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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 chain of mass m and length l is kept on a horizontal frictionless table, such that (1)/(4) th of the length of the chain is hanging out of the table. Find the work done to pull the hanging part of the chain onto the table. |
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Answer» Solution :Mass per UNIT LENGTH of the chain `(m)/(l)` `:.` Mass of hanging part of the chain = `(m)/(l)xx(l)/(4)= (m)/(4)` `:.` Centre of gravity of this portion is at `(l)/(4xx2) =(l)/(8)` below the top of the table. `:. ` Work to be done to LIFT the hanging part of the chain `=(m)/(4)xxgxx(l)/(8)=(mgl)/(32)`. |
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| 2. |
A diatomic molecule has atoms of masses m_(1) and m_(2). The potential energy of the molecule for the interatomic separation r is given by V(r )=-A+B(r-r_(0))^(2), where r_(0) is the equilibrium separation, and A and B are positive constants. the atoms are compressed towards each other from their equilibrium positons andreleased. what is the vibrational frequency of the molecule? |
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Answer» |
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| 3. |
For the planet-sun system identify the correct statement |
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Answer» the angular momentum of the planet is not conserved |
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| 4. |
A 2kg block lying on smooth table which is connected by a body of mass 1 kg by a string which passes through a pulley. Take 1 kg mass hanging vertically. Find the acceleration of block & tension in string. |
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| 5. |
A pendulum bob is hanging from the roof of an elevator with the help of a light string. When the elevator moves up with uniform acceleration 'a' the tension in the string is T_(1).When the elevator moves down with the same acceleration, the tension in the string is T_(1).If the elevator were stationary, the tension in the string would be |
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Answer» `(T_(1)+T_(2))/(2)` |
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| 6. |
In case of pulling & pushing minimum forces required are (w sin alpha)/(cos(theta-alpha)_(r )) & (w sin alpha)/(cos(theta + alpha)) then accelerations are |
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Answer» `a=G, a=g//mu` |
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| 7. |
Which of the following is/are likely to affect the time period of a simple pendulum? (i) length (ii) amplitude of vibration (iii) mass of the bob |
| Answer» Solution :(i) Length. If the MASS INCREASES WITHOUT CHANGING the centre of mass it will not AFFECT the period. | |
| 8. |
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tension, densities, lengths and diameter of the two wires are in the ratio 8:1, 1:2, x:y and 4:1 respectively. If the note of the higher pitch has a frequency of 360 Hz and the number of beats produced per second is 10, then the value of x:y is |
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Answer» `36 : 35` |
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| 9. |
In the equation y ( x, t ) = Asin (omega t- kx + phi ), what does the terms ( omega t - kt + phi ) represent ? |
| Answer» Solution :The terms `( OMEGA t - kx + phi ) ` is CALLED as ARGUMENT or PHASE of the wave. | |
| 10. |
A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 xx 10^(24) kg, radius = km, G = 6.67 xx 10^(-11)Nm^(-2)//kg^(-2). |
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Answer» Solution :Gravitational potential at height hfrom the surface of earth is `V = -(GM)/((R+h)) = (-6.67 xx 10^(-11) xx (6 xx 10^(24)))/((6.4 xx 10^(6) + 36 xx 10^(6))) = -9.4 xx 10^(6) J//kg`. |
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| 11. |
A body travels uniformly a distance of (20.0 +- 0.2)min time (4.0 + -0.04)s. The velocity of the body is |
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Answer» `(5.0 +- 0.4) MS^(-1)` |
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| 12. |
Explain the principle of homogeneity of dimensions. What are its uses ? Illustrate by giving one example of each. |
| Answer» SOLUTION :1 (d). 7 & 1 (d). 8 | |
| 13. |
The distance travelled by a body moving along a line in time t is proportional to t^3. The acceleration time (a, t) graph for the motion of the body will be |
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Answer»
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| 14. |
Four rods of equal length l and mass each form a square as shown in figure. The moment of inertia of the system about the axis 3 is |
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Answer» `2/3ml^(2)` |
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| 15. |
A 2kg block is placed over a 5kg block and both are placed on a smooth horizontal surface. The ceofficient of friction between the blocks is 0.10. Find the acceleration of the two blocks if a horizontal force of 14 N is applied to the upper block (g=10 ms^(-2)). |
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Answer» SOLUTION :Consider the motion of 2kg BLOCK. The forces on it are (i) gravitational force, `2g = 2xx10=20 N`. Vertically downwards, (ii) normal REACTION N by the 5 kg block, vertically upwards, (iii) force of friction `f = mu` N to the left and (iv) applied force 14 N.  In the vertical direction, there is no acceleration. `therefore N=20 N` In the horizontal direction, the acceleration of the 2 kg block is a. `therefore 14 - mu N = 2a ""` (`because` Resultant force = ma) `14-0.10xx20=2a "" (mu = 0.10)` `14-2=2a, a=6 ms^(-2)`. Consider the motion of 5KG block. The forces on it are (i) gravitational force `5 g = 5xx10=50N`, vertically downwards, (ii) normal reaction N of the horizontal surface, vertically upwards (iii) force of friction `f = mu` N to the right by Newton.s third law of motion and (iv) normal reaction, N downwards by 2kg block. In the vertical direction, there is no acceleration. `therefore N = 50 N`. In the horizontal direction, the acceleration of the 5kg block is .a.. `mu N= 5a.`. `rArr 0.10xx20=5a., a.=0.4 ms^(-2)`. |
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| 16. |
A particle of mass is made to move with uniform speed v along the perimeter of a regular polygon of 2n sides. The magnitude of impulse of impulse appliedat each corner of the polygon in amv sin pi//bn. Then a/b = |
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Answer» 1 |
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| 17. |
Maximum wavelength of radiation from black body at 1640 K temp. is 1.75 mum. If moon is considered as perfect black body and maximum wavelength radiation from it is 14.35 mum, then what should be the temperature of moon ? |
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Answer» 100K `lamda_(m_(1))T_(1)=lamda_(m_(2))T_(2)` `:.T_(2)=T_(1)XX(lamda_(m_(1)))/(lamda_(m_(2))` `=1640xx(1.75)/(14.35)=200K` |
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| 18. |
Two small sphere of masses 10kg and 30 kg are joined by a rod of length 0.5 m and of negligible mass. The M.I. of the system about a normal axis through centre of mass of the system is |
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Answer» `1.875" KGM"^(2)` |
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| 19. |
A gas undergoes a change of state during which 100 J of heat is supplied to it and it does 20 J of work. The system is brought back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process ? |
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Answer» SOLUTION :In a CYCLIC PROCESS `DeltaU=0` `:.DeltaQ=DeltaW"" (100-20)=20+W_(2)` `:.W_(2)=+60J` |
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| 20. |
A bullet of mass in travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be |
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Answer» zero `mv + M(0) = (M + m)V " or " V = (mv)/((M + m))` The heat generated in the process `= 1/2 mv^2 - 1/2 (M + m) V^2 = 1/2 mv^2- 1/2 (M + m)((mv)/(M + m))^2` ` = 1/2 mv^2 - 1/2 (m^2v^2)/((M + m)) = 1/2 mv^2 (1 - (m)/((M + m)))` `= 1/2 mv^2 ((M)/(M + m)) = (mMv^2)/(2(M + m))` . |
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| 21. |
Calculate the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to a vertical height of 10 cm. |
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| 22. |
The work done by a force F=(hati+2hatj+3hatk) N ,to displace a body from positionAto position B is [The position vector of A is r_(1)=(2hati+2hatj+3hatk)m] |
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Answer» Solution :(a) Work done ,`W=F.(r_(2)-r_(1))` position `r_(2)-r_(1)=hati-hatj+2hatk` `W=(hati+2hatj +3hatk).(hati-hatj+2hatk)=1-2+6=5J` |
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| 23. |
Two masses, initially at rest, attract each other with a constant force. If there is no external force acting on the masses, prove that the centre of mass of the system remains stationary. |
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Answer» Solution :Let the line joining the two masses be taken as X-axis. Initial POSITION of mass `m_(1) is x_(1)` and that of mass `" "m_(2) ""is" " x_(2) (x_(2) gt x_(1))`. Hence, the initial position of the centre of mass is `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` Suppose a force F acts on the first mass along the positive x direction. Hence a force F will act on the second mass too along the negative direction of the x-axis as per Newton.s third law of motion. Hence, acceleration of the first mass `a_(1)= (F)/(m_(1))` and DISPLACEMENT in time `t=(1)/(2)a_(1)t^(2)`. So, the position of the first one after a time t, `x_(1).=x_(1)+(1)/(2)a_(1)t^(2)=x_(1)+(1)/(2)(F)/(m_(1))t^(2)` Similarly, the position of the second one after a time t, `x_(2).=x_(2)+(1)/(2).((-F))/(m_(2))t^(2)=x_(2)-(1)/(2)(F)/(m_(2))t^(2)` Hence, the position of the centre of mass after time t, `x._("cm")=(m_(1)x_(1).+m_(2)x_(2).)/(m_(1)+m_(2))=(m_(1)x_(1)+(FT^(2))/(2)+m_(2)x_(2)-(Ft^(2))/(2))/(m_(1)+m_(2))` or, `x._("cm")=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=x_("cm")` Hence, the centre of mass remains stationary. 
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| 24. |
Suppose a smooth tunnel is dug along a straight line joining two points on the surface of the carth and a particle is dropped from rest at its one end. Assume that mass of earth is uniformly distributed over its volume. Then: |
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Answer» the particle will emerge from the other end with velocity `sqrt((G M_e)/(2R_e))` where `M_e` and `R_e` are earth's mass and radius respectively. |
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| 25. |
A bomber flying horizontally with constant speed releases a bomb from an aeroplane. (a) The path of bomb as seen by the observer on the ground is parabola (b) The path of the bomb as seen by a pilot is a straight line. (c) The path of the aeroplane with respect to bomb is a straight line (d) The path of the bomb as seen by pilot observed as parabola. |
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Answer» a is correct |
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| 26. |
A ball of mass m moving with a horizontal velocity v strikes the bob of the pendulum at rest. Moss of the bob of the pendulum is also m. During this collision the ball sticks with the bob. The height to which the combined mass rises |
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Answer» `(v^2)/(4g)` |
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| 27. |
A plane undamped harmonicwave progates in a medium. Find the mean space density of energy becomes equal to W_(0) at an instant t=t(0)+T//6, where t_(0) is the instant when amplitude is maximum at this location and T is the time period of oscillation. |
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Answer» <P> Solution :Let us consider the wave`Y=A cos(omegat-kx)`, then its energy density (energy per unit VOLUME)is GIVEN by `W=pA^(2)omega^(2) sin ^(2) (omegat-kx)` `[`where `p` is of medium, for string waves, `mu =pS]`. Let us consider `x=0`, `t_(0)=0` at which amplitude is maximum At `t=t_(0)+(T)/(6)` And `t=A cos[(omegaT)/(6)]` and the energy density is `w=pA^(2)omega^(2)sin^(2)[(omegaT)/(6)]=pA^(2)omega^(2)sin^(2)(PI)/(3)` `W=(pA^(2)omega^(2))(3)/(4)` From given DATA, `W=W_(0)` `impliespA^(2)omega^(2)=(4)/(3)W_(0)=(2W_(0)/(3)` |
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| 28. |
"Heat cannot itself flow from a body at lower temperature to the body at higher temperature" is a statement or consequence of |
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Answer» SECOND LAW of THERMODYNAMICS |
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| 29. |
A bar of length 3.77m is fixed between two rigid ends. It has a crack at its centre which buckles as result of temperature rise of 32°C. If the coefficient of linear expansion of the bar is 25 xx 10^(-6)//^(@) C the rise of the centre is |
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Answer» `2.5 XX 10^(-2)` m |
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| 30. |
A thin rod having length L_(0) at 0^(@)C and coefficient of linear expansion alpha has its two ends maintained at temperature theta_(1) and theta_(2) respectively. Find its new length. |
Answer» Solution :The temperature in rod changed by going linearly from its one end to another end and temperature at midpoint is `theta`. In thermal steady state heat current `=(dQ)/(dt)=` constant.  `:.KA(theta_(1)-theta)/((L_(0//2)))=(KA(theta-theta_(2)))/((L_(0//2)))` where K is thermal conductivity, `:.theta_(1)-theta=theta-theta_(2)` `:.theta_(1)+theta_(2)=2THETA` `:.theta=(theta_(1)+theta_(2))/(2)` temperature of midpoint Now, its length INCREASES with increase in temperature, `:.L=L_(0)(1+alpha theta)` `:.L=L_(0)[1xxalpha((theta_(1)+theta_(2))/(2))]`which is new length. |
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| 31. |
A polished metal plate with a rough black spot on it is heated to about 1400K and quickly taken to a dark room.Which of the following statements is ture? |
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Answer» The SPOT will APPEAR brighter than the PLATE. |
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| 32. |
Which of the following is true for a satellite in a circular orbit |
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Answer» it is a freely FALLING body |
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| 33. |
The string , the spring and the pulley shown in the figure are light . Find the time period of the mass m |
| Answer» SOLUTION :`2PI SQRT(m/k)` | |
| 34. |
Two identical spheres of gold are in contact with each other. The gravitational force of attraction between them is |
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Answer» directly PROPORTIONAL to the SQUARE of their radius |
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| 35. |
A faulty thermometer has its fixed points marked as 3^(@)and102^(@). The temperature of a body as measured by the faulty thermometer is 80^(@). Find the correct temperature of the body on Celsius scale. |
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Answer» `36.5^(@) C` |
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| 36. |
Is negative temperature possible on Kelvin scale ? Why ? |
| Answer» Solution :No. Because absolute ZERO TEMPERATURE is MINIMUM temperature on Kelvin SCALE. | |
| 37. |
A spring is stretched by 3 cm, its potential energy is U_1. If the spring is stretched by 6 cm, its potential energy be that is stored in it is U_(2)="……"U_(1). (Fill in the blank). |
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Answer» Solution :`U_(1) = (1)/(2)ky_(1)^(2)= (1)/(2)k(3)^(2)= (1)/(2)KXX 9` `U_(2)= (1)/(2)ky_(2)^(2)= (1)/(2)k(6)^(2)= (1)/(2)kxx 36` `therefore (U_2)/(U_1)= (36)/(9)= 4` `therefore U_(2) = 4U_(1)`. |
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| 38. |
If m_(1) = 10kg, m_(2)= 4kg, m_(3)= 2kg, the acceleration of system is |
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Answer» `5g//2` `T_(2)-m_(2)g=m_(3)a` |
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| 39. |
An alpha- particle is moving along a circle of radius R with a constant angular velocity omega. Point A lies in the same plane at a distance 2R from the center. Point A records magnetic field produced zero magnetic fields is 't', then the angular speed omega in terms of t is |
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Answer» `(2PI)/(t)` |
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| 40. |
A plane surface is incilined making an angle thetawith the horizontal. From the bottom of this inclined plane, a bullet is fired with velocity'v' . Find the maximum possible range of the bullet on the inclined plane |
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Answer» `(V^(2))/(G)` |
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| 41. |
A: Apparent weight of 1 kg of iron is equal to that of 1 kg of cotton in air R: Buoyant force on 1 kg of iron is lesser than Buoyant force on 1 kg of cotton. |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 42. |
A bullet is fired from a rifle anf the rifle recoils. Kinetic energy of rifle is |
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Answer» less than K.E. of BULLET |
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| 43. |
A wire elongates 2mm when a stone is suspended from it. When the stone is completely immersed in water, the wire contracts by 0.6 mm. Find the density of stone. |
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| 44. |
You are given the following group of particles, n_i represents the number of molecules with speed v_i n_(i)" 24863" V_i(ms^(-1))" 1.02.03.04.05.0" Calculate (i) average speed (ii) rms speed (iii) most probable speed. |
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Answer» SOLUTION :(i) average speed `V_(ANG)=(n_(1)v_(1)+n_(2)v_(2)+n_(3)v_(3)+n_(4)v_(5)+n_(5)v_(5))/((n_(1)+n_(2)+n_(3)+n_(4)+n_(5))` `=((2 xx 1)+(4 xx 2)+(8 xx 3)+(6 xx 4)+(3 xx 5))/((2+4+8+6+3))` `=(73)/(23)=3.17 ms^(-1)` (ii) root mena square speed is `v_(rms) =((n_(1)v_1^2+n_2v_2^2+n_3v_3^2+n_4v_4^2+n_5v_5^2)/(n_1+n_2+n_3+n_4+n_5))^(1//2)` `=((2 xx 1+4 xx 4+8 xx 9 +6 xx 16 +3 xx 25)/(2+4+8+6+3))^(1//2)=3.36ms^(-1)` (ii) Most probable speed 3m/s by which largest molecules are moving at any TIME. |
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| 45. |
Which of the examples represent simple harmonic motion and which represent periodic but not simple harmonic motion ? Motion of an oscillating mercury column in a U tube , |
| Answer» SOLUTION : It is S.H.M. | |
| 46. |
Which of the examples represent simple harmonic motion and which represent periodic but not simple harmonic motion ? Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower must position. |
| Answer» SOLUTION : It is S.H.M. | |
| 47. |
Which of the examples represent simple harmonic motion and which represent periodic but not simple harmonic motion ? The rotation of earth about its axis. |
| Answer» SOLUTION : It is PERIODIC but not S.H.M. because it is not to and FRO motion about a fixed point. | |
| 48. |
Which of the examples represent simple harmonic motion and which represent periodic but not simple harmonic motion ? General vibrations of a polyatomic molecule about its equilibrium position. |
| Answer» SOLUTION :It is a periodic but not S.H.M. A POLYATOMIC gas MOLECULE has a number of NATURAL frequencies and its GENERAL motion is the resultant of S.H.Ms. of a number of different frequencies. The resultant motion is periodic but not S.H.M. | |
| 49. |
A gas is taken in an enclosure. The pressure of the gas is reduced by pumping out some gas. Will you expect the temperature of the gas to decrease by Charles. law ? |
| Answer» Solution :No, since CHARLES. LAW is applicable for the VARIATION of the pressure of a GAS of constant mass. | |
| 50. |
In an experiment it was found that when a sonometer in its fundamental mode of vibration and a tunning fork gave 5 beats when length of wire is 1.05 metre or 1 metre. The velocity of transverse waves in sonometer wire when its length is 1m |
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Answer» `400m//s` `n prop (1)/(l) IMPLIES(n_(1))/(n_(2))=(l_(2))/(l_(1))=(1)/(1.05)impliesn_(1)=(100)/(105)n_(2)` `impliesn_(2) gt n_(1)` Here FORK GIVES `5` beats/sec with each length and hence `n_(2) gt n gt n_(1)` `implies{:(n_(2)-n=5.....(1)),(ul(n-n_(1)=5).....(2)),(ul(n_(2)-n_(1)=10)):}` |
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