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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. |
A ring and a disc of same mass roll without slipping along a horizontal surface with same velocity If the K.E. of ring is 8J then that of disc is |
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Answer» 2J |
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| 2. |
(A) : If earth suddenly stops rotating about its axis, then the value of acceleration due to gravity will become same at all the places. (R) : The value of acceleration due to gravity is independent of rotation of earth. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 3. |
Viscous force exerted by the liquid flowing between two plates in a streamline flow depends upon the |
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Answer» Velocity gradient in the DIRECTION PERPENDICULAR to the plates |
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| 4. |
Time taken by a body to slide down a rough surface inclined at an angle theta , is n time that required by the body to slide down the same distance on a smooth surface inclined at theta also. Find the coefficient of sliding friction between the body and the plane. |
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Answer» |
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| 5. |
The driver of a three - wheeler moving with a speed of 36 km/hr sees a child standing in the middle of the road and brings his vehicle at rest in 0.4 s just in time to save the child. The average retarding force on the vehicle is (The mass of the three - wheeler is 400 kg and the mass of the driver is 65 kg). |
| Answer» Answer :B | |
| 6. |
A spring balance reads W_(1) when a ball is suspended from it. A weighing machine reads W_(2) when a tank of liquid is kept on it. When the ball is immersed in the liquid, the spring balance reads W_(3) and the weighing machine reads W_(4). Then which of the following are not correct? |
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Answer» `W_(1) lt W_(3)` |
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| 7. |
A particle is moving with a contant velocity along a line parallel to positive X-axis. The magnitude of its angular momentum with respect of the origin is |
| Answer» Answer :D | |
| 8. |
One endof a lightrod isclamped in sucha mannerthat rodcan be rotatedin a verticalplane . A blockof mass mis attached to the otherenedof rodandtheblockstaysat rest at its lowest position . What minimum speed should beimparted to thebockso that itmay complete itscircle ? Assume length of rod to be l . |
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Answer» Solution :This is a case of rodhencethere is no chance of losingcircularpath dueto slackness as in case of STRING . KINETICENERGY at the bottom SHOULDBE enough that block just reaches the top most position which is at a height 2l from bottom POINT . Force applied by the rod remains along teh LENGTH displacementsi perpendicularto that hence no workis done by this force . Only workdone by weight is involvedso systemis conservative .Let v be the speedof blockat lowest point. Loss in kineticenergy = gain gravitational potential energy `rArr"" 1/2 mv^(2) = mg(2l)` `rArr "" v = sqrt(4gl)` |
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| 9. |
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of ther river is 2 km, the time taken to cross the river is (in hours) |
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Answer» `(2)/(27)` |
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| 10. |
Which one of the following represents simple harmonic motion ? |
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Answer» ACCELERATION = kx |
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| 11. |
A mass kg is suspended by a weightless string. The horizontal force required to hold the mass at 60^(@) with the vertical is |
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Answer» MG |
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| 12. |
A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough-surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals eta = (1)/(3) of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table ? |
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Answer» |
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| 13. |
A dry clean needle of diameter 'd' and density p when carefully placed on the surface of water remains floating. If T is the surface tension of water, then maximum value for the diameter 'd' of the needle for enabling it to float will be |
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Answer» `d= sqrt( ( 8rho pi)/( T g))` |
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| 14. |
The centre of mass of a non uniform rod of length Lwhose mass per unit length Klambda = (Kx^(2))/L where K is aconstant and x is the distance from one end is |
| Answer» ANSWER :A | |
| 15. |
CHOOSE THE INCORRECT PAIR: |
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Answer» Amplitude - Displacement |
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| 16. |
The tension and diameter of a sonometer wire of fundamental frequency n are doubled and density is halved.Its fundamental frequency will be |
| Answer» ANSWER :C | |
| 17. |
The centre of gravity of a body is a point where………..and…………on the body is…………… . |
| Answer» SOLUTION :the weight of the body ACTS , total gravitational torque , ZERO. | |
| 18. |
An object has a displacement from position vector vec(r )_(1)= (2hat(i) + 3hat(j))m " to " vec(r )_(2)= (4hat(i) + 6hat(j))m under a force vec(F) = (3x^(2) hat(i) + 2y hat(j))N. Find the work done by this force |
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Answer» Solution :`W= underse(r_(1))OVERSET(vec(r )_(1))INT vec(F).d vec(r )= underset(vec(r )_(1))overset(vec(r )_(2))int (3x^(2) hat(i) + 2y hat(J)).(DX hat(i) + dy hat(j) + dz hat(k))` `=underset(2)overset(4)int 3x^(2) dx + underset(3)overset^(6)int 2y dy - [x^(3)]_(2)^(4) + [y^(2)]_(3)^(6)= 83J` |
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| 19. |
Find the time period of small oscillations of the following systemA uniform square plate of edge suspended through a corner . |
| Answer» SOLUTION :`2PI SQRT((sqrt(8)a)/(3G))` | |
| 20. |
A body of mass 10 gm moving the velocity 5ms^(-1) collides with another body of mass 15gm moving with 3ms^(-1) in the same direction. After collision if they stick and move together, the loss of kinetic energy of the system is |
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Answer» 12 J |
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| 21. |
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, he has to swim in the direction |
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Answer» `30^(0)` with the RIVER FLOW |
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| 22. |
A simple pendulum has a time period 'T' in vacuum. Its time period when it is completely immersed in a liquid of density one eighth of the density of the material of the bob is |
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Answer» `sqrt((7)/(8))` |
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| 23. |
50 grams of ice at 0°C is mixed with 40 gm of water at 60°C. Then the ratio of ice and water at equilibrium is |
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Answer» `2:7` |
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| 24. |
A disc of radius R is spun to an angualr speed omega_(0) about its axis and then imparted a horizontal velocity of magnitude (omega_(0)R)/(4). The coefficient of friction is mu. The sense of rotation and direction of linear velocity are shown in the figure. The disc will return to its initial position. |
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Answer» if the value of `mult0.5`  About bottommost point angular momentum `L=I_(C)omega_(0)-mv_(0)R` (anticlockwise) `=(1)/(2)mR^(2)omega-m(omega_(0)(R)/(4))R`  `=(1)/(4)mR^(2)omega_(0)` `=+` ve or anticlockwise during slip friction acts about bottommost point. So, its torque is zero or angular momentum about bottommost point should also REMAIN anticlockwise when pure rolling STARTS. so, figure should be as shown below. so, the DISC will return to its initial position for all value of `mu`. |
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| 25. |
Define position vector. |
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Answer» <P> SOLUTION :A vector that DENOTES the position of a particle at any instant of time. With respect to origin of coordinate system .The position vector`VEC(r)`of the particle at a point P is given by`vec(r) = xhat(i) + y hat(j) + z hat(K)`. |
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| 26. |
A uniform round object of mass M, radius R and moment of inertia about its centre of mass I_(cm) has a light, thin string wrapped several times around its circumference. The free end of string is attaced to the celling and the object is released from rest. Find the acceleration of centre of the object and tension n the string. [ Take (I_(cm))/(MR^(2))=k] |
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Answer» Solution :When the CLOCK keeps the correct time, it completes one revolutions in 60 5. So the number of revolutions made per second is 1/60 RPS. i.e., `n_(0) = 1//60 rps = 0.017` rps Final frequency `=n=50/(60 xx 60) = 0.014` rps Time t = 7 minutes `=7 xx 60 = 420` s Acceleration `alpha = ?, omega = omega_(0) = omega_(0) + alpha t` `alpha = (omega - omega_(0))/t = (2PI n - 2pi n_(0))/t = (2pi (0.014 - 0.017))/420` `=-4.48 xx 10^(-5) rad//s^(2)` |
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| 27. |
A hemispherical bowl of radius 0.1m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity omega. A particle of mass m = 10^(-2) kgplaced inside the bowl also the particle from the bottom of the bowl is h, find the relation between h and omega. |
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Answer» SOLUTION :The particle is height h from the bottom of the bowl. With the rotation of the bowl, the particle also rotates about the vertical axis along a circular path of radius r In this situation, the weight mg of the particle of mass m, the CENTRIFUGAL force `m OMEGA^(2)`r and the normal force (R) on the particle by the surface of the bowl KEEP the particle in equilibrium. So,Rsin`theta = m omega^(2) r "" and " " Rcos theta = ` mg `therefore"" tan theta = (omega^(2) r)/(g)` If the radius of the bowl is a , then from we get tan`theta = (r )/(a- h)" " therefore (r)/(a - h)= (omega^(2)r)/(g) " " or, " " a - h = (g)/(omega^(2))` or, `"" h = a - (g)/(omega^(2)) = 0.1 - (9.8)/(omega^(2)) = 0.1 (1 - (98)/(omega^(2)))`m. 
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| 28. |
A ring of mass m and radius R is placed on rought inclined plane so that it rolls without slipping Match the following table. {:("Table-1","Table-2"),("(A) Linear acceleration of centre of mass","(P) is directly proportional to m"),("(B) Angular acceleration","(Q) is inversely proportional to m"),("(C) Rotational kinetic energy at any instant","(R) is directly proportional to "R^(2)),("(D) Translational kinetic energy at any instant","(S) is inversely proportional to R"),(,"(T) independent of mass m"):} |
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Answer» |
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| 29. |
The velocity of a particle under going SHM at the mean position is 4ms^(-1). Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude. |
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Answer» `2sqrt(3)MS^(-1)` |
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| 30. |
A moving particle of mass .m. makes head-on elastic collision with a particle of mass 2m. which is initially at rest. What is the fraction of K.E. lost by colliding particle ? |
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Answer» |
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| 31. |
An ideal material for making cooking vessels must have |
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Answer» Small CONDUCTIVITY and learge HEAT CAPACITY |
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| 32. |
The lengths and radii of two metal rods are in the ratio 3:2 and 2:3 respectively. When those two are heated from 0^(0)C to 100^(0)C, find the ratio of their linear expansions? |
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Answer» Solution :Expansion `Deltal=l alpha Delta t` `Deltal alpha l` [Both are STEEL RODS, so `alpha` is some] `(Deltal_(1))/(Deltal_(2))=(l_(1))/(l_(2))=(Deltal_(1))/(Deltal_(2))=(2)/(3)` |
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| 33. |
In the equation b=a^(2) cos^(2)((2pi beta gamma)/(alpha)). If the unitsof a, alpha, and beta are m, s^(-1) and (ms^(-1))^(-1). respectively. The unit of b and Y...... |
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Answer» `m and ms^(-2)` `:.` Unit of b= unit of`a^(2)` `:.` unit of `(2pi beta gamma)/(alpha)`= unitless `:.` UNITS of `gamma`= unit `(alpha)/(beta) [ :.` unitless] `:.` Units of `gamma=(s^(-1))/((ms^(-1))^(-1))=s^(-1)xxms^(-1)` `:.` Unit of `gamma==(m)/(s^(2))=ms^(-2)` |
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| 34. |
What is the time period of rotation of the earth around its axis so that the objects at the equator becomes weightless ? (g=9.8m//s^2, Radius of earth = 6400 km) |
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Answer» Solution :g at the equator is `g_(0) = g -g_0 = g - Romega^2` If bodies are to become WEIGHTLESS at the equator , `g_(0)=0` `0 = g - Romega^2implies Romega^2=g` `omega=sqrt(g/R)` Time PERIOD of ROTATION , `T = (2pi)/omega=2pisqrt(R/g)` ` R= 6400 xx10^3 m, "" g = 9.8 m//s^2` `T = 2pisqrt((6.4xx10^(6))/(9.8))=5078s=84` minute 38s. |
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| 35. |
The least count of a vernier callipers is 0.01 cm. Then, the error in the measurement is |
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Answer» `GT 0.01 CM` |
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| 36. |
Pick the odd one out from the following: |
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Answer» ATOMISER |
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| 37. |
A simple pendulum has a time period T_1 when on the earth's surface and T_2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T/T, is: |
| Answer» Answer :A | |
| 38. |
When a ceiling fan is switched on, it makes 10 revolutions in the first 3 seconds. Assuming a uniform angular acceleration, how many rotations it will make in the next 3 seconds? |
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Answer» 10 Using , `theta = omega_(0) t + (1)/(2) alpha t^(2)` `therefore 2 pi xx 10 = 0 + (1)/(2) alpha xx 3^(2) = (9)/(2) alpha "" …. (i)` For the rotationof fan in next three seconds , the total time of rotation = 3 + 3 = 6 s Let total NUMBER of rotation = N Then angle of rotation , `theta = 2 pi N` `therefore 2 pi N =0 + (1)/(2) alpha xx 6^(2) = 18 alpha "" .... (II)` Dividing (ii) by (i) , we get N = 40 No. of ROTATIONS in last three seconds = 40 - 10 = 30 rotations |
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| 39. |
The component of vec(A) along vec(B) is sqrt(3) times that of the component of vec(B) along vec(A). Then A : B is |
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Answer» `1:SQRT(3)` |
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| 40. |
A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of the pendulum will |
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Answer» INCREASE |
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| 41. |
A mild steel wire of length 2L and crosssectional area A is stretched, well within elastic limit, horizontally between two pillars as shown in figure. A mass m is suspended from the mid point of the wire, the mid point lower by distance x. Then strain in the wire is ...... |
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Answer» `(x ^(2))/(2L ^(2))`  `therefore DELTA L = 2 BO - 2 BD [ because BO = OC, BD= DC]=2 (BO -BD)` `=2 [ (x ^(2) + L ^(2)) ^(1//2) - L]` `=2L [ ((x ^(2))/( L ^(2)) + 1 ) ^(1//2) -1 ]` `=2L [ (1 + (x ^(2))/( L ^(2))) ^(1//2) -1 ]` `=2L [(1 + (1)/(2) (x ^(2))/( L ^(2)) + ...)-1 ] [because x lt lt L] ` By binomial theorem UPTO two terms, `= 2L xx 1/2 (x^(2))/( L ^(2))` `Delta L = (x ^(2))/( L )` `therefore` Strain `= (Delta L )/( 2L ) = (x ^(2) // L )/( 2 L ) = (x ^(2))/( 2 L ^(2))` |
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| 42. |
A particle was at x =0 at t = 0 . It moves with velocity v along the positive x -axis . If v changes with respect to x following the rule v = k sqrt(x) , how do its velocity and acceleration change with time? |
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Answer» |
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| 43. |
A body weighs w_(1) in air w_(2) in water and w_(3) in a liqud. The density of the liquid is given by |
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Answer» `(W_(1))/(W_(2)-W_(3))` |
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| 44. |
Obtain an expression for the time period T of a simple pendulum. [The time period T depend upon (i) mass l of the bob (ii) length m of the pendulum and (iii) acceleration due to gravity g at the place where pendulum is suspended. Assume the constant k=2pi] |
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Answer» Solution :`T ALPHA m^(a) L^(b) g^( c)` `T= k.m^(a) l^(b) g^(c )` Here k is the dimensionless constant. Rewriting the above equation with DIMENSIONS. `[T^1] = [M^a][L^b][LT^(-2)]^c` `[M^0L^0T^1] = [M^aL^(b+c)T^(-2c)]` Comparing the powers of M,Land T on both sides, `a=0,b+c=0,-2c=1` SOLVING for a,b and c `a=0,b=(1)/(2)" and "c= -(1)/(2)` From the above equation `T=k.m^(0) l^(1/2) g^(-(1)/(2))` `T= k k(l/g)^(1/2)= ksqrt((l)/(g))` Experimentally `k=2pi " hence "T= 2pi sqrt((l)/(g))`. |
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| 45. |
A simple pendulum of length lm is oscillated at a place where g = 9.8m//s^(2). Find the maximum velocity of the bob of the simple pendulum if the amplitude of oscillation is 3cm. |
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Answer» `5.291 CMS^(-1)` |
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| 46. |
Mention a pair of solid and liquid for each of the following cases where the angle of contact is (i) 90^@ (ii) less than 90^@ (iii) more than 90^@. |
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Answer» SOLUTION :(i) In the case of silver and WATER, the angle of contact is `90^@`. (ii) In the case of glass and water, the angle of contact is LESS than `90^@`. (iii) In the case of glass and mercury, the angle of contact is more than `90^@`. |
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| 47. |
Some amount of gas is compressed at STP in an (i) adiabatic process. What will be the final temperature and pressure of the gas in the case? (gamma = 1.4 for the gas) |
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Answer» |
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| 48. |
A string is wrapped around a wheel of radius 'r'. The axis of the wheel is horizontal and its M.I. about the axis is I. Weight 'mg' is ties to free end of the string which is released to fall down from rest position. The angular velocity of the wheel after it has fallen through distance 'h' will be |
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Answer» `((2gh)/(I+mr^(2)))^(1/2)` |
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| 49. |
Two particles of masses m_(1) andm_(2) in projectile motion have velocities vecv_(1) and vecv_(2) respectively at time t= 0. They collide at time t_(0). Their velocities become vecv_(1) and vecv_(1) at time 2t_(0) while still moving in air. The value of |(m_(1)vecv_(1)^(1)+m_(2)vecv_(2)^(1))-(m_(1)vecv_(1)+m_(2)vecv_(2))| is |
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Answer» ZERO `m_(1)v_(1)^(1)+m_(2)v_(2)^(1)=m[v_(1)+2g t_(0)]+m[v_(2)+2g t_(0)]` |
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| 50. |
The following are the p-V diagrams for cyclicprocesses for a gas. In which of these processes heat is not absorbed by the gas? |
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