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 stone of mass (m) is attached to one end of a small wire of length (l) and cross-sectional area (A) suspended vertically. The stone is now rotated in horizontal plane such that the wire makes an angle .theta. with vertical. Find the increase in length of wire if its Young.smodulus is Y. |
Answer» Solution : From fig. `Tcos theta =MG RARR T=(mg)/(cos theta)` `T SIN theta=m R omega^(2)` `e=(Fl)/(AY)=(Tl)/(AY)` `e=(mgl)/(AY cos theta)` If .`theta`. was not GIVEN but R, w and m was given then in such case `T^(2)cos^(2)theta+T^(2)sin^(2)theta=(mg)^(2)+(mRw^(2))^(2)` `T=sqrt((mg)^(2)+(mRw^(2))^(2)) " and usee"=(Tl)/(Ay)` |
|
| 2. |
Two particles executing SHM possess the same frequency. When the first particle just passes the mean position of its path, the second particle moving in the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm, when is the phase difference of the two particles? |
|
Answer» SOLUTION :If A is the amplitude and `theta` is the phase angle, then displacement, `x=Acostheta` For the first particle, `x_(1)=A_(1)costheta_(1)or,0=A_(1)costheta_(1)` or, `costheta_(1)=0or,theta_(1)=pm90^(@)` For the second particle, `x_(2)=A_(2)costheta_(2)or,costheta_(2)=x_(2)/A_(2)=3/6=1/2` or, `theta_(2)=pm60^(@)` At the given instant, the TWO particles are in motion in the ssame direction. So, (i) `theta_(1)=90^(@),"then "theta_(2)=60^(@),` phase difference = `theta_(2)-theta_(1)=-30^(@)` or, (II) If `theta_(1)=-90^(@),"then "theta_(2)=-60^(@),` phase difference `theta_(2)-theta_(1)=+30^(@)` `therefore" "` The required phase difference is `pm30^(@)`. |
|
| 3. |
A boat is containing leaves. When they are thrown into the water, the level of water in the lake |
|
Answer» REMAINS same |
|
| 4. |
Internal force can change |
|
Answer» KINETIC ENERGY |
|
| 5. |
Cars and aeroplanes are streamlined. Why? |
| Answer» Solution :To reduce the viscous drag FORCE exerted by the AIR close to the car. Also this shape provides a LIFT which HELPS to increase the SPEED of the car. | |
| 6. |
A tuning fork of frequency 550 Hz produces resonance with air column of a pipe, closed at one end , at one end, at 14 . 5 cm and 44 . 5 cm . Determine the velocity of sound in air and the end error . |
|
Answer» |
|
| 7. |
A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2m on a smooth horizontal circular tube of radius r. The time in which the next collision wll take place is equal to |
|
Answer» Solution :Relative VELOCITY of separation =relative velocity of approach = v `THEREFORE` TIME of NEXT collision = `(2pi R)/(v)` |
|
| 8. |
A body is thrown up with a velocity 29.23ms^(-1). The distance traveled in lastbut one second of upward motion is |
|
Answer» 14.7 m |
|
| 9. |
Which quality can be added with a velocity vector? |
|
Answer» Displacement |
|
| 10. |
On what factors, does the stopping distance depend ? |
| Answer» SOLUTION :Stopping distance depends on INITIAL velocity of vehicle and capacity of applied BRAKES on vehicle. | |
| 11. |
A given mass of gas is compressed isothermally untal its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restoreand its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately |
|
Answer» `1.20` For ISOTHERMAL compression, `P_(1) = P = V_(1) = V, P_(2) = 2P, V_(2) = ?` As `P_(1)V_(1) = P_(2)V_(2) :. PV = 2PV_(2)` or `V_(2) = (PV)/(2P) = (V)/(2)` For ADIABATIC expansion, `P_(1) = 2P, P_(2) = 0.75P, V_(1) = (V)/(2) , V_(2) = V` As `P_(1)V_(1)^(GAMMA) = P_(2)V_(2)^(gamma) :. (2P) ((V)/(2))^(gamma) = (0.75P)(V)^(gamma)` or `0.375 = ((1)/(2))^(gamma)` Solving, we get `gamma = 1.41` |
|
| 12. |
A small ball of mass m suspended from the celling at a point O by a thread of length l moves along a horizontal circle with a constant angular velocity omega. . |
|
Answer» linear momentum of system (BALL + rod) is conserved (A) For angular momentum about `A`, it will have angular momentum perpendicular to `l` That will have TWO component one along `CO` which is constant and another in the plane which canges direction. (B) Angular momentum about `C` is constant `(mvr)` along `OC`. |
|
| 13. |
Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle theta with respect to the horizontal direction. |
|
Answer» SOLUTION :(i) Consider an OBJECT thrown with initial VELOCITY `vecu` at an angle `theta` with the horizontal. (ii) Since ACCELERATION due to gravity acts vertically downwards, velocity along the horizontal x - direction `u_(x)` doesn't change throught the motion. Whereas velocity along the y-direction `u_(y)` is changed.  The path of the projectile : (a) Motion along x - direction : (i) The horizontal distance travelled by the projectile at a point P after a time t can be written as, `s_(x)=u_(x)t+(1)/(2)a_(x)^(2)` (ii) Here, `s_(x)=x, u_(x)=u cos theta, and a_(x)=0,` Therefore, `x=u cos.t` `t=(x)/(u cos theta)"....(1)"` (b) Motion along y - direction : (i) The downward distance travelled by the projectile at a point P after a time t can be written as, `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)` (ii) Here, `s_(y)=y, u_(y)=u sin theta, and a_(y)=-g` Therefore, `y=u sin theta t-(1)/(2g t^(2)` (iii) Substituting equation (1) , weget, `y=usin theta.(x)/(ucos theta)-(1)/(g)g((x)/(ucos theta))^(2)` `y=x tan theta -(1)/(2)g(x^(2))/(u^(2)cos^(2)theta)` (iv) Thus, the path travelled by the projectile is an inverted parabola. MaximumHeight : `(h_("max"))` (i) The maximum vertical distance travelled by the projectile during its JOURNEY is called maximum height. (ii) For the vertical part of the motion, `v_(y)^(2)=u_(y)^(2)+2a_(y)s_(y)` (iii) Here, `v_(y)=0, s_(y)=h_("max"), u_(y)=u sin theta and a_(y)=-g` Therefore. `0=u^(2)sin^(2)theta-2gh_("max")` `h_("max")=(u^(2)sin^(2)theta)/(2g)` Time of flight : `(T_(f))` (i) The time of flight `(T_(f))` is the total time taken by the projectile to hit the ground after thrown. (ii) The downward distance travelled by the projectile at a time t can be written as, `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)` (iii) Here substituting the values `S_(y)=0, t=T_(f)` `u_(f)=u sin theta, and a_(y)=-g` we get, `0=u sin thetaT_(f)-(1)/(2)gT_(f)^(2)` Therefore,`T_(f)=(2usintheta)/(g)` Horizontial range : (R) (i) The horizontal range (R) is the maximum horizontal distance between the point of projection and the point where the projectile hits the ground. (ii) The horizontal distance travelled by the projectile at a time t can be written as, `s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)` (iii) Here, `S_(x)=R, u_(x)=u cos theta, a_(x)=0 and t=T_(f)` `R=u cos.T_(f)` `R=u cos theta.((2usin theta)/(g))=(2u^(2)sin theta cos theta)/(g)` `""[because T_(f)=(2u sin theta)/(g)]` (iv) Therefore,`""R=(u^(2)sin 2theta)/(g)` `[because 2 theta=2sin theta.cos theta]` (v) For maximum range, `sin 2theta=1` `2theta=(pi)/(2)` `theta=(pi)/(4)` The maximum range is, `R=(u^(2))/(g)` |
|
| 14. |
A mark at the bottom of a liquid appears to rise by 0.1m The depth of the liquid is 1m. The refractive index of the liquid is |
|
Answer» 1.33 |
|
| 15. |
A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to the plane with n angular velocity omega another disc of one forth mass and same dimension is gently placed over it coaxially, the angular speed of the composite disc will be ............ |
|
Answer» `(5)/(4)OMEGA` `I_(1)omega_(1)=I_(2)omega_(2)` but `I_(1)=(1)/(2)MR^(2)` where M = mass, R = radius) and `I_(2)=(1)/(2)MR^(2)+(1)/(2)((M)/(4)R^(2))` `=(1)/(2)MR^(2)((5)/(4))=(5MR^(2))/(8)` and `omega_(1)=omega_(2)` `therefore (1)/(2)MR^(2)omega=(5)/(8)MR^(2)omega_(2)` `therefore omega_(2)=(4)/(5)omega` |
|
| 16. |
The unit of coefficient of kinetic friction is/has |
|
Answer» Nm |
|
| 17. |
(A) : Air in a cylinder is suddenly compressed by a piston. If the volume after compression is halfthe original volume the final pressure is twice the initial pressure. (R ): Suddenly compressed process is isobaric process |
|
Answer» Both (A) and(R ) are true and (R ) is the CORRECT explanation of (A) |
|
| 18. |
Which law forms the basis of equation of continuity? |
| Answer» SOLUTION :LAW of CONSERVATION of MASS. | |
| 19. |
The period of a simple pendulum measured inside a stationary lift is T. If the lift starts moving upwards with a acceleration g/3. What will be the period |
|
Answer» `T//3` |
|
| 20. |
Match the column: |{:(,"Column -1",,"Column -2"),((A),theta=37^(@),(P),f" is upwards"),((B),theta=45^(@),(Q),f" is downwards"),((C),theta=53^(@),(R),f" is static"),(,,(S),f" is kinetic"):}| |
|
Answer» |
|
| 21. |
It is more important to have beauty in the equations of physics that to have them agree with experiments. The great British physics P.A.M. Dirac held the view. Criticize this statement. |
| Answer» Solution :(i) The two equations relating force `vecF` and `vecr`, one for work `W=vecF.vecr` and another for torque `vec(tau)=vecrxxvecF`. (Details you will study in the coming chapters). The two equations express entirely different concepts. (ii) The Maxwell equations relating the electric fiedl `VECE` and the magnetic FIELD `vecB`. (iii) The de Broglie wave equation expressing the particle nature and wave nature of MATTER. `lamda=h/ mv`, where h is the Plank.s constnat, m is the mass of a particle moving with a velocity v. (iv) And the most beautiful equation is of course, Einstein.s `E=mc^(2)`, the rest ENERGY associated with a BODY of rest mass m and c is the velocity of light. | |
| 22. |
A uniform rod is placed at rest on a rough horizontal floor such that it is vertical. Due to a gentle disturbance it started failing down and friction is sufficient to avoid slipping. Then before the rod hits the floor |
|
Answer» The ACCELERATION of centre of the rod is never EQUAL to g |
|
| 23. |
Force constant of spring is 0.5 Nm^(-1). The force necessary to increase the length of 10 cm of spring will be……….. (Fill in the blank). |
|
Answer» Solution :`F= K TRIANGLE x` `=0.5 xx 10xx 10^(-2)` `= 0.05 N`. |
|
| 24. |
A coin is placed on a horizontal platform AB which undergoes SHM about mean point O in a vertical plane. The coin does not slip on the platform. The force of friction acting on the coin is F. Then |
|
Answer» F is always directed towards one end of PLATFORM |
|
| 25. |
A body is kept on a rough horizontal surface of angle of friction 'alpha' and moved by applying a force of magnitude 'F' making angle 'theta' with horizontal. Arrange the following cases in the order of increasing magnitude for F.(i) F is pulling force and theta = alpha(ii) F is pulling force and theta = 0(iii) F is pulling force and alpha = theta = 0(iv) F is pushing force and theta = alpha |
|
Answer» IV, III, II and i |
|
| 26. |
A spinning ball through air takes a curved path than a ball that does not have spin. What forces come into play for a spinning ball. |
| Answer» SOLUTION :DYNAMIC LIFT and VISCOUS DRAG | |
| 27. |
One mole of an ideal gas at an initial temperature of TK does 6R joules of work adiabatically. If theratio of specific hats of this gas a constant pressure and at constant volume is 5/3, the final temperature (T_f)of gas will be : |
| Answer» Answer :D | |
| 28. |
Distinguish between translational and rotatinal equilibrium. |
Answer» SOLUTION :
|
|
| 29. |
The temperature of an ideal gas is increased from 21°C to 927°C. The rms speed of its molecules willbecomes. |
|
Answer» Twice |
|
| 30. |
Three identical rings, each of mass M and radius R are placed in the same plane touching each other such that their centers form the vertices of an equilateral triangle. The M.I of the system about an axis passing through center of one of the rings and perpendicular to its plane is |
|
Answer» `(MR^(2))/(2)` |
|
| 31. |
A stone is dropped from to of a tower 300 m high and at the same time another is projected vertically upward with a velocity of 100 ms^(-1) from the root of the tower. Find when and where the two stones meet. |
|
Answer» 2S, 200.9 m |
|
| 32. |
A cylinder witih fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20^(@) C is ________ J. [Given that R = 8.31 J mol^(-1) K^(-1)] |
| Answer» Solution :`C _(V) = (3//2) R,` No of MOLES `=N = 67.2//22.4 = 3. Delta Q = n C _(V) DT =3 xx (3//2) xx R xx 30 = 1121.85J` | |
| 33. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end A. |
Answer» Solution :In the figure, the point A and B, 10 cm apart, are the the extreme positions of theparticle in SHM, and the point O is the MEAN position. The direction from A to B is POSITIVE, as indicated. The direction of velocity, displacement and acceleration from A to B is positive.  At the ene A (at extreme positive), velocity is zero, acceleration and force are DIRECTED TOWARDS B and are positive (0, +, +). |
|
| 34. |
An object falls from a height h (hltltR).the speed of the object when it reaches the ground is |
|
Answer» `(1)/(2)"gt"^(2)` |
|
| 35. |
A particle moving with uniform acceleration has its final velocity (vartheta=sqrt(150+8x))m//s where x is the distance travelled by the body. Thenthe acceleration is |
|
Answer» `+4m//s^(2)` |
|
| 36. |
A body is moving undirectionally under the influence of a source of constant power. It displacement in time t is proportional to |
|
Answer» `t^(1/2)` |
|
| 37. |
A bimetallic strip of total thickness 'x' is unbent (straight) at temperature theta_0. The coefficient of linear expansion of the two metals are alpha and 2 alpha respectively. Its radius of curvature (R), when its temperature is raised to theta_0 + Delta theta is |
|
Answer» `(x)/( 4) [ (2 + 3 ALPHA ( Delta theta ) )/( alpha ( Delta theta ) ) ]` |
|
| 38. |
A uniform rod of mass m , length l moving with a velocity v ( perpendicular to its length ) on a smooth horizontal plane, encounters a fixed peg P at adistance l/4 from its nearer end. The rod collides with the peg. Theduration of impact Delta t is very small. The average force exerted by the peg on the rod during the impact is of magnitude. |
|
Answer» `(4mv)/(7 DELTAT)` |
|
| 39. |
A Trolley containing a liquid slides down a smooth inclined plane of angle alpha with the horizontal. Find the angle of inclination of the free surface with the horizontal |
| Answer» SOLUTION :`theta=alpha` | |
| 40. |
The minimum number of non coplanar forces that can keep a particle is equilibrium is |
|
Answer» 1 |
|
| 41. |
A block of mass 20 kg is pushed with a horizontal force of 90N. If the co-efficient of static & kinetic friction are 0.4 & 0.3, the frictional force acting on the block is (g=10 ms^(-2)) |
| Answer» Answer :C | |
| 42. |
Calculate the net acceleration produced in the arrangement shown below. |
|
Answer» Solution :(a) Net pulling force =3 mg-mg=2MG Total mass =3m+m=4M `a=(2mg)/(4m)=(g)/(2) ` (B) Pulling force= 2mg Total mass =3m `a=(2 mg)/(3m)=(2g)/(3)` (c ) Net pulling force= 3mg-2mg=mg Total mass =3m+m+2m=6 m `a=(mg)/(6mg)=(g)/(6)` (d) Net pulling force= 3 mg +mg-2 mg= 2 mg Total mass =3m+m+2m=6m `a=(2g)/(6mg)=(g)/(5)` (e) Net pulling force = 4 mg `sin 30^(@)-mg`=mg Total mass =4m+m= 5m `implies a=(mg)/(5m)=(g)/(5)` |
|
| 43. |
A carpet of mass .m. made of inextensible material is rolled along its length in the form of a cylinder of radius .R. and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a very small push is given to it. Calculate the horizontal velocity of the axis of cylindrical part of the carpet when its radius to R/2. |
|
Answer» Solution :Lose in PE due to unrolling when RADIUS changes from R to `(R )/(2)` is MGR - `((MgR)/(8))=(7)/(8)(MgR)` The lose in PE is equal to the gain in KE which is `K=K_(T)+K_(R )=(1)/(2)MV^(2)+(1)/(2)I OMEGA^(2)` If .V. is the velocity when half the carpet has unrolled then as `V=(R omega)/(2), "" M to (M)/(4)` and `I=(1)/(2)((M)/(4))((R )/(2))^(2)` `therefore (7)/(8)MgR = (3)/(16) MV^(2)` on solving `V=((14gR)/(3))^((1)/(2))` |
|
| 44. |
A sphere is performing pure rolling as shown in figure. The radius of curvature of trajectory of point A in the shown portion is (OA = R/2) |
|
Answer» `9/4R` |
|
| 45. |
What is the difference between wave velocity and particle velocity? |
| Answer» Solution :wave velocity is constant for a given MEDIUM `(V = vlamda)` while particle velocity changes harmonically with TIME. it is maximum at the mean position and zero at the extreme position. | |
| 46. |
A solid whose volume does not change with temperature, floats in a liquid. Fractions f_1 and f_2 of its volume remain submerged fortwo different temperatures t_1 and t_2 of liquid. The volume coefficient of expansion of liquid is |
|
Answer» `( f_1 + f_2)/( f_1 t_1 + f_2 t_2)` |
|
| 47. |
A person wears eyeglasses of focal length -80cm. Where is the far point of the person? |
|
Answer» 80CM from the eye |
|
| 48. |
The emissivity of a perfect black body is |
|
Answer» O |
|
| 49. |
Select the correct statement from the following |
|
Answer» The angular momentum of a comet revolving around a star remains conserved. |
|