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. |
Prove that result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^(2)=(2gh)/((1+(k^(2))/(R^(2)))) using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. |
Answer» Solution :Suppose a BODY of mass M and radius R is rolling on inclined plane of height h, length l and angle of inclination `theta` is show as in figure. The forces acting on body are shown as in figure.  Suppose a is acceleration in downward direction. Equation of motion this body `N-Mgcostheta=0` and `F=Ma-f` `therefore F=Mgsintheta-f....(1)` Torque NECESSARY against friction force f `tau=Rxxf` `Ialpha=Mk^(2)((a)/(R))` where `alpha=(a)/(R )` `therefore RF=(Mk^(2))/(R)a` `therefore f=(Mk^(2))/(R^(2))a` `therefore` From eqn. (1), `Ma=Mgsintheta-(Mk^(2))/(R^(2)).a [F=Ma]` `(M+(Mk^(2))/(R^(2)))a=Mgsintheta` `therefore a=(Mgsintheta)/(M((R^(2)+k^(2))/(R^(2))))` `therefore a=(gsintheta)/((R^(2)+k^(2))/(R^(2)))=(gsintheta)/(1+(k^(2))/(R^(2)))` Let the height of SLOPE is h and its length is l. `therefore` From figure `(h)/(l)=sinthetaimpliesl=(h)/(sintheta)` The velocity at the bottom of slope is v and at maximum height is `u=0` `therefore` In eqn. `v^(2)-u^(2)=2al,u=0 and a=(gsintheta)/(1+(k^(2))/(R^(2)))` `v^(2)-0=2xx(gsintheta)/(1+(k^(2))/(R^(2)))XX(h)/(sintheta)` `therefore v^(2)=(2gh)/(1+(k^(2))/(R^(2)))` is proved. |
|
| 2. |
A body of mass m is kept on an inclined plane . The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is mu, what is the inclination of the plane ? |
|
Answer» |
|
| 3. |
A body is projected horizontally from the top of a tower with a velocity of 20 m/s. After what times the vertical component of velocity is four times the horizontal component of velocity ? (g = 10 m//s^(2)) |
|
Answer» 16 sec |
|
| 4. |
.n.number of liquids of masses m,2m, 3m, 4m, ....... having specific heats s. 2s,3s,4s, .... are at temperatures t, 2t, 3t, 4t, .... are mixed. The resultant temperature of mixture is |
|
Answer» `(3n)/((2N+1))t` |
|
| 5. |
A simple pendulum of length . l . has a bob of mass .m.. If it is released when the string makes an angle 60° with the vertical, the tension in the string at the lowest position is, |
| Answer» ANSWER :C | |
| 6. |
If vecA = (2hati + 3hatj)and vecB =(hati - hatj)then component of vecA perpendicular to vector vecBand in the same plane is: |
|
Answer» `5/2(HATI + HATJ)` |
|
| 7. |
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand, so that the spring is neighter stretched nor comprssed. Suddenly the support of the hand is removed. The lawest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand. |
Answer» Solution :When SUPPORT of the hand is REMOVED, the body of mass m oscillate about mean position.  Suppose, a body reaches at lower extreme point and the maximum extension of spring is x, The decrease in potential energy of body = mgx Increase in elastic potential energy of spring `=(1)/(2) kx^(2)` Now mechanical energy is conserved `therefore mgx = (1)/(2) kx^(2)` `therefore x= (2mg)/(k)"""........."(1)`  When the forces below and above the ball suspended at the end of spring are equal, the position of ball BECOMES meanposition. Suppose the support of hand is removed, the extension of spring x. increases, the resultant force will be zero `therefore F= +kx.` `therefore mg= kx.` `therefore x.= (mg)/(x)"""........."(2)` Taking ratio of equation (1) and (2), `(x)/(x.)= 2` `therefore x= 2X.` but x= 4 CM (is given) `therefore 4= 2x.` `therefore x.= 2cm`, will be the amplitude of oscillation. |
|
| 8. |
A body moves along x-axis such that it's position is given by x=(t^(3))/(3)-(5t^(2))/(2)+6t+2(x is in meter, 't' is in seconds). Identify correct options |
|
Answer» The smallest value of x during the motion is `x=2` |
|
| 9. |
A 3cm cube of iron has one lace at 100^(@)C and the other in a block of ice at 0^(@)C. If k of iron=0.2 CGS units and L for ice is 80cal/gm then the amount of ice that melts in 10 minutes is (assume steady state heat transfer) |
| Answer» Answer :A | |
| 10. |
Two cars 1 & 2 starting from rest are moving with speeds V_(1) and V_(2) m/s (V_(1) gt V_(2)). Car 2 ahead of car 'I' by 'S' metres when the driver of car '1' sees car '2' .What minimum retardation should be given to car'I' avoid collision |
|
Answer» `(V_(1)-V_(2))/(S)` |
|
| 11. |
Three transverse progressive waves arez_(1) = A cos (kx - omega t) ,z_(2) = A cos (ks + omega t)andz_(3) = A cos (ky - omega t). How may these be superposed to generate(i) a stationary wave,(ii) a wave propagating in a direction inclined at an angle of 45^(@) withboth the positive x and y - axes ?In each case, find out the positions where the resultant intensity would always be zero . |
|
Answer» Solution :(i)The first and the SECOND waves are TWO identical but oppositely directed waves. So, they would generate a stationary wave. Theequation of the resultant wave would be ` Z = z_(1) + z_(2) = A [COS (kx - omega t) + cos (kx + omega t)]` `= 2 a cos kx . cos omega t ` The resultant intensity is zero , where 2 A cos kx = 0 ` :. cos kx = 0 or , x = ((2n + 1)pi)/(2k) [ n = 0 , 1 , 2 , 3, .....]` (ii)The first wave directed along positive x - axis and the third wave directed along positive y - axis are identical waves.So the resultantwave propagates in a direction which is inclined at `45^(@)`with both the x are the y-axes.Theequation of the resultant wave would be ` z = z_(1) + z_(3) = A [cos (kx - omega t) + cos (ky - omega t)]` `= 2 A"cos"(k(x+y)-2omegat)/(2)*"cos"(k(x-y))/(2)` The resultant intensity is zero, where 2 a cos ` (k(x - y))/(2) = 0 ` `:."cos"(k(x-y))/(2)=0` or, `(k(x - y))/(2) = ((2n + 1) pi) /(2)[ n = 0 , 1 , 2 , 3 , .....]` or,` x - y = ((2 n + 1)pi)/(k)`. |
|
| 12. |
Define the velocity gradient and give its units. |
| Answer» SOLUTION :The rate of vriation of VELOCITY (v) with perpendicular distance (x) to the direction of FLOW, is called velocity gradient, `((DV)/(DX))`. Its unit is S. | |
| 13. |
Weight of a body of mass.m. decreases by 1% when it is raised to a height .h. above the earth.s surface. If the body is taken to a depth .h. in a mine, the change in its weight is |
|
Answer» DECREASES by 2% |
|
| 14. |
The length of a smooth inclined plane of incilination 30^(@) is 5 m. The work done is moving a 10kg mass from the bottom of the inclined plane to top is |
|
Answer» 245 j |
|
| 16. |
What is a heat pump ? |
| Answer» SOLUTION :A heat pump is a mechanical device that removes thermal energy from a REGION at a LOWER TEMPERATURE to a region at higher temperature. | |
| 17. |
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will suffer (1) greater deviation (2) no deviation (3) same deviation as before (4) total internal reflection |
Answer» Solution :Figure (a) is part of an EQUILATERAL prism of figure (b) as shown in figure. Which is a magnified image of figure (c ). Therefore the RAY will suffer the same DEVIATION in figure (a) and figure ( c ). 
|
|
| 18. |
The x and y coordinates of the particle at any time are x = 5t - 2t^(2) and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is |
|
Answer» `-8m//s^(2)` |
|
| 19. |
The blades of an aeroplane propeller are 2m long and rotate at 300 rpm. Calculate (i) frequency (ii) period of rotation (iii) angular velocity (iv) linear velocity of a point 0.5 m from the top of the blades. |
|
Answer» |
|
| 20. |
Which one of the following statements is incorrect? |
|
Answer» Frictional force opposes the relative motion |
|
| 21. |
The kinetic energy of a satellite is 2 MJ. What is the total energy of the satellite? |
| Answer» SOLUTION :KINETIC ENERGY, | |
| 22. |
The number of degrees of freedom for a diatomic gas molecule is ……………. |
|
Answer» 2 |
|
| 23. |
The two lenses of a compound microscope are of focal lengths 2cm and 5cm . IF an object is placed at a distance of 2.1cm from the objective of focal length 2cm the final image forms at the least distance of distinct vision of a normal eye, Find the distance between the objective and eyepiece. |
|
Answer» 46.17cm |
|
| 24. |
A particle of mass m moves in a conservative force field along x axis where the potential energy U varies with position coordinate x as U = U_(0)(1- cos ax),U_(0) and a being positive constants. Which of the following statement is true regarding its motion. Its total energy is U_(0) and starts from x=0 |
|
Answer» the ACCELERATION is constant |
|
| 25. |
Figure 2 (EP). 13 shows the orientation of two vectorsvec u and vec v in the (XY) plane. Ifvec u = a hat i + b hat j andvec v = p hat i+ q hat j which of the following is correct ? . |
|
Answer» a and p are positive while B and Q are NEGATIVE |
|
| 26. |
The process of measurement is basically |
|
Answer» a PROCESS of comparison |
|
| 27. |
A wedge is moving rightwards on which a block of mass 10kg is placed on it. Friction coefficient between the wedge and the block is 0.8.[ takes g=10m//s^(2)]. Select correct alternative (s) among the following options. |
|
Answer» If WEDGE is moving with constant velocity then friction ACTING on block is `64N`. `=(0.8)(10)(10)(4//5)=64N`.  but `mgsintheta=60N` so required friction is `60N`. So net force is zero . `(C ) a=2hat (i) f=mg sin 37^(#)-m a cos 37^(@)=44N` `(D) f=mg sin 37^(@)-ma cos 37^(@)=-20N` |
|
| 28. |
At the top of a mountain a thermometer read 7°C and barometer reads 70cm of Hg. At the bottom of the mountain the barometer reads 76cm of Hg and thermometer reads 27°C. The density of air at the top of the mountain is _______times the density at the bottom. |
|
Answer» `0.99` |
|
| 29. |
It is found that |A +B| = |A| . This necessarily implies : |
|
Answer» B = 0 |
|
| 30. |
In the above problem, direction of friction force is |
|
Answer» towards left if `F_(1)` is applied |
|
| 31. |
Trajectories are shown in figure for three kicked footballs. Initial vertical and horizontal velocity components are u_(y) and u_(x) respectively. Ignoring air resistance, choose the correct statement from column-2 for the value of variable in column-1. |
|
Answer» |
|
| 32. |
An object of mass 2 kg attached to a spring is moved to a distance x = 10 m from its equilibrium position. The spring constant k= 1Nm^(-1) and assume that the surface is frictionless (a) When the mass crosses the equilibrium position, what is the speed of the mass? (b) What is the force that acts on the object when the mass crosses the equilibrium position |
|
Answer» Solution :(a) Since the spring force is a conservative force, the total energy is constant. At `x=10m`, the total energy is purely potential `E=U=1/2KX^(2)=1/2xx(1)xx(10)^(2)=50J` When the mass crosses the equilibrium position (x = 0), the potential energy `U=1/2xx1xx(0)=0J` The entire energy is purely kinetic energy at this position `E=KE=1/2mv^(2)=50J` The SPEED `v=sqrt((2KE)/m=sqrt((2xx50)/2)=sqrt(50)ms^(-1)=7.07ms^(-1)` (b) Since the restoring spring force is F LOR, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When object is at `x=+10m` (elongation), the force `F=-kx` `F=-(1)(10)=-10N`. Here the negative sign implies that the force is towards equilibrium ie, towards negative x-axis and when the object is at `x=-10` (compression), it experiences a forces `F=-(1) (-10)=+10N`. Here the positive sign implies that the force points towards positive x-axis. The object comes to momentary rest at `x=10m` EVEN THOUGH it experiences a maximum force at both these points |
|
| 33. |
Assertion: A uniform circular motion is an acceleration motion. Reason: Direction of acceleration is parallel to velocity vector. |
|
Answer» |
|
| 34. |
The equation of state of some gases can be expressed as (P + a/(V^2)) (V-b)=RT . Here, P is the pressure,Vthe volume, T the absolute temperature and a, b, R are constants. Find the dimensions of a. |
|
Answer» |
|
| 35. |
A steel ball of mass 'm' falls in a viscous liquid with a terminal velocity 4ms^(-1). If another steel ball of mass '8 m' falls through the same liquid then its terminal velocity is |
| Answer» Answer :C | |
| 36. |
Electromotive force and Electric potential differ in the dimensions of |
|
Answer» Mass |
|
| 37. |
Three moles of an ideal gas (C_(p)=7//2R) at pressure p_(A) and temperature T_(A) is isothermally expanded to twice its initial volume. It is then composed at constant pressure to its original volume. Finally the gas is compressed at constant volume to its original pressure p_(A) (i) Sketch p-V and p-T diagrams for the complete process (b) Calculate the net work done by the gas and net heat supplied to the gas a during the complete process. |
|
Answer» |
|
| 38. |
One man takes 1 minute to raise a box to a height of 1 metre and another man takes 1/2 minute to do so. The energy of the two is |
|
Answer» different |
|
| 39. |
A sphere has a elastic obique collision with another identical sphere which is initially at rest. The angle between their velocities after the collision is |
|
Answer» `30^(@)` |
|
| 40. |
Estimate the total number of air molecules in a room of capacity 25m^(3) at a temperature of 27^(@)C. |
|
Answer» SOLUTION :`T=27^(@)C+273=300K,k_(B)=1.38xx10^(-23)JK^(-1),V=25M^(3)` As Boltzmann.s CONSTANT, `k_(B)=R/Nimplies Rk_(B)N` Now , `PV=nRT=nk_(B)NT` The number of molecules in the room, `NN=(PV)/(k_(B)T)=(1.013xx10^(5)xx25)/(1.38xx10^(-23)xx300)` `=(25.325xx10^(5))/(414xx10^(-23))=0.06117xx10^(28)` `nN=6.1117xx10^(26)"molecules"` |
|
| 41. |
The expansion of an ideal gas of mass ma t a constant pressur p is given by the straight line B. Then, the expansion of the same ideal gas of mass 2m at a pressure 2p is goven by the straight line. |
|
Answer» C Slope of V-T graph `propm/p` SINCE `m/p` ratio in both case in same. STRAIGHT LINE will be B. |
|
| 42. |
A simple pendulum has a time period T_(1) when on the earth.s surface and T_2 when it taken a height .R. above the earth.s surface, where .R. is the radius of the earth. The value of T_(2)//T_(1) is: |
| Answer» Answer :D | |
| 43. |
A moter car is moving towards a vertical wall. The driver abserves that the frequency of the sound ofhis horn changes from 440 Hz to 480 Hz when . If the velocity of sound is 330 ms^(-1) , what is the speed of the car ? |
|
Answer» |
|
| 44. |
Kinetic energy of the body is always |
|
Answer» zero |
|
| 45. |
A particle is acted simultaneouosly be mutually perpendicular simple harmonic motion x= a cos omega t" and "y= a sin omega t. The trajectory of motion of the particle will be……… |
|
Answer» an ellipse Squaring and adding the both, `x^(2)+y^(2) = a^(2) [ cos^(2) omega t+ sin omega^(2) t]` `THEREFORE x^(2)+y^(2)= a^(2)` is a general FORMULA of circle, its radius is a. |
|
| 46. |
1 fermi is equal to |
|
Answer» `10^(-7)` micron |
|
| 47. |
A brass sphere is hung from one end of a massless and inextensible thread. When the sphere is set into oscillation, it oscillates with a time period T. If now the sphere is dipped completely into a non-viscous liquid, then what will be the time period of its oscillation? (density of the liquid is 1/10th of that of brass) |
|
Answer» Solution :LET VOLUME of the sphere be V, density of brass be `RHO`, density of the LIQUID be `rho.`. `therefore` Apparent weight of the sphere when immersed in the liquid = real weight - weight of displaced liquid = `Vrhog-Vrho.g=VG(rho-rho.)` `therefore` Apparent acceleration due to gravity of the sphere immersed in the liquid, `g.=("apparent weight")/("mass")=(Vg(rho-rho.))/(Vrho)=g(1-(rho.)/rho)` According to the problem, `(rho.)/rho=1/10`, hence `g.=g(1-1/10)=9/10g`. In the case of a simple pendulum, `Tprop1/sqrtg`, so, if the time period of oscillation of the sphere, when immersed in the liquid, is T., then `(T.)/T=sqrt(g/(g.))=sqrt(10/9)or,T.=sqrt10/3T`. |
|
| 48. |
A bullet is fired with a velocity of 900 km/h in the horizontal direction from a gun kept 20 m above the ground . Take g=10 m//s^2.Find out the horizontal range. |
|
Answer» |
|
| 49. |
A vehicle of 100 kg is moving with velocity of 5 m/s. How much force will be required to stop in 1/10 sec ? |
|
Answer» Solution :`m= 100 KG` `v_(0) = 5m//s` `t = (1)/(10) SEC` `=100 (-50) ` `=- 5000 N` `a= (V-v_(0))/( t )= (6-5)/((1)/(10))` `=-50` Negtive SIGN indicateretardingforce |
|