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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. |
Two cylinders A and B of equalcapacity are connected to each other via a stopcock. A containsa gas at standard temperature and pressure. B is compeltely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :(a)What is the final pressure of the gas in A and B? (b)What is the change in internal energy of the gas (c ) What is the change in the temperatureof the gas (d) Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface? |
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Answer» Solution :The gas undergoes FREE expansion. During this PROCESS the work done dw = 0 (zero). As the system is thermally insulated dQ = 0 (zero). Hence there is no change is internal ENERGY or temperature. (a) Initial state of gas P, V final after expansion `P_(1)`, 2V from Bolyer law `PV=P_(1)2V` `:. P_(1)=(P)/(2)"":.`Pressure is halved (b) dU = 0 (c )dT = 0 (d) During the intermediate stages, the parameters change quickly and the gas does not attain equilibrium. So the gas does not obey gas equation hence they do not appear on P, V, T surface. |
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| 2. |
Steam at 100^(@)C is passed into 22 grams of water at 20^(@)C. When resultant temperature is 90^(@)C, then weight of the water present is |
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Answer» 27.33g |
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| 3. |
Taking that earth revolving round the earth in circular orbitsof radius 15xx10^(10^(m)) with a time period of 1 year . The time taken by another planet which is at a distance of 50xx10^(10m) to revolve round the sun in circulation orbit once will be |
| Answer» Answer :A | |
| 4. |
Find the time in which a layer of ice 3 cm thick on the surface of a pond will increase in thickness by 1 mm when the temperature of the surrounding is -20°C .the thermal conductivity of ice is 2.1 W/m K, latent heatof ice 3.36 10^5J/kg and density is 900 kg/m^-3. |
| Answer» SOLUTION :3 MIN 36 s | |
| 5. |
A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m . The height of the table from the ground is 0.8 m . If the angular speed of the particle is 12 rad s^(-1) . Find the magnitude of its angular momentum about a point on the ground right under the centre of the circle . |
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Answer» Solution :Here , m = 2 kg , `OMEGA = 12 rad s^(-1)` `R = SQRT((0.8)^(2) + (0.6)^(2)) = 1` m Angular momentum of the particle about point O , `L = mvr sin 90^(@)` `= m XX (0.6 omega) r` `=2 xx 0.6 xx 12 xx 1 = 14.4 kg m^(2) s^(-1)` . 
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| 6. |
Two droplets merge with each other and forms a larger droplet .In this process…. |
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Answer» energy is liberated. |
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| 7. |
Let V and E denote the gravitational potential and gravitational field at a point. It is possible to have (a) V = 0 and E = 0(b) V = 0 and E != 0 (c) V != 0 and E = 0(d) V != 0 and E != 0 |
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Answer» only a, B are TRUE |
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| 8. |
Assuming the earth to be a uniform sphere ofmass M and radium R. which of the following graphs represents the variation of acceleration due to gravity with distance .r. from the centre of the earth. |
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Answer»
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| 9. |
A glass of radius 10^(-3) and density 2000 kg m^(-3) fall in a jar filled with oil of density 800 kg m^(-3) .The terminal velocity is found to be 1 cm/s. Calculate the coefficient of viscosity of oil |
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Answer» |
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| 10. |
A metal ball of mass 2 kg moving with speed of 36 km h^(-1) has a head on collision with a stationary ball of mass 3 kg. If after collision, both the balls move together, then the loss in kinetic energy due to collision is |
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Answer» Solution :Applying principle of conservation of linear momentum , `m_1 u_1 + m_2u_2 = (m_1 + m_2) v` `2 xx 10 + 3 xx 0 = (2 + 3) v :. v = 4 ms^(-1)` Initial KE, `K_1 = 1/2 m_1u_1^2 = 1/2 xx 2 xx 10^2 = 100 J` Final KE, `K_2 = 1/2 (m_1 + m_2) v^2 = 1/2 (2 + 3) xx 4^2 = 40 J` `:. ` Loss in KE = `K_1 - K_2 = 100 - 40 = 60 J`. |
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| 11. |
Write the short note on radius of gyration. |
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Answer» Solution :As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, a new parameter called radius of gyration is define. It is related to the moment of inertia and the total mass of the body. A rigid body rotating about given axis its each particle having mass m suppose it is MADE of n particles, then total mass of rigid body `M=nm`.  The moment of inertia about given axis, `I=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)+....m_(n)r_(n)^(2)` `=mr_(1)^(2)+mr_(2)^(2)+mr_(n)^(2) [because m_(i)=m]` `=m[r_(1)^(2)+r_(2)^(2)+....r_(n)^(2)]` `=mn[(r_(1)^(2)+r_(2)^(2)+....r_(n)^(2))/(n)]` [`because` Dividing and multiplying by n] `I=Mk^(2)[because mn=Mandk^(2)=(r_(1)^(2)+r_(2)^(2)+.....r_(n)^(2))/(n)]` but `k^(2)` = average `r^(2)(ltr^(2)gt)` HENCE, `k^(2)` denotes the MEAN of squares of perpendicular distance of the particles of the body from the given axis. k is called the radius of gyration. By ANOTHER way, from formula of the moment of inertia of different body comparing with `I=Mk^(2),k^(2)` is obtained and its square root is k. k is a geometric property of the body and axis of rotation. The radius of gyration of a body about a axis defined as the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment ofinertia is equal to the momet of inertia of the body about the axis. The unit of radius of gyration is the unit of length m and its dimensional formula is `[L^(1)]`. |
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| 12. |
Which one of the following forces is a pseudo force? |
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Answer» FORCE of friction |
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| 13. |
The capillary rise of water in a tube depends on |
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Answer» the OUTER RADIUS of the tube |
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| 14. |
The moment of inertia l of a solid sphere having fixed volume depends upon its volume V as |
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Answer» `I PROP V` |
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| 15. |
The elongation of a spring of length 'L' and of negligible mass due to a force is 'x'. The spring is cut into two pieces of length in the ratio 1: n. The ratio of the respective spring constants is |
| Answer» ANSWER :A | |
| 16. |
We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature consist till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state ? (Hydrogen molecules can be considered as spheres of radius 1Å). |
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Answer» Solution :No. In the FINAL state, one is not justifiedin assuming the ideal gas law as detailed below: Number of MOLECULES in 0.5 G of hydrogen= `(6.023 XX 10^(23))/(2) xx 0.5 = 1.5 xx 10^(23)` Volume of each molecules = `4/3 pi r^(3) = 4/3 xx 22/7 (10^(-10))^(3) = 4 xx 10^(-30) m^(3)` `:.` TOTAL volume of the molecules contained, `V_(1) = 1.5 xx 10^(23) xx 4 xx 10^(-30)m^(3) = 6 xx 10^(-7)m^(3)` Final volume available , `V_(1) = 1.5 xx 10^(23) xx 4 xx 10^(-30) m^(3) = 6 xx 10^(-7)m^(3)` As `V_(2) = V_(1)`, the intermolecular forces cannot be ignored as done in ideal gas situation. |
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| 17. |
What is the ratio of angular speed of hour hand and minute hand of a clock? |
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Answer» Solution :ANGULAR SPEED of hour hand : `omega_(1)=(2pi)/(12xx3600)" rad/s"` For minute hand angular speed : `omega_(2)=(2pi)/(3600)" rad/s"` `THEREFORE (omega_(1))/(omega_(2))=(1)/(12)` `therefore omega_(1):omega_(2)=1:12` |
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| 18. |
A drop of kerosene oil placed on water |
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Answer» REMAINS as a drop |
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| 19. |
The length of a uniform ladder of massm is 1. It leans against a smooth wall making an angle theta with the horizontal. What should be the force of friction between the floor and ladder so that the ladder does not slip ? |
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Answer» Solution :Let the reaction at P and Q be `N_(1)` and `N_(2)` respectively as shownin fig. For the equilibrium of the ladder, `N_(2)=mg `and `N_(1)=F` where F is the force of friction between the ladder and the floor. Taking MOMENTS about Q. `N_(1)xxPQ sin theta=mg XX(PQ)/(2) costheta` `N_(1)=(1)/(2)mgcot theta` Force of friction `F=N_(1)=(1)/(2)mg cot theta` 
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| 20. |
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km ^(-1) in 10 minute. Find its acceleration. |
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Answer» `1ms ^(-2)` `16= v _(0) + 8a ""…(1)` Now, `d = 9M, t =3s` `9= 3v_(0) + (9a)/(2) THEREFORE 18 = 6v_(0) + 9a ""…(2)` `therefore 48=12v_(0) + 24 a ` (eq. (i) is multiplied by 3) `36=12v _(0) + 18 a ` (eq. (ii) is multiplied by 2) `{:(-,-,-), (12=6a,,),(therefore a = 2ms ^(-2),,):}` |
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| 21. |
A father and his seven year old son are facing each other on ice skates. With their hands, they push off against one another. Regarding the forces that act on them as a result of this and the accelerations they experience, which of the following is correct ? |
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Answer» FATHER EXERTS more force on the son and experiences less ACCELERATION |
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| 22. |
Select the correct reason for,Small droplets of a liquid are usually more spherical in shape than larger drops of the same liquid because: |
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Answer» FORCE of SURFACE tension is equal and opposite to the force of gravity. |
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| 23. |
One end of a uniform rod of length l and mass m is hinged at A. It is released from the rest from horizontal position AB as shown in figure. The force exerted by the rod on the hinge when it becomes verticle is |
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Answer» `3/2mg` |
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| 24. |
Define 'G'. What are the units and dimensions of 'G' ? |
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Answer» SOLUTION :If `m_(1)=m_(2)=1andr=1` F = G Universal gravitational constant is defined as the FORCE of ATTRACTION between two bodies of unit mass each and are placed unit DISTANCE apart. Unit of G: `Nm^(2)kg^(-2)` Dimension of G:`M^(-1)L^(3)T^(-2)` |
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| 25. |
When the tense wire of sitar is pulled slightly from the middle and then released which type of waves are produced ? |
| Answer» Solution :Machanical and transvers WAVES in the wire of SITAR which INDUCES mechanical and LONGITUDINAL waves in the surrounding AIR medium. | |
| 26. |
(A): The carnot cycle is useful in understanding the performance of heat engines.(R ): The Carnot cycle provides a ways of determining the maximum possible efficiency achievable with reservoirs of given temperatures. (2006) |
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Answer» Both (A) and(R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 27. |
A stone is thrown upwards with initial velocity 'u' from top of a tower and it reaches to ground with velocity '3u' then height of the tower is ...... |
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Answer» `(3u ^(2))/(g) ` `(because v = 3u, u=u, a = g and x=h)` `THEREFORE 9u^(2) -u^(2) =2gh` `therefore h = (8u^(2))4u^(2))/(g)` |
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| 28. |
The contact angle between a solid and a liquid is a property of |
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Answer» the material of the SOLID |
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| 29. |
Figure shows the position-time graph of particle moving along a straight line. Match the entries of Column I with the entries of Column II {:("Column-I"," Column-II"),("A) The particle A is ","P) Accelerating"),("B) The particle B is ","Q) Decelerating"),("C)The particle C is ","R) Speeding up"),("D) The particle D is ","S) Slowing down"):} |
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Answer» |
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| 30. |
An ideal gas compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? |
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Answer» Solution :Work done on the gas = AREA under the curve `W_("ADIABATIC") GT W_("ISOTHERMAL") gt W_("isobaric")` `W_("isochoric")` is OBVIOUSLY zero because in an isochoric process there is no change in Volume. 
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| 31. |
The position x of a particle at time t is given by x=(V_(0))/(a)(1-e^(-at)), where V_(0) is constant and a gt 0. The dimensions of V_(0) and a are |
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Answer» `[M^(0)LT^(-1)]` and `[M^(0)L^(0)T^(-1)]` `:. [a] = (1)/([t])= (1)/([T])= [T^(-1)]= [M^(0)L^(0)T^(-1)]` ALSO `[x]= ([V_(0)])/([a])` or `[V_(0)]= [x][a]= [L][M^(0)L^(0)T^(-1)] = [M^(0)LT^(-1)]` |
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| 32. |
The circular motion of a particle with uniform speed is………… |
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Answer» PERIODIC and SIMPLE HARMONIC |
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| 33. |
A pendulum beating seconds at one place (g=985cm//sec^(2)) is taken ot another place (g=981cm //sec^(2)). If it is to beat seconds at another place, its length is to decreases by |
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Answer» `0.1cm` |
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| 34. |
Find the massof 1 cubicmetre of moist air at 27^(@)Cs and 759.2mm pressure, the dew point being 10^(@)C. (Saturated vapour pressure at 10^(@)C=9.2mm. Density of dry air at STP =1.293 kg m^(-3). Density of moisture at any temperature and pressure =5/8 of density of dry air at the same temperature and pressure.) |
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Answer» |
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| 35. |
What is the efficiency of a carnot engine working between ice point and steam point ? |
| Answer» SOLUTION :`ETA = (T _(1) - T _(2)) //T_(1) = (373 - 273 ) // 373 = 0.268 = 26.8%` | |
| 36. |
Find the gravitational potential energy of system consisting of uniform rod AB of mass M, length l and a point mass m as shown in figure. |
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Answer» |
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| 37. |
The random error in the arithmetic mean of 50 observations is 'a', then the random error in the arithmetic mean of 200 observations would be ............ |
| Answer» SOLUTION :`a/4` | |
| 38. |
An aluminium sphere is dipped into water at 10^(@)C. If the tem.perature is increased, the force of buoyancy |
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Answer» Will increase |
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| 39. |
Maximum and minimum magnitudes of the resultant of two vectors of magnitudes P and Q are in the ratio 3:1. Which of the following relations is true |
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Answer» `P=2Q` |
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| 40. |
(A) : When bubble comes from the bottom of a lake to the top Its radius increases. (R ) : When bubble rises to top of a lake pressure decreases. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 41. |
A wooden block of 100 kg is about to be pushed on floor of coeficient of friction 0.4. The magnitude of the force of riction on the wooden block when it is just pushed is |
| Answer» ANSWER :A | |
| 42. |
If the radius of the earth is doubled without changing its mass, what will be the change in weight of body of m on the surface of earth ? |
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Answer» SOLUTION :`IMPLIES` Weight`W = mg = (GM_(E)m)/(R_e^2)` Radius of the earth is doubled, Weight W. = mg. `= (GM_(e) m)/(4R_e^2)=W/4 ` becomes `1//4^(TH)` of original weight . |
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| 43. |
Velocity and acceleration of a particle at time t=0 are vecu =(2hati + 3hatj)m//s s and a = (4hati +2hatj) m//s^(2)respectively. Find the velocity and displacement of particle at t = 2s. |
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Answer» Solution :Here, acceleration `vec(a) = (4hat(i) + 2HAT(j))m//s^(2)` is contant. So, we can apply `vec(V) = vec(u) + vec(a)t` and `vec(s) = vec(u) t + (1)/(2) vec(a) t^(2)` substituting the proper values, we get `vec(v) = (2hat(i) + 3hat(j)) + (2)(4hat(I) + 2hat(J)) = (10hat(I) + 7hat(J))m//s` and `vec(s) = (2)(2hat(i) + 3hat(j)) + (1)/(2)(2)^(2)(4hat(I) + 2hat(J)) = (12 hat(I) + 10hat(J))m//s` Therefore, VELOCITY and DISPLACEMENT of particle at t = 2s are `(10 hat(i) + 7hat(j))m//s` and `(12hat(i) + 10 hat(j))m` respectively. |
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| 44. |
The maximum and minimum values of the resultant of two forces acting at a point are 7 N and 3 N respectively, the smaller force is equal to |
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Answer» 4 N |
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| 45. |
Moment of inertia of a body depends upon |
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Answer» DISTRIBUTION of MASS of the body |
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| 46. |
A uniformpressure.P.is exertedonall sidescubeattemperaturet^(0)C. Bywhatamountshould the temperatureof the cube beraisedin orderto bring its volumeback to the volume it hadbeforethe pressurewas applied, if the bulk modulusand coefficientof volume expansion of the materialare B and gamma respectively. |
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Answer» <P> Solution :A by definitionfo bulk MODULUS`B= - V((DeltaP)/(DeltaV))`, withincreasein pressure decreasein volumeof the CUBE will be given by `-DeltaV=(VP)/(B)`, (as `DeltaP=P`) Now withrise in temperaturedue to thermalexpansion, volumeincreasesso if .`Deltatheta`. is the RISE in temperature then `DeltaV=V gamma Delta THETA ("as "gamma=(DeltaV)/(V Deltatheta))` As the volume of the cube remainsconstant `(VP)/(B) =Vgamma Delta theta rArr Delta theta =(P)/( gammaB)` |
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| 47. |
Two blocks are in contact on a frictionless table. A horizontal force is applied to one block as shown in fig. If m_1=10kg and m_2=5kg and F=15N. Find the force of contact between the two bodies. |
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| 48. |
A uniform wooden plank of mass 280 kg and length L is floating on still water with its one end at the origin O and its length lying along +Ve x-axis. A man of mass 84 kg standing at the end O, slowly walks to the other end. {:(,"Column - I",,"Column - II"),((A),"Displacement of man relative to water",(P),L),((B),"Displacement of centre of mass of plank relative to wate",(Q),(3)/(13)L),((C ),"Displacement of man relative to plank",(R ),(10)/(13)L),((D),"Final distance of centre of mass of system from O",(S),(5)/(13)L):} |
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| 49. |
The resistance of the wire of a Callendar and Barnes apparatus is 10 ohms at 20^(@)C. A cell of steady volate 2.2 volts and internal resistance 1 ohm was connected to it. A liquid was slowly and steadily forced through it and the temperatures of the incoing and outgoing flow of liquid were found to b 18^(@)C and 22^(@)C, respectively in the steady state. The liquid collected in 40 minutes was 120g. Find the specific heat capacity of the liquid. Neglect loss of heat by radiation. |
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Answer» |
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| 50. |
Steam at 100^(@)C is passed into 20 g of water at 10^(@)C, then water acquires a temperating of 80^(@)C, the man of water present will be [Take specific heat of water = 1 cal g^(–1) ""^(@)C^(–1) and Latent heat of steam = 540 cal g^(–1)] |
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Answer» 24 g `20 xx 1 xx (80 –100) = m xx 540 + m xx 1 × (100 – 80)` 1400=560 m `m= (1400)/(560)` =2.5 g Total MASS of water = 20 + 2.5 = 22.5 g |
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