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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.
| 6551. |
Match the laws given in Column I to the respective correct statement given in Column II. {:("Column I","Column II"),((a)"Wien's displacement law explains" ,(p) "why days are hot and nights are cold in deserts"),((b) "Planck's law explains",(q) "why a blackened platinum wire, when gradually heated, appears first dull red and then blue"),((c ) "Kirchhoff's law explains",(r ) "the distribution of energy in blackbody spectrum at shorter as well as longer wavelengths"),((d ) "Newton's law of cooling explains that ",(s ) "the rate of cooling of body is proportional to the excess temperature of the body over the surroundings"):} |
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| 6552. |
The resistances in the two arms of the mater bridge are 5Omega and ROmega, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6l_(1) . The resistance 'R' is |
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Answer» `15OMEGA` |
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| 6553. |
A bar magnet 8 cm long is placed in the magnetic merdian with the N-pole pointing towards geographical north . Two netural points separated by a distance of 6 cms are obtained on the equatorial axis of the magnet . If horizontal component of earth's field =3.2xx10^(-5)T, then pole strength of magnet is |
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Answer» 0.5 A m |
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| 6554. |
A spark plug in a bike or a car is used to ignite the air - fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure. To create the spark an electric field of magnitude 3xx10^(6)Vm^(-1) is required. (a) What potential difference must be applied to produce the spark ? ( b) If the gap is increased , does the potential defference increase decrease or remains the same the same ? ( c) find the potential difference if the gap is 1 mm . |
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Answer» Solution :Separation gap between two electrodes , d= 0.6 mm d=`0.6 xx10^(-3)m` Magnetude of electric field `E= 3xx10^(6)Vm^(-1)` Electric field `E= (V)/(d)` (a) APPLIED potential differenceV=E. d `= 3xx10^(6)xx0.6 xx10^(-13) = 1.8 xx10^(3)` V= 1800 V (b) From equation V= E. d If the gap (DISTANCE ) between the electrodes increased the potential difference also increases. ( c) Gap between the electrodes `"" d =1mm =1 xx10^(-3)` m Potential difference `"" ` V= E.d `= 3xx10^(6)xx1 xx10^(-3) = 3xx103 ` V = 3000 V |
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| 6555. |
For a thin convex lens when the heights of the object is double than its image, its object distance is equal to ....... Focal length of a lens is f. |
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Answer» f Accepting that IMAGE is inverted m =`-1/2` `m=V/u=-1/2implies v=(-u)/(2)` For mirror `1/f=(-1)/(u)+1/v` `implies -u=3f` (As per sign CONVENTION object distance is NEGATIVE) |
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| 6556. |
In Fig (a) , the V_(BB) supply can be varied from 0 V to 5.0 V . The Si transistor has beta_(dc) = 250 and R_(B) = 100 kOmega , R_(C) = 1 k Omega , V_(C C) = 5.0 V . Assume that when the transistor is saturated , V_(CE) = 0Vand V_(B E) = 0.8 V. Calculate (a) the minimum base current , for which the transistor will reach saturation . |
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Answer» Solution :Given at saturation `V_(CE) = 0 V , V_(BE) = 0.8` V `V_(CE) = V_(C C) - I_(C) R_(C)` `I_(C) = (V_(C C))/(R_(C)) = (5.0)/(1.0) = 5.0` mA a `therefore I_(B) = (I_(C))/(beta) = (5.0 mA)/(250)= 20 mu A` b. The input voltage at which the transistor will go into saturation is given by `V_(IH) - V_(B B) = I_(B) R_(B) + V_(BE) = 20 mu xx 100 + 0.8 = 2.8 V` The value of input voltage below which the transistor remains cutoff is given by `V_(R) = 0.6 V , V_(IH) = 2.8 V` c . Between 0.0 V and 0.6 V , the transistor will be in the .SWITCHED off. state . Between 2.8 V and 5.0 V , it will be in switched on state . Note that the transistor in active state when `I_(B)` varies from 0.0 mA to 20 mA . In this range , `I_(C) = beta I_(B)` is valid . In this saturation range `I_(C) le beta I_(B)` . |
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| 6557. |
When light is refracted from air into glass |
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Answer» Its WAVELENGTH and frequency both increase |
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| 6558. |
An object of mass 0.4 kg is rotating in horizontal circle of radius 2 m. If the breaking tension in the string is 31.54 N, the maximum number of revolutions per minute is, |
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Answer» 30 rev/min |
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| 6559. |
A wave is given by the equation Y = 10^(-4)sin (60t + 2x) where x and y are in metres and t in second, then wavelength of the wave is : |
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Answer» `pi` metre Here W = 60t and K =2 Now k = `(2pi)/(LAMBDA)` `rArr lambda = (2pi)/(2) = pi ` metre Hence correct choice is (a). |
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| 6560. |
In the arrangement shown, the 2 kg block is held to keep the system at rest. The string and pulley are ideal. When the 2kg block is set free, by what amount the tension in the string changes? [g=10m//s^(2)] |
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Answer» INCREASE of 12 N |
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| 6561. |
(i) Draw a circuit diagram to study the input and output characteristics of an n-p transistor in its common emitter configuration. Draw the typical input and output characteristics. (ii) Explain, with the hepl of a circuit diagram, the working of n-p transistor as a common emitter amplifier. |
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Answer» Solution :(i) The circuit arrangement to obtain the characteristics of a n-p transistor is shown in figure. Input characteristic : Input characteristics are graph between `V_(BE) and I_(B)` at constant `V_(CE)` Output characteristics : Output characteristics are graph between `V_(CE) and I_(C)` at constant `I_(B)`  (ii) Transistor amplifier is based on the principle that a WEAK input signal given to base region produces an amplified (LARGE) output signal in the collector region.  A simple amplifier circuit in common emitted configuration using a n-p-n transistor has been shown in Fig. The input signal `v_(i)` is CONNECTED between base and emitter through a capacitor `C_(1)`, which blocks d.c. voltage VBB from going towards the a.c. input source. The output is taken from the collector resistance RC. The capacitor `C_(2)` blocks the d.c. voltage VCC from the output signal `v_(0)` Due to input signal the base current changes by a value `i_(B) = (v_(i))/(R_(B))` and correspondingly collector current changes by `i_(C) = beta i_(B) = beta (v_(i))/(R_(B))` `:.` Output voltage ACROSS RC will be `v_(0) = R_(C).i_(C) = beta (R_(C))/(R_(B)) v_(i) or " Voltage gain " A_(v) = (v_(0))/(v_(i)) = beta.(R_(C))/(R_(B))` The input and output wavefronts have been shown in fig.(b) and (c), respectively. The output is amplified but is in opposite phase i.e., the input differ in phase by `180^(@) or pi` radian. |
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| 6562. |
The incorrect feature of nuclear forces among following is |
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Answer» NUCLEAR FORCES are CHARGE dependent |
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| 6563. |
A boy of mass 60 kg standing on a platform of mass 40 kg placed over a smooth horizontal surface. He throw, a stone of mass 1 kg with velocity u = 10 m/s at on angle of 45^(@) w.r.t the ground. The displacement of platform (with boy) on the horizontal surface when the stone lands on the ground is (g=10 m//s^(2)) |
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Answer» SOLUTION :`1 u cos 45^(@)=(60+40)v , v=(1)/(10sqrt(2))` Time of FLIGHT `= T_(f)=(2U sin theta)/(g)=sqrt(2)sec`. `S= v XX T_(f)=0.1 m` |
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| 6564. |
When an electricfield is applied across a semiconductor |
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Answer» electrons move from lower ENERGY level to higher energy level in the conduction band `RARR` Option (A) is correct. SIMILARLY, at room temperature, bound electrons in valence band of crystalline structure of semiconductor, move from one to another bound state. When electric field is applied they also get accelerated and they also get energy. Hence they move within the valence band, from low to high energy level. But we KNOW that motion of hole seems to be opposite to that of bound electron. Hence, hole seems to be moving from higher to lower energy level. `rArr ` Thus option (C ) is also correct. |
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| 6565. |
The radius of a spherical soap bubble is 0·5 mm. If the surface tension of the soap solution be 30 dyne/ cm, the excess pressure is : |
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Answer» `1500` dyne/c`m^(2)` `P_(i)-P_(0)=(4T)/r` `=(4xx30)/(0.5xx10^(-1))=2400` dyne/c`m^(2)` So the correct choice is (c). |
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| 6566. |
An infinitely long rod lies along the axis of a concave mirror of focal length 'f'. The near end of the rod is at a distance u > f from the mirror. Its image will have a length. |
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Answer» `(UF)/(U-F)` |
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| 6567. |
Four point charges q_(A) = 2 muC, q_(B)=-5muC, q_(C)=2mu C and q_D=-5 muC are located at the corners of a square ABCD of side 10cm. What is the force on a charge of 1muC placed at the centre of the square? |
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Answer» Solution :Forces on the charge `1 muC` at O due to the CHARGES `-5muC and -5muC` are EQUAL and OPPOSITE. So they cancel each other. SIMILARLY the forces due to `+2muC and +2 muC" on "1muC` at O, also cancel each other. So the force on a charge of `1 muC` placed at the centre of the square is zero. 
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| 6568. |
A particle executes SHM with amphtude 0.2 m and time period 24s. The time required for it to move from the mean position to a point 0.1 m from the mean position is : |
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Answer» 2 s `:.""t=(24)/(12)=2S` So CORRECT CHOICE is (a). |
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| 6569. |
A solid sphere rolls without sliding with constant velocity. What fraction of total K.E of sphere is rotational K.E. ? |
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Answer» `1//5` IMPLIES T.E. = `(1)/(2)mv^(2)+(1)/(5)mv^(2)=(7)/(10)mv^(2)` implies T.E. `=(1)/(2)mv^(2)+(1)/(5)mv^(2)=(7)/(10)mv^(2)` `therefore (E_("Rotational"))/(E_("TOTAL"))=((1)/(5)mv^(2))/((7)/(10)mv^(2))=(2)/(7)` |
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| 6570. |
In a double experiment , instead of takingslits of equal widths , one slit is made twice as wide as the other . Then in the interference pattern : |
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Answer» the INTENSITIES of both the maxima and MINIMA increase |
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| 6571. |
What phase difference between two identical traveling waves, moving in the same direction ulung a stretched string, results in the combined wave having an amplitude 0.852 times that of the common amplitude of the two combining waves! Express your answer in (a) degrees (b) radiants, and (c) wavelengths |
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| 6572. |
Net charge on an isolated system always remains constant.This is called as the law of. ___________ |
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| 6573. |
Using B_(0)=mu_(0)H_(0) formula find ratio of (E_(0))/(H_(0)) for plane electromagnetic wave. |
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Answer» `(C )/(mu_(0))` `THEREFORE (E_(0))/(mu_(0)H_(0))=c rArr (E_(0))/(H_(0))= mu_(0)c` |
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| 6574. |
Give the logic symbol, Boolean expression and truth table of a NOR gate. |
Answer» SOLUTION :
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| 6575. |
Eye can see different objects at different distance by changing the_____of the lens .It is called_____. |
| Answer» SOLUTION :FOCAL LENGTH, POWER of ACCOMMODATION | |
| 6576. |
A thick conducting slab is introduced between the plates of an isolated charged capacitor to fill the free space partially and without touching any plate. Which of the following is wrong statement regarding the points inside the capacitor |
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Answer» Electric field at some points will CHANGE and at some points remains unchanged |
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| 6578. |
Two conducting coils are kept parallel to each other so that they have a common axis as shown in the figure. Now a bar magnet moves with velocity vecv towards coil (2) as shown in figure, then _____ |
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Answer» the NORTH POLE is INDUCED on the FACE of coil (1) towards the magnet. |
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| 6579. |
Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed omega. The magnetic moment of the sphere is |
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Answer» `1/3 qomegar^(2)` `THEREFORE M/(2/5mr^(2)omega) = q/(2m) or M = 1/5qomegar^(2)` |
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| 6580. |
Which of the following statement related to the hysteresis loop is incorrect? |
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Answer» The curve of B against H for a ferromagnetic material is called hysteresis loop |
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| 6581. |
How is a transistor biased to be in active state ? |
| Answer» SOLUTION :TRANSISTOR is said to be in ACTIVE state when its emitter-base JUNCTION is (suitably) forward BIASED and base-collector junction is suitably) reverse biased. | |
| 6582. |
A uniform electric field E axisbetweentwo charged plates as shownin Fig. What would be work done in movinga charge q alongthe closedrecetangualr path ABCDA ? |
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Answer» Solution :When a capacitor is charged by a battery, WORK is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy. Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates = zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates V =`q/C`. Now work done in giving an additional infinitesimal charge dq to capacitor `dW=Vdq=q/C dq`. The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works, which may be obtained by integration. Therefore total work `W=int_0^QV dq=int_0^Q q/C dq` If V is the final potential difference between capacitor plates, then Q = CV `W=(CV)^2/(2C)=1/2CV^2=1/2QV` This work is stored as electrostatic potential energy of capacitor i.e., Electrostatic potential energy, `U=Q^2/(2C)=1/2CV^2=1/2QV` If V is the findal potential difference between capacitor plates, then Q = CV Energy density: Consider a parallel PLATE capacitor consisting of plates, each of area A, separated by a distance d. If space between the plates is filled with a medium of dielectric constant K, then Capacitance of capacitor, `C=(K epsilon_0A)/d` If `sigma` is the surface charge density of plates, then electric field strength between the plates `E=sigma/(K epsilon_0) rArr sigma=K epsilon_0 E` Charge on each plate of capacitor `Q=sigmaA=K epsilon0 EA` `therefore` Energy stored by capacitor, `U=Q/(2C)=((Kepsilon_0EA))/(2(Kepsilon_0A1d))=1/2Kepsilon_0E^2Ad` But Ad = volume of space between capacitor plates Energy stored, `U=1/2K epsilon_0 E^2 Ad` Electrostatic Energy stored per unit volume, `UE = U/(Ad)=1/2K epsilon_0 E^2` This is expression for electrostatic energy density in medium of dielectric constant K. In air or free space (K = 1), therefore energy density, `u_e=1/2Kepsilon_0E^2` |
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| 6583. |
a vertical wall is in South north direction. a current carrying where is kept in the world such that to the west of the wall magnetic field due to wire is towards south then the wire where should be |
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Answer» VERTICAL and current in downwards |
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| 6584. |
What are coherent sources of light ? |
| Answer» Solution :light SOURCES which emit light waves of same WAVELENGTH (or frequency) having EITHER zero or a constant originating phase DIFFERENCE are called coherent sources of light. | |
| 6585. |
A solid sphere of mass 10 kg and diameter 5 cm rolls without slipping on a smooth horizontal surface with velocity 5 cm/s. Its total kinetic energy is |
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Answer» `175 xx 10^(-4) J` |
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| 6586. |
A relativistic particle with rest mass m cillides with a stationary particle of mass M and activities at reaction leading to formation of new particles: m+Mrarrm_(1)+m_(2)+….., where the rest masses of newly formed particles are written on the right-hand side. Making use of invariance of the quantity E^(2)-p^(2)c^(2), demonstrate that the threshold kinetic energy of the particle m required for this reaction is defined by Eq.(6.7c). |
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Answer» <P> Solution :With PARTICLE masses standing for the names of particles, the reaction is`m+Mrarrm_(1)+m_(2)+…..` On R.H.S. let the energy momenta be `(E_(1),cvec(p)_(1)),(E_(2),cvec(p)_(2))` etc. On the left the energy momentum of the particle `m is (E,cvec(p))` and that of the particle is `(Mc^(2),VEC(O))`, where OFCOURSE, the usual relations `E^(2)-c^(2)vec(p)^(2)=m^(2)c^(4)eta` hold. From the conservation of energy momentum we see that `(E+Mc^(2))^(2)-c^(2)vec(p)^(2)=(SigmaE_(i))^(2)-(SigmaE_(i))^(2)-(Sigmacvec(p)i)^(2)` Left hand side of `m^(2)c^(4)+M^(2)c^(4)+2Mc^(2)E` We evaluate the R.H.S. in the frame where `Sigmavec(pi)=0` (`CM` frame of the decay product). Then R.H.S `=(SigmaE_(i))^(2)=(SigmaE_(i))^(2) ge(Sigmam_(i)c^(2))^(2)` beacuse all energies are `+ve`. THEREFORE we have the result `E ge((Sigmam_(i))^(2)-m^(2)-M^(2))/(2M)c^(2)` M or Since `E=mc^(2)+T`, we see that `T geT_(th)` where `T_(th)=((Sigmam_(i))^(2)-(m+M)^(2))/(2M)c^(2)` |
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| 6587. |
A rod of length 10.0 cm lies along the principal axis of a concav mirror of focal length in such a way that the end closer to the pole is 20.0 cm away from it. The length of the image is : |
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Answer» 10.0 cm  For end M, u = - 30 cm `therefore (1)/(u) + (1)/(V) = (1)/(f)` `therefore (1)/(v) = (1)/(u) - (1)/(u) = (-1)/(10) = (-1)/(15)` `therefore` v = - 15 cm = distance of image of end M. , Thus lenght of image of rod = 20 - 15 = 5 cm. |
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| 6588. |
A hydrogen atom in the ground state absorbe 12.09 eV of energy. The change in the orbital angular momentum of the electron is: |
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Answer» `+1.05 XX 10^(-34)Js` `E_(3)=-1.51eV for n=3` `E_(3)-E_(1)=[(1.51-(-13.6)]=12.09eV` `triangleE=12.08eV for trianglen=3-1=2` Now `L=n.(h)/(2pi)` `TRIANGLEL=trianglen.(h)/(2pi)` here `trianglen=2` `triangleL=2 .(h)/(2pi)=h/pi` `triangleL=(6.6 xx 10^(-34))/(3.14)=2.11 xx 10^(-34)Js` |
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| 6589. |
Assume that the earth moves around the sun in circular orbit of radius R and there exists a planet which also moves around the sun in a circular orbit with an radius of the orbit of the planet is |
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Answer» <P>`2^(-2//3)R` As we KNOW, `T_(e )=(2pi)/(omega_(e ))` and `T_(p)=(2pi)/(omega_(p))` `:. (T_(e ))/(T_(p))=(omega_(p))/(omega_(e ))""`...(i) Again, from Kepler.s law, we get `(T_(e )^(2))/(T_(p)^(2))=(R_(e )^(3))/(R_(p)^(3)) rArr ((T_(e ))/(T_(p)))^(2)=((R_(e ))/(R_(p)))^(3)` `rArr (omega_(rho))/(omega_(e ))=((R_(e ))/(R_(rho)))^(3//2)""`[from eq. (i)] `rArr (R )/(R_(rho))=((2omega_(e ))/(omega_(e )))^(2//3)=2^(2//3)""[.: R_(e )=R]` `rArr R_(rho)=2^(-2//3)R` |
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| 6590. |
A baseball is hit straight up and is caught by the catcher 2.0 s later, at the same height at which it left the bat. The maximum height of the ball during this interval is : |
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Answer» 4.9 m |
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| 6591. |
For a long sighted person, least distant of distinct vision is x meter. He wants to read news paperplaced atmeter from him. He uses spectacle having lenses of power xD. What is x :- |
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Answer» `0.5` f=1m |
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| 6592. |
Figure given below shows four progressive waves A, B, C and D with their phases expressed withrespect to the wave A. It can be calculated from the figure that :- |
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Answer» the WAVE C is ahead by a phase angle of`(PI)/(2)`and the wave B LAGS behind by a phase angle of `(pi)/(2)` angleof `(pi)/(2) `, waveDhasphasedifferenceof `pi` |
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| 6593. |
A convex lens of focal length 15cm and a concave mirror of focal length 30cm are kept with their optic axes PQ and RS paralledl but separated in vertical direction by 0.6 cm as shown in Figure. The distance between the lens and mirror is 30cm. An upright object Ab of height 1.2 cm is placed on the optics axis PQ of the lens at a distance of 20 cm from the lens. If A^(')B^(')is the image after refraction from the lens and reflection from the mirror, find the distance of A^(')B^(') from the pole of the mirror and obtain its magnification. Also, locate positions of A^(')and B^(') with respect to the optic axis RS. |
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Answer» Solution :For convex lens using sign convectionof coordinate geometry, `u=+20cm, f=-15cm` So, `(1)/(f)=(1)/(v)-(1)/(u)rArr -(1)/(15)=(1)/(v_(1)) -(1)/(20)` `rArr (1)/(v_(1)) =(1)/(20)-(1)/(15)=(3-4)/(60)rArrv_(1)=-60cm` i.e., IMAGE is formed at a distance 60 CM to the left of lens L. Magnification, `m_(1)=(v_(1))/(u_(1))=-(60)/(20)=-3` This image is real and inverted. It is intercepted by the mirror. For concave mirror, `u_(2)=-60+30=-30cm, f_(2)=+30cm` So, `(1)/(f)=(1)/(v)+(1)/(u)` gives `(1)/(30)=(1)/(v_(2))-(1)/(30)rArr (1)/(V_(2))=(1)/(30)+(1)/(30)=(2)/(30) rArr v_(2)=15cm` Magnification, `m_(2)=-(v_(2))/(u_(2))=-(15)/(-30)=+(1)/(2)` `:.` Net magnification, `m=m_(1)xxm_(2)=(-3)XX((1)/(2)) =-1.5 ` Size of image `A^(')B^(') = -1.5XX1.2cm=-18cm`. Magnification of mirror is HALF and image of B fomred by convex lens is 0.6 cm above RS, so the length of image will be 1.5 cm below RS. Thus, `B^(')` will be 0.3 cm above RS and `A^(')`will be 1.5 cm below RS.  
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| 6594. |
The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 5 cm, then the object distance for objective will be in cm) |
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Answer» `1.8` |
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| 6595. |
A triangle has sides of length 6, 7 and 8. The line through its incenter parallel to the shortestside is drawn to meet the other two sides at P and Q. The length of the segment PQ is |
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Answer» `12/5`  Let AD=h , ID=x AREA of triangle ABC=R.s `=r((6+7+8)/2)` `=21/2r` …(1) Area of triangle `ABC=1/2BCxxh` `=1/2(6)xxh=3h` …(2) From (1) and () `"2lr"/2=3h` `r/h=6/21=2/7` TriangleABC and triangle APQ are similar `therefore (PQ)/(BC)=(h-r)/h` `(PQ)/6=1-r/h` `(PQ)/6=1-2/7` `rArr PQ=30/7` |
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| 6596. |
Which semiconducting material can be used for constructing LED if it is to emit light in the visible range? |
| Answer» Solution :We can use that semiconducting material for which forbidden ENERGY gap `(E_(G))` is 1.8 eV or more so that light PHOTONS emitted lie in visible light range. | |
| 6597. |
A particle has a charge of +1.5 mu C and moves from point A to point B, a distance of 0.20 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPE_(A) - EPE_(B)=+9.0 xx 10^(-4) J. Find the magnitude and direction of the electric force that acts on the particle. |
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Answer» `3.0 xx 10^(-3)` N, from A TOWARD B |
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| 6598. |
In 1954, ............... invented the first digitally operated programmable robot called unimate. |
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Answer» EDWARD purcell |
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| 6599. |
If the electric flux entering and leaving a closed surface are 6 xx 10^(6) and 9 xx 10^6 S.I. units respectively, then the charge inside the surface of permittivity of free space epsi_(0) is |
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Answer» `epsi_(0) XX 10^(6)` |
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| 6600. |
In the above problem what force should the man exert on the rope to get his correct weight on the machine ? |
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Answer» Solution :`T-CANCEL(Mg)+cancel(Mg)=Ma` `(because R=Mg)a=(T)/(M)` `T-Mg-mg=ma` `T-m(T)/(M)=(m+M)g` `T=((m+M)Mg)/(M-m)=((30+60)60xx10)/(60-30)` `=(""^(3)cancel(90)xx600)/(cancel(30))=1800 N` 
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