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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.
| 27451. |
A circuit has 1000 turns enclosing a magnetic circuit 20cm2 in section. With 4A, the flux density is 1.0 Wb/m2 and with 9A, it is 1.4 Wbm2. Find the mean value of the inductance between these current limits and the induced e.m.f. if the current falls from 9 A to 4 A in 0.05 seconds. |
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Answer» L = N(dΦ/dl) = N(d/dI)(BA) = NA(dB/dI) henry = 1000 × 20 × 10−4 (1.4 − 1)/(9 − 4) = 0.16 H Now, eL = L.dI/dt ; dI = (9 − 4) = 5 A, dt = 0.05 s ∴ eL = 0.16 × 5/0.05 = 16 V |
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| 27452. |
An iron rod, 2 cm in diameter and 20 cm long is bent into a closed ring and is wound with 3000 turns of wire. It is found that when a current of 0.5 A is passed through this coil, the flux density in the coil is 0.5 Wb/m2 . Assuming that all the flux is linked with every turn of the coil, what is (a) the B/H ratio for the iron (b) the inductance of the coil ? What voltage would be developed across the coil if the current through the coil is interrupted and the flux in the iron falls to 10 % of its former value in 0.001 second ? |
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Answer» H = NI/l = 3000 × 0.2 = 7500 AT/m B = 0.5 Wb/m2 (a) Now, B/H = 0.5/7500 = 6.67 x 10-5 H/m. Also μr = B/μa H = 6.67 × 10−5 /4π × 10−7 = 53 (b) L = NΦ/I = (3000 x π (0.02)2 x 0.5)/(4 x 0.5) = 0.94H eL = N(NΦ/dt) volt; dΦ = 90 % of original flux = (0.9 x π (0.02)2 x 0.5)/4 = 0.45π x 10-4Wb dt = 0.001 second ∴ eL = 3000 × 0.45π × 10−4 /0.001 = 424 V |
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| 27453. |
A flux of 0.5 mWb is produced by a coil of 900 turns wound on a ring with a current of 3 A in it. Calculate (i) the inductance of the coil (ii) the e.m.f. induced in the coil when a current of 5 A is switched off, assuming the current to fall to zero in 1 milli second and (iii) the mutual inductance between the coils, if a second coil of 600 turns is uniformly wound over the first coil |
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Answer» (i) Inductance of the first coil = NΦ/I = (900 x 0.5 x 10-3)/3 = 0.15H (ii) e.m.f. induced e1 = L(di/dt) = 0.15 x (5 - 0)/(1 x 10-3) = 750V (iii) M(N2Φ1/I1) = (600 x 0.5 x 10-3)/3 = 0.1H (ii) Second Method for M We will now deduce an expression for coefficient of mutual inductance in terms of the dimensions of the two coils. Flux in the first coil Φ1 = N2I1/(l/μ0μrA) Wb; Flux/ampere = Φ1/I1 = N1/(l/μ0μrA) Assuming that whole of this flux (it usually is some percentage of it) is linked with the other coil having N2 turns, the weber-turns in it due to the flux/ampere in the first coil is M = N2Φ1/I1 = N2N1/(l/μ0μrA) ∴ M = ((μ0μrAN1N2)/l)H Also M = N1N2/(l/μ0μrA) = N1N2/reluctance = (N1N2/S)H |
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| 27454. |
the solution of the differential equation `(dy)/(dx)+(x(x^(2)+3y^(2))/(y(y^(2)+3x^(2)))=0` isA. `x^(4)+y^(4)+x^(2)y^(2)=c`B. `x^(4)+y^(4)+3x^(2)y^(2)=c`C. `x^(4)+y^(4)+6x^(2)y^(2)=c`D. `x^(4)+y^(4)+9x^(2)y^(2)=c` |
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Answer» Correct Answer - C put `y=vx` Differential equal becomes `int(4(v^(3)+3v))/(v^(4)+6v^(2)+1)dv=-4int(dx)/(x)` `ln(v^(4)+6v^(2)+1)=-4lnx+lnc` `impliesx^(4)+y^(4)+6x^(2)y^(2)=c` |
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| 27455. |
The spectra of radiation emitted by two distant stars are shown below. The ratio of the surface temperature of star A to that of star B, `T_(A):T_(B),` is approximately-A. `2:1`B. `4:1`C. `1:2`D. `1:1` |
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Answer» Correct Answer - 1 `(T_(1))/(T_(2))=(lamda_(2))/(lamda_(1))` |
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| 27456. |
A five digits number divisible by 3 is to be formed using the number 0,1,2,3,4 and 5 without repetition. The number of such numbers are `m^(3)` then m is equal to |
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Answer» Correct Answer - 6 since a five digits no. if formed using the digits {0,1,2,3,4 and 5} divisible by 3 i.e., only possible Case-1 using digits 0,1,2,4,5 no. of ways `=4.4.3.2.1=96` Case-2 using digits 1,2,3,4,5 no of ways `5,4,3,2,1=120` Total numbers formed `=120+96=216=6^(3)` `=120+96=216=6^(3)` |
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| 27457. |
`int(xcostheta+1)/((x^(2)+2xcostheta+1)^((3)/(2)))dx` isA. `sqrt(xcostheta+1)+C`B. `(1)/(sqrt(x^(2)2xcostheta+1))+C`C. `(x)/(x^(2)+2xcostheta+1))+C`D. `sqrt(x^(2)+2xcostheta+1)+C` |
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Answer» Correct Answer - C `int(xcostheta+1)/((x^(2)+2xcostheta+1)^((3)/(2)))dx=` `int((1)/(x^(2))costheta+(1)/(x^(3)))/((1+(2)/(x)costheta+(1)/(x^(2)))^((3)/(2)))dx` let `1+(2)/(x)costheta+(1)/(x^(2))=t^(2)` `(-(2)/(x^(2))costheta-(2)/(x^(3)))dx=2tdt` `1=-int(t)/(t_(3))dt=(1)/(t)+C=(1)/(sqrt(1+(2)/(x)costheta+(1)/(x^(2)))+C` |
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| 27458. |
If `alpha` and `beta` are roots of the equation `x^(2)-3x+1=0` and `a_(n)=alpha^(n)+beta^(n)-1` then find the value of `(a_(5)-a_(1))/(a_(3)-a_(1))` |
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Answer» Correct Answer - 8 `alpha+beta=3, alphabeta=1` `because alpha^(2)beta^(2)=7becausea_(1)=2,a_(2)=6` Also `alpha^(n)-3alpha^(n-1)+alpha^(n-2)=0` and `beta^(n)-3beta^(n-1)+beta^(n-2)=0` `because(alpha^(n)+beta^(n))-3(alpha^(n-1)+beta^(n-1))+(alpha^(n-2)+beta^(n-2))=0` `because(alpha^(n)+beta^(n)-1)-3(alpha^(n-1)+beta^(n-1)-1+(alpha^(n-2)+beta^(n-2)-1)=1` `becausea_(n)=1+3a_(n-1)-1_(n-2)` `becausea_(3)=17,a_(4)=46,a_(5)=122,because(a_(5)-a_(1))/(a_(3)-a_(1))=8` |
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| 27459. |
If `[.]` denotes the greatest integer less than or equal to x and (.) denotes the least integer greater than or equal to x, then domain of the function `f(x)=sin^(-1){x+[x]+(x)}` isA. `x in {0}`B. `x in [0,1]`C. `x in (-1,0]`D. `x in phi` |
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Answer» Correct Answer - A `-1lex+[x]+((x)le1` case 1: if `x in I` `-1le3xle1,-(1)/(3)lexle(1)/(3)` `x=0` case ii: if `x cancel(in)I` `(x)=[x]+1` `becausex+[x]+(x)=x+2[x]+1` `because-1lex+2px]+1le1` `-2ltx+2[x]le0` if `xlt0` `x+2[x]lt-2` if `xge0,x+2[x]ge0` No solution |
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| 27460. |
Let `S_(n)=1+2+3+…+n` and `P_(n)=(S_(2))/(S_(2)-1).(S_(3))/(S_(3)-1).(S_(4))/(S_(4)-1)…..(S_(n))/(S_(n)-1)`. Where `nepsilonN(nge2)` `lim_(ntoprop) P_(n)` is equal to |
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Answer» Correct Answer - 3 `S_(n)=(n)/(2)(n+1),S_(n)-1=((n+2)(n-1))/(2)` `(S_(n))/(S_(n)-1)=(n(n+1))/((n+2)(n-1))=(n)/((n-1))((n+1))/((n+2))` `P_(n)=((2)/(1),(3)/(2)…(n)/(n-1))((3)/(4).(4)/(5)…(n+1)/(n+2))` `P_(n)=((n)/(1))((3)/(n+2))impliesunderset(ntoinfty)(lim)P_(n)=3` |
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| 27461. |
A Body intially at rest, starts moving along `x-`axis in such a way so that its acceleration vs displacement plot is as shown in figure. The maximum velocity of particle is `:-` A. `1 m//s`B. `6 m//s`C. `2 m//s`D. none |
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Answer» Correct Answer - A Area of `a-x` graph `=intadx=int(v(dv)/(dx))dx=intvdv` `rArrint_(0)^(v_(m))vdv=(1)/(2)xx1xx1` `rArr(V_(m)^(2))/(2)=(1)/(2)rArrv_(m)=1m//s` |
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| 27462. |
A wedge `B` of mass 2 m is placed on a rough horizontal suface. The coefficient of friction between wedge and the horizontal surface is `mu_(1)`. A block of mass m is placed on wedge as shown in the figure. The coefficient of friction between block and wedge is `mu_(2)`. The block and wedge are released from rest. Q. Suppose the inclined surface of the wedge is at `theta=37^(@)` angle from horizontal and `mu_(2)=0.9` then the wedge:A. will remain in equilibrium if `mu_(1)=0.5`B. will accelerate towards left if `mu_(1)=0`C. will acceleration toward left if `mu_(1)=0.25`D. will remain in equilibrium if `mu_(1)=0.3` |
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Answer» Correct Answer - A::D Since `mu_(2)gttantheta` so block will remain at rest, so net content force onblock will be vertically upward. Net contact force on wedge due to block will be vertically downward. There is no tendency of sliding of wedge so wedge will remain in equlibrium for any value of `mu_(1)` |
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| 27463. |
A particle is shifted from `A` to `B` and then from `B` to `C` where `A`,`B` and `C` are the midpoints of the corresponding faces of a cube of side `2m`. If a force `vec(F)=(3hat(i)+4hat(j)-5hat(k))N` is continously acting on the particle, then select incorrect alternative `:-` A. work done form `A` to `B` is `7J`B. work done from `B` to `C` is `1J`C. work done `A` to `C` is `8J`D. force `vec(F)` is non`-`conservative force . |
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Answer» Correct Answer - D `vec(OA)=hat(i)+hat(k)`, `vec(OB)=2hat(i)+hat(j)+hat(k)`, `vec(OC)=hat(i)+2hat(j)+hat(k)rArr``vec(AB)=vec(OB)-vec(OA)=hat(i)+hat(j)` `vec(BC)=vec(OC)-vec(OB)=hat(j)-hat(i)`, `vec(AC)=vec(OC)-vec(OA)=2hat(j)` `W_(AB)=vec(F).vec(AB)=7`, `W_(BC)=vec(F).vec(BC)=1` `W_(AC)=vec(F).vec(AC)=8` A constant force is always a conservative force. |
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| 27464. |
A particle is moving along `x-`axis whose position is given by `x=4-9t+(t^(3))/(3)` then choose the CORRECT statement for thes motion`-`A. Direction of motion is not changing at any of the instantsB. For `0lttlt2s`, the particle is speeding upC. Direction of motion is changing at `t=3`secD. For `0lttlt3s` the particle is speeding up. |
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Answer» Correct Answer - C `x=4-9t+(t^(3))/(3)rArrv=(dx)/(dt)=(-9+t^(2))` and `a=((dv)/(dt))=2t`. Therefore (`vlt0` and `agt0` for `0lttlt3s`) |
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| 27465. |
Which of the following forces in non conservation one ?A. `3hat(i) +4hat(j)`B. `4xhat(i) + 3yhat(i)`C. `3x^(2) hat(i) + 4y^(2) hat(j)`D. `y^(2)hat(i) + x^(2)hat(j)` |
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Answer» Correct Answer - D for conservation force `F_(x)=(-dU)/(dx), F_(y) = (-dU)/(dy)` `(deltaF_(x))/(deltay)=(deltaF_(y))/(deltaX)=(delta^(2)U)/(deltaxdeltay)` |
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| 27466. |
If a ball is projected vertical upward from ground such that total distance traveled is 3.6 times of distance travel by the ball in 1st second, then find time of flight of ball? |
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Answer» Let initial velocity be u Total height,H= u2/2g Total distance the ball travel is u2/g Total time of flight,T=2u/g Distance in 1 second s=u-(g/2) As given in question 3.6s=2H 3.6 [u-(g/2)]=2×u2/2g 3.6 [u-5]=u2/10 36u-180=u2 When we solve the above equation we get u=30 or 6 T=2×30/10=6s Or T=2×6/10=1.2S From the given options the correct answer is T=6s |
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| 27467. |
A projectile is projected from ground, such that 2 second before it reaches the highest point, it makes an angle 37 degrees with vertical, then the velocity of projectile at highest point is (1 ) 5 m/s (2) 10 m/s (3) 15 m/s ( 4) 20 m/s |
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Answer» Velocity 2 seconds before it reaches maximum height=velocity 2 seconds after it reaches maximum height Vy= uy-gt = -2g = 20 m/s The angle made by the vertical is 37 but to find tan we need angle with the horizontal so the needed angle is 90 - 37 = 53. So, tan 53=Vy/Vx 4/3=20/Vx Vx=20×3/4 =15 m/s |
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| 27468. |
Which of the following quantities has the units `Kg m^(2) s^(-3) A^(-2)`?A. ResistanceB. InductanceC. CapacitanceD. Magnetic flux |
| Answer» Correct Answer - A | |
| 27469. |
The mutual inductance between the rectangular loop and the long straight wire as shown in figure is M. A. M=ZeroB. `M=(mu_(0)a)/(2pi) In (1+c/b)`C. `M=(mu_(0)b)/(2pi)In((a+c)/b)`D. `M=(mu_(0)a)/(2pi)In (1+b/c)` |
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Answer» Correct Answer - D `phi=int(mu_(0)i)/(2pix)xadx M_(x)i` `M=(mu_(0)a)/(2pi) ln. ((b+c))/c` |
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| 27470. |
The magnetic field shown in the figure consists of two uniform regions. The width of the first part is 5cm and the magnetic induction here is `0.001` T. The width of the other part is also 5cm, with the direction of the induction being opposite in direction and `0.002` T in magnitude. What should be the minimum speed of the electron arriving from the direction indicated in the figure so that it can pass through the magnetic field? Mass of electron `=9xx10^(-31)`kg A. `8/9xx10^(7) m//s`B. `4/9xx10^(7) m//s`C. `16/9xx10^(7) m//s`D. none |
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Answer» Correct Answer - A Suppose in 1^(st) region radius of circular path is r_(1) & in region 2 this is r_(2). `:.r_(1)lgt 5 & r_(2)gt5` `r=mv//(qB)` so,`v_(min)=(rqB_(min))/m` `:.v_(min)=(5xx10^(-2)xx1.6xx10^(-19)xx0.01)/(9xx10^(-31))=8/9xx10^(7)m//s` |
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| 27471. |
ABCDA is a closed loop of conducting wire consisting of two semicircular sections, the part ABCD lying in the XY plane and the part CDA lying in the YZ plane, both the parts having the centre at origin O (see the diagram). This loop is placed in a uniform field `vec(B)` which varies with time. Radius of the semicircular section is 0.5 m If `vec(B)` be directed along the vector `(vec(i) - vec(j))` and decreases at the rate of `10^(-2) Tesla/second`. The magnitude and sense of the induced emf in the loop as seen along the positive x-axis will beA. zeroB. `(5pi)/(4sqrt(2))` mV in the counterclockwise senseC. `(5 pi)/(4sqrt(2))` mV in the clockwise senseD. `(5pi)/(2sqrt(2))`mV in the clockwise sense. |
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Answer» Correct Answer - C It is given that `vec(B)=(hat(i)=hat(j))/(sqrt(2)) |B|` This direction is perpendicular to z-axis. Hence there will be no flux in the semicular section ABC. The induced emf in the other semiculaer section ADC will be `e=-(d phi)/(dt)=(d)/(dt)[vec(B)*vec(A)]` `=-(d)/(dt)[B(pia^(2))/(dt)*cos(pi)/(4)]` Since the field makes an angle of `pi//4` with the yz plane `e=(pi a^(2))/(2sqrt(2))*(dB)/(dt)=(pia^(2))/(2sqrt(2))*10^(-2)V` Putting `a=0.5m e, (pi(0.5)^(2))/(2sqrt(2))10^(-2)V=(5 pi)/(4sqrt(2))mV` This will be in the clockwise sence looking along positive x-axis ie,`vec(CDA)`. |
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| 27472. |
Why the Indian Constitution was amended keeping in view of MLAs and MPs? |
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Answer» The procedure of amendment in the constitution is laid down in Part XX (Article 368) of the Constitution of India. This procedure ensures the sanctity of the Constitution of India and keeps a check on arbitrary power of the Parliament of India. |
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| 27473. |
state the various functions political parties perform in a democracy |
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Answer» Some functions of political parties: 1. Breaking faith of common people 2. Fighting with other parties 3. Full violence 4. Not following what they promised to people 5. Benefitting their relatives 6
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| 27474. |
You can keep your personal files/folders in_________A. My folderB. My DocumentsC. My filesD. My text |
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Answer» Correct Answer - B B. My Documents
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| 27475. |
System software:-A. Allows the user to diagnose and troubleshoot the deviceB. is a programming languageC. is part of a productivity of softwareD. helps the computer manage internal resources |
| Answer» Correct Answer - D | |
| 27476. |
What utility to you use to transfer files and exchange messages?A. Web browsersB. WWWC. EmailD. Hpertext |
| Answer» Correct Answer - C | |
| 27477. |
Files are organised by storing them inA. tablesB. databasesC. foldersD. graphs |
| Answer» Correct Answer - C | |
| 27478. |
You organize files by storing them in________A. ArchivesB. ListsC. IndexesD. folders |
| Answer» Correct Answer - D | |
| 27479. |
In a slider crank mechanism, the length of the crank and connecting rod are 150 mm and600 mm respectively. The crank position is 60° from inner dead centre. The crank shaft speedis 450 r.p.m. (clockwise). Using analytical method, determine: 1. Velocity and acceleration ofthe slider, and 2. Angular velocity and angular acceleration of the connecting rod |
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Answer» Given : OC = 150 mm = 0.15m ; PC = 600 mm = 0.6 m ; CD = 150 mm = 0.15 m ; N = 450 r.p.m. or ω = 2π × 450/60 = 47.13 rad/s 1. Velocity & Accceleration of the slider: Vp= ω x OM=6.834m/s; ap= ω2 x NO=124.4m/s2 2. Angular velocity and angular acceleration of the connecting rod: ωPC= Vpc/PC=6.127rad/s; ωPC=at PC/PC=481.27rad/s2 |
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| 27480. |
The crank and connecting rod of a petrol engine, running at 1800 r.p.m. are 50 mm and 200 mm respectively. The diameter of the piston is 80 mm and the mass of the reciprocating parts is 1 kg. At a point during the power stroke, the pressure on the piston is 0.7 N/mm2, when it has moved 10 mm from the inner dead centre. Determine: 1. Net load on the gudgeon pin, 2. Thrust in the connecting rod, 3. Reaction between the piston and cylinder, and 4. The engine speed at which the above values become zero. |
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Answer» Given : N = 1800 r.p.m. or ω = 2π x 1800/60 = 1888.52 rad/s; r = 50 mm = 0.05 m; l = 200mm; D = 80 mm; mR = 1 kg; P = 0.7 N/mm2; x = 10 mm 1. Net load on the gudgeon pin We know that load on the piston When the piston has moved 10 mm from the inner dead centre i.e when P1P = 10 mm the crank rotates from OC1 to OC through an angle θ By measurement we find that θ = 33°.We know that ratio of lengths of connecting rod and crank n = l/r = 200/50 = 4and inertia force on the reciprocating parts We know that net load on the gudgeon pin FP = FL - F = 3520 - 1671 = 1849 N 2. Thrust in the connecting rod We know that thrust in the connecting rod 3. The reaction between the piston and cylinder We know that reaction between the piston and cylinder 4. Engine speed at which the above values will become zero A little consideration will show that the above values will become zero if inertia force on the reciprocating parts F1 is equal to the load on the piston FL. Let ω1 be the speed in rad/s at which FI = FL |
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| 27481. |
An object O is placed in front of a partially reflecting plane mirror M1 and a concave mirror M2 of focal length f. The distance between O and M1 is x and the distance between M1 and M2 is y. The two images of O formed by partial reflection in M1 and reflection in M2 (of rays passing through M1) coincide. The magnitude of f is(a) (y2 - x2) / 2y(b) (y2 + x2) / 2y(c) y - x(d) y2 + x2) / y - x |
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Answer» Correct Answer is: (a) (y2 - x2) / 2y |
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| 27482. |
an infinitely long thin conductor, shaped as shown, carries current. Each section is of the same length. The magnetic field at the point P due to the secion from `-oo` to A is B. The field at P due to the entire conductor isA. zeroB. BC. `sqrt(2)B`D. `2B` |
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Answer» Correct Answer - D |
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| 27483. |
An infinitely long thin conductor, shaped as shown, carries current. Each section is of the same length. The magnetic field at the point P due to the section from -∞ to A is B. The field at P due to the entire conductor is (a) zero (b) B (c) √2B (d) 2B |
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Answer» Correct Answer is: (d) 2B The sections up to A and the sections beyond A form pairs which produce equal magnetic fields at P. |
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| 27484. |
Two infinitely long conductors carrying equal currents are shaped as shown. The short sections are all of equal lengths. The point P is located symmetrically with respect to the two conductors. The magnetic field at P due to any one conductor is B. The total field at P is (a) zero (b) B (c) √2B (d) 2B |
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Answer» Correct Answer is: (a) zero The sections of the conductors on either side of P form pairs which produce equal and opposite magnetic fields at P. |
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| 27485. |
A particle is moving with constant acceleration starting from origin. Following graph are given for the motion of particle. Which of the following combination of graph are correct for the motion - |
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Answer» The correct option is (1) A, B, C. |
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| 27486. |
The arrangement shown consists of three elements. (i) A thin rod of charge `-3.0 mu C` that froms a full circule of radius `6.0 cm`. (ii) A second thin rod of charge `2.0 mu C` that forms a circular are of radius `4.0 cm` and concentric with the full circle, subtending an angle of `90^(@)` at the centre of the full circle. (iii) an electric dipole with a dipole moment that is perpendicular to a radial line and has magnitude `1.28xx10^(-21) C-m`. Find the net electric potential in volts at the centre. |
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Answer» Potential due to dipole at the centre of the circle is zero. Potentials due to charge on circle `=V_(1)=(K.(-3xx10^(-6)))/(6xx10^(-2))` Potential due to are `V_(2)=(K.(2xx10^(-6)))/(4xx10^(-2))` Net potential `=V_(1)+V_(2)=0` |
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| 27487. |
An inducatane L and a resistance R are connected in series with a battery of emf epsilon. Find the maximum rate at which the energy is stored in the magnetic field.A. `(E^(2))/(4R)`B. `(E^(2))/(R )`C. `(4E^(2))/( R)`D. `(2E^(2))/(R )` |
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Answer» Correct Answer - A The graph of current is given by : `i=i_(0)(1-e^(-1//tau)) rArr (di)/(dt)=(i_(0))/(tau)e^(i//tau)` Energy stored in the form of magnetic field energy is: `U_(B)=(1)/(2)Li^(2)` `therefore` Rate of increase of magnetic field energy is : `R=(dU_(B))/(dt)=Li(di)/(dt)=(Li_(0)^(2))/(tau)(1-e^(-1//tau))e^(-1//tau)` This wiill be maximum when `(dR)/(dt)=0` `rArr=e^(-1..tau)=1//2` `R_("max")=(Li_(0)^(2))/)tau)=(Li_(0)^(2))/(tau)(1-(1)/(2))((1)/(2))=(Li(0)^(2))/(4tau)` `=[(L(E//R)^(2))/(4(L//R))]=(E^(2))/(4R)` |
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| 27488. |
The ratio of density of nuclei O16 and Ca40 is -(1) 3 : 4 (2) 1 : 2 (3) 1 : 3 (4) 1 : 1 |
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Answer» The ratio of density of nuclei O16 and Ca40 is 1 : 1. |
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| 27489. |
Figure shows cross section of two large parallel metal sheets carrying electric currents along their surface. The current in each sheet is `10//pi A//m` along the width, Consider two points A and B, as shown in the figure with their positions. A. Megnetic field at A is `4mu T` along x-direction.B. Magnetic field at A is `4 mu T` along negative x-direction.C. Magentic field at B is zero.D. Magnetic field at B is `2 mu T` along x-direction. |
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Answer» Correct Answer - A::C Field due to each plate `=1/2 (mu_0)K=2(mu)K=2(mu)T` At A, field add up, being in the same direction whereas at B, fields cancle out due to opposite directions. |
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| 27490. |
Calculate the modulus of rapture of a beam of breadth 0.2m and depth 0.5m given that the failure load under centre-point loading (CPL)of span 0.7m is 2000N |
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Answer» Modulus of rapture σ = 3Fl/2bd2 where, σ = modulus of rapture F = Breaking load l = length b = breadth d = depth. σ \(=\frac{3\times2000\times0.7}{2\times0.2\times(0.5)^2}\) σ = 4200/0.1 σ = 42000 N/m2 |
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| 27491. |
Assertion (A) : The number of electric lines of force emanating from 1μc charge in vacuum is 1.13× 105. Reason (R) : This follows from Gauss’s theorem in electrostatics.(a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false |
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Answer» Option : (A) From Gauss’s theorem in electrostatics : ϕ = \(\frac{q}{∈_0}\) = \(\frac{1\times 10^{-6}}{8.85\times 10^{-12}}\) = 1.13 × 105 |
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| 27492. |
Differentiate between trading business and service business. |
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Answer» A trading business does not manufacture a good or product but only facilitates the act of bringing the finished goods from the manufacturing unit to the buyer or customer. Any business activity that is intangible, which cannot be seen and felt, but is for the benefit of a buyer is called a service. Services do not have a fixed time and it is flexible as per the demands of the customers. Services do not even need a shop to sell their products. |
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| 27493. |
The difference between the GNP and the NNP is equal to the (a) Consumer expenditure on durable goods (b) Direct tax revenue (c) Indirect tax revenue (d) Capital depreciation |
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Answer» The difference between the GNP and the NNP is equal to the Capital depreciation. |
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| 27494. |
Investment is equal to (a) Gross total of all types of physical capital assets (b) Gross total of all capital assets minus wear and tear (c) Stock of plants, machines and equipments (d) None of these |
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Answer» Investment is equal to Gross total of all capital assets minus wear and tear. |
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| 27495. |
What are two major differences between qualities and values of an entrepreneur and employee? |
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Answer» Belief in Self/Customer Focus/Decision Making and Responsibility/Belief that Environment can Change |
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| 27496. |
Industry means, which is connected with conversion of raw materials into finished goods. It may be of various types. Explain: (a) Analytical industry (b) Genetic industry and (c) Construction industry? |
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Answer» 1. Analytical Industry: It analyses and separates different elements from the same materials, as in the case of oil refinery. 2. Genetic Industries: These industries remain engaged in breeding plants and animals for their use in further reproduction. The seeds, nursery companies, poultry, diary, piggery, hatcheries, nursery, fisheries, apiary etc are classic examples of genetic industries. 3. Construction Industries: These industries are involved in the ,construction of building, dams, bridges, roads, as well as tunnels and canals. |
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| 27497. |
Physical activity carried out under an agreed set of universal rules for competitions is known as? |
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Answer» Sports Physical activity carried out under an agreed set of universal rules for competitions is known as |
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| 27498. |
Between Games and Play, which form of activity don’t follow any set of rules? |
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Answer» The activity don’t follow any set of rules is Play |
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| 27499. |
Describe any two sectors of green economy. |
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Answer» Two sectors of Green Economy: 1. Agriculture: Agriculture refers to growing crops for our food. It is the largest part of our economy and the most important because it provides us with food. 2. Water Management: Water is one of our most important resources. Billions of people worldwide lack access to clean drinking water or improved sanitation services and population growth is making the problem more serious |
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| 27500. |
Describe any three federal features of Indian democracy |
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Answer» (i) Division of powers between the centre and states – (ii) There are three lists: Union List, State List, Concurrent List. (iii) Residuary subjects (iv) Control of union territories with Centre |
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