (ie., where the area of cross-section of the tube is smaller), the flow speed is larger due to mass conservation. Consequently, pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible. |
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| 5956. |
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmospheric pressure. |
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Answer» p = hρg, h = \(\frac{P}{ρg}\)= \(\frac{1.01×10^5}{984×9.8}\)m = 10.47m. |
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| 5957. |
Two thin evacuated (one end closed) glass take A and B are carefully immersed in a beaker containing mercury such a way that there is no chance to get air in to the tubes. A is stand vertically and B is making an angle θ with the vertical.1. Is any rise of mercury in the tubes?2. Is any height difference of mercury levels in tube A and B? Justify your answer.3. When the doctors are measuring body pressure, it is advisable to lie on a table. Why? |
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Answer» 1. Yes 2. No. Pressure is same at same level. To get same pressure, height of mercury becomes same. 3. When we lie on the table, the pressure of our body will be same at all points. |
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| 5958. |
Find the depression of the miniscus in the capillary tube of diametre `0.4mm` dipped in a beaker containning mercury (density of mercury `=13.6xx10^(3)kg//m^(3)` and surface tension of the mercury is `0.49 N//m` and angle of contact is `135^(@)`).A. `0.025cm`B. `0.021 cm`C. `0.020 cm`D. `0.027 cm` |
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Answer» Correct Answer - A `T=(hrdg)/(2costheta)` |
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| 5959. |
A particle of mass `1.5` kg moves along x-axis in a conservative force field. Its potential energy is given by `V(x)=2x^(3)-9x^(2)+12x,` where all quantities are written in SI units. The plot of this potential energy is given below. It is seen that the particle can be in stable equilibrium at a point on x-axis, x_(0). When it is displaced slightly from this equilibrium position, It executes SHM with time period T. What is the range of total mechanical energy of the particle for which its motion can be oscillatory about a pointA. `Elt5J`B. `Elt8J`C. `Elt12 J`D. `E lt 9 J` |
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Answer» Correct Answer - A `V_(max)at(dv)/(dx)=0` `6x^(2)-18x+12=0` `x^(2)-3x+2=0` `x=1 , 2` `(d^(2)v)/(dx^(2))=2x-3` `=-1atx=1Rightarrow maximum` `=1at x=2Rightarrow minimum` Motion can be oscillatory about x=2 `EltE_(max)` `E_(max)=2xx1^(3)-9xx1^(2)+12=5j` |
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| 5960. |
A particle of mass `1.5` kg moves along x-axis in a conservative force field. Its potential energy is given by `V(x)=2x^(3)-9x^(2)+12x,` where all quantities are written in SI units. The plot of this potential energy is given below. It is seen that the particle can be in stable equilibrium at a point on x-axis, x_(0). When it is displaced slightly from this equilibrium position, It executes SHM with time period T. What is the time period of SHM mentioned in the paragraph?A. `pi sec`B. `2pi sec`C. `pi/2 sec`D. `pi/4 sec` |
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Answer» Correct Answer - A `a=F=-6(x^(2)-3x+2)` `a=-4(x^(2)-3x+2)=-4((2+y)^(2)-3(2+y)+2)` `=-4(4+y^(2)+4y-6-3y+2)` `a=-4y` `omega-2` `T=(2pi)/2=pi sec` |
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| 5961. |
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be :(A) √3 years(B) 3 years(C) 9 years(D) 3√3 years |
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Answer» Correct option is (D) 3√3 years T' = T\((\frac{3R}R)^{3/2}=3√3T\) |
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| 5962. |
The distance of a geostationary satellites from the centre of earth (Radius R=6400km) is nearest to |
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Answer» We know that distance of geostationary satellite from the surface of earth is 36000 km. The radius of earth is 6400km. Therefore, distance of geostationary satellite from the centre of earth = 36000+6400=42400km Therefore, 6400xD = 42400 or, D = 42400/6400 = 6.6R or, D = 7R |
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| 5963. |
Two satellites revolve around the ‘Sun’ as shown in the figure. First satellite revolves in a circular orbit of radius R with speed `v_(1)` . Second satellite revolves in elliptical orbit, for which minimum and maximum distance from the sun are `(R)/(3)` and `(5R)/(3)` respectively. Velocities at these positions are `v_2` and `v_3` respectively. The correct order of speeds is A. `v_(2)gtv_(3)gtv_(1)`B. `v_(3)ltv_(2)ltv_(1)`C. `v_2gtv_(1)gtv_(3)`D. `v_(2)gtv_(3)=v_(1)` |
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Answer» `V_(1)=sqrt((GM)/(R))` (orbital velocity in circular path) For elliptical orbit conservation of angular momentum `mV_(2)(R)/(3)=(5R)/(3)mV_(3)` conservation energy `-(GMm)/(R//3)+(1)/(2)mV_(2)^(2)=(-GMm)/(5R//3)+(1)/(2)mV_(3)^(2)` Solving `V_(2)=sqrt((5GM)/(R))` and `V_(3)=sqrt((GM)/(5R))` |
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| 5964. |
A vernier calipers has `1mm `marks on the main scale. It has `20` equal divisions on the Verier scale which match with `16` main scale divisions. For this Vernier calipers, the least count isA. `0.2 mm`B. `0.05 mm`C. `0.1 mm`D. `0.2 mm` |
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Answer» Correct Answer - D Least count of vernier callipers `LC=1MSD= -1VSD` `=("Smallest division on main scale")/("Number of divisions on vernier scale")` `20` divisions of vernier scale `=16` divisions of main scale `:. 1VSD=(16)/(20)mm=0.8mm` `LC=1MSD-1VSD` `=1mm-0.8mm` `=0.2mm` |
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| 5965. |
A motorcyclist wants to drive on the vertical surface of a wooden well of radius 5m with a minimum speed of 5√5 The minimum value of coefficient of friction between the tyres and the wall of the well must be? |
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Answer» Frictional force = f = mg Centripetal Force = Contact force = N = mv² / r μ = f / N = mg / (mv2 / r) = rg / v² = (5 × 10) / (5√5)² = 50 / 125 = 0.4 Minimum coefficient of friction between the tyres and wall of the well must be 0.4 |
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| 5966. |
A person wants to drive on the vertical suirface of a large cylindrical wooden `well` commonly known as `death well` in a circus. The radius of the well 2 meter, and the coefficient of friction between the tyers of the motorcycle and the wall of the well is `0.2` the minimum speed the motorcyclist must have in order to prevent slipping should beA. `10 m//s`B. `15 m//s`C. `20 m//s`D. `25 m//s` |
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Answer» `N=(mv^(2))/(R )` …….(`A`) `f_(max)=mg` `muN=mg` `mu(mv^(2))/(R )=mg` `v=sqrt((Rg)/(mu))=sqrt((2xx10)/(0.2))=10m//s` |
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| 5967. |
Two uniform solid cylinders A and B each of mass 1 kg are connected by a spring of constant `200Nm^(-1)` at their axles and are placed on a fixed wedge as shown in fig.There is no friction between cylinders and wedge. The angle made by the line AB with the horizontal, in equilibirum is A. `0^(@)`B. `15^(@)`C. `30^(@)`D. None of these |
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Answer» In equilibrium, there will not be any friction between the cylinder and the wedge. If `theta` be the required angle Cylinder `A : mg sin60^(@)=kx cos(60^(@)-theta)(-mg cos30^(@))` Cylinder `B : mg sin30^(@)=kx cos(30^(@)+theta)=kx sin(60^(@)-theta)` `:. cot30^(@)=cot(60^(@)-theta)rArrtheta=30^(@)`. |
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| 5968. |
The sum of the series`C 20()_0-C 20()_1+C 20()_2-C 20()_3+...-...+C 20()_(10)`is:(1) `-C 20()_(10)`(2) `1/2C 20()_(10)`(3) 0 (4) `C 20()_(10)`A. `.^(20)C_(10)`B. `1/2.^(20)C_(10)`C. `0`D. `.^(20)C_(10)` |
| Answer» Correct Answer - A | |
| 5969. |
If 10gm of water4 is added to 150gm of oleum (104.5%), then the find solution:A. Mass of `SO_(3)` left is 10gmB. Mass of `H_(2)SO_(4)` is 156.75gmC. No water will be leftD. Labelling of new solution is 102.25% |
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Answer» 100gm convert into maximum 104.5gm `H_(2)SO_(4)` 150gm convert into pure `(104.5)/(100)xx150=156.75gm H_(2)SO_(4)` and `w_(H_(2)O)` requried `=(4.5)/(100)xx150=6.75gm` we add water `gt6.75gm` thus no `SO_(3)` remain after 10gm water addition labelling of new solution `=100%` as no `SO_(3)` left in it. |
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| 5970. |
Two radioactive materials A & B have decay constant `3lamda` and `2lamda` respectively. At `t=0` the numbers of nuclei of A and B are `4N_(0)` and `2N_(0)` respectively then,A. Their number of radioactive nuclei will be equal at `t=(ln2)/(lamda)`B. Their decay rate will be equals at `t=(ln4)/(lamda)`C. Their decay rate will be equal at `t=(ln3)/(lamda)`D. At `t=(ln4)/(lamda)` the decay rate of A will be greater than that of B. |
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Answer» Correct Answer - A::C When no. of nuclei or equal `4N_(0)e^(-3lamdat)=2N_(0)e^(2lamdat)` `impliest=(ln2)/(lamda)` when decay rate equal `3lamda4N_(0)e^(-3It)=2lamda2N_(0)e^(-2lamdat)` `t=(ln3)/(lamda)` at `t=(ln4)/(lamda)` decay rate of `A=(3lamda)(4N_(0))e^(-3lamda(ln4)/(lamda))` at `t=(ln4)/(lamda)` decay rate of `B=2lamda4N_(2)e^(-2lamda(ln4)/(lamda)` |
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| 5971. |
A uniform rope of mass M=0.1kg and length L=10m hangs from the celling. `[g=10m//s^(2)]` :-A. time taken length by the transverse wave to travel the full length of the rope is 2 secB. Speed of the transverse wave in the rope increases linearly from bottom to the topC. Speed of the transverse wave in the ropedecreases linearly from bottom to the topD. Speed of the transverse wave in the rope remain constant along the length of the rope |
| Answer» Correct Answer - A | |
| 5972. |
A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` isA. `(10)/(pi)`B. (5)/(pi)`C. `(100)/(pi)`D. `(50)/(pi)` |
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Answer» Correct Answer - B Potential energy `U = mV` `rArr U = (50 x^(2) + 100) 10^(-2)` `F = -(dU)/(dx)=-(100x)10^(-2)` `rArr m omega^(2)x=-(100 xx10^(-2))10^(-2)` `10 xx 10^(-3) omega^(2)x=100 xx 10^(-2)x` `rArr omega^(2) =100,omega =10` `rArr f =(omega)/(2pi)=(10)/(2 pi)=(5)/(pi)`. |
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| 5973. |
The string in fig. is passing over small smooth pulley rigidly attached to trolley A. If the speed of trolley is constant and qual to `v_(A)` towards right, speed and magnitude of acceleration of block B at the instant shown in figure are A. `V_(B)=V_(A)`B. `a_(B)=(16 V_(A)^(2))/(125)`C. `a_(B)=0`D. `V_(B)=4/5 V_(A)` |
| Answer» Correct Answer - B | |
| 5974. |
In which process maximum energy is released ?A. `S_(g)^(-)+e^(-)rarrS_(g)^(-2)`B. `F_(g)^(-)+e^(-)rarrF_(g)^(-)`C. `N_(g)^(-)+e^(-)rarrN_(g)^(-)`D. `S_(g)^(-)+e^(-)rarrS_(g)^(-)` |
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Answer» Correct Answer - 2 `EA_(1) of F lt EA_(1) of S` |
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| 5975. |
Green house gases are-A. CFC, `O_(3)` and water vapoursB. Peroxy aeetyl niotrate, CFC and`O_(3)`C. `NO_(2) CH_(4) andSO_(2)`D. peroxy acetyl nitrate, `NO_(2) and SO_(2)` |
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Answer» Correct Answer - 1 Fact based |
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| 5976. |
9+(5x)/(2)=4 |
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Answer» Subtract From Both Sides `(9 + 5x / 2) - 9 = 4 - 9` Simplify Left Side `5x / 2 = 4 - 9` Simplify Arithmetic `5x / 2 = -5` Multiply Both Sides By Inverse Fraction `5/2 x * 2/5 = -5 * 2/5` Simplify Left Side `x = -5 * 2/5` Simplify Right Side `x = -2` |
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| 5977. |
Three of the following four are same in a certain way and hence form a group. Find out the one which does not belong to that group?1. Ego2. Anger3. Miser4. Brave |
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Answer» Correct Answer - Option 3 : Miser Given: Some words Calculation: ⇒ The "Miser" is different from another because means "A person who hoards wealth and spends as little money as possible" and the remaining word represents the states of mind. ∴ The required result will be " Miser". |
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| 5978. |
Were you singing when I came in? Yes, I ________ A) sang B) sung C) was D) were |
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Answer» Correct option is C) was |
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| 5979. |
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semi-circles |
| Answer» Required Area= Ar(ABCD) - Ar(APD) - Ar(BPC)=7*7 - 2*(0.5*22/7*7/2*7/2)=10.5 cm^2 | |
| 5980. |
A solid cube of each side 8 cm, has been Painted red, blue and black on pairs of Opposite faces. It is then cut into cubical blocks of each side 2 cm. How many cubes have no face painted? |
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Answer» Correct Answer - Option 3 : 8 Given: A solid cube of each side 8 cm Calculation: ⇒ Cube has no face painted = Inner cubes (No colour) ⇒ The total number of cubes without any colour on any side = (n - 2)3 ⇒ When n = 8/2 = 4 ⇒ The total number of cubes without any colour on any side = (4 - 2)3 = 23 = 8 ∴ The required result will be 8. |
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| 5981. |
Find the probability that the month of June may have 5 Mondays in a year.(A) \( \frac{2}{7} \)(B) \( \frac{1}{6} \)(C) \( \frac{5}{6} \)(D) \( \frac{1}{7} \) |
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Answer» → Total number of days in June = 30 → Therefore, there must be 4 Mondays, 4 Tuesdays, 4 Wednesdays, ..... 4 Sundays. → Rest days = 30 - 7 x 4 = 30 - 28 = 2. → These 2 days may be (i) Sunday & Monday, (ii) Monday & Tuesday, (iii) Tuesday & Wednesday, (iv) Wednesday & Thursday, (v) Thursday & Friday, (vi) Friday & Saturday, (vii) Saturday & Sunday. → Hence, total outcomes for these 2 days. (Rest 2 days) is n(S) = 7 → Outcomes favourable to 5 Mondays are Monday & Tuesday and Sunday & Monday. → Number of outcomes favourable to get 5 Mondays is n(E) = 2. → ∴ Probability of getting 5 Mondays in month of June is P(E) = n(E)/n(S) = 2/7. |
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| 5982. |
We often hear her ___ at concerts. A) sings B) singing C) sang D) to sing E) have sung |
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Answer» Correct option is B) singing singing option b |
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| 5983. |
2+4+6= |
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Answer» Simplify Arithmetic `12` |
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| 5984. |
[6.67times10^(-17)times6times10^(26)],[6.4times10^(6)] |
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Answer» Cancel Terms `(67 * 10 ^ -17 * 6 * 10 ^ 26) / (4 * 10 ^ 6)` Cancel Terms `(67 * 10 ^ (-17 + -6) * 6 * 10 ^ 26) / 4` Cancel Terms `(67 * 10 ^ (-17 + -6) * 3 * 10 ^ 26) / 2` Collect And Combine Like Terms `(201 * 10 ^ (-17 + -6) * 10 ^ 26) / 2` Simplify Arithmetic `(201 * 10 ^ -23 * 10 ^ 26) / 2` Simplify Arithmetic `(201 * 1e-23 * 10 ^ 26) / 2` Collect And Combine Like Terms `(2.01e-21 * 10 ^ 26) / 2` Simplify Arithmetic `(2.01e-21 * 1e+26) / 2` Simplify Arithmetic `201000 / 2` Simplify Fraction `100500` |
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| 5985. |
What is a stack? Write an algorithm for PUSH() and POP() operations. |
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Answer» Algorithm for PUSH operation: Algorithm for POP operation: Step 2: ITEM = STACK[POP] |
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| 5986. |
Using recursion, write a C program to reverse a given number. |
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Answer» #include<stdio .h> #include<conio .h> void main() { int n,r; clrscr(); printf("enter an integer"); scanf("%d",&n); rev(n); getch(); } rev (int n) { if (n>0) { printf ("%d",n%10); rev(n/10); } } |
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| 5987. |
Infinite recursion leads to(A) Overflow of run-time stack(B) Underflow of registers usage(C) Overflow of I/O cycles(D) Underflow of run-time stack |
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Answer» Correct option- (A) Overflow of run-time stack |
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| 5988. |
Differentiate between pointer (*) and address (&) operator using examples. |
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Answer» The indirection operator (*) gets the value stored in the memory location whose address is stored in a pointer variable. The address of (&) operator returns the address of the memory location in which the variable is stored. The output of the following example shows the difference between * and &. //difference between * and &. #include<conio .h> void main() { int k; int *ptr; clrscr(); k=10; ptr=&k; printf("\n Value of k is %d\n\n",k); printf("%d is stored at addr %u\n",k,&k); printf("%d is stored at addr %u\n",*ptr,ptr); *ptr=25; printf("\n Now k = %d\n",k); getch(); } Output: Value of k is 10 10 is stored at addr 65524 10 is stored at addr 65524 Now k=25 |
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| 5989. |
Write an algorithm for PUSH operation in stack. |
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Answer» Algorithm for PUSH operation |
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| 5990. |
Differentiate between recursion and iteration. |
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Answer» Recursion and iteration: The factorial of a number, written as n!, can be calculated as follows: n! = n * (n-1) * (n-2) * (n-3) * ... * (2) * 1 This approach is called iteration, that is step by step calculation. However, we can also calculate n! like: n! = n * (n-1)! Going one step further, we can calculate (n-1)! using the same procedure: (n-1)! = (n-1) * (n-2)! We can continue calculating recursively until the value of n is 1. This approach is called recursion. It refers to a situation in which a function calls itself either directly or indirectly. Indirect recursion occurs when one function calls another function that then calls the first function. |
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| 5991. |
The potential energy `varphi`, in joule, of a particle of mass `1kg`, moving in the x-y plane, obeys the law `varphi=3x+4y`, where `(x,y)` are the coordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t=0`, thenA. The particle has constant accelerationB. The work done by external forces from the position of rest to the instant the particle crossing x-axis is `25J`C. The speed of the particle when it crosses the y-axis is `10ms^(-1)`D. The coordinates of the particle at time `t = 4s` are `(-18, -28)` (in m) |
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Answer» Correct Answer - A::B::C::D `phi = 3 x + 4y` `vec(F) = (- delta phi)/(delta x) hat(i) - (delta phi)/(delta y) hat(j)` `vec(F) = - 3 hat(i) - 4 hat(j)` Potential energy at (6, 4) `phi = 3 (6) + 4 (4) = 18 + 16 = 34` Joule Here force is conservative so its work done will not depends upon the path choosen. (A) acceleration of particle `a = (F)/(m) rArr a = (-3 hat(i) - 4 hat(j)) m//sec^(2)`. (It is constant) `vec(u) = 0 hat(i) + 0 hat(j)` `t = 4` sec `s_(x) = (s_(x))_(0) + u_(x) t + (1)/(2) a_(x) t^(2)` `s_(x) = 6 + 0 - (1)/(2) (3) xx (16) rArr s_(x) = 6 - 24 rArr (s_(x) = - 18)` `s_(y) = (s_(y))_(0) + u_(y) t (1)/(2) a_(y) t^(2)` `rArr s^(y) = 4 + 0 - (1)/(2) (4) (16)` `rArr s_(y) = 4 + 0 - (1)/(2) (4) (16)` `rArr s^(y) = 4 - 32` `rArr (s_(y) = - 28)` (C) Speed of particle when it crosses y - axis i.e. at that time its x - coordinate will be zero. `s_(x) = (s_(x))_(0) + a_(x) t + (1)/(2) a_(x) t^(2)` ` 0 = 6 + 0 - (1)/(2) (3) (t)^(2) rArr (3)/(2) t^(2) = 6` `rArr t = 2` sec `v_(x) = u_(x) + a_(x) t rArr v_(x) = - 3 (2) rArr v_(x) = - 6 m//sec` `v_(y) = u_(y) + a_(y) t rArr v_(y) = 0 - 4 (2) rArr v_(y) = - 8 m//sec` Speed of particle `= sqrt((-6)^(2) + (-8)^(2)) = 10 m//sec` (B) When the body is crossing x-axis than that time `s_(y) = 0` `s_(y) = (s_(y))_(0) + t + (1)/(2) xx a_(y) t^(2)` `0 = 4 + 0 + (1)/(2) xx (-4) t^(2)` `2t^(2) = 4 rArr t = sqrt2 sec` `v_(x) = a_(x) t rArr v_(x) = - 3 sqrt2 m//sec` `v_(y) = - 4 sqrt2 m//sec` `KE = (1)/(2) mv_(x)^(2) + (1)/(2) mv_(y)^(2)` `KE = (1)/(2) xx 1 xx (3 sqrt2)^(2) + (4 sqrt2)^(2)` `rArr KE = 25` Joules |
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| 5992. |
Examine whether the following logical statement pattern is tautology, contradiction or contingency.[(p → q) ∧ q] → p |
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Answer»
In the above truth table, the entries in the last column are a combination of T and F. ∴ [(p → q) ∧ q] → p is contingency |
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| 5993. |
A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation `v = (5t^(2) - 3t + 2)m//s` . Now give the answer of following questions : Velocity of particle when acceleration is zero is :A. `1 m//s`B. `(35)/(20)m//s`C. `(31)/(20)m//s`D. `38m//s` |
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Answer» Correct Answer - C Velocity of ……………. `a = 10t - 3 = 0` `t = (3)/(10)` `V = ((3)/(10))(2)/(3)` `V_((t= 3//10))= 5((3)/(10))^(2)-3 ((3)/(10))+2 = (31)/(20)m//s` |
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| 5994. |
Using examples from your area ,compare and contrast the activities and funtions of private and public sectors |
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Answer» Private Sector: (i) Ownership of assets and delivery of services is in the hands of private individuals or companies. (ii) This sector is guided by the motive to earn more and more profits. (iii) Services of this sector can be obtained only on payment hence, the poor people can not afford services of this sector. (iv) This sector does not render services of public utility. Public Sector : (i) The government owns assets and takes responsibility to deliver services. (ii) This sector is not guided by profit motive. (iii) It invests funds to construct structures and infrastructure of public utility e.g. construction of roads, bridges, railways, harbours, generation of electricity, construction of dams, school and college buildings etc. (iv) All citizens of a country use and obtain services produced by this sector unconditionally. (v) It collects funds for investment indirectly through taxes on people having certain income and capable to pay them. |
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| 5995. |
A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation `v = (5t^(2) - 3t + 2)m//s` . Now give the answer of following questions : Velocity of particle at t = 3 sec. is :A. `30 m//s`B. `38 m//s`C. `22 m//s`D. `36 m//s` |
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Answer» Correct Answer - B Velocity of …………. `V=(5t^(2)-3t+2)m//s` `V_((t=3))=5(3)^(2)-3(3)+2=45-9+2=38m//s`. |
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| 5996. |
The slope of the curve `y = x^(3) - 2x+1` at point x = 1 is equal to : |
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Answer» Correct Answer - B The slope ………… `y=x^(3)-2x+1` `(dy)/(dx)= 3x^(2)-2` slope at `x = 1 is (dy)/(dx)|_(x=1) = 3(1^(2))-2=1` |
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| 5997. |
The position vector of a particle is given by `vec r = (2t hati+5t^(2)hatj)m` (t is time in sec). Then the angle between initial velocity and initial acceleration isA. zeroB. `45^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - C The position ………… `(vecV)= (dvecr)/(dt) = 2hati+10t hatj` `veca = 10 hatj` `vecV(t = 0) = 2hati` `veca(t=0)=10hatj` `implies vecV _|_veca(theta = 90^(@))` |
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| 5998. |
Suppose that water drops are released from a point at the edge of a roof with a constant time interval `Deltat` between one water drop and the next. The drops fall a distance h to the ground. If `Deltat` is very short ie the number of drops falling through the air at any given instant is very large then the CM of the drops is very nearly at a height (above the ground ) ofA. `h//2`B. `h//3`C. `2h//3`D. `3h//4` |
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Answer» Correct Answer - C The droplets fall a distance `(1)/(2)g t^(2)` in time t, and the number of the droplets and hence their mass is proportional to `dt`. Computing the CM of the droplets using the definition, we get the result. |
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| 5999. |
A bi-convex lens is placed between a light source and a concave mirror as shwon such that image of the light source coincides with itself. Then, A. light, after being refracted, may fall normally on the mirrorB. light, after being refracted, may fall at the pole of the mirrorC. source and its image coincide for two positive of sourceD. none of the above |
| Answer» Normal incidence `rArr` normal reflection | |
| 6000. |
Which section of front office provide the communication facilities and service like STD, ISD, E-mail, internet, fax, xerox etc to guests? |
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Answer» Business Center |
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