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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A block of mass m and surface area A just begins to slide down an inclined plane when the angle of inclination is pi//5. Keeping the mass of the block same, if the surface area is doubled, the inclination of the plane at which the block starts sliding will be |
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Answer» `pi//5` |
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| 2. |
1 g of steam at 100^(@)C is passed into a calorimeter of water equivalent 10 g containing water of mass 90 g. Find the resultant temperature of the mixture if the initial temperature of water is 27^@C. |
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| 3. |
A stone is thrown vertically up from a bridge with velocity 3ms^(-1). If it strikes the water under the bridge after 2s, the bridge is at a height of (g=10ms^(-2)) |
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Answer» 26m |
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| 4. |
In the diagram shown, no friction at any contact surface. Initially, the spring has no deformation. What will be the maximum deformation in the spring ? Consider all the strings to be sufficiency large. Consider the spring constant to be K. |
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Answer» In `C` frame `/_\K=0` `W_(|peseudo|)=W_(s)+W_(1F)=W_(2F)=/_\K=0`…….i  Work done by pseudoframe in `C`-frame is zero. `W_(1F)=Fx_(1C)` `W_(2F)=2Fx_(2C)` `0-1/2kx_(max)^(2)+FX_(1C)+2Fx_(2C)=0`.........ii `1/2k_(max)^(2)=Fx_(1C)+2Fx_(2C)` `x_(max)=x_(1C)+x_(2C)` From the concept of centre of MASS `Mx_(1C)=2Mx_(2C)` `X_(1C)=2x_(2C)` `:. x_(1C)+2X_(2C)=x_(max)` `:. 3x_(2C)=x_(max)` `x_(2C)=x_(max)/3` and `x_(1C)=(2x_(max))/3` Puttng the valuesin eqn ii `1/2kx_(max)^(2)=F((2x_(max))/3)+2F(x_(max)/3)` `1/2kx_(max)=(2F)/3+(2F)/3` `x_(max)=(8F)/(3K)` |
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| 5. |
energy supplied to convert unit mass of substance from solid to liquid state at its melting point is called |
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Answer» LATENT heat of fusion |
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| 6. |
Find the absolute error in the parallel combination of R_(1) and R_(2) |
| Answer» SOLUTION :`DeltaR_(P)=R_(P)^(2)XX(DeltaR_(1))/(R_(1)^(2))+R_(P)^(2)xx (DeltaR_(2))/(R_(2)^(2))` | |
| 7. |
Find the PE of three objects of masses 1kg, 2kg and 3 kg placed at the three vertices of anequilateral triangle of side 20 cm. |
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Answer» 50Gj |
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| 8. |
What is the moment of inertia of a disc about one of its diameter? |
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Answer» Solution :Suppose a DISC of mass M and RADIUS R in XY plane as shown in figure. Z is perpendicular axis to the plane PASSING through the centre of disc. Moment of inertia of disc about Z-axis `I_(Z)=(MR^(2))/(2)` ACCORDING to perpendicular axis theorem `I_(z)=I_(x)+I_(y)""....(1)` Since disc is symmetrical with axis x and y, `I_(x)=I_(y)` `therefore` From eqn. (1), `I_(z)=2I_(x)""........(2)` but `I_(z)=(MR^(2))/(2)` `(MR^(2))/(2)=2I_(x)` `therefore I_(x)=(MR^(2))/(4)andI_(x)=I_(y)` `therefore I_(y)=(MR^(2))/(4)` |
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| 9. |
Diameter of a glass bottle with a flat base is 4 cm and the bottle has mass 30 g. Centre of gravity of the empty bottle is 10 cm above the base. Up to what height the bottle should be filled with water such that , when it floats vertically in a tank of water such that when it floats vertically in a tank of water, its centre of gravity will be at the mid-point of the immersed portion of the bottle? |
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| 10. |
An elastic string carrying abody of mass 'm' extends be 'e'. The body rotation in a vertical circle with critical velocity. The extension in the string at the lowest position is |
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Answer» `2e` |
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| 11. |
A hollow cylinder of mass M and radius R is rotating about its axis of symmetry and a solid sphere of same mass and radius is rotating about an axis passing through its centre. If torques of equal magnitude are applied to them, then the ratio of angular accelerations produced is |
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Answer» `(2)/(5)` Moment of inertia of SOLID sphere about an axis passing through its centre`I_(2) = (2)/(5) MR^(2)` Let `alpha_(1)` and `alpha_(2)` be ANGULAR acceleration produced in the cylinder and the sphere respectively on applying same torque `tau` in each CASE . then `alpha_(1) = (tau)/(I_(1)) (As rau = I alpha) and alpha_(2) = (tau)/(I_(2))` Their corresponding ratio is `(alpha_(1))/(alpha_(2)) = (I_(2))/(I_(1)) = ((2)/(5) MR^(2))/(MR^(2)) = (2)/(5)` |
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| 12. |
A unifromsphereof massm andradiusR is placed ona rough horizontalsurface(figure).thesphere isstuuck horizontally at a heighth fromthefloor . Match the following |
Answer» Solution :Considerthe diagramwherea sphere ofm andradius ,R, struckhorizontally atheighthabove thefloor  the spherewillrollwithoutsippingwhen `omega=(v)/( R)` , WHEREV islinear velocityand `omega` is angular velocity of thesphere Now , angular momentumof sphere, aboutcentreof mass ( [We areapplyingconservationof angular momentumjustmomentumjustbeforeandafter struck ] `mv(h-R)=Iomega=((2)/(5)mR^(2))((v)/(R))` ` impliesmv(h-R)=(2)/(5)MVR` `h-R=(2)/(5)Rimplies h=(7)/(5) R` Thereforethesphereeill rollwithout sippingwitha constantvelocityandhence, noloss ofenergy,so`(D) to (i)` Torquedueto appliedforce,F aboutcentre ofmass For `tau=0,h=R` ,spherewillhaveonlytranslational motion. itwould loseenergy byfriction hence, `(B ) to (iv)` thesphere willspinclokwisewhen`taugt 0implies hgt R` therefore , ` (C ) to (II)` the spherewill spinanti- clockwisewhen`tault 0implies hltR,(a) to (iii)` |
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| 13. |
The surface tension of a liquid |
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Answer» INCREASES with area |
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| 14. |
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 Gpa. What is the vertical deflection of this face? |
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| 15. |
Each question contains statements given in two columns, which have to be matched. Statements in column-I are labelled as A and D whereas statements in column-II CP are labelled as p, q, r and S. Match the entries of DP S column-I with appropriate entries of column-II. Each entry in column-I may have one or more than one correct option from column-II. The answers to these questions have to be appropriately bubbled as illustrated in the given example, if the correct matches are A to (q,r),B to (p,s), C to (r,s) and D to (q). |
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| 16. |
The mean distance of mass from the sun is 1.524 times that of earth from the sun. Find the number of years required for mass to make one revolution about the sun. |
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Answer» SOLUTION :GIVEN that `R_(m) = 1.524 R_(e) and T_(e)=1` year According to Kepler.s third law , we can write `T^(2) prop R^3` `implies ((T_s)/(T_e))^(2) = ((R_s)/(R_e))^3 implies T_s=Te((R_S)/(R_e))^(3//2)` `implies T_(s)=(1) ((1.524R_e)/R_e)^(3//2) implies T_(s) = (1) (1.524)^(3//2) = 1.88` YEARS |
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| 17. |
A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of |
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Answer» `SQRT2:1` |
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| 18. |
The high thermal conductivity of metal is due to free electrons. The relevant electron property is : |
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Answer» its being charged |
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| 19. |
The bob of pendulum is released from a horizontal position . If the length of the pendulum is 1.5 m,what is the speed with which the bob arrives at the speed with whichthe bob arrivesat the lowermost point , given that it dissipated 5 % of its intial energy against air resistance ? |
Answer» Solution : KINETIC ENERGY at point B = 95 % POTENTIAL energy at point A , `1/2 MV^(2) = 0.95 xx mgh` ` :. 1/2 mv^(2) = 0.95 xx mgh` ` :. V^(2) = 2XX 0.95 xxgh` ` :. v = sqrt(2xx0.95 xx9.8 xx1.5) = 5.285 ms^(-1)= 5.3 ms^(-1)` |
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| 21. |
A uniform thin rod of mass m and length l is standing on a smooth horizontal suface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling.Find the velocity of centre of mass of the rod at the instant when it makes an angle theta with horizontal. |
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Answer» Solution :The table is frictionless. The centre of MASS of the scale moves vertically downward as the scale falls. Initial kinetic energy K is zero and the potential energy U is `mg l/2`. Total energy of the scale`E = K + U` `E =0 + mg l/2` The position of the scale after certain time is shown in Fig. 7.2.54 using the DOTTED line. LET `theta`be the angle through which the scale has rotated. Let the angular velocity be m and the linear velocity of the centre of mass be v.  `K =1/2 Iomega^(2) + 1/2mv^(2)` `U = mg(1/2-s)` Total energy is conserved as there is no friction `1/2Iomega^(2)+ 1/2mv^(2) + mg(l/2-s) = mg l/2` `1/2 I omega^(2) + 1/2mv^(2) - mgs =0`........(i) From Fig. `s=l/2 -l/2 cos theta =l/2(1-cos theta)` Differentiating `(ds)/(dt) =-l/2 xx sin theta xx (dtheta)/(dt)` `v=l/2 sin thetaomega, omega = (2v)/(l sin theta)` Also `I = (ml^(2))/12` Substituting in eqns (i) `1/2 xx (ml^(2))/12 xx (4V^(2))/(l^(2) sin^(2) theta) + 1/2 mv^(2) -mgs =0` `v^(2)/2 [1/(3 sin^(2)theta)+1] =gs, v^(2)[(1+3 sin^(2)theta)/(3 sin^(2)theta)]=2gs` `v=[(6gs sin^(2)theta)/(1+3 sin^(2)theta)]^(1/2)` |
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| 22. |
A rocket of mass 6000 kg is set for vertical firing. If the exhaust speed is 1000m/sec. The rate of mass of gas ejected to supply a thrust needed to give the rocket an initial upward acceleration of 20.2 m//s^(2) is |
| Answer» Answer :D | |
| 23. |
The velocity of a particle under going SHM at the mean position is 2ms^(-1). Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude. |
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Answer» `1.234ms^(-1)` |
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| 24. |
(A) : All physically correct equations are dimensionally correct. (R) : All dimensionally correct equations are physically correct. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 25. |
If veca=2hati-hatj-5hatk and vecb=hati+hatj+2hatk, then find scalar and vector product. |
| Answer» SOLUTION :SCALAR product = `-9`, vector product `=3(hati-3hatj+hatk)` | |
| 26. |
Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air … |
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Answer» decreases by a factor 10 `therefore (v _(1))/( v _(2)) = (lamda _(1))/( lamda_(2)) implies (350 )/( 3500) =(lamda_(1))/( lamda_(2)) implies lamda_(2) = 10 lamda_(1)` `therefore` Wavelength becomes 10 times its initial VALUE. |
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| 27. |
An object of mass 5kg falls from rest through a vertical distance of 20 m and reaches a velocity of 10 m/s. How much work is done by the push of the air on the object ? (g = 10 m//s^(2)) |
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| 28. |
Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass 2m placed at the centroid G of the triangle? (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = CG = 1 m |
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Answer» Solution :(a) The angle between GC and the positive x-axis is `30^(@)` and so is the angle between GB and the negative x-axis. The individual forces in vector notation are `F_(GA) = (Gm(2m))/(1) hatj` `F_(GB) = (Gm(2m))/(1) ( - hati cos 30^(@) - hatj sin 30^(@))` `F_(GC) = (Gm(2m))/(1) ( + hati cos 30^(@) - hatj sin 30^(@))` From the principle of superposition and the law of vector addition, the resultant gravitational force `F_(R)` on (2m) is `F_(R) = F_(GA) + F_(GB) + F_(GC)` `F_(R) = 2Gm^(2) hatj + 2Gm^(2) ( - hati cos 30^(@) - hatj sin 30^(@))` `+ 2Gm^(2) (hati cos 30^(@) - hatj sin 30^(@)) = 0 ` Alternatively. one expects on the basis of symmetry that the resultant force ought ot the zero. (b) Now if the MASS at vertex A is doubled then `F_(GA)= (G2m.2m)/(1) hatj = 4Gm^(2) hatj` `F_(GB) = F_(GB) and F_(GC) = F_(GC)` `F_(R)= F_(GA) + F_(GB) + F_(GC)` `F_(R) = 2 GM^(2) hatj` For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction. We have to add up these forces vectoriallyfor all the point masses in the extended object to get the total force. This is easily done using calculus. For two SPECIAL cases, asimple law results when you do that : (1) The force of attraction between a hollow SPHERICAL shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell. Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell LEAVING only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above. (2) The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely 
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| 29. |
Consider the equation T = 2pi sqrt(l/g) and check whether it is correct or not. |
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Answer» SOLUTION :We know, dimensional. Formulae of T, l and g are `[T], [L], [LT^(-2) ]` respectively. `THEREFORE` Dimensional formula of `sqrt((l)/( g)) = sqrt((L)/( LT^(-2) ) )= sqrt(T^(-2) ) = [T]` Dimensional formula of LEFT side is also `[T]`. As `2pi` is a dimensionless QUANTITY, the dimensional formula of left hand side of the equation is equal to dimensional formula of right hand side. Hence we conclude that the given equation is a CORRECT one. |
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| 30. |
Two particles P and Q perform SHM with amplitude A and time period T along same line. At time t=0, P is at +A/2 moving along+x axis. Particle Q is at phase difference . The time at which they collide is 5T/12: |
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Answer» this notion when described in a circular motion in circle of RADIUS A and TIME period T the angle made by particle with horizontal diameter will be `30^@` in clockwise direction |
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| 31. |
The distance covered by a body in time (5.0 pm 0.6)sis (40.0 pm0.4)m . The percentage error in the speed is |
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Answer» 0.13 |
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| 32. |
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much highabove the ground can the cricketer throw the same ball ? |
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| 33. |
Find the bulk modulus of an oil of volume 1.6 m^(3) when a pressure of 250 atomosphere is applied on it causing a change of volume by 8l. K=50xx 1.013xx10^(8)=50.65xx10^(8)=5xx10^(7)N//m^(2) (approximately) |
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| 34. |
The thermodynamic process in which the volume of the system remains constant is called |
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Answer» ISOTHERMAL |
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| 35. |
A certain amount of gas is taken through acyclic process (A B C D A) that has two isobars, one isochore and one isothermal. The cycle can be represented on a P-Vindicator diagram as : |
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Answer»
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| 36. |
What are the factors affecting the surface tension of a liquid? |
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Answer» SOLUTION :(i) The presence of any cotamination orimpurities. (ii) the presence of dissolved SUBSTANCES. (III) electrification (iv) temperature |
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| 37. |
A Carnot engine having an efficiency of eta = (1 / 10) as a heat engine is used as a refrigerators. If the work done on the system is 10 J, what is the amount of energy absorbed from the reservoir at lowest temperature ? |
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Answer» Solution :As HEAT ENGINE, LET Q the heat energy ABSORBED from the SOURCE and `Q_(2)` be the energy rejected to the sink. Then `Q_(1)=W+Q_(2)=10+Q_(2)"…(1)"` `"Given :"eta=1-(Q_(2))/(Q_(1))=(1)/(10)` `(Q_(2))/(Q_(1))=1-(1)/(10)=(9)/(10)` `Q_(1)=(10)/(9)Q_(2)"...(2)"` From equation (1) and (2), `10+Q_(2)=(10)/(9)` `10=((10)/(9)-1)Q_(2)` `therefore""Q_(2)=90J` |
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| 38. |
Block A of mass M = 2 kg is connected to another block B of mass m= 1 kg with a string and a spring of force constant k = 600 N//m as shown in Fig. Initially, spring is compressed to 10 cm and whole system is moving on a smooth surface with a velocity v = 1 m//s. At any time, thread is burnt, the velocity of block A, when B is having maximum velocity w.r.t. ground, is |
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Answer» zero  CONSERVATION of momentum, `2v_(1)+1v_(2)(2+1)xx1implies2v_(1)+v_(2)=3`…..i Conservation of energy `1/22v_(1)^(2)+1/21v_(2)^(2)=1/2600(10/100)^(2)=1/23(1)^(2)implies2v_(1)^(2)+v_(2)^(2)=9` From EQN i and ii `v_(1)=2m//s, v_(2)=-1m//s` or `v_(1)=0, v_(2)=3m//s` The second solution is possible. |
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| 39. |
Assertion :when a springs of force constatn k is cut into two equal halves , the force cosntant of each half become K//2 Reasion : when an elastic springs is elongated by length x work done is (1)/(2)kx^(2) |
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Answer» If the assertion and reason are CORRECT and reason is a correct EXPLANATION of the assertion |
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| 40. |
Can a quantity have dimensions but still has no units? |
| Answer» SOLUTION :No, a quantity having DIMENSIONS must have some UNITS of its MEASUREMENT. | |
| 41. |
Which of the following planets has two moons phobos and deimos ? |
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Answer» Jupiter |
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| 42. |
A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinctic energy is lost by impact, the value of coefficient of restitution is |
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Answer» `1/(2sqrt(2))` Here, `u_2 = 0 and v_1 = 0` `K_i = 1/2 m u_1^2 + 1/ m u_2^2 = 1/2 m u_1^2` `K_f = 1/2 mv_1^2 + 1/2 mv_2^2 = 1/2 mv_2^2` Loss of K.E. = `1/2 m u_1^2 - 1/2 m v_2^2` According to questions, `1/2 (1/2 m u_1^2) = 1/2 m u_1^2 - 1/2 mv_2^2` (as half of KE is LOST be IMPACT) `:. u_1^2 = 2v_2^2 " or " v_2 = (u_1)/(sqrt2) :.e = (v_2 - v_1)/(u_1 - u_2) = (v_2)/(u_1) = 1/(sqrt2)`. |
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| 43. |
A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness (t)/(4) . Then the relation between the moment of inertia I_(x) and l_(y) is |
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Answer» `I_(y) = 32 I_(X)` where `rho` = DENSITY of material of disc. `therefore I_(X) = (1)/(2) m_(X) R^(2) = (1)/(2) piR^(2) t rho R^(2)` or `I_(X) = (1)/(2) pi rho t R^(4)` Mass of disc (Y) , `m_(Y) = pi (4R)^(2) (t)/(4) ""rho = 4 pi R^(2) t rho` `therefore I_(Y) = (1)/(2) m_(Y) (4R)^(2) = (1)/(2) 4 pi R^(2) t rho (16R^(2)) = 32 pi t rho R^(4)` `therefore (I_(Y))/(I_(X)) = (32 pi t rho R^(4))/((1)/(2) pi rho t R^(4)) = 64 or I_(Y) = 64 I_(X)` |
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| 44. |
A piece of copper having a rectangular cross -section of 15.2 mm xx 19.1 mm is pulled in tention with 44,500 N force, producing only elastic deformation. Calculate the resulting strain ? |
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| 45. |
A ball is thrown vertically down with velocity of 5m//s. With what velocity should another ball be thrown down after 2 seconds so that it can hit the 1^(st) ball in next 2 second. |
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Answer» `40m//s` |
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| 46. |
The mass of a spool of wire in the form of a uniform solid cylinder is m and its radius is r. The wire is unwound under a constant force F. Assume that the cylinder does not slip, find (i) the acceleration of the centre of mass, (ii) the force of friction, (iii) what is the speed attained by the centre of mass after the cylinder has rolled through a distance, assume that the cylinder starts from rest and it rolls without slipping ? |
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Answer» Solution :Mass of the cylinder =m Radius =r Force = F Due to the external force APPLIED F, the cylinder starts rotating in the CLOCKWISE direction. The force of friction/acts towards right as shown in Fig. 7.2.53. For the linear MOTION of the centre of mass, if a is its acceleration, then, `F +f = ma`......(i) The torque CAUSING rotational motion is: `tau = I alpha` `I =1/2 mr^(2), alpha = a/r` `Fr - fr =1/2 mr^(2) xx a/r` `F -f = (ma)/2`......(ii) From eqns. (i) and (ii) `2F =ma + (ma)/2 = (3ma)/2, a = (4F)/(3m)` Substituting the value of a in equation (i) `F + f = m xx (4F)/(3m)` `f = (4F)/3 - F = F/3` `v^(2) = u^(2) + 2AS, v^(2) = 2 xx (4F)/(3m) xx s, v = sqrt((8Fs)/(3m))` 
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| 47. |
Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components p_(x), p_(y) and p_(z). Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. |
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Answer» Solution :If position vector of a particle is `vecr` and LINEAR momentum `VECP` than angular momentum `vecl=vecrxxvecp` but `vecr=xhati+yhatj+zhatk` and `vecp=p_(x)HATI+p_(y)hatj+p_(z)hatk` are the respectively COMPONENTS of `vecr and vecp` `vecl=l_(x)hati+l_(y)hatj+l_(z)hatk` `:.vecl=[xhati+yhatj+zhatk]xx[p_(x)hati+p_(y)hatj+p_(z)hatk]` `l_(x)hati+l_(y)hatj+l_(z)hatk=|(hati,hatj,hatk),(x,y,z),(p_(x),p_(y),p_(z))|` `=hati(yp_(z)-zp_(y))+J(zp_(x)-xp_(z))+hatk(xp_(y)-yp_(x))` From this `l_(x)=yp_(z)-zp_(y)` `l_(y)=zp_(x)-xp_(z)` and `l_(z)=xp_(y)-yp_(x)` The particle moves in the XY plane hence `Z=0 and p_(z)=0` `:.vecl=0hati+0hatj+(xp_(y)-yp_(x))hatk` `:.vecl=(xp_(y)-yp_(x))hatk` Therefore, when the particle is confined to move in the XY-plane, the direction of angular momentum `(vecl)` is along the Z-direction. |
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| 48. |
A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens out. Then he descends with a net retardation of 2m//sec^(2). His velocity when he reaches the ground is 8m/sec. Find te height at which he gets out of the aeroplane? |
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Answer» SOLUTION :Distance he falls before the parachute OPENS is `1/2g xx 100=490m` Then his velocity `=gt=98.0 m//s=u` Velocity on reaching GROUND =8=V retardation =2 `v^(2)-u^(2)=2as""8^2-(98)^(2)=2 xx (-2)S` `S=(106 xx 90)/(4)=2385m` Total distance =2385+490 =2875m=height of aeroplane. |
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| 49. |
A particle of mass 'm' is executing oscillations about the origin on the x - axis. Its P.E is U(x) = K |x|^(3), where 'k' is a positive constant. If the amplitude of oscillation is 'a', then how does its time period 'T vary with amplitude. |
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Answer» Solution :`U(x) = K |x|^(3) "" F= (DU)/(dx)= 3kx^(2)` At extreme POSITIONS , `x=-a` `F= maomega^(2)= 3KA^(2)` `omega prop sqrt(a):. T prop (1)/(sqrt(a))` HENCE`T alpha ("amplitude")^(-1//2)` |
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| 50. |
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is - 20^@ C, the temperature of the surroundings to which it rejects heat is _____ |
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Answer» `21^@C` `=T_L/(T_H-T_L)` where `T_L` = lower temperature in K `T_H`=HIGHER temperature in K `therefore 5=T_L/(T_H-T_L)` `therefore 5T_H-5T_L=T_L` `therefore 5T_H=6T_L` `=6/5xx(273-20)` `=6/5xx253` =303.6 K `therefore T_H`=303.6-273=`30.6^@C approx 31^@C` |
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