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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 is projected up along a rough incline with a velocity of u = 10 m/s. After 4 s the block was at point B at a distance of 5 m from the startingpoint A and was travelling down at a velocity ofv = 4 m/s. (a) Find time after projection at which the block came to rest. (b) Find the coefficient of friction between the block and the incline. Take g = 10 m//s^(2) |
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Answer» (B) 0.18 |
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| 2. |
A uniform cube of side 10 cm weighs 880gxxg. It is floating in saline water (specific gravity of saline water = 1.1). What will be the thrust on each face of the cube? |
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Answer» Solution :Density of the material of the cube = `880/10^(3)=0.88g*cm^(-3)`. Since the density is less than the density of saline WATER, some part of the cube remains above water. Let the lower SURFACE of the cube be at a depth of x cm from the upper surface of saline water. So, according to the CONDITION of floatation, weight of the cube = weight of the saline water DISPLACED by the cube or, `880xxg=x xx10xx10xx1.1xxgor,x=8cm`  So, the area of the part of a side of the cube that remains immersed in the saline water, `A=8xx10=80cm^(2)` and the average depth that of the surface = `(0+x)/2=x/2`. Hence, the thrust on each lateral surface of the cube `=x/2xx1.1xxgxxA=8/2xx1.1xx981xx80` `=3.45xx10^(5)DYN`. |
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| 3. |
Four particles, each of mass 1kg, are placed at the corners of a square of side one meter in the X - Y plane. If the point of intersection of the diagonals of the square is taken as the origin, the coordinates of the centre of mass are |
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Answer» (1,1) |
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| 4. |
(A): The bridges declared unsafe after a long use.(R) : Elastic strength of bridges losses with time due to elastic fatigue. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 5. |
A uniform stick of length L is pivoted at one end on a horizontal table. The stick is held forming an angle theta_(0) with the table. A small block of mass m is placed at the other end of the stick and it remains at rest. The system is released from rest (a) Prove that the stick will hit the table before the block if cos theta_(0) ge sqrt((2)/(3)) (b)Find the contact force between the block and the stick immediately before the system is released.Take theta_(0) = cos^(-1) (sqrt((2)/(3))) (c) Find the contact force between the block and the stick immediately after the system is released if theta_(0) cos ^(-1) (sqrt((2)/(3))). |
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Answer» (C) ZERO |
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| 6. |
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each gm then error in the value of resistance of the wire is : |
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Answer» 0 |
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| 7. |
Statement I: During free fall of a person one feels weightlessness because his weight becomes zero. Statement II: He falls with an acceleration of g. |
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Answer» Statement I is true, statement II is true, statement II is a CORRECT EXPLANATION for statement I. |
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| 8. |
An aluminium wire of length 60 cm and cross sectional area10^(-2) cm^(2)is joined with a steel wire of the same cross sectional area. A 10 kg mass is used to produce tension in the conjoined wires. The length of the steel wire is 86.8 cm . A transverse wave is produced in the wire with the help of an external source . Find the minimum frequency of vibration in the wire for which a stationary wave is produced with a node formed at the junction of the two wires.The densities of aluminium and steel are 2.6 g*cm^(-3) and 7.8 g*cm^(-3), respectively . |
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Answer» |
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| 9. |
The rate of work done is |
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Answer» energy |
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| 10. |
When a soapwater bubble is given a positive charge it expands. If it is given a negative charge then it |
| Answer» Answer :A | |
| 11. |
Theelectricfieldin a regions is givenby vec(E) = E_(0) (x)/(L) hat(i) . Findthechargecontainedinsidea cubical volume boundedbythe surfacex =0,x=L,y = 0,z = 0and z=L. |
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Answer» SOLUTION :At x = 0,E = 0 andat `x = L , vec(E) = E_(0) hat(i)` Thedirection of the field is alongthe X - AXIS ,so it will cross the yz-faceof thecube. Thefluxof thisfield .  `PHI =phi_("left face") + phi_("RIGHT face") = 0 + E_(0)L^(2) = E_(0)L^(2)` By Gauss.s law ,`phi = (q)/(in_(0)) "" therefore q = in_(0) phi= in_(0)E_(0)L^(2)` |
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| 12. |
A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then aconstant force F starts acting on the block of mass M to pull it. Find the force on the block of mass m |
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Answer» `(MF)/M` |
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| 13. |
The volume of a sphere is 1.76cm^(3). The volume of 25 such spheres will be correctly represented by |
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Answer» `0.44 xx 10^(2) CM^(3)` |
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| 14. |
Determine the elastic potrntial energy stored in stretched wire. |
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Answer» Solution :When a wire is stretched, work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire. Consider a wire of length L and area of cross section A. Let deforming force F be the applied on the wire and l be the increase in length of the wire. `Y = (FL)/(ADeltaL)` here U is the Young.s modulus of wire. `DELTA l = l` Let work dW is needed to increase in small length dl. `therefore dW=F XX dl OR Yal (dl)/(L)` Work W is done for INCREAE the length of wire from L to `L + l.` It is work for `l =0` to `l =l` `W = int _(0)^(l) (YAl)/(L) dl (YA )/(2) (l ^(2))/(L)` `W = 1/2 xx Y xx ((l )/(L)) ^(2) xx AL` `1/2xx` Young modulus `xx("Strain")^(2) xx` Volume of wire `W =1/2 xx ` Strees `xx ` Stress ` xx` Volume of wire Hence, work stored in wire it the elastic potential energy. So the stored elastic potential energy per unit is obtained from `u = 1/2 SIGMA epsi` SI unit of elastic potential density is `Nm ^(-12) or `Pa and dimensional formula `[M^(1) L ^(-1) T ^(-2)].` Elastic potential energy in eleastic body in equal to the enclosed area under the applied force F to the body versus change in length. But enclosed area under the graph of stress versus strain is equal to the elastic potnetial energy per unit voluem (energy) (density) of the body. |
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| 15. |
In Doppler effect, when a source moves towards a stationery observer, the apparent increase in frequency is due to |
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Answer» INCREASE in WAVELENGTH of sound RECEIVED by OBSERVER |
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| 16. |
Cooking gas containers are kept in a lorry moving with uniform speed. What will be the effect on the temperature of the gas molecules? |
| Answer» Solution :As there is no change in the translation KINETIC ENERGY of the gas MOLECULE, the temperature of gas will REMAIN the same. | |
| 17. |
A planet is revolving round the sun . Its distance from the sun at apogee is r_A and that at perigee is r_P.The masses of plancet and sun are .m. and M respectively , v_A is the velocityof planet at apogee and V_P is at perigee respectively and T is the time period of revolution of planet round the sun, then identify the wrong answer. |
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Answer» `T^2=pi^2/(2GM)(r_A+r_P)^3` |
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| 18. |
A spherical soap bubble of radius 1 cm is formed inside another of radius 4 cm. The radius of single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is --------------- cm. |
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Answer» 1 |
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| 19. |
Two identical satellites are orbiting are orbiting at distances R and 7R from the surface of the earth, R being the radius of the earth. The ratio of their |
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Answer» KINETIC energies is 4 `r_(1)=R+R=2R and r_(2)=7R+R=8R` (`because` R is the radius of the earth) Now, `K=(GMm)/(2r)rArr(K_(1))/(K_(2))=(r_(1))/(r_(2))=4` `U=-(GMm)/(r),(U_(1))/(U_(2))=(r_(2))/(r_(1))=4` and `E=-(GMm)/(2r)or(U_(1))/(U_(2))=(r_(2))/(r_(1))=4`. |
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| 20. |
A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as shown in figure. A horizontal acceleration of magnitude a_(0) is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is |
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Answer» `sqrt((4R sin alpha)/(a_(0)))` |
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| 21. |
A constant power 'P' is applied to a particle of mass 'm'. The displacement of the particle when its velocity increases from v_(1) " to " v_(2) is (ignore friction) |
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Answer» Solution :Power P= F.v (ma) v `a= (P)/(MV) rArr v.(dv)/(ds)= (P)/(mv)` `v^(2).dv= (P)/(m)ds " " (P)/(m) UNDERSET(0)overset(s)INT ds= underset(v_(1))overset(v_(2))int v^(2)dv` `(P)/(m).s = (1)/(3) (v_(2)^(3)-v_(1)^(3)) THEREFORE` 
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| 22. |
In a private school a physics teacher was asked by her student how to understand first law of thermodynamics in an easier way by using hints or any examples. The teacher was also thinking for a long time and when she was taking classes for 12^(th) standard students about Einstein's photoelectric equation, she was striked with an idea for clearing doubt of her 11^(th) std students. Can you guess her idea? (ii) Derive and Give expression for C_(P)-C_(V)&'gamma' for Monoatomic molecule. |
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Answer» Solution :The lady teacher explained first law of thermodynamics comparing Einstein's PHOTO electric equation. When an photo of energy (of some joule) incident on a metal surface, a part of the energy of the photon will be USED to eject electrons from each atm in the metal surface (i.e. SYSTEM) and remaining part of energy of the photon will be used to impart kinetic energy to the ejected electrons. Here the energy of photon is compared with heat energy `(DeltaQ)`. supplied to the gas (system). A part of the energy of `DeltaQ` will be used in increasing the internal energy of the gas by increasing its temperature. This is compared with the kinetic energy in the case of photoelectric equation and after gaining temperature. The molecules that are excited expands and does some work on the system. This is compared with the work function in photoelectric equation i.e. `DeltaQ=DeltaU+DeltaW` (ii) Mono Atomic molecule : Average K.E. of a molecule `=(3)/(2)KT` Total energy of a mole of gas `(3)/(2)KTxxN_(A)=(3)/(2)RT` For one mole, the molar specific heat at CONSTANT volume `C_(V)=(dU)/(dT)` `=(d)/(dT)[(3)/(2)RT]` `C_(V)=(3)/(2)R` `C_(P)=C_(V)+R` `C_(P)=(5)/(2)R` The ratio of specific heats, `gamma=(C_(P))/(C_(V))=((5)/(2)R)/((3)/(2)R)=(5)/(3)` `gamma=1.67` |
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| 23. |
In a private school a physics teacher was asked by her student how to understand first law of thermodynamics in an easier way by using hints or any examples. The teacher was also thinking for a long time and when she was taking classes for 12^(th) standard students about Einstein's photoelectric equation, she was striked with an idea for clearing doubt of her 11^(th) std students. Can you guess her idea? (i) Define degrees of freedom? |
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Answer» Solution :The lady teacher explained first law of thermodynamics comparing Einstein's photo electric EQUATION. When an photo of energy (of some joule) INCIDENT on a metal surface, a part of the energy of the photon will be used to eject electrons from each atm in the metal surface (i.e. system) and remaining part of energy of the photon will be used to IMPART kinetic energy to the ejected electrons. Here the energy of photon is compared with heat energy `(DeltaQ)`. supplied to the gas (system). A part of the energy of `DeltaQ` will be used in increasing the internal energy of the gas by increasing its temperature. This is compared with the kinetic energy in the case of photoelectric equation and after gaining temperature. The molecules that are excited expands and does some work on the system. This is compared with the work function in photoelectric equation i.e. `DeltaQ=DeltaU+DeltaW` (i) Degrees of freedom : The minimum number of independent coordinates needed to specify the POSITION and CONFIGURATION of a thermodynamical system in space is called degrees of freedom of the system. Ex : A particle moving in space has 3 degrees of freedom. |
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| 24. |
A body .x. with a momentum .p. collides with another identical stationary body .y. one dimensionally. During the collision .y. gives an impulse .J. to the . body .x.. Then the coefficient of restitution is: |
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Answer» <P>`(2J)/p -1` |
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| 25. |
A spring of force constant 'K' is cut into two pieces such thal one piece is double the length of the other. Then the long piece will have a force constant of |
| Answer» Answer :B | |
| 26. |
Why a cricket player lowers his hands while catching a cricket ball? Explain. |
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Answer» Solution :We know, impulse= force`XX` time `""` change in linear momentum. Force not only depends on the change in momentum but also how fast the change is brought about. The same change in momentum brought about in shorter time needs GREATER force. by lowering his hands while catching the CRICKET ball, the player allows a longer time for the momentum change, thereby PREVENTING his hands from getting INJURED. |
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| 27. |
Mathch Column -I with Column-B: |
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Answer» `therefore (a)rarr(q,r)` (b) As velocity `(v=72//6=12ms^(-1))` REMAINS the same, the motion is univorm linear motion. Also x-t graph is a straight line inclined to x-axis. `therefore (b)rarr(p,s)` (c) When body is thrown VERTICALLY upward, the velocity decreases (while going upward) and thenincreases (while coming downward). Therefore, the motion is non-uniform linear motion. The motion is under the ACTION of force of gravity, the `therefore (c)rarr(q,r)` (d) When a bu llet is fired into air, it is an example of non-linear motion. Also,linear momentum does not remain CONSERVED as the motion is under the action of force of gravity. `therefore (d)rarr(q,r)` |
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| 28. |
An iceberg of density 900 kgm^(-3) is floating in water of density 1000 kgm^(-3). What is the percentage of volume of iceberg outside the water ? |
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Answer» Solution :Fraction of volume INSIDE water = Relative DENSITY of the BODY `(v.)/v = 900/1000 = 0.9` Fraction of volume outside water = `1- 0.9 = 0.1` PERCENTAGE of volume outside water = `0.1 xx 100 = 10%`. |
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| 29. |
Where does the acceleration due to gravity is larger, equator or poles of the earth? Why? |
| Answer» Solution :`IMPLIES` The acceleration of GRAVITY g is larger at poles of earth because RADIUS of earth near poles is LESS than the radius near equator by about 21 km and acceleration DUE to gravity `g prop 1/R_(e)^2` andhence g is larger at poles than at the equator. | |
| 30. |
A pendulum that beats seconds on the surface of the earth were taken to a depth of (1//3) th the radius of the earth. What will be its time period of oscillation? |
| Answer» Answer :A | |
| 31. |
A cylinder is released from rest from the top of an incline of inclination theta and length 'l' . If the cylinder roles without slippling its speed at the bottom |
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Answer» `SQRT((4glsintheta)/(3))` |
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| 32. |
A steel wire is 1m long and 1mm^2in area of cross-section. If it takes 200 N to stretch this wire by 1mm how much force will be required to stretch the wire of the same area and of same material having length of 10m to 1002 cm |
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Answer» 100N |
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| 33. |
The torque (tau) due to a force gives us the turning effect of the force about a fixed point/axis. It is measured by the product of magnitude of force (F) and perpendicular distance (r ) of the line of action of force form the axis of rotation. overset rarr(tau) = overset rarr(r ) xx overset rarr(F) = rF sin theta hatn where thete is smaller angle between overset rarr(r ) and hatn is unit vector along overset rarr(tau). Read the above passage and answer the following questions : (i) What is the significance of torque ? (ii) How do you determine the direction of torque ? (ii) What is the implication of this concept in day to day life ? |
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Answer» Solution :(i) Torque plays the same role in rotational motion as FORCE plays in linear motion. (II) The direction of torque, `overset rarr(tau)` is given by right handed screw rule. `overset rarr(tau) _|_ overset rarr(r ) and overset rarr(tau) _|_overset rarr(F)`. (iii) From `oversetrarr(tau) = overset rarr(r ) xx overset rarr(F) = rF sin THETA`, we find that when `theta = 0^(@) = zero`. In day to day life, it IMPLIES that when you align yourself along the force (i.e., your boss), the turning effect on you becomes zero, ity, say GOD, i.e., when you merge your wishes with the wishes of God, you are stable. |
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| 34. |
When an ideal diatomic gas is heated atconstant pressure, the fraction of heat energy supplied which is used in doing work to maintain pressure constant is |
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Answer» `5//7` |
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| 35. |
A heavier sphere moving eastward with a certain velocity .v. collides with a lighter sphere at rest. If it is perfect elastic head on collison, then after collision |
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Answer» heavier sphere MOVES WEST ward with same speed |
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| 36. |
Calculate the temperature at which rms velocity of SO_2 is the same as that of oxygen at 27^@C |
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Answer» Solution :For `O_2,v_(RMS)=sqrt((3RT)/M)=sqrt((3Rxx300)/32)` For ` SO_2, V'_(rms)=sqrt((3RT')/M')=sqrt((3RT')/64)` As v' = v `:. sqrt((3RT')/64)=sqrt((3Rxx300)/(32))` (or) `T'=600K = 600-273=327^@C` |
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| 37. |
A block A of mass 3kg and another block B of mass 2 kg are connected by a light inextensible string as shown in figure. If the coefficient of friction between the surface of the table and A is 0.5. What maximum mass C is to be placed on A so that the system is to be in equilibrium ? |
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Answer» 3kg |
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| 38. |
What will be the power incident normally on unit area of spherical surface which is at distance R from the outermost surface of Sun of radius r at t^(@)C ? sigma= Stefan - Boltzmann's constant |
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Answer» `(r^(2)SIGMA(t+273^(4)))/(4piR^(2))` `P=AesigmaT^(4)` `:.P=4pir^(2)sigmaT^(4)""[becauseA=4pir^(2),e=1]` POWER INCIDENT per unit area, `=(P)/(4piR^(2))=(4pir^(2)sigmaT^(4))/(4piR^(2))=(r^(2)sigma(t+273)^(4))/(R^(2))` |
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| 39. |
A wall clock uses a vertical spring mass system to measure the time. Each time the mass reaches an extreme position, the clock advances b a second. The clock gives correct time at the equator. If the clock is taken to poles it will |
| Answer» Answer :D | |
| 40. |
flat plate is separated from a large plate by a layer of glycerine of thickness 3 xx 10^(-3)m. If the coefficient of viscosity of glycerine is 2 Ns//m^(2)what is the force required to keep the plate moving with a velocity of 6xx10^(-2)m//s. Area of the plate is4.8 xx 10 ^(-3) m |
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Answer» |
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| 41. |
Mercury is used in liquid thermometers because it has |
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Answer» HIGH SPECIFIC HEAT and high conductivity |
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| 42. |
By neglecting the air resistance is a heavy and light body each have same acceleration due to gravitation ? |
| Answer» Solution :`implies` YES, it does not DEPEND on the MASS of the BODY. | |
| 43. |
Why absolutely pure simple harmonic motion is not possible? |
| Answer» Solution :ABSOLUTELY pure SIMPLE harmonic motion is not possible because, pure simple harmonic motion is an ideal but from definate conditions SYSTEMS executing simple harmonic motion can be SEEN. | |
| 44. |
If a friction of 200 N exerted oppositeto the motion of a objectdragged up to 10m , then what will be the work done by the friction force on the road ? |
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Answer» SOLUTION :`W = Fdcos theta ""[ :. d = 0 , theta =PI]` ` :. W = 0` Joule |
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| 45. |
A band music at a frequency f is moving towards a wall at a speedv_(b) . A motorist is following the band with a speedv_(m). If v is the speed of sound , obtain an expression for the beat frequency heard by the motorist . |
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Answer» SOLUTION :Two separate sounds will be heard by the motorist one is DIRECT from band and other is the echo the WALL . Apparent frequency of direct sound . `f_(1) = f ((V + v_(m))/(v+ v_(b))) ` Apparent frequency of the echo , `f_(2) = f ((v + v_(m))/(v+ v_(b))) ` Therefore , the beat frequency heard by the motorist, ` n= f_(2) - f_(1)= f(v + v_(m)). [(1)/(v - v_(b)) - (1)/(v+ v_(b))]` `= (2fv_(b) (v + v_(m)))/(v^(2)- v_(b)^(2))` |
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| 46. |
Figure shows a box of mass m is placed on a wedge of mass .M. on a smooth surface. How much force .F. is required to be applied on .M. so that during motion .m. remains at rest on its surface . |
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Answer» Solution :The force action on the bodies m and M are shown in the figure along with free BODY diagrams of mass m and M. As the two bodies move together, we can find the acceleration of system towards right directly as `a=F/(m+M)`  Here the condition is , the small block of mass .m. should remain at rest on the incline SURFACE of the wedge block. Look at the FBD of m, the force ACTING on it towards left is .ma. . It is teh pseudo force on it as its reference frame is the wedge . As wedge . As wedge is moving with an acceleration , we consider m relative to it . Now with respect to wedge block m is at rest or in EQUILIBRIUM , we can balance all the forces along the tendency of motion of body (i.e., inclined plane) and perpendicular to it shown in FBD of it. For m to be at rest , from FBD of m , along the plane MG `sin theta = ma cos theta "or" a =g tan theta "or" F/(m+M)=g tan theta ,F=(m+M) g tan theta` |
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| 47. |
Passage - I: An instrument known as triple point cell ( at a temperature of 273.16 K and pressure of 4.6mm of mercury) in which triple point is attained for water. It consists of a U-tube containing ice-cold water to evacuate air by means of an exhaust pump and sealed. It is then lowered into a Dewar's flask containing ice-water mixture. The space corresponding to the inner portion of the vessel is filled with a freezing mixture and this space is referred as "thermometric well". As a result, the layer of the water freezes into ice. Thus the space in the U-tube contains ice, water. water-vapor all the three in equilibrium and the corresponding coordinates of pressure and temperature the "triple point of water" 1. Evacuated Cell 2. Inner U-tube 3. Ice-Water Mixture 4. Freezing Mixture 5. Water-frozen into ice 6. Water Vapour 7. Dewar's Tube Triple Point cell is an experimental arrangement |
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Answer» To practically demonstrate /observe the triple POINT of water |
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| 48. |
Passage - I: An instrument known as triple point cell ( at a temperature of 273.16 K and pressure of 4.6mm of mercury) in which triple point is attained for water. It consists of a U-tube containing ice-cold water to evacuate air by means of an exhaust pump and sealed. It is then lowered into a Dewar's flask containing ice-water mixture. The space corresponding to the inner portion of the vessel is filled with a freezing mixture and this space is referred as "thermometric well". As a result, the layer of the water freezes into ice. Thus the space in the U-tube contains ice, water. water-vapor all the three in equilibrium and the corresponding coordinates of pressure and temperature the "triple point of water" 1. Evacuated Cell 2. Inner U-tube 3. Ice-Water Mixture 4. Freezing Mixture 5. Water-frozen into ice 6. Water Vapour 7. Dewar's Tube In the space within the U-tube as given in the description |
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Answer» The layer in contact with region ( 4) of FIGURE is frozen into ice |
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| 49. |
On which factors centre of mass of rigid body depends? |
| Answer» SOLUTION :CENTRE of mass DEPENDS upon the mass distribution and rigid BODY, shape and size of the body as well as the AXIS of system. | |
| 50. |
The bob of a simple pendulum is made of iron. Exactly below the equilibrium position of the pendulum, the pole of a strong magnet is placed. What will be the change in the time period of the pendulum? |
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Answer» SOLUTION :In addition to the GRAVITATIONAL force, MAGNETIC force of attraction of the magnet acts on the pendulum bob. If acceleration produced by the magnetic field on the bob is a, total acceleration of the pendulum bob g. = g + a `therefore` Time period of the simple pendulum `T=2pisqrt(L/(g+a))""...(1)` It is understood from equation (1) that the time period decreases due to the increase in acceleration of the pendulum bob. |
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