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. |
A uniform circular disc has a radius of 20cm. A circular disc of radius 10 cm is cut concentrically from the original disc. The shift in centre of mass is |
| Answer» Answer :D | |
| 2. |
An electrical bulb of volume 250 c.c was sealed off at pressure of 10^(-3) mm of Hg and at 270C. Calculate the number of air molecules now present in the bulb. |
|
Answer» |
|
| 3. |
Determine the 'effective focal length' of the combination of two lenses of focal lengths 30 cm and -20 cm if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ? (b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and convex lens is 40 cm. Determine the magnification produced by the two lens system and size of image. |
|
Answer» SOLUTION :(a) Here, `f_1 = 30 cm, f_2 = -20 cm,d= 8.0, f = ?` Let a parallel beam be incident on the convex lens first. If `2 nd` lens were ABSENT, then `:. mu_1 = oo and f_1 = 30 cm` As `(1)/(v_1)-(1)/(u_1)=(1)/(f_1)` :. `(1)/(v_1)-(1)/(oo)=(1)/(30)` or `v_1 = 30 cm` This image would now act as a VIRTUAL object for `2 nd ` lens. `:. u_2 = + (30 - 8)=+22 cm, v_2 = ? f_2 = -20 cm` As `(1)/(v_2)=(1)/(f_2)+(1)/(u_2)` :. `(1)/(v_2)=(1)/(-20)+(1)/(22)= (-11 + 10)/(220) = (-1)/(220)` `v_2 = -220 cm` `:.` Parallel incident beam would appear tom diverge from a point `220 - 4 = 216 cm` from the centre of the two lens system. (ii) Suppose a paralleol beam of light from the left is incident first on the concave lens. `:. u_1 = -oo, f_1 = -20 cm, v_1 = ?` As `(1)/(v_1)-(1)/(u_1)=(1)/(f_1) :. (1)/(v_1)=(1)/(f_1)+(1)/(u_1)=(1)/(-20)+(1)/(-oo)=(-1)/(20)` `v_1 = -20 cm` This image acts as a REAL object for the `2 nd` lens `u_2 = -(20 - 8) = -28 cm, f_2 = 30 cm, v_2 = ?` As `(1)/(v_2)-(1)/(u_2)=(1)/(f_2) :. (1)/(v_2)=(1)/(f_2)+(1)/(u_2)=(1)/(30)-(1)/(28)=(14 - 15)/(420)` `v_2 = -420 cm` `:.` The parallel beam appears to diverge from a point `420 - 4 = 416 cm`, on the left of the centre of the two lens system. From the above discussion, we observe that the answer depends on which side of the lens system the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be useful here. (b) Here, `h_1 = 1.5 cm, u_1 = -40 cm, m = ?, h_2 = ?` For the `1st` lens, `(1)/(v_1)-(1)/(u_1)=(1)/(f_1)` `(1)/(v_1)=(1)/(f_1)+(1)/(u_1)=(1)/(30)-(1)/(40)=(4 - 3)/(120) = (1)/(120)` Magnitude of magnification produced by first lens, `m_1 = (v_1)/(u_1) = (120)/(40) = 3` The image formed by `1st` lens acts as virtual object for the `2nd` lens `:. u_2 = 120 - 8 = 112 cm, f_2 = -20 cm, v_2 = ?` As `(1)/(v_2)-(1)/(u_2)=(1)/(f_2) :. (1)/(v_2)=(1)/(f_2)+(1)/(u_2)=(1)/(-20)+(1)/(112)=(-112 + 20)/(112 xx 20) = (-92)/(112 xx 20)` `v_2 = (-112 xx 20)/(92)` Magnitide of magnification produced by second lens `m_2 = (v_2)/(u_2)=(112 xx 20)/(92 xx 112) = (20)/(92)` Net magnification produced by the combination `m = m_1 xx m_2 = 3 xx (20)/(92) = (60)/(92) = 0.652` `:.` Size of image, `h_1 = mh_1 = 0.652 xx 1.5 = 0.98 cm`. |
|
| 4. |
Two satellites S_(1) and S_(2)are to beset into the orbits of heights R/4 and R/6 (above the surface of earth ) respectively. They revolve around the earth in coplanar circular orbits in opposite sense (where, R = radius of earth ). Which of the following statement (s) is /are correct ? |
|
Answer» The RATIO of their time period is `((15)/(14))^(3//2)` |
|
| 5. |
If no external force acts on the system (a) acleleration of centre of mass is zero (b) Velocity of centre of mass is zero (c) Centre of mass may be at rest (or) moving with constant velocity (d) Velocity and momentum of centre of mass remains constant |
|
Answer» a, B and C are TRUE |
|
| 6. |
The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. |
|
Answer» SOLUTION :(a) At t = 0, OP makes an angle of `45^(@) = pi//4` rad with the (positive direction of ) x-axis. After time t, it covers an angle `(2PI)/Tt` in the anticlockwise sense, and makes an angle of `(2pi)/Tt+pi/4` with the x-axis. The projection of OP on the x-axis at time t is given by, `x(t)=Acos((2pi)/Tt+pi/4)` For `T=4s`. `x(t)=Acos((2pi)/4t+pi/4)` which is a SHM of amplitude A, period 4 s, and an initial PHASE*=`pi/4`. (b) In this case at t = 0, OP makes an angle of `90^(@)=pi/2` with the x-axis. After a time t, it covers an angle of `(2pi)/Tt` in the clockwise sense and makes an angle of `(pi/2-(2pi)/Tt)` with the x-axis. The projection of OP on the x-axis at time t is given by `x(t)=Bcos(pi/2-(2pi)/Tt)` `=Bsin((2pi)/Tt)` For T = 30 s, `x(t)=Bsin(pi/15t)` WRITING this as x (t ) = B COS`(pi/15t-pi/2)`, and comparing with Eq. (14.4). We find that this represents a SHM of amplitude B, period 30 s, and an initial phase of `-pi/2`. |
|
| 7. |
Consider a rectangular plate of size (a x b) form which two circular holes of radii R_(1) and R_(2) have been cut. The separation between the holes is x. If a'. b'. R'. R_(2)' and x' are therespective valuesat higher temperature t. then |
|
Answer» `(R_1)/( R_1) = (R_2)/( R_2) = ( 1+ ALPHA t) ` |
|
| 8. |
When a small ice crystal is placed into supercooled water, it begins to freeze instantaneously. What amount of ice is formed from 1 kg of water supercooled to -8^(0)C? |
|
Answer» `m XX 80 = 1000 xx 1 xx 8 ; m = 100 G`. |
|
| 9. |
Assuming that a car whose CG is at a height b from the rear wheel and at a distance c from the front wheel and at a height h above the pavement, has sufficient power and there is sufficient friction, find the maximum acceleration that it would be able to develop without tipping over backward. |
|
Answer» |
|
| 10. |
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system? |
|
Answer» Solution :When 0.02 kg mass is added, the spring STRETCHES by 7 cm As `MG=kx` `therefore k=(mg)/(x)=(0.02xx10)/(7XX10^(-2))=(20)/(7)NM^(-1)` When 0.02 kg mass is removed, the period of vibration will be `T=2xsqrt((m)/(k))=2pisqrt((0.2)/(20//7))=2pisqrt((7)/(100))=(2pixx2.645)/(10)=1.66s` |
|
| 11. |
In the figure shown a uniform rod of length l and mass m is kept at rest in horizontal position on an elevated edge. The value of x is such that the rod will have maximum angular acceleration alpha, as soon as it is set free. |
|
Answer» x is equal to `(L)/(2 sqrt(3))` |
|
| 12. |
One face of a copper plateof area 5m^(2)and thickness 0.01m isin contact with water boilingat 100^(@)C, the other face is at 0^(@)C. Findthe heatconducted in one second (K ofcopper = 400 Wm^(-)K^(-1)) |
|
Answer» Solution :`A = 5m^(2) , l = 0.01 m , K = 400 Wm^(-1) K^(-1)` `theta_(1)= 100^(@)C , theta_(2) = 0^(@) C andt = 1s ` from `Q = (KA(theta_(1)-theta_(2)t))/(l)= (400 xx 5 xx (100 -0)xx1)/(0.01)= 2 xx 10^(7)J` `rArr (2K_(2)A_(1)(theta_(1)- theta))/(l_(1)) = (K_(1)A(theta - theta_(2)))/(l_(1)) rArr 2 (theta_(1) - theta) = (theta- theta_(2))` But `(theta_(1) - theta_(2)) = (theta_(1) + theta) + (theta - theta_(2)) rArr (theta_(1) - theta_(2)) = 3(theta_(1) - theta)` `rArr (theta_(1) - theta) = ((theta_(1)- theta_(2))/(3))= (36)/(3) = 12^(@)C"" therefore `i.e., temperature differenceacross the LAYES P is `12^(@)C` |
|
| 13. |
Three particles A, B and C of mass m, 2m and 3m respectively lie on a smooth horizontal table at the vertices of an equilateral triangle. A and B as well as B and C are connected by light inextensible strings. C is given a velocity v_(0)parallel to AB as shown in figure. Show that Aeventually begins to move with a velocity (2 v _(0))/(19). |
|
Answer» |
|
| 14. |
A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence 45^@. The ray undergoes total internal reflection IF mu is the refractive index of the medium with respect to air, select the possible value(s) of mu from the following |
|
Answer» Solution :For total INTERNAL reflection to OCCUR, `I gt ` CRITICAL angle c or SIN `I gt sin C` `or sin 45^@ gt 1/mu or 1/sqrt2 gt 1/mu or mu gtsqrt2 or mu gt 1.414` Possible values of `mu` can be ( c) 1.5 or (d) 1.6 |
|
| 15. |
An object A is dropped from rest the top of a 30 m high building and at the same moment another object B is projected vertically upwards with an initial speed of 15 m/s from the base of the building. Mass of the object A is 2 kg while mass of the object B is 4 kg. Find the mximum height above the ground level attained by the centre of mass of the A and B system (take g = 10 m//s^(2)) |
|
Answer» SOLUTION :`m_(1)=4kg, m_(2)=2kg` Initially 4 KG is on the ground `therefore x_(1)=0` 2 kg is on top of the building `therefore x_(2)=30m` `X_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `=(0+2xx30)/(4+2)=10m` `therefore` Initial height of CM = 10 m. Initial velocity of CM, `u_(cm)=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))` `u_(cm)=(4xx15+0)/(4+2)=10` m/s UPWARDS `therefore` Maximum height attained by CM from initial position, `h_(cm)=(u_(cm)^(2))/(2g)=(10^(2))/(20)=5m` `therefore` Maximum height attained by CM of 4 kg and 2 kg from the ground `= 10+5=15 m` 
|
|
| 16. |
Two particle of equal mass m at A and B are connected by a rigid light rod AB lying on a smooth horizontal table. An impulse J is applied at A in the plane of the table and perpendicular at AB. The velocity of particle at A is |
| Answer» Answer :B | |
| 17. |
Two simple pendulums of equal length cross each other at mean position. What is their phase difference ? |
| Answer» SOLUTION :`PI` RADIAN or `180^@` | |
| 18. |
The escape velocity of a body on the earth.s surface is V . A body is thrown vertically up with a speed of (kV_(e)(k lt 1)). The maximum height reached by the body above the earth is |
|
Answer» `R(K^(2)/(1-k^(2)))` |
|
| 19. |
Assertion : Gravitational potential energy increases if we move away from the surface of earth. Reason : Gravitational potential energy is zero at infinity. |
|
Answer» If both assertion and REASON are correct and reason is a correct EXPLANATION of the assertion. |
|
| 20. |
To which group of error do zero error, back lash error and end error in metre bridge belong? |
| Answer» SOLUTION :INSTRUMENTAL ERROR. | |
| 21. |
A sphere of steel of mass 1 kg moving with a velocity of 12m/s ,along X-axis collides elastically with a stationary sphere after the collision is 8 m/s and is moving at an angle of 45^(@) with X -axis , find the magnitude and directionof the second sphere after the collision . |
|
Answer» Solution :Momentum is conserved in COLLISION , ` :. M_(1)vec(v_(1)) +m_(2)vec(v_(2)) =m_(1)vec(v_(1))+m_(2)vec(v_(2))` TAKING X - components and `vec(v_(2))=0` `m_(1)v_(1) =m_(1)v_(1)costheta_(1)+m_(2)v_(2)cos theta_(2) ` ` :. 180 = 120 xx1/(sqrt(2)) +20 v_(2)cos theta_(2)` ` :. 180 = 120 xx1/(sqrt(2))+20v_(2)cos theta_(2)` ` :. 9- 3sqrt(2)=v_(2) cos theta_(2)""......(1)` Taking Y - components , ` 0 =m_(1)v_(1) sin theta_(1)- m_(2) v_(2) sin theta_(2)` ` :. 0 = 15 xx 8 xx sin 45^(@) -20 v_(2) sin theta_(2)` ` :. 0 = 15 xx 8 xx sin 45^(@) - 20 v_(2) sin theta_(2)` ` :. 0 = 120 xx 1/(sqrt(2)) -20 v_(2) sin theta_(2)` ` :. 0 = 3sqrt(2) -v_(2) sin theta_(2)` ` :. 4.2425 =v_(2) sin theta_(2) ""........(2)` Takingthe RATIO of equation (2) and (1) , ` :. (v_(2)sintheta_(2))/(v_(2)costheta_(2)) =(4.2425)/(4.758)` ` :. tan theta_(2) =0.8916` ` :. tan theta_(2) =0.8916` ` :. theta_(2) =tan^(-1) (0.8916)` ` :. theta_(2) =41^(@)43` ` :. v_(2)sin (41^(@)43)=4.2425` ` :. v_(2) =(4.2425)/(sin (41^(@)43)) =(4.2425)/(0.6654)` ` :. v_(2) = 6.37 ms^(-1)` |
|
| 22. |
In a molecule, the potential energy between two atoms is given by U(x)= (a)/(x^(12))-(b)/(x^(6)). Where'a' and 'b' are positive constant and 'x' is the distance between atoms. Find the value of 'x' at which force is zero and minimum P.E at that point |
|
Answer» Solution :Force is zero `rArr (DU)/(dx) = 0` i.e., `a(-12) x^(-13) -b (-6) x^(-7)== 0` `(-12a)/(x^(13)) + (6B)/(x^(7)) = 0 rArr (12a)/(x^(13))= (6b)/(x^(7))` `rArr x^(6)= (2a)/(b) THEREFORE x= [(2a)/(b)]^(1//6)` `U_("min")= a ((b)/(2a))^(12//6)- b((b)/(2a))^(6//6)` `rArr U_("min") = (ab^(2))/(4A^(2))-(b^(2))/(2a) rArr U_("min") = (-b^(2))/(2a)` |
|
| 23. |
The Young's modulus of steel is more than that of rubber. For a given strain, which one will have a greater tensile stress? |
|
Answer» STEEL |
|
| 24. |
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E. which is Translational is |
| Answer» Answer :C | |
| 25. |
A convex lens forms a real image of a point object placed on its principal axis. IF the upper half of the lens is painted black (A) the image will be shifted downward (B) the image will be shifted upward ( C) the image will not be shifted (D) the intensity of the image will decrease |
|
Answer» (A) (C ) and (D) |
|
| 26. |
A river is 150 m wideand has a current of 1 m*s^(-1). A boat is moving directly across the river at a speed of 2 m*s^(-1) . How long will theboat taketo cross the river ? |
|
Answer» |
|
| 27. |
The dimensional formula for potential energy is |
|
Answer» `M^(2)L^(2)T^(-2)` |
|
| 28. |
A uniform circular disc of radius 50cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0rads^(-2). Its net acceleration in ms^(-2) at the end of 2.0 s is approximately |
|
Answer» `7.0` |
|
| 29. |
Derive an expression for the potential energy of an elastic stretched spring. |
|
Answer» Solution :(i) The potential energy possessed by a spring due to a deforming FORCE which streatches or compresses the spring is termed as elastic potential energy. The work done by the applied force against the restoring force of the spring is stored as the elastic potential energy in the spring. (ii)Consider a spring mass system. Let us assume a mass, m lying on a SMOOTH horizontal table as shown in FIGURE. Here, x = 0 is the equilibrium position. One end of the spring is attached to a RIGID wall and the other end to the mass.  (iii) As long as the spring remains in equilibrium position, its potential energy is zero. Now an external force `vec(F)_(a)` is applied so that it is stretched by a distance (x) in the direction, of the force. (iv) There is a restoring force called spring force `vec(F)_(s)` developed in the spring which tries to bring the mass back to its original position. This applied force and teh spring force are equal in magnitude but opposite in direction i.e., `vec(F)_(a)=-vec(F)_(s)`. According Hooke's law, the restoring force developed in the spring is `vec(F)_(s)=-k vec(x)` (v) The negative sign in the above expression implies that the spring force is always opposite to that of displacement `vec(x)` and k is the force constant. Therefore applied force is `vec(F)_(a)= + k vec(x)`. The positive sign implies that the applied force is in the direction of displacement `vec(x)`. The spring force is an example of variable force as it depends on the displacement `vec(x)`. Let the spring be stretched to a small distance `d vec(x)`. The work done by the applied force on the spring to stretch it by a displacement `vec(x)` is stored as elastic potential energy. `U = int vec(F)_(a). d vec(r )= int_(0)^(x)|vec(F)_(a)||d vec(r )|cos theta` `= int_(0)^(x)F_(a)dx cos theta` (vi) The applied force `vec(F)_(a)` and the displacement `d vec(r )` (i.e., here dx) are in the same direction. As, the initial position is taken as the equilibrium position or mean position, x = 0 is the lower limit of integration. `U= int_(0)^(x)kxdx , U = k [(x^(2))/(2)]_(0)^(x)` `U=(1)/(2)kx^(2) ""`....(1) (vii) If teh initial position is not zero, and if the mass is changed from position `x_(i)` to `x_(f)`, then the elastic potential energy is `U=(1)/(2)k(x_(f)^(2)-x_(i)^(2))""` ....(2) From equations (1) and (2), we observe that the potential energy of the stretched spring depends on the force constant k and elongation or compression x. |
|
| 30. |
A block of steel of size 5 cm x5 cmx5cm is weighted in water. If the relative density of steel is |
|
Answer» `6xx5xx5xx5gf` |
|
| 31. |
According to Newton's law of gravitation , everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and the inversely proportional to the square of the distance between their centres, i.e., F prop (m_1m_2)/(r^2) or F=G(m_1m_2)/(r^2) where G is universal gravitational constant =6.67xx10^(-11)N*m^2*kg^(-2). Read the above passage and answer the following questions : (iii) What values of life do you learn from it ? |
| Answer» SOLUTION :EVERY BODY in the UNIVERSE attracts every other body with a certain amount of force. | |
| 32. |
Explain the isobaric process and derive the work done in this process. |
|
Answer» <P> Solution :Isobaricprocess : thisis a thermodynamicprocessthat occursat constantpressure.eventhrough pressureis CONSTANTIN this process , temperature, volumeand internalenergyare not constant. Fromthe idealgas equationwe have` V= ((mu R )/( P ))T `...(1) Here`(mu R) /( P)= ` constant In an isobaricprocessthe temperatureis directly proportionalto volume. ` V porpT ` ( Isobaricprocess ) THisimpliesthat fora isobaricprocess, theV-Tgraphis astraightlinepassingthroughthe origin. IF agasgoesfroma state`(V_i , T_i ) `to` (V_f , T_f)`at constantpressurethen thesystemsatisfiesthefollowinngequation ` (T_f )/(V_f )=(T_i )/(V_i)` Examplesfor Isobaricprocess : ( i ) whenthe gasheated andpushes thepiston so thatit exertsa forceequivalent toatmosphericpressureplusthe forcedueto gravity then this process is isobaric . (ii )Mostof thecookingprocesses in ourkitchenare isobaricprocesses , whenthe foodis cookedin an openvesselthepressureabovethefoodis alwaysat atmosphericpressure The PVdiagramfor anisobaricprocessis a horizontallineparallelto volumeaxisfigure(a)representsisobaricproceesswheredecreases figure(b)representsisobaricprocesswherevolumeincreases  The workdonein an isobaricprocess : workdoneby thegas ` W= int_(v_1)^(v_f)pd V ` In an isobaricprocess the pressureis constant, soP comesoutoftheintegral ` W= P int_(v_i)^(v_f) DV ` ` W=P[V_f -V_i]=P Delta V ` Where`Delta V `denoteschange in thevolume.IF`Delta V ` isnegative ,Wis alsonegativethisimpliesthat theworkis doneon thegas.If ` Delta V `is positive, Wis positiveW isalsopositiveimplying that workis done by thegas . Theequation(6)can ALSOBE rewrittenusingtheidealgasequation. fromidealgasequation ` PV = mu RTand V =( mu RT )/(P )` Substitutingthis inequation( 6)we get `W = mu RT_f(1- (T_i )/(T_f ))` inthePVdiagram, areaunderthe isobariccurveis equalto theworkdoneis isobaricprocess .  Thefirstlaw ofthermodynamics forisovaricprocessis givenby ` DeltaU =Q- P Delta V ` ` W = P DeltaV ,Delta U = Q- mu RT _f [ 1-(T_i )/(T_f)]` |
|
| 33. |
A body of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed vin mutually perpendicular directions. Total energy released in the process is |
|
Answer» `mv^2` As two fragments of mass m each are moving with speed V each at RIGHT angles, so `|vec(p_1) + vec(p_2)| = msqrt(v^2 + v^2) = sqrt(2) mv` `:. |vecp_3| = |vecp_1 + vecp_2| = sqrt(2) mv` The mass of third fragment is 2m. `:. ` KINETIC energies of three fragments are `K_1 = (p_1^2)/(2m) = 1/2 mv^2 , K_2 = (p_2^2)/(2m) = 1/2 mv^2` and `K_3 = (P_3^2)/(2(2m)) = 1/2 mv^2` Total energy released during explosion = `K_1 + K_2 + K_3 = 3/2 mv^2` . |
|
| 34. |
Explain Laplace's correction. |
|
Answer» Solution :Laplace.s correction: In 1816, Laplace satisfactorily corrected this DISCREPANCY by assuming that when the sound PROPAGATES through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling EFFECT due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson.s law (not Boyle.s law as Newton assumed), which is `PV^gamma` = Constant....(1) where ` gamma = (C_p)/(C_v)`which is the RATIO between specific heat at constant pressure and specific heat at constant volume. Differentiating EQUATION (1) on both the sides, we get `V^(gamma) dP + P (gamma^(V^(gamma - 1)) dV) = 0` or` gamma P = - V (dp)/(dV) = B_A` where, `B_A`is the adiabatic bulk modulus of air. Now, substituting equation (2) in equation `V = sqrt(B/rho)`the speed of sound in air is ` v_A = sqrt( (B_A)/(rho) ) = sqrt( (gamma P)/(rho) )= sqrt(gamma v_T)` |
|
| 35. |
A 1000 c.c of air at 50^(@)C is cooled down to 0^(@)C at constant pressure. If the volume coefficient is 0.00366/""^(@)C, Calculate its volume. |
|
Answer» |
|
| 36. |
Assertion : An object has given two velocities vecv_(1) and vecv_(2) has a resultant velocity vecv = vecv_(1) + vecv_(2). Reason : vecv_(1) and vecv_(2) should be velocities with reference to some common reference frame. |
|
Answer» |
|
| 37. |
What is the condition for two different metal rods to have the same change in their lengths at all temperatures? |
| Answer» SOLUTION :`L_(1)alpha_(1)=L_(2)alpha_(2)` | |
| 38. |
Particles of masses m, 2m, 3m, …….. nm grams are placed on the same line at distances l, 2l, 3l, …… nl cm from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeter is |
|
Answer» `((2n+1)l)/(3)` |
|
| 39. |
If coefficient of cubical expansion is x times coefficient of superficial expansion, then the value of x is |
|
Answer» 1.5 |
|
| 40. |
Four particles each of mass 100g are placed at the corners of a square of side 10cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system. |
|
Answer» Solution :mass `=m=100g`, RADIUS `r=5sqrt(2)cm` MOMENT of inertia `I=4mr^(2)` `=4xx100xx(5sqrt(2))^(2)=2XX10^(-3)kgm^(2)` `k=SQRT((I)/(M))=sqrt((2xx10^(-3))/(0.4))=0.07m=7cm` |
|
| 41. |
The periods of a pendulum on two planets are in the ratio 3: 4. The acceleration due to gravity on them are in the ratio |
|
Answer» (:16 |
|
| 43. |
Calculate the temperature which has same numeral value on celsius and Fahrenheit scale. |
|
Answer» Solution :`(""^(@)F-32)/(180)=(""^(@)C)/(100)` SUPPOSE, `F=C=Q` `:.(Q-32)/(180)=Q/100` `:.Q-32=(180)/(100)Q` `:.Q-9/5Q=32` `:.-(4Q)/(4)=32` `:.Q=-(32xx5)/(4)` `:.Q=-40` `:.Q=-40" "^(@)F` or `Q=-40" "^(@)C` |
|
| 44. |
When a glass prism of refracting angle 60^@ is immersed in a liquid its angle of minimum deviation is 30^@. The critical angle of glass with respect to the liquid medium is |
| Answer» Answer :B | |
| 45. |
An air bubble is glass is situated at a distance 3cm from a convex surface of diameter 10cm as shown in figure. At what distance from the surface will the bubble appear? |
|
Answer» SOLUTION :In CASE of refraction from curved SURFACE `mu_2/v-mu_1/u=(mu_2-mu_1)/R` Here `mu_1=1.5 mu_2=1, R=-5cm` and `u=-3cm` So `1/v-((1.5))/((-3))=(1-1.5)/((-5))`i.e., `v=-2.5 cm` |
|
| 46. |
A body of mass .m. and radius .r. rolling on a horizontal floor with velocity .y., rolls up an inclined plane up to vertical height (3V^2)/(4g). Find the moment of inertia of body and comment on its shape |
|
Answer» Solution :The TOTAL KE of the body `K=K_(T)+K_(R)=(1)/(2)mv^(2)[1+(I)/(MR^(2))]` When ROLLS up an inclined plane of height `h=(3V^(2))/(4g)` its KE is converted into PE. So `(1)/(2)mv^(2)[1+(I)/(mr^(2))]=mg((3v^(2))/(4g))` on simplification `I=(mr^(2))/(2)` hence the body is either a disc or CYLINDER. |
|
| 47. |
For a particle oscillating along x-axis according to equation x = A sin omega t The mean value of its velocity averaged over first 3/8 of the period is |
|
Answer» |
|
| 48. |
Define orbital velocity. |
| Answer» Solution :ORBITAL velocity is the velocity REQUIRED to put the SATELLITE into its orbit around the EARTH. | |
| 49. |
What is meant by temperature radiation ? |
| Answer» SOLUTION :Radiation whose intensity depends UPON the TEMPERATURE of the BODY. | |
| 50. |
What is phase of SHM? |
|
Answer» Solution :(i) The phase of a vibrating particle at any instant completely specifies the state of the particle. (ii) It expresses the position and direction of MOTION of the particle at that instant with RESPECT to its mean position. (iii) where `omegat+varphi=varphi` is CALLED the phase of the vibrating particle. `y=Asin(omegat+varphi_(0))` (iv) The displacement acquired by an oscillating body when time t = 0 s (initial time) the phase `vaphi=varphi_(0))` is called epoch (initial phase) where `phi_(0)` is called the angle of epoch. |
|
