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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. |
At what temperature is the rms speed of gaseous hydrogen molecules equal to that of oxygen molecules at 47^(@) C |
| Answer» Answer :A | |
| 2. |
A metal rod has a length of Im at 30^(@) C . 'alpha'of metal is 2.5 xx 10^(-5)//^(@) C. The temperature at which it will be shortened by 1mm is |
| Answer» Answer :C | |
| 3. |
Uniform circular motion. |
Answer» Solution :When a point object is moving on a circular path with a constant speed, it covers equal distances on the circumference of the circle in equal intervals of time. Then the object issaid to be in uniform circular motion. This is shown in Figure  In uniform circular motion, the velocity is ALWAYS changing but speed remains the same. Physically it implies thatmagnitude of velocity vectorremains constant and only the direction changes continuously. If the velocity changes in both speed and direction during the circular motion, we get non uniform circular motion.  As seen already, in uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in Figure. Note that the length of the velocity vector(blue) is not CHANGED during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towardsthe center of the circle. This is called centripetal acceleration. It always points towards towards the center of the circle. This is shown in the Figure.  The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors  (i) Let the DIRECTIONS of position and velocity vectorsshift through the same angle `theta` in a SMALL interval of time `trianglet`. (ii) For uniform circular motion, `r=|vecr_1|=|vecr_2|" and " v=|vecv_1|=|vecv_2|`. If the particle moves from position vector `vecr_1 " to " vecr_2`, the displacement is given by `triangle vecr = vecr_2 - vecr_1` and the change in velocity from `vecv_1" to "vecv_2` is given by `triangle vecv= vecv_2 - vecv_1`. (III) The magnitude of the displacement `triangle r" and of "triangle v` satisfy the following relation. `(triangle r)/(r )= -v((triangle r)/(r ))` Then, `a=(triangle v)/(triangle t)=(v)/(r )((triangle v)/(triangle t))= -(v^2)/(r )` For uniform circular motion `v= omega r`, where `omega` is the angular velocity of the particle about the center. Then centipetal acceleration can be written as `a= -omega^2 r`. |
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| 4. |
A uniform ladder of length L and mass M leans against a frictionless vertical wall. The ladder makes an angle theta with ground and the coefficient of the static friction between the ground and the foot of the ladder is mu. What maximum height a man of mass m can climb up the ladder so that the ladder will not skid? |
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Answer» SOLUTION :The value of limiting friction betweenthe ground and the LADDER is `mu`n Fig. Let the man can climb a maximum height x up the ladder. The point P in figure denotes the position of the man. Here frictional force between the ladder and ground. f = `mu`n where n = contact or NORMAL force on ladder by ground = (Mg +mg) and the contact force on ladder by the wall is n. . At equilibrium NET force on the ladder is zero, i.e., n.- f=0 or, n. = `mu` (M+m) g Also the moment about the point B is zero n.L sin`theta -Mg"" (L)/(2)"cos"theta-mg.xcos theta =0` or, mxcos`theta = mu (M+m) L sin theta - (ML)/(2) cos theta` `:. x = (mu(M+m)Lsintheta-(ML)/(2)costheta)/(mcostheta)` 
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| 5. |
Distance between the centres of two stars is 10a. The masses of these stars are M and 16M and their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? |
| Answer» SOLUTION :`SQRT((45GM)/(4A))` | |
| 6. |
Conversion of heat into electrical energy can be achieved by Accroding to newton's law of cooling (provided the difference of temperature is small ), the rate of loss of heat is proportional to |
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Answer» transistor |
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| 7. |
A simple harmonic oscillator consist of a particle of mass m and an ideal spring with spring constant k. The particle oscillates with a time period T. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be........... |
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Answer» `T//2` If the SPRING is CUT into TWO equal parts, the force constant of each part becomes 2K. Therefore, `T.=2pisqrt((m)/(2k))=(T)/(sqrt(2))` |
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| 8. |
If water drops are falling at regular time intervals from ceiling of height H, then position of drop from the ceiling is (when nth drop falling from the ceiling and rth drop is in its way) |
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Answer» `((n-1)^2)/((n-r)^2)H` |
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| 9. |
1 "cm"^3 of water at 1 atm pressure is converted into steam, volume of steam becomes 1671 "cm"^3 hence increase in internal energy will be ____ |
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Answer» 2087 J =1X 540 =540 cal Now from the first law of THERMODYNAMICS, `DeltaQ=DeltaU+DeltaW` `DeltaQ=DeltaU+PDeltaV` `THEREFORE DeltaU=DeltaQ-PDeltaV` `=540-10^5(1671-1)xx10^(-6)` `=540-1670xx10^(-1)` =540-167 =373 cal |
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| 10. |
When a stone falling from the top of vertical tower has fallen a distance x m, another is let fall from a point y m below the top. If the fall from rest and reach the ground together, then the height of the tower is |
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Answer» `((X+y)^2)/(4X) m` |
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| 11. |
A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane |
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Answer» The HOLLOW SPHERE reaches the bottom first |
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| 12. |
(A) : Higher is the accuracy of measurement, if instrument have smaller least count. (R) : Smaller the percentage error, higher is the accuracy of measurement. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 13. |
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collison occur? |
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Answer» `2/3` |
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| 14. |
A cyclic process for an ideal monatomic gas (C_v = 12.5 J mol^1 K^1)is represented in the figure. The temperatures at 1,2 and 3 are 300 K and 455 K, respectively. Compute the values of triangleQ,triangleU and triangleW for each of the processes. The process from 2 to 3 is adiabatic. |
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Answer» `triangle W = intpd V= 0` (volume remains constant) `triangleQ = overset(T_1)UNDERSET(T_2)int(C_v)dT=C_(v)(T_(2)-T_(1))` `triangleQ = underset(T_1)overset(T_2)intC_(v)dT = C_(v)(T_(2)-T_(1))` 12.5(600-300)=3750joules By the FIRST law of thermodynamics `triangleQ=triangleU+triangleW or triangleU=triangleQ-triangleW` 3750-0=3750 JOULE In the process 2 to 3 `triangleQ=0` (Since the process is adiabatic) `triangleW = (R(T_2-T_3))/(gamma-1)` `C_v = (T_2-T_3) (becauseC_v=(R)/(gamma-1))` `=12.5(600-455)=12.5xx145=-1812.5` joules `thereforeU=triangleQ`- `triangleW=0-1812.5=-1812.5` joules In the process from 3 to 1, `triangleW= underset(V)overset(v_1)intpdV=p(V_1-V_3)=pV_(1)-pV_(2)` or `triangleW=R(T_(1)-T_(3))` (becausepV=RT) `=8.31(300-455)=-1288 joules` `triangleQ=undersetT_(3)oversetT_(1)intC_(p)dT=C_(p)T_(1)-T_(3))=1.67xx12.5xx(300-455)`(because`gamma=(C_(p))/(C_(v))` `=-3235.6 joules` By the first law of thermodynamics `triangleQ=triangleU+triangleW` `therefore``triangleU+triangleQ-triangleW`=(-3235.6)-(-1288)=1989.1 joules |
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| 15. |
When stationary waves are set up, pick out the correct statement from the following |
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Answer» all the particles in the medium are in the same PHASE of vibration at all times and distances |
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| 16. |
A 1 m long rod having a constant cross sectional area is made of four materials. The first 0.2 m are made of iron, the next 0.3 m of lead, the following 0.2m of aluminium and the remaining part is made of copper. Find the centre of mass of the rod. The densities of iron, lead, aluminium and copper are 7.9xx10^(3)kg//m^(3),11.4xx10^(3)kg//m^(2),2.7xx10^(3)kg//m^(3) and 8.9xx10^(3)kg//m^(3) respectively. |
Answer» SOLUTION : mass (m) =volume (v) `xx` density (d) m= Area (A) `xx` length (1) `xx` density (d) m=Ald Mass of iron part, `m_(1)=Axx0.2xx7.9xx10^(3)` `=1.58xx10^(3)A` Mass of lead part, `m_(2)=Axx0.3xx11.4xx10^(3)` `=3.42xx10^(3)A` Mass of aluminium part, `m_(3)=Axx0.2xx2.7xx10^(3)` `=0.54xx10^(3)A` Mass of copper part, `m_(4)=Axx0.3xx8.9xx10^(3)` `=2.67ccxx10^(3)A` Co-ordinate of iron part from end "O" of the ROD, `x_(1)=0.1m` Co-ordinate of lead part from end "O" of the rod, `x_(2)=0.35m` Co-ordinate of aluminium part from end “O” of the rod, `x_(3)=0.6m` Co-ordinate of copper part from end "O" of the rod, `x_(4)=0.85m` `:.` Centre of mass of the rod, `X_(c)=(m_(1)x_(1)+m_(2)x_(1)+m_(3)x_(3)+m_(4)x_(4))/(m_(1)+m_(2)+m_(3)+m_(4))` `impliesX_(CM)=(((1.58xx10^(3)xx0.1+3.42xx10^(3)xx0.35+),(0.54xx10^(3)xx0.6+2.67xx10^(3)xx0.85))A)/((1.58xx10^(3)+3.42xx10^(3)+0.54xx10^(3)+2.67xx10^(3))A)` `impliesX_(cm)=0.481m` from the end ..O.. of the rod. |
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| 17. |
A glass u-tube is such that the diameter of one limb is 3.0mm and the other limb is 6.0mm. The tube is inverted vertically with open ends below the surface of beaker. What is the difference between the heights to which water rises in the two limbs? T=0.07N//m" "theta=0^(@) |
| Answer» Solution :Apply : EXCESS PRESSURE concept `2T{1/r_(1)-1/r_(2)}=HPG` | |
| 18. |
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^(3) in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump? |
| Answer» SOLUTION :`=43.6 KW` | |
| 19. |
If force F is reprsented in different, ways then which of the following is correct and which are not correct ? M= mass, a = acceleration r= radius, v= velocity d= distance t=time. (a) F=(mv^(2))/(r), (b) F=ma, (c) F=(ma)/(d) (d) F=ma+(1mv^(2))/(r) (e) F=(md^(2))/(t^(2)) |
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Answer» |
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| 20. |
An ideal gas heat engine operates in a carnot cycle between 227^@ C and 127^@ C. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to _____ |
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Answer» 1.2 `T_1`=227+273=500 K `T_2`=127+273 = 400 K `ETA=(500-400)/500=1/5` `THEREFORE` Output work = `(eta)xx` INPUT heat = `1/5xx6` =1.2 kcal |
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| 21. |
A particle with total energy E is moving in a potential energy region U (x). Motion of the particle is restricted to the region when |
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Answer» `U(x)LT E` |
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| 22. |
A thin square plate of side 5cm is suspended vertically from a balance so that lower side just dipsinto water with side parallel to surface. When the plate is clean (theta= 0^@ ), it appears to weigh 0.044N. But when the plate is greasy ( theta= 180^@ ), it appears to weigh 0.03N. The surface tension of water is sigma (10^(-2) Nm^(-1) ). Find sigma |
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Answer» |
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| 23. |
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiments is 7.30 xx 10^(-2) "N/m". 1 atmospheric pressure =1.01 xx 10^(5) Pa density of water =7000" kg/m"^(3), g="9.80 ms"^(-2). Also calculate the excess pressure. |
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Answer» Solution :Given, `R=(2.00 mm)/(2)=1.00 mm` `=1.00 xx 10^(-3)m` `h=8.00 cm =8 xx 10^(-2) m` `sigma=7.30 xx 10^(-2) N//m` `P_("atm")=1.01 xx 10^(5) Pa` `p=1000 kg//m^(3), g=9.80 m//s^(2)` EXCESS pressure inside the bubble is `p=(2sigma)/(r)=(2 xx 7.30 xx 10^(-2))/(1 xx 10^(-3)) =146 Pa` Pressure required in the TUBE=Atmospheric pressure +Pressure due to water column `=P_("atm")+HPG` `=1.01 xx 10^(5)+8 xx 10^(-2) xx 10^(-3) xx 9.80` `=1.01 xx 10^(5) +0.00784 xx 10^(5)` `=1.01784 xx 10^(5) Pa` Pressure inside the bubble =Pressure outside the tube +Exces pressure `=1.01784 xx 10^(5)+146` `=1.0193 xx 10^(5) Pa or` `=1.02 xx 10^(5) Pa` |
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| 24. |
If for damped oscillations if (k)/(m) = ((b)/(2m))^(2), then is the oscillation exist? |
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Answer» Solution :No, because angular frequency fo damped oscillator is `OMEGA.= SQRT((k)/(m)-(b^2)/((2M)^2))" if "(k)/(m)= ((b)/(2m))^(2)" then "omega.=0` Periodic TIME will be infinite. Oscillation does not exist. |
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| 25. |
Givethe equationof statefor anisothermalprocess. |
| Answer» Solution :It is a process in which the temperature REMAINS CONSTANT but the pressure and volume of a thermodynamic system will CHANGE. The IDEAL GAS equation is `PV=muRT` | |
| 26. |
The time period of a mass suspended by a spring (force constant K) is T. If the spring is cut into three equal pieces, what will be the force constant of each part? If the same mass be suspended from one piece what will be the periodic time? |
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Answer» Solution :CONSIDER the SPRING be made of a combination of three springs in SERIES each of spring constant k. The effective spring constant K is given by ` 1/K = 1/k + 1/k + 1/k = 3/k` or`K = k/3 " or" k = 3k` `therefore ` Time period of VIBRATION of a body attached to the endof this spring, ` T = 2K sqrt(m/k) = 2pi sqrt( (m)/(k/3) ) = 2pi sqrt( (3m)/(k) `.....(1) When the spring is cut into three pieces, the spring constant = k, time period of vibration of a body attached to the end of this spring, `T_1 = 2pi sqrt(m/k)` From eqns (1) and (2) `(T_1)/(T)1 = (1)/(sqrt3) " or " T_1 = (T)/(sqrt3)` |
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| 27. |
A setellite osrevolving very close to a planet of density p. The period of revolutionof satellite is |
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Answer» `sqrt((3pirho)/G)` |
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| 28. |
One Angstrom is……. |
| Answer» Solution :`10^(-10)` m | |
| 29. |
A metal cube is placed in empty vessel. When water is filled in the vessel so that the cube is completely immeresed in the water, the force on the bottom of the vessel in contact with the cube |
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Answer» Will increase |
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| 31. |
A body executing SHM at a displacement 'x' its PE is E_(1) at a displacement 'Y' its PE is E_(2) The P.E at a displacement (x + y) is |
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Answer» `SQRT(E)= sqrt(E_(1))-sqrt(E_(2))` |
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| 32. |
If veca and vecb are the two vectors vecR resultant of veca, vecb as shown |
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Answer» a-1, B-2, c-3, d-4 |
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| 33. |
The angular frequency of the damped oscillator is given by omega=sqrt((k/m-(b^(2))/(4m^(2)))) where k is the spring constant, m is th mass of the oscillatory and b is the damping constant. If the ratio (b^(2))/(mk) is 8%, the change in time period compared to the undamped oscillator isapproximately as follows: |
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Answer» INCREASES by 1% |
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| 34. |
A ballon with mass M is decending with an acceleration a (lt g) What mass m of its contents must be removed so that is starts moving up with an acceleration a ? |
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Answer» Solution :Let the retarding force acting on the ballon in vertically upwards direction be `F` When the ballon is descending down with acceleration its equation of MOTION can be WRITTEN as `Ma = M G - F or F= mg - Ma` Let a mass m be removed from the balloon As it moves up with an acceleration a as shown in its equation of motion can be written as `(M - m) a =F - (M - m) g` using (i) `Ma - ma = Mg - Ma - M g + mg` `2 M a = m (g + a)` `m = ( 2Ma)/(g + a )`  .
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| 35. |
A body of mass m is accelerated uniformaly from rest to a speed v in a time. T. The instantaneous power delivered to the body as a function of time, is given by |
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Answer» `(MV^(2))/(T^(2))t` |
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| 36. |
Assertion : The amplitude of oscillation can never be infinite.Reason:The energy of oscillator is continuously dissipated. |
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Answer» If both assertion and reson are true and reason is the CORRECT explanation of assertion From above equation, amplitude of oscillation, in ABSENCE of damping force (b=0), that the steady state amplitude approaches infinity as `omegatoomega_(0)`. That is, if there is not resistive force in the system and then it is possible to DRIVE an oscillator with sinusoidal force at the RESONANCE frequency, the amplitude of motion will build up without limit. this does not occur in practice because some damping is always present in real oscillation. due to presence of various dissipative force in the system, the amplitude of oscillation can grow to a large value only but cann never be infinite  The graph showing amplitude as function of frequency for the forced oscillator with varying resistive force. |
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| 37. |
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 xx 10^(8)m cdot s^(-1)) |
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Answer» `10.1` GHz `therefore ` Doppler effect in relevant with the theory of relativity as follows: `V.=vsqrt((1+beta)/(1-beta))` `beta = (" speed of observer " ) / (" speed of light " ) = 1/2` `therefore v. = 10 SQRT((1+1/2)/(1-1/2))=10sqrt(3) = 10 xx 1.73` `=17.3` GHz The OPTION (c) is CORRECT. |
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| 38. |
A cylinder of volume 10L contains nitrogen gas at 27^@C, and 50 standard avogadro pressure. The cylinder is kept in vaccum and a 1cm^2 hole is made on it. Calculate the time after which the cylinder will be empty. k=1.38 times 10^-23 J.K^-1, N=6.023 times 10^23 and molecular mass of nitrogen =28. |
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Answer» |
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| 39. |
Assertion : A stone and a feather dropped from the same height do not reach the ground at the same time. Reason : Acceleration due to gravity is dependent on the mass of the object. |
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Answer» If both assertion and reason are true and reason is the CORRECT EXPLANATION of assertion. |
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| 40. |
Is there any specific reason for using Cesium-133 atom for defining one second? |
| Answer» SOLUTION :YES Tis atom emits electromagnetic radiation of a PRECISE and UNVARYING frequency. | |
| 41. |
Three liquids A, B and C having same specific heat and mass m , 2m and 3m have temperatures 20^(@) C,40^@Cand 60^@C respectively. Temperature of the mixture when : {:("Column - I","Column - II"),("A) A and B are mixed","P) " 33.33^@C),("B) A and C are mixed","Q) "52^@C),("C) B and C are mixed","R) "50^@C),("D) A , B and C all three are mixed","S) "46.66^@C):} |
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Answer» <P> |
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| 42. |
If a platinum wire is stretched by 0.5% what is the strain on the wire? |
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Answer» Solution :Fractional increase in the LENGTH `= (E)/(L)=` strain ` THEREFORE ` The strain `=0.5%= (0.5)/(100)=0.005` |
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| 43. |
What is the values of frequency in the differential equation of SHM (d^(2)x)/(dt^(2))+100x =0? |
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Answer» Solution :Comparing `(d^(2)x)/(dt^(2))+100X= 0` to general equation `(d^(2)x)/(dt^(2))+omega^(2) x=0` `omega^(2)= 100` `THEREFORE omega = 10` `therefore 2pi F= 10` `therefore f= (10)/(2pi)HZ`. |
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| 44. |
A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to |
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Answer» `r^5` `(DQ)/(dt)=P=F. v=6pi eta r v^2 or,v = kr^2` `therefore P=6pi eta r(kr^2)^2 = 6pi k^2 eta r^5` ` P prop r^5` |
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| 45. |
A simple pendulum starts with its bob at position X.The bob is pulled aside to Y and then released. It swings from Y to Z and back to Y. |
| Answer» SOLUTION :WORK,PE,KE,CONSTANT | |
| 46. |
There were two fixed points in the original Celsius scale as mentioned above which were assigned the mimber 0^(@) C " and " 100^(@) C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is ass ned the number 273.16 K What is the other ftsed point on this (Kelvin) scale ? |
| Answer» SOLUTION :The other FIXED POINT is the ABSOLUTE ZERO itself. | |
| 47. |
Is potential energy the property of a single object? Justify. |
| Answer» Solution :No, Potential energy is a property of a SYSTEM rather than of a SINGLE object DUE to its physical POSITION. Because gravitation potential energy depends on relative position. So, a reference level at which to set the potential energy equal to ZERO. | |
| 48. |
Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the joining them where the gravitational field is zero. |
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Answer» Solution :Let Thepoint be the position when the gravitational FIELDIS zero . ` (Gm)/(x^(2)) = (G(4M))/((r-x)^(2))` ` 1/(x^(2)) = 4/((r-x)^(2))` Square root on the sides ,`1/x = 2/((r-x)) ` ` x = r/3 ` The POINTP is distance`r/3` from mass .m. and`(2R)/3` frommass .4m. Gavitational potential` V = (-Gm)/((r/3))- (G(4m))/(((2r)/3))` ` = (-3(Gm))/4 - (3G(4m))/(2r)= (-3Gm)/r - (12Gm)/(2r) rArr V = (-9Gm)/r ` |
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| 49. |
In which process (isothermal or adiabatic) there is a greater change in internal energy ? |
| Answer» SOLUTION :ADIABATIC | |
| 50. |
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 Nm^(-2), what is the maximum load the cable can support? |
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Answer» Solution :Maximum stress `= ("maximum force")/("cross sectional area")` `therefore` Maximum force `F_(MAX) =` Maximum stress `xxA` `= 10^(8) xx PI R ^(2)` `= 10^(8) xx 3.14 xx (1.5 xx 10 ^(-2)) ^(2)` `= 3.14 xx 2.25 xx 10 ^(8) xx 10 ^(-4)` `=7.-065 xx 10 ^(4) N` `~~7.07 xx 10 ^(4) N` |
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