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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. |
Principle of conservation of linear momentum: |
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Answer» Solution :(i) When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force `vec(F)_(21)` on particle 2 and particle 2 exerts an exactly equal and opposite force `vec(F)_(12)` on particle 1, according to Newton's third law. `""vec(F)_(21)=-vec(F)_(12)` (ii) The force on each particle (Newton's second law) can be written as, `vec(F)_(12)=(dvec(p)_(1))/(DT) " and " vec(F)_(21)=(dvec(p)_(2))/(dt)` (iii) Here `vec(p)_(1)` is the momentum of particle 1 which changes due to the force `vec(F)_(12` exerted by particle 2. Further `vec(p)_(2)` is the momentum of particle 2. This changes due to `vec(F)_(21)` exerted by particle 1. `(dvec(p)_(1))/(dt)=-(dvec(p)_(2))/(dt)` `(dvec(p)_(1))/(dt)+(dvec(p)_(2))/(dt)=0` `d/(dt)(vec(p)_(1)+vec(p)_(2))=0` (IV) It implies that `vec(p)_(1)+vec(p)_(2)=` constant vector (always). (v) `vec(p)_(1)+vec(p)_(2)` is the total linear momentum of the two particles `(vec(p)_("tot")=vec(p)_(1)+vec(p)_(2))`. It is also called as total linear momentum of the system. Here, the two particles constitute the system. (vi) If there are no external forces acting on the system, then the total linear momentum of the system `(vec(p)_("tot"))` is always a constant vector. |
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| 2. |
Assertion :Damped oscillation indicates loss of energy. Reason : The energy loss in damped oscillation my be due to friction, air resistance. |
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Answer» Both are ture and the reason is the CORRECT EXPLANATION of the assertion. |
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| 3. |
Solve the following with regard to the correct significant figures (i) sqrt(4.8-2.35) (ii) 2.35xx10^(3)-1.2xx10^(3)//2.0xx10^(2) |
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| 4. |
Explain the relation in phase between displacement, velocity an acceleration in SHM, graphically as well as theortically. |
Answer» SOLUTION :
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| 5. |
A wall consists of alternating blocks of length 'd' and coefficient of thermal conductivity k_(1) and k_(2) respectively as shown in figure. The cross sectional area of the blocks are the same. The equivalent coefficient of thermal conductivity of the wall between left and right is. . . . . |
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Answer» `(k_(1)+k_(2))/(2)` `K=(sumk_(i)A_(i))/(sumA_(i))` `=(k_(1)A+k_(2)A+k_(1)A+K_(2)A+k_(1)A+k_(2)A)/(6A)` `=(3(k_(1)+k_(2)A))/(6A)=(k_(1)+k_(2))/(2)` |
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| 6. |
Show that the centre of gravity of three equal weights suspended from three vertices of a triangle coincide with the centre of mass of the triangle. |
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Answer» Solution :Let a median of the triangle ABC be AD Fig. Identical weights W are hung from each of the VERTICES A,B and C. RESULTANT of the weights suspended from B and C = 2W and it acts from the point D. Resultant of the forces 2W at D and W at A is 3W. SUPPOSE this resultant acts at G. Hence `WxxAG=2WxxDG` or, AG=2DG THUS, G is the point of intersection of the medians of the triangle, which is the centre of mass. 
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| 7. |
An ideal gas is trapped between a mercury column and the closed lower end of a narrow vertical tube of uniform bore. The upper end of the tube is open to the atmosphere. (Atmospheric pressure is 76 cm of mercury). The length of the mercury and the trapped gas columns are 20 cm and 43 cm, respectively. What will be the length of the gas column when the tube is tilted slowly in a vertical plane through an angle of 60^(@). Assume the temperature to be constant. |
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Answer» SOLUTION :Boyle.s LAW HOLDS good because the temperature is constant. So, `P_(1)V_(1)=P_(2)V_(2)orP_(1)l_(1)=P_(2)l_(2)`, since the bore is uniform `P_(1)=76+20=96cm` of HG `l_(1)=43cm,P_(2)=76+hcostheta=76+20cos60^(@)` = `76+10=86cm` of Hg `l_(2)=?i.e., 96xx43=86l_(2)rArrl_(2)=48cm` of Hg |
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| 8. |
Explain the acceleration. |
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Answer» Solution :The time rate of change of velocity is called acceleration. Let a particle be moving in a straight line andat time `t_(1) and t_(2)` its velocities are `v_(1) and v_(2)` respectively.Thus, the change in velocity of the particle in time interval `Delta t = t _(2) - t _(1) is v _(2) - v _(1).` According to definition of average acceleration, Average acceleration`= ("change in velocity")/(“time")` `therefore LT a gt = (v_(2) -v_(1))/(t _(2) - t _(1)) = (Delta t)/(Delta t)` Average acceleration is a vector quantity and its direction is in the direction of change in velocity `(Deltav)` . The UNIT of acceleration is `ms^(-2)` From average acceleration we cannot know how the velocity of particle changes with time. Taking `lim _(Delta t to 0)`in equation then we get instantaneous acceleration a at time t. `therefore a = lim _(Delta t to 0) (Delta v )/(Delta t) = (dv)/(dt)` Now, `v = (dx)/(dt)` `therefore a = (dv)/(dt) = (d)/(dt) ((dx)/(dt))` `therefore a = (d ^(2) x)/(dt ^(2)) = bar x` In other words second DERIVATIVE of position with respect to time is acceleration of a particle. If `(dv)/(dt)` is positive, acceleration is along the positive X-axis and if it is negative the acceleration is along the negative X-axis. |
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| 9. |
A disc of mass M and radius R is rotating with angular velocity omega_(0) about a vertical axis passing through its centre (O). A man of mass (M)/(2) and height (R)/(2) is standing on the periphery. Theman gradually lies down on the disc such that his head is at a distance (R)/(2) from the centre and hisfeet touching the edge of the disc. For simplicity assume that the mancan be modelled as a thin rod of length (R)/(2). Calculate the angular speed (omega) of the platform after the man lies down. |
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| 10. |
Consider the motion of the tip of the minute hand of a clock. In one hour (a) the displacement is zero (b) the distance covered is zero (c) the average speed is zero (d) the average velocity is zero |
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Answer» a & B are correct |
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| 11. |
A: A body is thrown vertically upwards with an initial speed 25 m/s from a position 1. It falls back to position 1 after some time. During this time duration, total change of velocity of the body is zero. R : Average acceleration of the body during this time is zero |
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Answer» If both A and B are true and R is the CORRECT explanation of A |
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| 12. |
Three vectors of magnitude 3,6 and 6 units are acting along the sides of an equilateral trinangle taken in order the magnitude of their resultant is |
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| 13. |
Find out the center of mass for the given geometrical structures. (a) Equilateral triangle (b) Cylinder (c) Square |
Answer» Solution : (a) For equilateral triangle, centre of mass lies at its centroid. (b) For CYLINDER, centre of mass lies at its geometrical centre. (c) For SQUARE, centre of mass lies at the POINT where the diagonals MEET |
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| 14. |
A body executes SHM.The PE and KE and total energy (TE) are measured as functions of displacement x.Which of following statements is true? |
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Answer» KE is MAXIMUM,when X=0 |
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| 15. |
Passage -II The figure shows the variation of potential energy of a particle as a function of x, the X-coordinate of the region. It has been assumed that potential energy depends only on X. For all other values of x, U is zro, i.e. for x lt -10 and x gt15,U=0. If total mechanical energy of the particle is 25J, then it can be found in region |
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Answer» `-10ltxlt-5` and `6lexlt15` |
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| 16. |
If A is the areal velocity of a planet of mass M, its angular momentum is |
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Answer» M/A |
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| 17. |
The relative error in the determination of the surface area of a sphere is alpha . Then the relative error in the determination of its volume is : |
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Answer» `(3)/(2) ALPHA` |
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| 18. |
Shown below is a distribution of charges. Find the flux of electric field due to these charges through the surface S. |
| Answer» SOLUTION :`2Q// epsi_(0) ` | |
| 19. |
The tank in figure discharges water at constant rate for all water levels above the air inlet R. Find the height above datum to which water would rise in the manometer tubes M and N. |
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| 20. |
A uniform circular disc of radius R is placed on a frictionless horizontal plane. Another identical disc rotating with velocity omega_(0) is gently placed on top the first disc.The coefficient of friction between the discs is mu. The time in which both disc acquire same angular velocity is nRomega_(0)//8mu g. Find the value of n. |
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| 21. |
The bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring . A third body C of mass m moves with a velocity v_0 along the line joining A and B and collides elastically with A as shown in fig1.18.At a certain instant of time t_0 after the collision, it is found that the instantaneous velocities of A and B are the same. Further,at this instant the compression of the spring is found to be x_0. FInd out the force constant of the spring. |
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Answer» Solution :According to the principle of conservation of ENERGY, `1/2mc_(0^2)=1/2mv^2+1/2times2mv^2+1/2kx_(0^2)` or,`mv_(0^2)=3mv^2+kx_(0^2)` [k=force constant of the spring] or, `mv_(0^2)=3M(v_0/3)^2+kx_(0^2)or,k=2/3(mv_(0^2))/x_(0^2)` |
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| 22. |
Inner and outer radii of a spool are r And R respectively. A thread is wound over ita ineer surface and placed over a rough horizontal surface. Thread is pulled by a force F as shown in the figure. Then in case of pure rolling. |
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Answer» thread unwinds, spool ROTATES ANTICLOCKWISE and FRICTION leftwards |
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| 23. |
What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator ? |
| Answer» SOLUTION :`(a_("MAX"))/(v_("max"))= OMEGA`. | |
| 24. |
The bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring . A third body C of mass m moves with a velocity v_0 along the line joining A and B and collides elastically with A as shown in fig1.18.At a certain instant of time t_0 after the collision, it is found that the instantaneous velocities of A and B are the same. Further,at this instant the compression of the spring is found to be x_0. FInd outthe common velocity of A and B at time t_0and (ii) the force constant of the spring. |
Answer» Solution :After elastic collision, C will same to rest and A will GAIN in velocity `v_0` (as their mass are the same).  Let us assume that the COMMON velocity of A and B is v, at time `t_0` after the collision. According to the principle of CONSERVATION of MOMENTUM, `mv_0=mv+2mv or,v=v_0/3` |
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| 25. |
Dielectric constant has the same dimensions as |
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Answer» Stress |
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| 26. |
From the top of a towerbody A is thrown up vertically with velocity u and another body B is thrown vertically down with the same velocity u.If V_(A) and V_(B) are their velocities when they reach the ground and t_(A) and t_(B) are their times of flight ,then |
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Answer» `v_(A)=v_(B) and t_(A)=t_(B)` |
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| 27. |
The resultant of two vecP and vecQ is vecR. If the magnitude of vecQ is doubled, the new resultant becomes perpendicular to vecP, then the magnitude of vecR is |
| Answer» Answer :C | |
| 28. |
The potential energy of a particle varies with distance x from a fixed origin as V= (A sqrtx)/(x + B)where A and B are constants. The dimensions of AB are |
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Answer» `M^1L^(5//2)T^(-2)` |
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| 29. |
At the top of a mountain, a thermometer reads 7^(0)Cand a barometer reads 70 cm of Hg. At the bottom of tlie ,nowuain, they read 27^(0)C and 76 cm of Hg respectively.Compare the density of air at tlu top with that at the bottom of the mountain. |
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Answer» Solution :At the top of the mountain, PRESSURE `P_(1) = 70` cm of HG Temperature `T_(1) = 7 + 273` = 280 K DENSITY of the air = `d_(1)` Atthe bottom of the mountain, pressure `P_(2) ` = 76 cm of Hg Temperature `T_(2) = 27 + 273 = 300 ` K Density of the air `d_(2)` From the gas equation, `(P_(1))/(d_(1)T_(1)) = (P_(2))/(d_(2)T_(2)) " (or)" (d_(1))/(d_(2)) = (P_(1))/(T_(1)) xx (T_(2))/(P_(2)) = (70)/(280) xx (300)/(76)` = 0.9868. |
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| 30. |
A vertical water jet flows out of a round hole. One of the horizontal sections of the jet has a diameter d = 20mm while the other section located 1 = 20cm lower has the diameter which is n = 1.5 times less. The volume of the water flowing from the hole each second is found to be 9 xx 10^(-n) cm^3/s. What is the value of n ? (Surface tension T = 0.073 N/m, and density of water = 10^3 kg/m^3). |
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| 31. |
Find the distance covered by a particle from time t=0 to 1=6 sec, when the particle follows the movement according to y = a cos ((pi)/(4)) t: |
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Answer» a |
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| 32. |
Assertion : Space rockets are usually launched in the equatorial line from west to east The acceleration due to gravity is minimum at the equator . |
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Answer» Both ASSERTION and REASON are true and Reason is the correct EXPLANATION of Assertion |
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| 33. |
A U-tube is supported with its limbs vertical and is partly filled with water. If internal diameters of the limbs are 1xx10^(-2)mand1xx10^(-4)m respectively, what will be the difference in height of water in the two limbs ? Surface tension of water is 0.07Nm^(-1) ? |
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Answer» Solution :Surface tension, `sigma=0.07Nm^(-1)`, Density, `rho=1000kgm^(-3),g=9.8ms^(-2)` Angle of contact `theta=0^(@)`, radius, `r_(1)=0.5xx10^(-2)m`, Radius, `r_(2)=0.5xx10^(-4)m`, Let `h_(1)` be the height of water in the limb of radius `r_(1)`. Then, `h_(1)=(2sigmacostheta)/(r_(1)rhog)=(2xx0.07xxcos0^(@))/(0.5xx10^(-2)xx1000xx9.8)m=2.86xx10^(-3)m` Let `h_(2)` be the height of water in the limb of radius `r_(2)`. Then, `h_(2)=(2sigmacostheta)/(r_(2)rhog)=(2xx0.07xxcos0^(@))/(0.5xx10^(-4)xx1000xx9.8)m=2.86xx10^(-1)m` Difference in heights = `h_(2)-h_(1)=2.68xx10^(-1)m-2.86xx10^(-3)m=(0.286-0.00286)m=0.283m` |
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| 34. |
A hot ball of iron weighing 200 g is dropped into 500 g water at 10°C .The resulting temperature is 22.8°C .Calculate the temperature of the ball |
| Answer» SOLUTION :422.8°C | |
| 35. |
O' is the centre of an equilateral triangle ABC F_(1),F_(2) and F_(3) are three force acting along the sides AB, BC and AC as shown in figure. What should be the magnitude of F_(3),so that the total torque about 'O' is zero ? |
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Answer» `(F_(1)+F_(2))/(2)` |
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| 36. |
Which of the following is the most accurate measurement ? |
| Answer» Solution :`200xx 10^(-4)` m | |
| 37. |
Angular velocity of a smooth sphere A moving on a frictionless horizontal surface is omega and the velocity of its centre of mass is v. When it undergoes elastic head on collision with another identical sphere B at rest then the angular velocities of the two spheres become omega_(A) and omega_(B) respectively. If friction is neglected, the relation between omega_(A) and omega_(B) will be |
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Answer» `omega_(A) ltomega_(B)` |
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| 38. |
Draw a graph showing the variation of volume of a given mass of water with temperature from 0^@C . In the graph mark the temperature at which water has maximum density. |
Answer» SOLUTION :
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| 39. |
A shell of mass of mass .m. is projected with a velocity .v. at an angle 60° to horizontal When it reaches the maximum height, its angular momentum with respect to point of projection is |
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Answer» `SQRT(3)(mv^(3))/(8G)` |
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| 40. |
A monoatomic gas molecule has |
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Answer» THREE DEGREES of freedom |
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| 41. |
How does the real coefficient of expansion of water change between 0^(@)C and 8^(@)C? |
| Answer» Solution :It is NEGATIVE as the volume DECREASES from `0^(@)C `to `4^(@)C.` But for temperature more than `4^(@)C,` it is positive. | |
| 42. |
One molecule of a hydrogen and one molecule of oxygen are at the same temperature. The rms speed of hydrogen and oxygen molecules are in the ratio |
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Answer» 1:1 |
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| 43. |
A physical quantity is given is y = (a b^(3))/(c^(2)) . "If" Deltaa, Deltab, Deltac are absolute errors, the possible fractional error in y is ......... |
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Answer» `(Deltay)/(y) =(DeltaaDeltab)/(2Deltac)` |
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| 44. |
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. |
| Answer» Solution :`a = -2.5 MS^(-2)` , Retarding force ` = 465 XX 2.5 = 1.2 xx 10^3 = N` | |
| 45. |
In deriving Bernoulli's equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of the blood flow in an artery of diameter 2 times 10^-3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively? |
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Answer» Solution :The pressure drop is greater (B) More important with increasing FLOW VELOCITY |
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| 46. |
There are three liquids A, B and C having equal masses. Their temperature are respectively 14^(@)C, 24^(@)C and 40^(@)C. When A and B are mixed, the temperature of the mixture becomes 20^(@)C, When B and C are mixed, the temperature of the mixture becomes 34^(@)C. What would be the temperature of the mixture when all the three liquids are mixed? |
| Answer» SOLUTION :`30^(@)C` | |
| 47. |
Generation, propagation and detection of electromagnetic waves is the basis of |
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Answer» COMPUTERS |
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| 48. |
A satellite launching station should be |
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Answer» NEAR the EQUATOR REGION |
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| 49. |
A satellite is a at height of 25, 600 km from the surface of the earth. If its orbital speed is 3.536 km/s find its time period. (Radius of the earth = 6400 km) |
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Answer» Solution :`R = 6400 KM = 6.4 xx 10^(6)m` `h = 25, 600 km = 25.6 xx 10^(6)m`, `v_(0) = 3.536 xx 10^(3) m//s` The time PERIOD `T = (2pi(R+h))/(V_(0)) = (2pi(256.6 xx 6.4)10^(6))/(3.536 xx 10^(3))` `= (2pi xx 32)/(3.536) xx 10^(3) = 56.870 s` |
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| 50. |
The escape velcoties on the surface of two planets of masses m_(1) and m_(2) and having same radius are v_(1) and v_(2) respectively. Then |
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Answer» `(v_(1))/(v_(2)) = SQRT((m_(1))/(m_(2)))` |
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