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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. |
From a uniform disc of radius R, a small disc of radius (R)/(2)is cut and removed as shown in the diagram. Find the center of mass of the remaining portion of the disc. |
Answer» Solution :Let us consider the MASS of the uncut full disc be M. Its center of mass would be at the geometric center of the disc on which the origin coincides. Let the mass of the small disc cut and REMOVED be mn and its center of mass is at a position `(R)/(2)` to the RIGHT of the origin as shown in the figurr.  Hence, the remaining portion of the disc should have its center of mass to the left of the origin, say, at a distance x. We can write from the principle of moments, `(M-m)x=(R)/(2)` `x=((m)/(M-m))(R)/(2)` If `sigma` is the surface mass density (i.e., mass per unit surface area). `sigma (M)/(PIR^(2))` of small disc is,surface mass density surface area m= surface mass density `xx` surface area `m= sigma xx pi ((R)/(2))^(2)` `m=((M)/(pi R^(2)))pi((R)/(2))^(2)=(M)/(pi R^(2)) pi(R^(2))/(4)=(M)/(4)` substituting m in the expression for x `x=((M)/(4))/((M-(M)/(4)))xx(R)/(2)=((M)/(4))/(((3M)/(4)))xx(R)/(2)` `x=(R)/(6)` he center of mass of the remaining portion is at a distance `(R)/(6)` to the left from the center of the disc. |
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| 2. |
A vessel contains one mole of O_2 gas (molar mass 32) at a temperature T. The pressure of the gas 1s.What will be the pressure of one mole of He (molar mass 4) at a temperature 2t in an identical vessel? |
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Answer» <P> Solution :Pressure of oxygen at temperature T isLet the pressure of Helium be `P_("He")`at a temperature 2 T Number of molecules of Helium `n_("OXY")=32/16=2` Number of molecules of Oxygen `n_(He) = 4/2=2` It is GIVEN that `(V_("OXY"))/(V_("He)) = 1` From standard GAS equation `P=(nRT)/V` We get , `P_("He")/(P_("OXY"))=(n_(He))/(n_("OXY"))xxT_(He)/(T_("OXY"))xxV_("OXY")/(V_(He))` `P_("He")/(P_("OXY"))=2/2xx(2T)/Txx1` `:. (P_(He))/(P_("OXY"))=2` `:.`Pressure of 1 mole of Helium =2 `xx` Pressure of 1 mole of oxygen |
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| 3. |
If .g. on the surface of the earth is 9.8 ms^(-2), find its value at a depth of 3200km (radius of the earth = 6400km) |
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Answer» SOLUTION :with d=3200 KM `g_d=g[1-d/R] RARR g_d=4.9 MS^(-2)` |
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| 4. |
An 8 kg metal block of dimensions 16 cm xx 8 cm xx 6 cm is lying on a table with its face of largest area touching the table. If g = 10 m s^(-2), the minimum amount of work done in making it stand with its length vertical is |
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Answer» Solution :When face of the largest AREA is touching the table, height of centre of gravity above the table `h_1 = 6/2 = 3 cm` With its length vertical, height of centre of gravity would become `h_2 = 16/2 = 8 cm` Minimum work required `W = (P.E.)_(2) - (P.E.)_(1) = MG(h_2 - h_1)` `:. W = (8 xx 10(8 - 3))/(100) = 4J` . |
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| 5. |
Two 3kg masses have velocities barv_(1)=2hat(i)+3hat(j)m//s and barv_(2)=4hat(i)-6hat(j)m//s. Find a) velocity of centre of mass, b) the total momentum of the system, C) The velocity of centre of mass 5s after application of a constant force bar(F)=24hat(i)N, d) position of centre of mass after 5s if it is at the origin at t=0 |
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Answer» Solution :(a) `vecv_(c )=(m_(1)vecv_(1)+m_(2)vecv_(2))/(m_(1)+m_(2))`, `vecv_(c )=(3(2hati+3hatj)+3(4hati-6hatj))/(6)` `:.` Velocity of centre of mass `vecV_(c )=3hati-1.5hatjms^(-1)` (b) The momentum of the system `=Mv_(c )=6kg(3hati-1.5hatj)ms^(-1)=18hati-9hatjkgms^(-1)` (c ) To FIND the velocity of centre of mass after 5s of application of the force `vecF=24hatiN` we first find the acceleration of the centre of mass. It is given by `veca_(c )=(vecF)/(M)=(24hati)/(6)=4hatims^(-2)` The velocity of centre of mass before the force is applied is `vecv_(c )` and from the equation `vecv_(c ).=vecv_(c )+veca_(c ).t` `vecv_(c ).=(3hati-1.5hatj)+(4hati)5=(3hati-1.5hatj+20hati)` `vecV_(c ).=(23hati-1.5hatj)ms^(-1)` (d) From the equation of the position vector `vecr=vecr_(0)+vecv_(0)t+(1)/(2)vecat^(2)` where `vecr_(0)=VEC0` (origin at `t=0`) , `vecv_(0)=vecv_(c )`, `veca=veca_(c )` and `t=5s` `vecr=(3hati-1.5hatj)5+(1)/(2)(4hati)25`, `vecr=(15hati-7.5hatj+50hati)` `:.vecr=(65hati-7.5hatj)m` The coordinates of the centre of mass after 5s of application of the force `vecF` are `(65m,-7.5m)` |
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| 6. |
A person travels along a straight road for the half time with a velocity v_(1) and the next half time with a velocity v_(2). The mean velocity v of the man is |
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Answer» `2/v - 1/v_(1) +(1)/(v_(2))` |
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| 7. |
If 484 J of energy is spent in increasing the speed of wheel from 60 rpm 360 rpm, the M.I. of the wheel is. |
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Answer» `1.6" KG m"^(2)` |
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| 8. |
An object of mass 100 g is thrown with initial velocity vecV=5(hati+hatj)ms^(-1) with respect to the ground. Neglect the effect of air on the motion of mass and take g=10 ms^(-1). What is the impulse transferred by the mass when it hits the ground. |
Answer» SOLUTION :
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| 9. |
The diagram show a force-extension graph for a rubber band. Consider the following statements a) It will be easier to compress this rubber than expand it b) Rubber does not return to its original length after it is stretched c) The rubber band will get heated if it is stretched and released Which of these can be deduced from the graph |
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Answer» C only |
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| 10. |
A particle makes SHM along a straight line and its velocity when passing through points 3cm and 4cm from the mean position is 16cm/sec and 12cm/sec respectively. Then its time period is |
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Answer» `PI/2` SEC |
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| 11. |
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 the height at which he gets out of the aeroplane? |
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Answer» Solution :Distance he FALLS before the parachute OPENS is `(1)/(2)gxx100=490m` Then his velocity = `g t=98.0m//s=u` Velocity on reaching ground = `8=vartheta` retardation = 2 `vartheta^(2)-u^(2)=2AS` `8^(2)-(98)^(2)=2xx(-2)SthereforeS=(106xx90)/(4)=2385m` TOTAL distance = `2385+490` = `2875m` = height of aeroplane |
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| 12. |
While taking a turn on a circular road with a velocity of 36 kmcdot h^(-1), a cyclist leans at an angle of 30^(@) with the vertical. What is the radius of the circular road? |
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Answer» |
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| 13. |
Can you think of a motion which is harmonic but not simple harmonic ? |
| Answer» SOLUTION :The MOTION of the SECONDS HAND in a CLOCK. | |
| 14. |
Explain how overtones are produced in a: (a) Closed organ pipe (b) Open organ pipe |
Answer» Solution :Closed organ pipesLookat thepictureof a clarin e t, showninfirgure. Itisa pipewith oneclosedandthe otherendopen. Ifone endof a pipeis closed , thewavereflectedat thisclosedend is `180^@ `out ofphasewiththeincomingwave. thusthereis nodisplacementof theparticlesat the closedend . THerefore, nodesare formedat theclosedend andantinodes areformed at openend .  (a)No motionof particleswhichleadsto nodesat closedend andantinodesat openand( fundamentalmode)(N-node, A - antinode) Let usconsiderthe simplestmodeofvibrationof the aircolumncalledthe fundamentalmode. Anti- nodeis formedat theopenand andnodeatclosedend .FROMTHE figure.LetLbe thelengthof thetubeand thewavelengthof the waveproduced. forthe fundamentalmodeof vibration,we have ` L = ( lamda_1 )/( 4 )orlamda_1= 4L` The frequencyof thenoteemittedis `f_1 = ( v)/( lamda_1) = ( v)/(4 L)` whichis calledthe fundamentalnote : Teh freqencieshigherfundamentalfrequencycan beproducedbyblowingairstronglyat openend .Suchfrequenciesare calledovertones . THefigure(B)showsthe secondmode of vibration havingtwonodes and twoantinodes ,  ` 4L = 3 lamda _2` `L=(3lamda _2)/(4 )or lamda_2= ( 4L )/(3) ` Thefrequencyfor this, ` f_2 =( V) /( lamda _2 )= ( 3v )/( 4 L)= 3 F_1` is calledfirstover, sincehere, thefrequencyis threetimesthefundamentalfrequencyit iscalledthirdharmonic . The figure(c )showsthirdmodeof vibrationhavingthreenodesand three anti- nodes  We have` 4L = 5 lamda_3` ` L= (5lamda_3) /(4 )orlamda_3= ( 4 L )/( 5 )` THefrequency`f_3 = ( V)/( lamda _3 ) = ( 5v )/( 4 L)= 5 F_1` is calledsecondovertoneand sincen=5here, thisis calledfifthharmonic, hencetheclosedorgranpipehasonlyoddharmonics and frequencyof thenthharmonicis` f_n= ( 2 n +1)f_1`: Thereforethefrequenciesof harmonicare in theratio ` F_1 : f_2: F_3: F_4 :...= 1:3:5:7:.............(3)` |
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| 15. |
What is resonance? Give example for it. |
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Answer» Solution :The frequency of exterrnal periodic FORCE (or driving force) matches with the frequency of the VIBRATING body (driven). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding VIBRATIONS are known as resonance vibrations. Example: The breaking of GLASS DUE to sound |
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| 16. |
A gun of mass M fires a bullet of mass m horizontally. The energy of firing is such that it is sufficient to project the bullet vertically to a height h. The velocity of recoil of the gun is |
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Answer» `SQRT((2m^2 GH)/(M(M+m)) )` |
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| 17. |
Derive the ratio of two specific heat capacities of monatomic, diatomic and triatomic molecules. |
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Answer» Solution :Monoatomic molecule : Average kinetic energy of a molecule `=[3/2kT]` (i) TOTAL energy of a mole of gas `=3/2 kT xxN_A = 3/2 RT` (ii) For one mole , the MOLAR specific heat at constant volume `C_V = (DU)/(dT) =d/(dT) [3/2RT]` `C_V = [3/2R]` `C_P = C_V + R =3/2 R+R` `=5/2R` (III) The ratio of specific heats, `gamma=C_P/C_V=(5//2R)/(3//2R)=5/3=1.67` Diatomic molecule : Average kinetic energy of a diatomic molecule at low temperature `5/2kT`. (i) Total energy of one mole of gas `=5/2kT xxN_A = 5/2RT` (Here, the total energy is purely kinetic) (ii) For one mole specific heat at constant volume `C_V=(dU)/(dT)=[5/2RT]=5/2R` But, `C_P = C_V + R = 5/2R + R = 7/2R` `:. gamma=C_P/C_V=(7//2R)/(5//2R)=7/5=1.40` (iii) Energy of a diatomic molecule at high temperature is equal to `7/2` RT. `C_V=(dU)/(dT)=[7/2RT]=7/2R` `:. C_P = C_v + R = 7/2 R+R` `C_P=9/2R` It is noted that the `C_V and C_P` are higher for diatomic molecules than the monoatomic molecules. It is implied that to increase the temperature of diatomic gas molecules by `1^@C` it require more heat energy than monoatomic molecules. `:.gamma=C_P/C_V=(9//2R)/(7//2R)=9/7=1.28` Triatomic molecule : (i) Linear molecule : Energy of one mole `=7/2 kTxxN_A = 7/2RT` `C_V=(dU)/(dT)=d/(dT)[7/2RT]` `C_V=[7/2R]` `C_P=C_V+R=7/2R+R` `=9/2R` `gamma=C_P/C_V=(9//2R)/(7//2R)=9/7=1.28` (ii) Non - Linear molecule : Energy of one mole `=6/2kTxxN_A=6/2RT=3RT` `C_V=(dU)/(dT)=3R` `C_P=C_V+R` `=3R+R =4R` `gamma=C_P/C_V=(4R)/(3R)=4/3=1.33` |
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| 18. |
Consider the quantities, pressure, power, energy, impulse, gravitationalpotential , electricalcharge, temperature , area. Out of these, the only vector quantities are |
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Answer» Impulse, pressure and area |
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| 19. |
Explain whether the escape speed depends on the mass of the object or not. |
| Answer» SOLUTION :`v_e` = `sqrt(2GM)/R`, so `v_e` does not DEPEND on the mass of OBJECT. | |
| 20. |
Boiling water is changing into steam.Under this condition the specific heat of water is ……………. . |
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Answer» zero |
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| 21. |
Figure shows a cyclic process ABCDBEA performed on an ideal cycle.If P_A =2 atm, P_B5 atm, and P_6 =6atm, V_E - V_A= 20 litre, find the work done by the gas in the complete process (latm, pressure = 1 xx 10^5Pa) |
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Answer» Solution :Evidently, the entire cycle can be visualised as MADE up of two CYCLES, i.e., cycle ABEA and cycle BCDB. Work DONE during the cycle ABEA, by the gas ` W_1=` area of the loop in the P-V diagram. `=1/2 (" (BASE)")/("(altitude) ")=1/2(V_E- V_A ) (P_B-P_A) =1/2( 20 xx 10^(-3)m^3) (5-2 ) xx 10^5Nm^(-2) = 3KJ` Work done in the cycle BCDB, by the gas `W_2 =`-( area of loop BCDB) Now, evidently, triangles ABE and BCD are similiar, the corresponding angles being equal ` therefore("Areaof " DeltaBCD)/( "Areaof ABE ") =((" lengthof normalfromB ONCD")/(" lengthof normalfromB ofEA"))^2=((P_c -P_B )/(P_B -P_A))^2 = ((6 atm- 5 atm)/( 5 atm -2atm ))^2 =1/9` `therefore` Areaof ` DeltaABCD= 1/9`Areaof `DeltaABE= 1/9 xx 3 KJ= 0.33KJthereforeW_2= 0.33KJ` ` therefore` Totalworkdoneby thegas ` DeltaW= W_1 +W_2= 3KJ- 0.33 KJ= 2.67KJ .` |
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| 22. |
Assertion: For motion in two orthree diemensions, velocity and acceleration vecotrs must have any angle between 0^(@) and 90^(@) between them. Reason: For such motion velocity and acceleration of an object is always in the opposite direction. |
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Answer» |
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| 23. |
A gas undergoes a process in which its pressure p and volume V are related as Vp^(n)=constant The bulk modulus for the gas in this process is |
| Answer» ANSWER :C | |
| 24. |
Heat of 30 kcal is supplied to a system and4200 J of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy. (J = 4200 J/kcal) |
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Answer» `1.302xx10^(5)J` |
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| 25. |
A mild steel wire of length 1.0 m and cross sectional area 0.50 xx 10 ^(-2) cm^(2)is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint. |
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Answer» Solution :Let AB a mild steel wire of length = 1 m and its cross sectional area `A =0.50 xx 10 ^(-2) cm ^(2)` Mass suspended `m = 100 g = 0.1 kg and C ` is mid-point of wire AB `therefore AC = CB =L = 1/2 and CD =x`  `therefore AD = sqrt (AC ^(2) + CD ^(2)) and DB = sqrt (L ^(2) + x ^(2))= sqrt(L ^(2) + x ^(2))` Increas in length `= AD +DB -2L` `Delta L = sqrt( L ^(2) + x ^(2)) + sqrt (L^(2) + x ^(2)) - 2L` `=2 sqrt ( L ^(2) + x ^(2)) - 2L` `=2L [ (1 + (x ^(2))/( L ^(2))) -1 ]` `= 2L [(1 + (x ^(2))/( 2L ^(2)) + ...)-1]` Only two terms taken from BIONOMIAL theorem.s expansion `=2L [1+ (x ^(2))/( 2 L ^(2)) -1 ]` `= 2L xx (x ^(2))/( 2 L ^(2))` `Delta L = (x ^(2))/(L) ""...(1)` Longitudinal strain `= (Delta L)/(2L) = (x ^(2))/(2L^(2)) ""...(2)` If T be the tension in the wire, then in equilibrium `2T cos theta = MG ` `therefore T= (mg )/( 2 cos theta )` `= ( mg )/( 2x (x )/( sqrt ( x ^(2) + L ^(2))))` `[because "From figure, " cos theta (CD)/(AD) = (x)/( sqrt (x^(2) +L^(2)))]= (mg sqrt (x ^(2) + L ^(2)))/( 2x)` `If x lt lt L implies x` neglected `T = (mgL)/(2x)""...(3)` `therefore "Stress" sigma = (T)/(A) = (mgL)/(2 x A) ""...(4)` `[because` Equation (2)] `therefore` YOUNG modulus `Y = ("Stress")/("Strain")` `= (mgL)/(2xA) xx (2 L ^(2))/(x ^(2))` `Y = ( mg L ^(3))/( A x ^(3))` `therefore x ^(3) = (mg L ^(3))/( AY)` `therefore x = L [ (mg)/(AY) ] ^(1/3)` `therefore x = 1/2 [ ( 0.1 xx 9.8 )/( 0.5 xx 10 ^(-6) xx 2 xx 10 ^(11) )] ^(1/3)` `therefore x = 1/2[(0.98)/(10^(5))]^(1/3)` `thereforex1/2 [ 9.8 xx 10 ^(-6) ] ^(1/3)` `log x = 1/2 [ (1)/(3) log 9.8 xx 10 ^(-6) ]` `= 1/2 {(1)/(3) [0.9912 +bar6.0000]}` `=1/2 { (1)/(3) [ bar (6).9912]}=1/2 [bar (2).3304]` By antilog, `x = 2.14 xx10 ^(-2)m` `thereforx=1. 07xx10 ^(-3)m` `thereforex ~~ 1 cm = 0.01 m` |
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| 26. |
A sphere of a radius r cm falls from rest in a viscous liquid. Head is produced due to viscous force. The rate of production of heat when the sphere attians its terminal velocity is proportional to ….. |
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Answer» `R^(2)` `=6pietarvxxv=6pietarv^(2)` `[v prop r^(2)]` TERMINAL VELOCITY `prop r^(5)` |
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| 27. |
A load of 981 N is suspended from a steel wire of radius 1 mm. what is the maximum angle through which the wire with the load can be deflected so that it does not break when the load passes through the equilibrium poision? Breaking stress is 7.85 xx 10^(8)Nm^(-2). |
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Answer» SOLUTION :Let `theta` be the maximum angular deflection of load and `upsilon` be the velocity ACQUIRED by load while passing through the EQUILIBRIUM POSITION A. PE of load at `B = KE ` of load at A `mg(L-l cos theta) = 1.2 m upsilon^(2)` …(i) If T is the tension in the string at the lowest position A, then `T - mg = (m upsilon^(2))/(l)` or `T = mg +(m upsilon^(2))/(l) = mg + 2 m (1- cos theta)` `= 3mg - 2 mg cos theta` For maximum value of `theta` stress = `(T//pi r^(2))` must be equal to breaking stress i.e., `(T)/(pi r^(2)) = 7.85 xx 10^(8)` or`(3 mg - 2mg cos theta)/(pi r^(2)) = 7.85 xx 10^(8)` or `(3 xx 981 - 2 xx 981 cos theta)/((22//7)xx(10^(-3))^(2)) = 7.85 xx 10^(8)` or `2943-1962 cos theta = (22//7) xx 10^(-6) xx 7.85 xx 10^(8)` `=2466.5` or `cos theta = (2943-2466.5)/(1962) = (476.5)/(1962)` `= 0.2429 = cos 75^(@) 57'` `theta = 75^(@) 57'`. 
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| 28. |
A lead bullet (specific heat = 0.032 "cal/g" ^@C ) is completely stopped when it strikes a target with a velocity of 300 m/s. The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be |
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Answer» `16.7^@C` |
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| 29. |
A cable is cut to half of its original length. What will be the effect on the increase in its length under a given load ? |
| Answer» Solution :Since `DELTA l prop l _(i),` The change in length is DIRECTLY PROPORTIONAL to the original length. The original length is halved the increase in length will be reduced to HALF. | |
| 30. |
Figure (a) shows a thin liquid film supporting a small weight =4.5xx10^(-2)N. What is the weight supported by a film of the same liquid at the same temperature in figure (b) and (c ) ? Explain your answer physically . |
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Answer» Solution :The weight SUPPORTED by both film (b)and (c ) is `W=4.5xx10^(-2)N`, because surface tension `S=(F)/(2L)`(Film has two FREE surfaces) According to F= W=2lS For all cases l=40cm and surface tension S are same . HENCE ,the total weight supported in each case (b) and (c ) is `4.5xx10^(-2)N`. |
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| 31. |
How thermal conduction is represented for given temperature difference for any material by rate of heat conduction ? |
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Answer» Solution :Consider a metallic bar of length L and uniform cross SECTION A with its two ends maintained at different temperatures. This can be done, for example, by putting the ends in thermal CONTACT with large reservoirs at temperatures, say `T_(C)` and `T_(D)` respectively. Let us assume the ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.  Here, as as `T_(C)gtT_(D)`, temperatures of different parts increase with time which is shown by dashed line in temperature. Distance graph :  After sometime, a steady state is reached , the temperature of the bar DECREASES uniformly with distance from `T_(C)` to `T_(D),(T_(C)gtT_(D))`. The reservoir at C supplies heat at constant rate, which transfers through the bar and is given out at the same rate to the reservoir at D. Now both `(dQ)/(dt)` and `(dT)/(dx)` remain constant with time. This condition of the rod is called as ..thermal steady state... It is found experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional to the temperature difference `(T_(C)-T_(D))` and the area of cross section A and is inversely proportional to the length L : `Hprop(A(T_(C)-T_(D)))/(L)` The constant of proportionality K is called the thermal conductivity of the material. The greater the value of K for a material, the more rapidly will it conduct heat. The SI UNIT of K is `S^(-1)m^(-1)K^(-1)` or `"W m"^(-1)K^(-1)`. Thermal conductivity of metals is high and wood, glasswool have vergy LESS thermal conductivity hence they are good insulators. Thermal conductivities of some materials : 
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| 32. |
A certain volume of a gas (diatomic) expands isothermally at 20^(@)C until its volume is doubled and then adiabatically until its volume is again doubled. Find the final temperature of the gas, given g = 1.4 and that there is 0.1 mole ofthe gas. Also calculate the work done in the two coses R = 8.3J mole^(-1) K^(-1) |
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Answer» We have`TV^(gamma-1)= constant//(273+20)V^(g-1)=(273+t)(2V)^(gamma-1)` or `273+t=(293)/(2^(1.4-1))=(293)/(2)^(0.4)` `log(273+t)=log293-0.4log2-0.4log2=log293-0.4xx0.3010` `=2.4669-0.1204` or `log(273+t)=2.3465` or `273+t=antilog2.3465` or `273+t=222.1` `t=-50.9^(@)C` (i) Work done in isothermal process `=nRTlog_(E)(V_(2))/(v_(1))=(8.3)/(10)xx293log_(e)(2V)/(V) therefore(n=(1)/(10))` `=0.83xx293xx2.3log_(10)^(2)(becauselog_(e)X=2.3log_(10)x)` `=0.83xx293xx2.3xx0.3010=1.684xx10^(2)J` (ii) Work done in adiabatic processs`= (nR(T-T'))/(gamma-1)=0.83(293-222.1)/(1.4-1)` `=(0 .83xx70.9)/(0.4)=1.47xx10^(2)J` |
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| 33. |
Number of significant digits in 2030........... |
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Answer» 1 |
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| 34. |
The angle between the normal force and the resultant force of normal force and maximum frictional force is |
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Answer» ANGLE of FRICTION |
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| 35. |
If the escape velocity on the earth.s surface is 11.2 "kms"^(-1) , its value on a planet having double the mass and double the radius that of the earth is |
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Answer» 5.6`"KMS"^(-1)` |
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| 36. |
Force and moton are in the opposite direction Given example |
| Answer» SOLUTION :In anobjectis thrownverticallyupwardthe DIRECTIONOF MOTIONIS upwardbutgravitationalforce is DOWNWARD . | |
| 37. |
A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is V_(0) and the particle has tangential acceleration of constant magnitude a, If V_(0) =0 then the distance covered by the stone till the string ruptures |
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Answer» `(R)/(2a)` |
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| 38. |
A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is V_(0) and the particle has tangential acceleration of constant magnitude a, If the breaking strength of the string is equal to weight of the stone. The time after which the string will break |
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Answer» `(1)/(a)(SQRT(gR)-V_(0))` |
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| 39. |
The gravitational field in a region is given by vecE = (5hati + 12hatj) (N)/(Kg) if the potential at the origin is taken to be zero. Then the ratio of potential at the points (12m, 0) and (0, 5m) is …… |
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Answer» |
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| 40. |
Give the magnitude and direction of the net force acting on a drop of rain falling down with a constant speed |
| Answer» SOLUTION : As the RAIN drop is falling with a CONSTANTSPEED , its acceleration a = 0 . Hecne net force F = ma = 0. | |
| 41. |
A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is V_(0) and the particle has tangential acceleration of constant magnitude a, After what time the tension of the string will be equal to T (gt (mv_(0)^(2))/(R)) |
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Answer» `(1)/(a)(sqrt((TR)/(m))-V_(0))` |
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| 42. |
A cyclest is riding a bicycle on a circular track in a horizontal plane with maximum possible speed on the track. Coefficent of friction between the track and cycle tyre is mu on seeing a cat on the track, the cyclist applies brakes so hard that the tyres of the cycle stop rotating abruptly. Let veca_(1) and veca_(2) be the acceleration of bicycle just befor and just after the brakes applied, then |bar(a_(2))-bara_(1)| is |
| Answer» ANSWER :D | |
| 43. |
A liquidtakes 5 minutes to cool from80^(@)Cto50^(@)C . Howmuchtimw will ittake to cool from 60^(@)Cto 30^(@)C? Thetemperatuer of thesurroundingis 20^(@)C . |
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Answer» SOLUTION :According to Newton.s law of cooling `(dT)/(dt) = k (T - T_(0))` If the BODYCOOLS by radiationfrom `T_(1)`to `T_(2)""^(@)C`in time T , then `(dT)/(dt) = (T_(1) - T_(2))/(t) and T = T_(ave) = (T_(1) + T_(2))/(2)` `therefore((T_(1) - T_(2))/(t)) = k[((T_(1) + T_(2))/(2))-T_(0)]` Substitutingthe values , we get . `((80 - 50)/(5)) = k[((80 + 50)/(2))-20] to (1) and ((60 -30)/(t)) = k [((60+30)/(2))-20 ] to (2) ` Solvingequations (1) and (2) , we gett = 9 MINUTES . |
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| 44. |
A satellite is launched in the equatorial pane in such a way that it can transmit signals upto 60^(@) latitude on the earth. The angular velocity of the satellite is sqrt((GM)/(nR^(3))). Find n. |
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Answer» |
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| 45. |
A lift is moving down with an acceleration equal to the acceleration due to gravity. A body of mass M kept on the floor of the lift is pulled horizontally. If the co-efficient of friction is m then the frictional resistance offered by the body is |
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Answer» mMg |
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| 46. |
The rates of fall of temperature of two identical solid spheres of different materials are equal at a certain temperature |
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Answer» Their specific heat CAPACITIES are EQUAL |
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| 47. |
Which of the following statement about the gravitational constant is true? |
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Answer» It is a force |
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| 48. |
A boat which has a speed of 5km hr^(-1) in still water crosses a river of with 1 km along the shortest possible path in 15 minutes. Calculate the velocity of the river water. |
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Answer» Solution :The SITUATION is shown in Fig. of previous example OA=A, represents velocity of the river WATER. OB=B , represents velocity of the boat in still water which is `5km hr^(-1)` The boat crosses the 1km wide river ALONG the SHORTEST possible PATH OC in 15 minutes. Hence OC=R represents resultant velocity of the boat which is `(1km)/(15mt)=4km hr^(-1)` Here , `OB=AC =5 km hr^(-1), OC=4 Km hr^(-1),OA=?` In`DeltaOAC, OA=sqrt((AC)^(2)-(OC)^(2))=sqrt((5)^(2)-(4)^(2))=3` Here the velocity of the river water is `3km hr^(-1)` |
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| 49. |
Triple point of water is 0.01^@C . Express this temperature in kelvin scale. |
| Answer» SOLUTION :`T_k\=t_c+273.15=0.01+273.15=273.16K` | |