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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 body of mass 2kg is projected from the ground with a velocity 20ms^(-1) at an angle 30^(@) with the verticle. If t_(1) is the time in seconds at which the body is projected and t_(2) is the time in seconds at which it reaches the ground, the change in momentum in kg ms^(-1) during the time (t_(2)-t_(1)) is |
| Answer» Answer :B | |
| 2. |
A piece of copper having an internal cavity weighs 260 gm in air and 221gm in water. The volume of the cavity is (Density of copper =8.8 gm/cc) |
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Answer» 43 cc |
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| 3. |
Under what conditions can a Carnot engine have 100% efficiency? |
| Answer» Solution :When TEMPERATURE of SINK is ABSOLUTE zero efficieny is 100%. | |
| 4. |
A spring of force constant 'k' is stretched by a small length 'x'. Find the work done in stretching it further by a small length 'y'. |
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Answer» Solution :Let `W_(1)` is the work DONE in STRETCHING a spring of FORCE CONSTANT .K. through a length ..x... Then `W_(1)= (1)/(2) Kx^(2)` Let `..W_(2)..` is the workdone is stretching the spring through a length `(x+y)`. Then `W_(2)= (1)/(2)K(x+ y)^(2)` `THEREFORE` Additional work done, to increase the elongation by ..y.. is `W= W_(2)-W_(1)` `W=(1)/(2) K (x+y)^(2) - (1)/(2) Kx^(2)` 
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| 5. |
A solid cylinder of mass M and radius R pure rolls on a rough surface as shown in the figure. Choose the correct alternative (s). |
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Answer» The acceleration of the centre of mass is `F/(M)` |
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| 6. |
The potential energy of a long spring when stretched by 2cm is U. If the spring is stretched by 8 cm the potential energy stored in it is………. |
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Answer» `8U` For `x=2, U= (1)/(2)k(2)^(2)= 4xx (k)/(2)= 2K"""……"(1)` For `x=8 cm," NEW "U= U.= (1)/(2) k(8)^(2)= (1)/(2)k(64)` `= 32k= 16(2k)` `THEREFORE U.= 16U""therefore " from equation (1) ")` |
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| 7. |
A small body of mass 10 gram is making simple harmonic oscillations along a straight line with a time period of pi//4 seconds and the maximum displacement is 10cm. Find the energy of oscillations and the average kinetic energy during one oscillation |
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Answer» `32 XX 10^(-2) J , 16 xx 10^(-2) J` |
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| 8. |
An elevator weighing 6000kg is pulled upward by a cable with an acceleration of 5 m//s^(2). Taking g to be 10ms^(-2), find the tension in the cable . |
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| 9. |
A ball of mass 2 kg hanging from a spring oscillates with tiem period 2pi seconds. Ball is removed when it is equilibrium position, then spjring shortens by what length? |
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Answer» Solution :`T=2pisqrt(m/k)` `:.k=(4pi^(2)m)/(T^(2))=((4pi^(2))(2))/((2)^(2))` Now `MG=kx_(0)` `:.x_(0)=(mg)/k=(2xxg)/2=g` metre or 10 m |
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| 10. |
The potential energy v in joule of a particle of mass 1 kg moving in xy plane, obeys the law v=3x+4y, when (x,y) are the co-ordinates of the particle in meter. If the particle is at rest at (6,4) at time t=0, then |
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Answer» The PARTICLE has constant acceleration |
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| 11. |
In an equilateral triangle ABC, AL, BM and CN are medians. Forces along BC and BA represented by them will have a resultant represented by |
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Answer» `2vec(AL)` |
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| 12. |
If a liquid does wet the surface ,its angle of contact is acute or obtuse ? |
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| 13. |
What is capillary action ? Derive the formula for rise of liquid in a capillary tube immersed vertically in liquid. |
Answer» Solution :The phenomenon, the surface tension of liquid plays an important role. Capillary rise , (a) SCHEMATIC picture of a narrow tube immersed water .(b) Enlarged picture near interface. The word Capilla means hair in Latin . If the tube were hair thin the rise WOULD be very lrge for liquid of angle of contact is acute (This type tubeis known as capillary tube) In figure , consider a vertical capillary tube of circular cross action (radius a) inserted into an open vessel of water . The contact angle between water and glass is acute .`(thetalt(90^(@))`. Thus the surface of water in the capillary is concave .This means that there is a pressure difference between in two SIDES of the TOP surface. `P_(i)-P_(0)=(2S)/(r)` where r is radius of meaniscus ..(1) From figure `costheta=(a)/(r)` `thereforer=(a)/(costheta)` ....(2) Putting value of equation(2) in (1), `P_(i)-P_(0)=(2s)/(a)costheta`....(3) Hence , the pressure `P_(i)` of the water inside the tube , justat the meniscus (air -water interface) is less thenthe atmospheric pressure `P_(0)`. According to figure (a) , consider two points A and and B . they must be at the same pressure, Hence , here `P_(B)=P_(A)` ....(4) but `P_(B)=` pressure on convex surface of meniscus + pressure of liquid column of height h. `thereforeP_(B)=P_(0)+hrhog` and `P_(A)=P_(i)"" (because` From equ .(4) and (5))Where `rho` is the density of liquid and h is the height of liquid column in capillary. `thereforeP_(i) -P_(0)=hrhog` From equation (3),(4), `hrhog=(2Scostheta)/(a)` `h=(2Scostheta)/(arhog)` As theradius a is small , the height of column is more . If meniscus is convex (for example mercury), means `costheta` is negative , liquid depressed in capillary. |
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| 14. |
A circular disc with a groove along its diameter is placed horizontally . A block of mass 1 kg is placed asshown . The coefficient of friction between the block and all surfaces of groove in contact is mu = 2//5 . The disc has an acceleration of 25 m//s^(2) . find the acceleration of the block with respect to disc . |
Answer»  N + ma SIN `THETA = "mg cos" theta` N = ( mg cos `theta - "ma sin" theta )` Acceleration of mass w.r. to disc ` a_(0) = (( "mg sin "theta + "ma cos" theta - muN)/(m))` |
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| 15. |
Particle is point like object with dimension. |
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| 16. |
A cyclic process consisting of two isobaric and two adiabatic processes is shown in the figure. If P_2 = nP_1 then prove that efficiency of thisprocess is eta=1-n^((1-1/gamma)) where gamma=C_P/C_V |
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Answer» Solution :For adiabatic process `TP^((1-gamma)/gamma)` = constant `therefore T_1P^((1-gamma)/gamma)=T_2P_2^((1-gamma)/gamma)` `therefore T_2/T_1=(P_1/P_2)^((1-gamma)/gamma)` But `P_2=nP_1 therefore P_1/P_2=1/n` `T_2/T_1=(1/n)^((1-gamma)/gamma)=(n)^((1-gamma)/gamma)=n^(1-1/gamma)` Now EFFICIENCY `eta=1-T_2/T_1` `therefore eta=1-(n)^(1-1/gamma)` (`because` PUTTING VALUE of `T_2/T_1`) |
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| 17. |
A ball at rest is dropped freely from a height of 20m. It loses 30% of its energy on striking the ground and bounces back. The height to which it bounces back is |
| Answer» ANSWER :A | |
| 18. |
The correct graph between surface tension T and temperature q in low temperature range is |
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Answer»
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| 19. |
Is there any advantage of washing clothes with warm water than cold water? Why? |
| Answer» Solution :Surface tension of WATER decreases with the INCREASE in temperature of water. HENCE warm water easily SPREADS over the clothes and cleans easily. | |
| 20. |
On a smooth table two blocks of masses 2.5 kg and 1.5 kg are placed one over the other as shown in figure. If the coefficient of static friction between two blocks is 0.2, the maximum horizontal force to be applied on the lower block so that the two blocks move together is (g=ms^(-2)) |
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Answer» 8N |
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| 21. |
The figure shows a uniform rod lying along the X-axis. The locus of all the points lying on the XY-plane, about which the moment of inertia of the rod is same as that about O is |
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Answer» an ellipse |
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| 22. |
A string of 7m length has a mass of 0.035 kg. If tension in the stringis 60.N, then speed of a wave on the string is |
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Answer» `77m//s` |
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| 23. |
When a copper wire of area of cross section 1 mm^(2)is stretched by a force of 10N. Then the work done per unit volume in stretching the wire is (Y=1.2xx10^(11)Nm^(-2)) |
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Answer» `416.7Jm^(-3)` |
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| 24. |
A man of mass 100 kg stand at the rim ofof radius and moment of inertia 4000 kg m^(2) mounted on a vertical frictionlessThe whole system is initially at rest. The man now walls along the outer edge of unable with a velocity of 1 m/s relative to the earth. |
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Answer» The angular VELOCITY of turn table is `0.05` RAD/s along the direction of MAN |
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| 25. |
Two particles of equal mass m go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is |
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Answer» `v=sqrt((1)/(GM))` i.e.,`(mv^2)/(R)=(GMxxm)/((2R)^(2))` `rArrv=(1)/(2)sqrt((GM)/(R))` 
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| 26. |
The gravitational potential at a point at height h from the surface of earth is -6.50 xx 10^(7) J kg^(-1), and acceleration due to gravity is 7.2 m//s^(2). Calculate the value of h if radius of earth is 6400 km. |
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Answer» 2,627 km `V = (-GM)/((R + h)) = -6.50 xx 10^(7) J kg^(-1)` …(i) `a = + (GM)/((R+h)^(2)) = 7.2 m//s^(2)` …(II) Dividing (i) by (ii) `(GM(R+h)^(2))/((R+h).GM) = (R+h) = (6.50 xx 10^(7))/(7.2) = 0.9027 xx 10^(7) m` (R +h) = 9,027 km Height of the point above earth surface `h = 9,027 km - 6,400 km` = 2,627 km |
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| 27. |
A small glass marble of mass .m. oscillates between the two edges, inside a hemispherical glass bowl of radius .r.. If .V. is the speed of marbal at the lowest position, the normal reaction at that position is |
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Answer» `(mv^(2))/(R )` |
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| 28. |
Is there any net work done by external forces on a car moving with a constant speed along a straight road ? |
| Answer» Solution : If the car moves CONSTANT speed, then there is no changes in its kinetic energy . It impliesthat if there is no change in die energy then there is so work done by the FORCE on the body PROVIDED is as REMAINS constant | |
| 29. |
Fill in the blanks using the words from the list appended with each statement For a fluid in a steady flow,the increase in flow speed at a constriction follows (conservation of mass/ Bernoulli's principle) |
| Answer» SOLUTION :CONSERVATION of MASS, Bernoulli.s EQUATION | |
| 30. |
Two simple harmonic motion are represented by the equation y_(1)= 0.1 sin(100 pi t+(pi)/(3)) and y_(2)= 0.1 cos pi t. The phase difference of the velocity of particle-1 with respect to the velocity of particle-2 is………… |
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Answer» `(pi)/(6)` `y_(1)= 0.1 sin [100pi t +(pi)/(3)]` `there v_(1) = 0.1xx 100 COS [100 pi t +(pi)/(3)]` and `y_(2)= 0.1 cos pi t` `therefore v_(2)= -0.1 sin(pi t+(pi)/(3))` The PHASE difference of velocity of particle-2 RELATIVE to particle-1 `delta= (pi t+(pi)/(3))-(pit +(pi)/(2))` `therefore delta = (pi)/(3)- (pi)/(3)= (2pi -3pi)/(6)= -(pi)/(6)rad`. |
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| 31. |
A paricle of mass m is projected with a velocity u at angle alpha with the horizontal. Work done by gravity during its descent from its highest point to, the position where its velocity vector makes an angle (alpha)/2 with the horizontal is |
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Answer» `1/2"MU"^(2)TAN^(2)ALPHA` |
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| 32. |
A simple pendulum from earth is shifted to moon (gm = ge/6). Foreign minute on earth, the time on the moon as measured by the pendulum is (For same number of oscillations) |
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Answer» `24.5 SEC` |
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| 33. |
In a given process of an ideal gas, dW=0 and dQ |
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Answer» the TEMPERATURE will decrease |
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| 34. |
If the velocity of a particle is v= At+Bt^2, where A and B constants, then the distance travelled by it between 1s and 2s is. |
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Answer» `3/2 A+4B` |
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| 35. |
A man can row a boat with 4 kmph in still water. If he is crossing a river where the current is 8kmph. The time taken to cross the river in shortest path is (k)/(sqrt(3)) hr. where 'k' is ( width of the river is 8km) |
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| 36. |
If there were a small alteration in gravitational force , then which of the following forces do you think would alter in some respect |
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Answer» Viscous FORCES |
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| 37. |
A constant force is acting on a particle and also acting perpendicular to the velocity of the particle. The particle describes the motion in a plane. Then |
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Answer» ANGULAR displacement is zero |
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| 38. |
A car without passengers moving with certain velocity on a level ground can be stopped in a distance of 10m. If the passengers add 25% of its weight, its stopped distance for the same braking force and velocity is (ignore friction) |
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Answer» 15m |
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| 39. |
In a vernier cliper, n divisions of vernier scale coincides with (n - 1) divisions of main scale. The least count of the instrument is............ |
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Answer» `1/N MSD` |
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| 40. |
The figure shows a siphon in action. The liquid flowing through the siphon has a density of 1.5 gm//cm^(3). Calculate the pressure difference between (a) Points A and D, (b) Points B and C |
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| 41. |
A ballon filled with helium rises against gravity increasing its potential energy . The speed ofthee ballonalso increases as it rises . How do you reconcile this with the law of conservation ofmechanical energy ? You can neglect viscous drag of air and assume that density of air is constant . |
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Answer» Solution :m = Mass of ballon V = Volumeof ballon `rho_("He") ` = Density of helium `rho_("air")` = Denity of air Volume V of ballon displaces volume V of air . So , `V(rho_(air)-rho_(He)) g = ma = m (dv)/(dt)""…(i)` Integrating with respect to t , ` rArr V(rho_("air") -rho_("He")) "gt" = mv ""...(ii)` `1/2 mv^(2) =1/2 m ((V^(2))/(m^(2))) (rho_(air)-rho_("He"))^(2) g^(2)t^(2)` `= 1/(2m) V^(2)(rho_(air)-rho_(He))^(2) g^(2)t^(2)` If the ballon rises to a HEIGHT h , from s = ut + `1/2 at^(2)` ` :. h = 1/2 "at"^(2)` `=1/2(V(rho_(air)-rho_(He)))/m "gt"^(2) ( :. U = 0 ) ....(iii)` From (iii) and (ii) `1/2 mv^(2) = [V (rho_(a)-rho_(He))"gt"^(2)][1/(2m)V(rho_(air)-rho_(He))"gt"^(2)]` ` =V(rho_(a)-rho_(He))gh` Therefore , `1/2 mv^(2) +V_(rho_(He))=gh =V_("pair") hg ` `KE_("ballon")+PE_("ballon") ` = Change in PE of air . So, as the ballon goes up , an equalvolume of air comes down , INCREASE in PE and KE of the ballon is EQUAL to decrease inPE of air . |
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| 42. |
A circular dise in rolling down in an inclined plane without slipping. The percentage of rotational energy in its total energy is |
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Answer» `66.61%` Translational `K.E.=(1)/(2) MV^(2)=(1)/(2)M(R omega)^(2)=(1)/(2) MR^(2) omega^(2)` Total kinetic energy `=E_("rot")+E_("trans")=(1)/(4)MR^(2) omega^(2)=(3)/(4) MR^(3)MR^(2) omega^(2)` `% "of" E_("rot") =(E_("rot"))/(E_("rot"))xx100%=33.33%` |
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| 43. |
Real gas deviats from perfect gas behaviour as its molecules a) Have definete size b) Attract each other c) Show brownian motion d) Are not spherical |
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Answer» a,B & C are TRUE |
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| 44. |
As shown in figure, a planet revolves round the sun in Elliptical orbit with sun at the focus. The shaded areas can be assumed to be equal. If t_1 and t_2represent the time taken for the planet to move from A to B and C to D respectively, then |
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Answer» `t_(1) LT t_2` |
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| 45. |
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8m//s^2 |
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Answer» 0.28m, 1.491 SEC |
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| 46. |
Select the correct statement of the following |
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Answer» 1 radian = ` 5 . 729^(@)` |
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| 47. |
If we consider the motion of all the molecules, the expression of mean free path is |
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Answer» `(1)/(N pi d^(2))` |
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| 48. |
If veca.vecb=0, then what is the angle between veca"and "vecb? |
| Answer» SOLUTION :`-pi//2` | |
| 49. |
A wooden plank of mass 20 kg is at rest on a smooth horizontal floor. A man of mass 60 kg starts moving from one end of the plank to the other end. The length of the plank is 10 m. Find the displacement of the plank over (he floor when the man reaches the other end of the plank |
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Answer» Solution :mass of MAN m = 60 kg , mass of the BOAT M = 20 kg , length of the plank = 10 m Distance moved by the boat `S =(ml)/(m+M) = (60 xx 10)/(60 + 20) = 7.5 m` |
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