Saved Bookmarks
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. |
What is the effect temperature on coeffieient of viscosity of a liquid ? |
| Answer» Solution :As temperature increase coefficient of VISCOSITY of liquid DECREASES, This is why oils of DIFFERENT viscosities are USED in different seasons , for lubrications. | |
| 2. |
Three point masses, each of mass m, are situated at the three corners of an equilateral triangle of each side l. The moment of inertia of this system about one of the sides of the triangle will be - |
|
Answer» `ml^(2)`  `I=(0)+(0)+m((sqrt(3))/(2)l)^(2)` `=(3)/(4)ml^(2)` |
|
| 3. |
What is meant by a fundamental frequency ( mode) or first harmonic or vibration of a system? |
| Answer» Solution :The lowest POSSIBLE natural FREQUENCY of a vibrating SYSTEM is called the fundamental MODE or FIRST harmonic. | |
| 4. |
What is the basis of cesium atomic clock ? |
| Answer» SOLUTION :The basis of caesium ATOMIC clock is the PERIODIC vibration of the ATOMS . | |
| 5. |
In natural convection, a heated portion of a liquid moves because |
|
Answer» its molecular MOTION BECOMES aligned |
|
| 6. |
A heavenly body is observed from two diametrically opposite points A and B on the earth. The angle substended at the heavenly body is 2.9 xx 10^(-4) rad. Given the diameter of Earth to be about 1.28 xx 10^(7) Compute the distance of the heavenly body from the earth. |
|
Answer» `2.267xx10^(-11) m` |
|
| 7. |
Four point masses 1 kg, 1 kg, 2 kg and 2 kg are placed at the corners of a square of side a m as shown in Fig. Find the centre of mass of thesystem. |
|
Answer» |
|
| 8. |
If a car covers 2/5th of the total distance with v_(1) speed and 3//5^(th) distance with v_(2) then average speed is |
|
Answer» `(1)/(2)SQRT(v_(1)v_(2))` |
|
| 9. |
A coin is kept at distance of 10 cm from the centre of a circular turn table. If mu=0.8the frequency of rotation at which the coin just begins to slip is |
|
Answer» 62.8 rpm |
|
| 10. |
A mercury drop shaped as a round tablet of radius R and thickness h is located between two horizontal plates, Assuming hltltR. Find the mass .m. of a weight which has to be placed on the copper plate to dimnish the distance between plates n-times (theta angle of contact) |
|
Answer» Solution :`Deltap=T{1/r_(1)+1/r_(2)}" now "Deltap=T{(2(costheta))/h}+1/R` `r_(1)=h/(2(costheta)),r_(2)=R" if "hltltRrArrDeltap=(2T(costheta))/h` then total force EXERTED by mercury on upper glass plate `F=(Deltap)` (A contact) = `({(2T(costheta))/h}piR^(2))/("if R is NEW radius of drop")` when `h^(1)=h/n` then CONSERVATION of volume `piR^(2)h=pi(R^(1))^(2)(h^(1))rArrR^(1)=sqrtnR" "f^(1)=n^(2)F` If .mg. is the weight placed on the upper plate, `F^(1)=f+mg` `rArrm^(1)=(F^(1)-f)/G=(2piR^(2)T(costheta))/(gh)(n^(2)-1)thereforem^(1)={(2piR^(2)T|costheta|)/(gh)}(n^(2)-i)` 
|
|
| 11. |
Does a vibrating body always produce sound ? |
| Answer» Solution :No. A vibrating body will produce sound only when its FREQUENCY of OSCILLATION LIES between 20 HZ to 20 KHz. | |
| 12. |
Iron, lead and brass spheres of equal masses are heated to a common temperature and are kept on a sheet of wax. The sphere that passes through the plate first is (lead 0.03cal//g//""^(@)C, iron 0.11cal//g//""^(@)C, brass 0.08cal//g//""^(@)C |
|
Answer» lead |
|
| 13. |
An ideal gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure P_1 . The same gas is expanded adiabatically to a volume 2V, the final pressure is P_A. In terms of the ratio of the two specific heats for the gas gamma , the ratio P_I//P_Ais : |
|
Answer» `T_2=T_1(V_1/V_2)^(GAMMA-1)` |
|
| 14. |
A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from his hole. a. The water level will rise up in the vessel b. The pressure at the surface of the water will decrease c. The force by the water on the bottom of the vessel will decrease d. The density of the liquid will decrease |
|
Answer» B and C are correct |
|
| 15. |
The cross project of the vectors (2i-3j+4k) and (i+4j-5k) is |
|
Answer» `-HAT(i)-14hat(j)+11 hat(K)` |
|
| 16. |
Two blocks of masses m and M connected by a spring are placed on frictionless horizontal ground. When the spring is relaxed, a constant force F is applied as shown. Find maximum extension of the spring during subsequent motion. |
Answer» Solution :If we use ground as INTERNAL frame as we usually do, solutiona of the problem BECOMES quite involved. Therefore, we prefer the center of mass frame, in which mass center remains at rest.  In the adjacent figure is SHOWN horizontal position of mass center (CM) by dashed line. It remains unchanged in center of mass frame. Mass center of TWO particle system divides separation between them in reciprocal ratio of the masses, therefore displacements `x_(1)` and `x_(2)` of the blocks must also be in reciprocal ratio of their masses. The extension x is sum of displacements `x_(1)` and `x_(2)` of the blocks as shown in the figure.When extension of the spring achieves its maximum value, both the block must stop receding away from the mass center, therefore, velocities of both the blocks in center of mass must be zero, nrgative work done by spring forces becomes equal to increase in potential energy and work done by applied force evidently becomes `Fx_(1)`. Using above fact in applying work energy theorem on the system relative to the cente of mass frame, we obtain `K_(i)+W_(irarrf)=K_(f) rarr0+W_(i rarr f,"springforce")+W_(i rarr f),F = 0` `-(0-1/2kx^(2)) + (FM_(x))/(m+m) = 0` `x=(2FM)/(k(m+m))` |
|
| 17. |
The speed of sound in oxygen is 332 m/s at STP. The speed of sound in hydrogen at STP will be ………….. |
|
Answer» 53/2 m/s `gamma` is same for both the gases , so , `(v_(H_(2)))/(v_(O_(2))) = sqrt((32)/(2)) = 4` `IMPLIES f_(H_(2)) = 4 xx 332 = 1328` m/s |
|
| 18. |
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along |
|
Answer» a line PERPENDICULAR to the PLANE of ROTATION |
|
| 19. |
A stone is projected vertically up to reach a maximum height 'h'. The ratio of its kinetic to potential energies at a height (4h)/(5) will be |
| Answer» ANSWER :C | |
| 20. |
A liquid of specific heat 0.5 at 60^(@)C, is mixed with another liquid of specific heat 0.3 at 20^(@)C. After mixing, the temperature of the mixture becomes 30^(@)C. In what proportion by weight are the liquids mixed ? |
|
Answer» Solution :Mass of liquid `A = m_(1)` , Mass of heat liquid `B = m_(2)` , Specific heat of `A(S_(1)) = 0.5` Specific heat of `B(S_(2)) = 0.3` , Temperature of `A(t_(1)) = 60^(0)C` , Temperature of `B(t_(2)) = 20^(0)C` Final temperature `(t) = 30^(0)C` , Fall in temperature of `A = 60 - 30 = 30^(0)C` RISE in temperature of `B = 30 - 20 = 10^(0)C` Heat lost by the liquid `A = m_(2)S_(1) (t_(1) - t) = m_(1) xx 0.5 xx 30` CAL Heat gained by the liquid `B = m_(2) S_(2) (t-t_(2)) = m_(2) xx 0.3 xx 10` cal Heat lost by the liquid A = heat gained by the liquid B `m_(1) xx 0.5 xx 30 = m_(2) xx 0.3 xx 10 (or) (m_(1))/(m_(2))=(0.3xx10)/(0.5xx30) = 1/5 = 1:5`. |
|
| 21. |
A wooden log of mass 120kg is floating on still water perpendicular to the shore. A man of mass 80kg is standing at the centre of mass of the log and he is at a distance of 30m from the shore. When he walks through a distance of 10m towards the shore and halts, now his distance from the shore is |
|
Answer» 20 m |
|
| 22. |
If a planet of given density were made larger (keeping its density unchanged) its force of attraction for an object on its surface would increased because of increased mass of the plane but would decrease because of larger separation between the centre of the planet and its surface. Which effect would dominate |
|
Answer» Increase in MASS |
|
| 23. |
The root mean square velocity of the molecules in a sample of helium is 5/7th that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is 0^(@)C , then the temperature of the helium sample is about |
| Answer» ANSWER :A | |
| 24. |
A small mass m is moved slowly from the surface of the earth to a height h above the surface. The work done (by an external agent) in doing this is (a) mgh for all values of h (b) mgh, for h lt lt R (c) (1)/(2) mgR , for h = R (d) -(1)/(2)mgR, for h = R |
|
Answer» only a & B are TRUE s |
|
| 25. |
Anvils made of single crystals of diamond, with theshape as shown in figure, are used to investigatebehaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameterof 0.50 mm, and the wide ends are subjectedto a compressional force of 50,000 N. Whatis the pressure at the tip of the anvil? |
|
Answer» SOLUTION : DIAMETEROF the narrow end (i.e., tip) of the anvil, `D=0.5 MM = 5XX10^(-4)m` Areaof the tip, `A=(piD^(2))/(4)=2xx10^(-7) m^(2)` Force, `F=5xx10^(4)N` Pressure at the tip, `P=(F)/(A) = 2.5xx10^(11) Pa`. |
|
| 26. |
An ideal spring with a spring constant 'K' is hung from the ceiling and a block of mass 'm' is attached to its lower end. The mass is released with spring initially unstretched. Then the maximum extension in the spring is |
|
Answer» `4 mg//K` |
|
| 27. |
A linear harmonic oscillator of force constant 2xx10^(-6)Nm^(-1) andamplitude 0.01 m has a total mechanical energy of 160J. Then find maximum and minilmum values o P.E and K.E? |
|
Answer» Solution :MAXIMUM `K.E.=1/2KA^(2)=1/2xx2xx10^(6)xx(0.01)^(2)=100J` since TOTAL energy is 160 J, Maximum P.E. is 160 J. From this it is understood that at the mean position potential energy of the simple harmonic OSCILLATOR is minimum which NEED not be zero. `PE_("min")=TE-KE_("max")=160-100=60J=KE_("min")=0` |
|
| 28. |
Two identical blocks each of mass ''M'' are tied to the ends of a string and the string is laid over a smooth fixed pulley. Initially the masses are held at rest at the same level. What fraction of mass must be removed from one block and added to the other, so that it has an acceleration of 1//5^("th") of the acceleration due to gravity |
|
Answer» `1//10` |
|
| 29. |
Refractive index of a flint glass (mu) was measured in an experiment and was found to be 1.655,1.667, 1.655, 1.659, 1.669" and "1.654. Find (i) the mean value of mu (ii) mean absolute error (iii) relative error (iv) percentage error. |
|
Answer» SOLUTION :Given, `mu_(1)=1.655, mu_(2)=1.667, mu_(3)=1.655, mu_(4)=1.659, mu_(5)= 1.669, mu_(6)=1.654` (i) Mean value of `mu` is `mu_(m)=(mu_(1)+mu_(2)+mu_(3)+mu_(4)+mu_(5)+mu_(6))/(6)` `=(1.655+1.667+1.655+1.659+1.669+1.654)/(6)=1.6598` `=1.660` (Rounded off to 4 significant figures) ABSOLUTE errors in the different measurments are `triangle mu_(1)= mu_(m)-mu_(1)` `=1.660-1.655=0.005` `triangle mu_(2)= 1.660-1.667= -0.007` `triangle mu_(3)= 1.660-1.655=0.005` `triangle mu_(4)= 1.660- 1.659= 0.001` `triangle mu_(5)= 1.660- 1.669= -0.009` `triangle mu_(6)= 1.660- 1.654= 0.006` Mean absolute ERROR `triangle bar(mu)= (sum|triangle mu_(i)|)/(6)` `=(0.005+0.007+0.005+0.001+0.009+0.006)/(6)` `=0.0055=0.006` (Rounded off to 1 significant figure) (iii) Relative error `triangle mu= (triangle bar(mu))/(mu_m)=(0.006)/(1.660)=0.0036` `=0.004` (iv) percentage error `= (triangle bar(mu))/(mu_m)xx100%= 0.4%`. |
|
| 30. |
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (a) Show p=p_(i)'+m_(i)V where pi is the momentum of the ith particle (of mass m_(i)) and p'_(i)=m_(i)v'_(i). Note v'_(i) is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass sump'_(i)=O (b) Show K=K'+1//2MV^(2) where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV^(2)//2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14. (c ) Show L=L'+RxxMV where L'=sumr'_(i)xxp'_(i) is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember – r'_(i)=r_(i)-R , rest of the notation is the standard notation used in the chapter. Note ′ L and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles. (d) Show (dL')/(dt)=sumr'_(i)xx(dp')/(dt) Further, show that (dL')/(dt)=tau'_(ext) where tau'_(ext) is the sum of all external torques acting on the system about the centre of mass. |
| Answer» Solution :USE the definition of centre of mass and third LAW of motion. Assume the INTERNAL forces between any two particles act along the LINE joining the particles. | |
| 31. |
In the system shown, the ratio of energy stored in the spring S_(1) to that of spring S_(2) in equilibrium is |
|
Answer» `(K_(1))/(K_(2))` |
|
| 32. |
Find the work done if a particle moves from position r_(1)="to a position"vecr_(1)=(3hati+2hatj-6hatk) "to position "vecr_(2)=(14hati+13hatj-9hatk)"under the effect of force " vecF=(4hati+hatj+3hatk)N |
|
Answer» SOLUTION :`vecr=vecr_(2)-vecr(1)11hati+11hatj-3hatk` `VECF=(4hati+hatj=3hatk)N` `W=vecF*vecr=46J` |
|
| 33. |
An unhappy mouse of mass mo moving on the end of a spring of spring constant p is acted upon by a damping force F_(x) =-btheta_(x). For what value of b the motion is critically damped ? |
|
Answer» `B=sqrt(p/m_(0))` |
|
| 34. |
Transverse waves are produced in a long string by attaching its free end to a vibrating tuning fork. Figure shows the shape of a part of the string . The points in phase are |
|
Answer» `A` and `D` |
|
| 35. |
The first two lengths of an air column , in a resonance column method , were found to be32.1 cm and 99.2 cm , respectively . Determine the end correction for the tube . If it is known that velocity of sound in the laboratory is332 m//s, then find the frequency of the vibrating tunning fork. |
|
Answer» SOLUTION :Here , `l_(1) = 32.1 cm and l_(2) = 99.2 cm` `:. Frome = (l_(2) - 3l_(1))/( 2) `, the end correction isgiven by ` e = ( 99.2 cm - 3 xx 32.1 cm)/( 2) = 1.45 cm` ALSO , from ` v = 2 f (l_(2) - l_(1))` , the FREQUENCY of the vibrating tunning fork is given by ` f = (v)/( 2( l_(2) - l_(1))) = ( 332 m//s)/( 2( 99.2 - 32.1 ) xx 10^(-2) m) = 247 .4 Hz` |
|
| 36. |
"Insulators also contain electrons, but they are not conductors." Why ? |
| Answer» SOLUTION :They are not FREE ELECTRONS. | |
| 37. |
A 60 cm metal rod M_(1) is joined to another 100 cm metal rod M_(2) to form an L shaped single piece. This piece is hung on a peg at the joint. The two rods are obeserved to be equallt inclined to the vertical. If the two rods are equally thick, the ratio of density of M_(1) to that of M_(2) is |
|
Answer» `5//3` |
|
| 38. |
A ball strikes a smooth horizontal floor obliquely and rebounds inelastically. |
|
Answer» The KINETIC energy of the ball just after hitting the FLOOR is equal to the additional potential energy of the ball at its maximum height after rebound. |
|
| 39. |
The V-t graphs of two objects make angle 30^(@) and 60^(@) with the time axis. Find the ratio of their accelerations. |
| Answer» Solution :`(a_(1))/(a_(2))=(tan30^(@))/(TAN60^(@))=(1//sqrt(3))/(SQRT3)=(1)/(3)=1:3` | |
| 40. |
Newton's second law gives |
|
Answer» `vecFprop (d VECP)/(DT)` |
|
| 41. |
In which type of collision mechanical energy is transformed into other forms of energy? |
| Answer» SOLUTION :INELASTIC COLLISION. | |
| 42. |
Answer the following questions based on the P-T phase diagram of carbon dioxide: What are the critical temperature and pressure for CO_(2)? What is their significance ? |
| Answer» Solution :The CRITICAL TEMPERATURE and pressure of `CO_(2) " are " 31.1 ^(@) C " and " 73.0 `atm, respectively. Above this temperature, `CO_(2)` will not LIQUEFY even if compressed to high PRESSURES. | |
| 43. |
A tank is filled with two immiscible liquids of densities 2rho and rho each of height h. Two holes are made to the side wall at (h)/(2) and (3h)/(2) from upper surface of the liquid, then find the ratio of velocity of efflux of the liquids through the holes |
Answer» Solution :According to Bernoullis theorem, to find `v_(1)`  `P+rhog((h)/(2))=P+(1)/(2)(rho)(v_(1)^(2))impliesv_(1)=SQRT(gh)` to find `v_(2),P+(rhogh)+(2rho)g((h)/(2))=(1)/(2)(2rho)v_(2)^(2)+P` `implies v_(2)=sqrt(2gh) therefore (v_(1))/(v_(2))=(1)/(sqrt(2))` |
|
| 45. |
Two horizontal discs of different radii are free to rotate about their central vertical axes. One is given some angular velocity, the other is stationary. Their rims are now brought in contact. There is friction between the rims |
|
Answer» the FORCE of friction between the rims will disappear when the discs rotate with EQUAL angular speeds. |
|
| 46. |
When object is projected in vertically upward direction with velocity u, its time of flight will be …... . |
|
Answer» |
|
| 47. |
A body of mass m moving with velocity u collides head on elastically with another body of mass 2m at rest. The ratio of kinetic energy of the colliding body before and after collision is |
|
Answer» `4 : 1` |
|
| 48. |
Duringforcedvibration,if the frequencyof free vibration of a body is equal to the frequency of the external periodic force,thenthe phenomenonthat occurs is called |
|
Answer» beats |
|
| 49. |
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313K. If the specific heat of air at constant volume is 0.718 k. j//kg K, find the amount of heat absorbed in kj and kcal (J=4.2 KJ// Kcal) |
|
Answer» 6.838 KJ |
|
| 50. |
Two small pegs (A and B) are at horizontal and vertical separation b and h respectively. A small block of mass M is suspended with the help of two light strings passing over A and B as shown in fig. The two string are always kept at right angles (i.e., lt APB = 90^(@)). Find the minimum possible gravitation potential energy of the mass assuming the reference level at location of peg A. [Hint: the potential energy is minimum when the block is at its lowest position] |
|
Answer» |
|