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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. |
The area of the surface of water in a vessel is 40cm^2. It is transferred into a cylindrical vessel of area of cross section 220cm^2. If the surface tension of water is 0.07Nm^(-1), the work done in the transfer of water is |
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Answer» 12.6 ERG |
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| 2. |
The volume of a certain mass of gas at a pressure of 5 xx 10 ^(4) Pais doubled adiabatically. What is the final pressure of the gas ? [ gamma = 1.4] |
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Answer» <P> Solution :`(V _(2))/(V _(1)) = 2 , P_(1) = 5 xx 10 ^(4) Pa , P_(2) = ?`` P _(1) V _(1) ^( gamma ) =P_(2) V _(2) ^( gamma)` `P _(2) = P_(1) = ((V _(1))/( V _(2))) = 5 xx 10 ^(4) ((1)/(2)) ^(1.4) = 1. 9 xx 10 ^(4) Pa ` |
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| 3. |
A ballon with mass 'm’ is descending down with an acceleration ‘a' (where a le g). How much mass should be removed from it so that is starts moving up with an acceleration ‘a ’ ? |
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Answer» `(2ma)/(G+a)`  from FIGURE (2) F- (m-m) g= (m-m) a wherem= m - m mg - ma - mg + mg = ma = m a mg+ma = ma m(g + a ) = 2 ma `m= (2ma )/( g+ a)` |
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| 4. |
In a simple pendulum the period of oscillation (T ) is related to the length of the pendulum (L) as |
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Answer» `L/T`=CONSTANT |
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| 5. |
Define intensity of sound and loudness of sound. |
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Answer» Solution :(i) The loudness of sound is defined as "the degree of sensation of sound PRODUCED in the ear or the PERCEPTION of sound by the LISTENER". (II) The intensity of sound is defined as "the sound power transmitted per uni"t area taken NORMAL to the propagation of the sound wave". |
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| 6. |
A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth its velocity must be increase …..... |
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Answer» `100%` `(mv_0^2)/R = (GM_em)/r^2` (r = distance of satellite from the centre of earth) `v_0^2=(GM_e)/r` `:.v_0 = (GM_e)/r ""...(1)` Escape speed for satellite `:. v_e = sqrt(2GM_e)/r ""...(2)` `:. v_e =sqrt2 sqrt((GM_e)/r)` `:. v_e = sqrt2 v_0` `:. v_e =1.414 v_0` `:. v_e/v_0 = 1.414` `:. (v_e-v_0)/v_0 =(1.414-1)/1 = 0.414` `:. (v_e-v_0)/v_0xx100% =41.4%` |
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| 7. |
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle theta without slipping and slipping down the incline without rolling is, |
| Answer» ANSWER :a | |
| 8. |
A block of mass 4kg is placed in contact with the front vertical surface of a lorry. The coefficient of friction between the vertical surface and block is 0.8. The lorry is moving with an acceleration of 15 m//s^2 (g = 10ms^(-2)). The force of friction between lorry and block is |
| Answer» ANSWER :C | |
| 9. |
A stone at the end of 1m long string is whirled in a vertical circle at a constant speed of 4ms^-1. The tension in the string is 6N when the stone is |
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Answer» (a) `10ms^-1` `(T_(max))/(T_(min))=4` (i) and `V_L=sqrt(V_H^2+4gL)` Tension at highest point, `T_(min)=(mv_H^2)/(K)-MG` Tension at lowest point, `T_(max)=mv_L^2//L+mg=(m(v_H^2+4gL))/(L)+mg` Now EQ. (i) can be WRITTEN as `m[(v_H^2+4gL)/(L)+g]=4xxm((v_H^2)/(L)-g)` `implies v_H=sqrt(3gL)=10ms^-1` |
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| 10. |
A stone is released from an elevator going up with an acceleration .a.. The acceleration of the stone after the release is |
| Answer» Answer :D | |
| 11. |
A block A of m ass rests on a horizontal table. A lig h t strin g connected to it passesover a frictionless pulley at the edge of table an d from its other end another block B of mass m_(2) is su sp en d e d . The co efficien t of k in eticfriction between the block and the table is mu_(k) hen the block A is sliding on the table, the ten sio n in the string i s : |
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Answer» `(m_(2)-mu_(K)m_(1)g)/(m_(1)+m_(2))` `m_(2) g - T -m_(2)a` `T-mu_(2)m_(1) g = m_(1) a` `RARR a= ((m_(2) -mu_(k)m_(1))g)/(m_(1) + m_(2))`  Forthe blockon MASS`m_(2)` `m_(2) g - T= m_(2) [(m_(2)- mu_(k) m_(1))/(m_(1) +m_(2))] g` `rArr = (m_(1)m_(2)(1+mu_(k))g)/( m_(1)+m_(2))` |
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| 12. |
An object initially at rest explodes, disintegrating into 3 parts of equal mass. Parts 1 and 2 have the same initial speed 'v', the velocity vectors being perpendicular to each other. Part 3 will have an initial speed of |
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Answer» <P>`sqrt2v` |
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| 13. |
When a loaded cargo ship enters a river from the sea, it sinks by a length x. When the ship is totallly unloaded, it rises by a length y. When the unloaded ship again enters the sea, it rises by z cm more. If the density of water of the river is rho_(w) and the body of the ship is vertical, show that the density of sea water is (yrho_(w))/((z-x+y)). |
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Answer» Solution :Let the WEIGHT of the loaded ship be w , the weight of cargo be w. , the base-area of the ship be A and the DENSITY of sea water be `rho`. Let US ALSO assume that the loaded ship sinks a LENGTH h in sea water. `therefore` In the case of floatation in sea water, `w=Ahrho""...(1)` In the case of floatation in river water, `w=A(h+x)rho_(w)""...(2)` When the ship is unloaded, `w-w.=A(h+x-y)rho_(w)""...(3)` When the unloaded ship enters the sea, `w-w.=A(h+x-y-z)rho""...(4)` From (1) and (2), we get `Ahrho=A(h+x)rho_(w)or,h+x=(hrho)/rho_(w)` From (3) and (4), we get `A(h+x-y)rho_(w)=A(h+x-y-z)rho` or, `h+x-y=(h+x-y-z)rho/rho_(w)` or, `(hrho)/rho_(w)-y=(hrho)/rho_(w)+(x-y-z)rho/rho_(w)or,y=(y+z-x)rho/rho_(w)` or, `rho=(yrho_(w))/((z-x+y))`. |
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| 14. |
The random error which exists invariably in screw gause |
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Answer» LEAST count ERROR |
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| 15. |
A person can see objects clearly from a distance 10cm at infinity. Then we can say that the person is |
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Answer» The PERSON with EXCEPTIONAL EYE having no eye defect |
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| 16. |
fig 3.36shownsthe velocity time graph of a rocket projeted upwards. Assumingthefirctionis neglibeladtermine from this graph (i) the time for which fuelbruns (ii) accendingand descerdingpart of therocket and (iii)maximumheightatteinedby the rocket . |
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Answer» |
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| 17. |
A wire PQ of length 5 cm which is free to move a and fro is attached to a rectangular dipped in a soap solution and raised. Find the force PQ in equilibrium. ( Surface tension of soap =4xx10^(-2)Nm^(-1)) |
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Answer» Solution :Length of the wire `L=5xm =5XX10^(-2)m` Surface tenson `S=4XX10^(-2)Nm^(-1)` When the wire PQ is pulled towards BC, due to surface tension (S) the force required to KEEP PQin equilibrium is `F=Sxx2L` `=(4xx10^(-2))xx(5xx10^(-2))xx2=40xx10^(-4)=4xx10^(-3)N` |
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| 18. |
Though India now has large base in science and technology, which is fast expanding, it is still a long way from realising its potential of becoming a world leader in science . Name some important factors which in your view have hindered the advancement of science in India . |
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Answer» Solution :Some of the important factors which have HINDERED the growth of science the india are given below: Lock of infrastructure and funds for quality research work in science. Science education is neither properly oriented nor directed. It needs specific direction depending on our requirements. The rural based science education is nearly non-existent so that majority of population is deprived of the benefits of advancements in science and technology. Poor PAY scales and other facilities to scientists as COMPARED to administrators. Indian society is full of superstitions and is highly traditional. Therefore, they are slow in adopting the new scientific trends. There is practically no co-ordination between the RESEARCHERS and the industrialists. The industrialists are the actual consumers of new research and technology. The industrialists of this country have little CONFIDENCE in the ability of the indian scientists. |
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| 19. |
A cubical vessel of side 10 cm is filled with Ice at 0^(@)C and is immersed in water bath at 100^(@)C. If thickness of walls of vessel is 0.2 cm and conductivity is 0.02 CGS units, then time in which all the Ice melts is (Density of Ice = 0.9 gm/cc) |
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Answer» 6 sec |
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| 20. |
A body of mass 2kg is projected from the ground with a velocity 20ms^(-1) at an angle 30^(@) with the verical. 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 kgms^(-1) during the time (t_(2)-t_(1)) is |
| Answer» Answer :B | |
| 21. |
The efficiency of an ideal heat engine working between the freezing point and boiling point of water is |
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Answer» `26.8%` FREEZINGPOINT of water `= 0^(@)C = 273 K` Boiling POINT of water `= 100^(C) = (100 + 273)K = 373 K` `T_(2)` : Sink temperature = 273 K `T_(1)` : Source temperature = 373 K `% eta = (1-(T_(2))/(T_(1))) xx 100 = (1-(273)/(373)) xx 100 = 26.8%` |
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| 22. |
A satellite is revolving near the earth's surface, Its orbit velocity is |
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Answer» 5.8 `"KMS"^(-1)` |
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| 23. |
In the absence of wind the range and maximum height of a projectile were R and H. If wind imparts a horizontal acceleration a = g/4 to the projectile then find the maximum range and maximum height. |
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Answer» Solution :`H^(1) = H (because u SIN THETA " REMAINS same")` `T^(1) = T` `R^(1) = u_(x)T + (1)/(2) a T^(2) = R + (1)/(2) (g)/(4)T^(2)` `= R = (1)/(8) g T^(2) = R + H` `R^(1) = R + H` `H^(1) = H` |
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| 24. |
Two wires are together end to end . The wires are made of the same material , but the diameter of one is twice that of the other . They are subjected to a tension of 4.60 N. The thin wire has a length of 40.0 cm and a linear mass density of 2.00 g//m. The combination is fixed at both ends and vibratedin such a waythat two antinodes are present , with the node between them being precisely at the weld . (a) What is the frequency of vibration ? (b) Find the length of the thick wire . |
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Answer» Solution :The mass per volume density , the tension , and the frequency must be the same for the TWO wires . The linear density , wave SPEED , wavelength and node - to node distance for it . a. Since the first node is at the weld , the wavelength in the thin wire is ` lmbda = 2 L = 80.0 cm` The frequency and tension are the same in both sections , so ` f = (v)/( lambda) = (1)/( 2L) sqrt ((T)/( MU)) = ( 1)/( 2(0.400 m)) sqrt (( 4.60 N)/( 0.00200 kg//m)) = 59.9 Hz` B. Since the thick wire is twice the diameter , it will have four times the cross - sectional area , and a linear density ` mu'` that is four times that of the thin wire . `mu' = 4 ( 2.00 g//m) = 0.00800 kg//m`. `L' = (lambda')/( 2) = (v')/( 2 f) = (1)/( 2 f) sqrt ((T)/( mu'))` `= (1)/( 2( 59.9 Hz)) sqrt(( 4.60 N)/( 0.00800 kg//m)) = 20.0 cm` Note that the thick wire is half the length of the thin wire . We could have reasoned the answer by noting that the wave speed on the thick wire is half as large , so the wavelength should be half as large for the same frequency . |
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| 25. |
As observed from the earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from the earth, this would |
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Answer» be similarly true (1) Relative MOTION between the earth andmercury. (2) Major gravitational force on mercury is due to the sun. |
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| 26. |
One cubic metre of an ideal gas is at a pressure of 10 N//m^2and temperature 300K. The gas is allowed to expand at constant pressure to twice its volume by supplying heat. If the change in internal energy in this process is 10^4 J, then the heat supplied is_ |
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Answer» `10^5J` |
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| 27. |
Why in winters birds swell their feathers? |
| Answer» Solution :When birds swell their feathers, they trap air in the feather. Air being a poor conductor PREVENTS loss of heat and KEEPS the BIRD warm. | |
| 28. |
A rod of mass M = 5kg and length L = 1.5 m is held vertical on a table as shown. A gentle push is given to it and it starts falling. Friction is large enough to prevent end A from slipping on the table. (a) Find the sum of linear momentum of all the particles of the rod when it rotates through an angle theta = 37^(@) (b) Find the friction force and normal reaction force by the table on the rod, when theta = 37^(@) (c) Find value of angle q when the friction force becomes zero. [tan 37^(@) = (3)/(4) " and " g = 10 m//s] |
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Answer» (b) `f = 9 N, N = 24.5 N` (C) `cos^(-1) ((2)/(3))` |
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| 29. |
n identical cubes each of mass .m. and side .l. are on the horizontal surface. Then the minimum amount of work done to arrange them one on the other is |
| Answer» Answer :C | |
| 30. |
Two lead balls of mass m and 2 m are placed at a separation d. A third ball of mass m is placed at an unknown location on the line joining the first two balls such that the net gravitational force experienced by the first ball is (6Gm^(2))/(d^(2) . What is the location of the third ball? |
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Answer» |
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| 31. |
Along the surface of a hemispherical container, a small ball is pushed down from a height h, such that the ball rises up to the oppositeedge. If theheight of the container is R , the ball must be pushed with a velocity |
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Answer» `SQRT(2HG)` |
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| 32. |
A body falling from rest has a velocity 'V' after it falls through a distance 'h' . The distance it has to fall down further for its velocity to become double is …. Times 'h'. |
Answer» Solution :Let a body STARTING from rest travels a distance of .H. m from A to B during which it acquires a velocity V as SHOWN in the figure . Its velocity becomes 2V at point C. `v^(2)-u^(2)=2as` Apply the above formula for both the cases we get, `V^(2)-0^(2)=2gh``(2V)^(2)-V^(2)=2gh` `RARR V^(2)=2gh`.....(1) `rArr 3V^(2)=2gh^(1)`.....(2) Eliminating the unconcerned terms by dividing equation (2)by equation (1) we get, `(2)/(1)=(3V^(2))/V^(2)=(2gh^(1))/(2gh)rArrh^(1)=3h`. |
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| 33. |
(A): Identical springs of steel and copper are equally stretched. More work will be done on the steel spring. (R) : Youngs modulus of steel is higher than copper. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct EXPLANATION of .A. |
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| 34. |
Two balls of different masses are thrown vertically upward with same initial velocity Maximum heights attained by them are h_(1) and h_(2)respectively, what is h_(1)//h_(2)? |
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Answer» SOLUTION :Same HEIGHT, `:.h_(1)//h_(2)=1` |
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| 35. |
A disc of radius R rolls on a rough horizontal surface. The distance covered by the point A in one revolution is |
| Answer» Solution :Path of A is cycloid and distance TRAVELLED in ONE rotation is 8R. | |
| 36. |
A lift is freely falling from 8th floor and is just about to stop at 4th floor. Taking ground floor as origin and positive direction upwards for displacement, velocity, acceleration. Which one of the following is correct? |
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Answer» `X lt 0, v lt 0, a GT 0` |
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| 37. |
A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle 45^(@) with the initial vertical direction is |
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Answer» Mg |
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| 38. |
The melting point of ice is 0^(0)C at 1 atm. At what pressure will it be -1^(0)C ? |
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Answer» Solution :Here `DeltaT = (- 1 - 0) = - 1, T = 273 + 0 = 273K and V_(2) - V_(1) = (1-1/(0.9))xx10^(-3)m^(3)` (given) L = 80 `cal//g` : We have,`(DELTAP)/(DeltaT)=L/(T(V_(2)-V_(1))) (or) (DeltaP)/(-1)=(80xx4.2xx10^(3))/(273(1-1/0.9)xx10^(-3))` `THEREFORE DeltaP = 132 xx10^(5) N//m^(2)=132 "atm" (or) P_(2)-P_(1)=132"atm" thereforeP_(2)=132 +P_(1)=133"atm"` |
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| 39. |
A duster weighs 0.5N. It is pressed against vertical board with a horizontal force of 11N. If the co-efficient of friction is 0.5 the minimum force that must be applied on the duster parallel to the board to move it upwards is |
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Answer» 0.4 N |
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| 40. |
Give two examples for angular harmonic oscillator. |
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Answer» Solution :EXAMPLES of angular harmonic oscillatio : (i) BALANCE wheel of a watch. (II) Torisonal pendulum. |
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| 41. |
A heavy partile slides under gravity download the inside of a smooth vertical tube dheld in vertical plane. It starts from the highest point with velocity sqrt(2ag) , where a is the radius of the circle. Find the angular position theta (as shown in figure) at which the vertical acceleration of the particle is maximum. |
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Answer» `v^(2)=v_(0)^(2)+2gh` where, `h=a(1-costheta)` :. `v^(2)=(sqrt(2ag))^(2)+2ag(1-costheta)` or `v^(2)=2ag(2-costheta)` ...(i) `N+mgcostheta=(mv^(2))/(a)` or `N+mgcostheta=2mg(2-costheta)` or `N=MG(4-3costheta)` NET vertical force, `F=Ncostheta+mg` `=mg(4costheta-3cos^(2)theta+1)` This force (or ACCELERATION) will maximum when `(dF)/(dtheta)=0` or `-4sintheta+6sinthetacos=0` So, EITHER `sintheta=0` , `theta=0^(@)` , or `costheta=(2)/(3)` , `theta=cos^(-1)((2)/(3))` `theta=0^(@)` is unacceptable Therefore, the desired position is at `thetacos^(-1)((2)/(3))` |
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| 42. |
The mean free path and rms velocity of a nitrogen molecule at a temperature 17^(@)C Are 1.2xx10^(-7)m and 5xx10^(2) m/s respectively.The time between two successivecollections |
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Answer» `2.4 XX 10^(-10)S` |
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| 43. |
A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0^(@) C " to " 100^(@) C, what is the longitudinal stress developed ? (y = 9 xx 10^(9) N//m , alpha = 10 xx 10^(-6) ""^(@) C^(-1)) |
| Answer» Answer :A | |
| 44. |
A swimmer is capable of swimming 1.65 ms^(-1) in still water. If she swims directly across a 180m wide river whose current is 0.85 m/s, how far downstream (from a point opposite her starting point) will she reach ? |
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Answer» 92.7 m |
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| 45. |
10N each , 6 forces are acting as different sides of a closed hexagon. Choose the correct options |
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Answer» Resultant is 10N, of the REMAINING vectors if one side is removed |
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| 46. |
A force barF=barVxxbarAis exerted on a particle in addition to the force of gravity, where barVis the velocity of the particle and barAis a constant vector in the horizontal direction. The minimum speed of a particle of mass 3kg be projected, so that it continues to move undeflected with a constant velocity is (xg)/A. Then 'x' is (consider the plane of the ground as xy plane) |
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Answer» |
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| 47. |
What are the sources of error in searlis experiment? How are the errors eliminated searlis experiment? |
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Answer» Solution :A. Source of errors Searle.s apparatus are 1) The supports may yield when a load is applied to the lower end of experimental wire. 2) While the EXPERIMENT is going on temperature may change. This cause thermal expansion is LENGTH. 3) Imporpoer adjustment of air bubble in spirit level. The errors can be MINIMISED by using a reference wire. The effect of yuield is equal on the both the wires Similarly thermal expansion is same for both the wires. So if another reference wire is used we will mesaure only erlative increase w.r.t. reference. So errors are minimised. |
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| 49. |
A solid sphere has linear velocity v_(0)=4 m/s and angular velocity omega_(0)=9 rad/s as shown. Ground on which it is moving , is smooth . Itcollides elastically with a rough wall of coefficient of friction mu . Radius of the sphere is 1m and mass is 2 kg . If the sphere after colliding with the wall rolls without slipping on opposite direction, then coefdicient of friction mu is |
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Answer» `(1)/(2)` |
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| 50. |
A solid sphere has linear velocity v_(0)=4 m/s and angular velocity omega_(0)=9 rad/s as shown. Ground on which it is moving , is smooth . Itcollides elastically with a rough wall of coefficient of friction mu . Radius of the sphere is 1m and mass is 2 kg . What is net linear impulse imparted by the wall on the sphere during impact ? |
| Answer» Answer :B | |
