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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. |
An object of mass m held against a vertical wall by applying horizonatal forceF as the minimumof the force is |
| Answer» Solution :greater than mg | |
| 2. |
A circular disc of radius R is removed from a bigger of the discs coincide. The centre of mass of the new disc is ar from the centre of the bigger disc. The value of alpha is |
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Answer» `(1)/(4)` `M = PI (2R)^(2) sigma` Mass of disc removed , `M_1 = pi R^2 sigma = 1/4 M` Mass of remaining disc , `M_2 = M - M_1 = M - 1/4 M = 3/4 M` Let centre of mass of remaining disc be at `C_2` where `OC_2 = x` As `M_1 xx O C_(1) = M_2 xx O C_(2)` `therefore M/4 R = (3M)/(4) ""x or x = (R)/(3) = alpha R therefore alpha = (1)/(3)` 
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| 3. |
A bridge over a canal is in the form of an arc of a circle of radius 20 m. Calculate the maximum speed at which a motor cycle can pass over the bridge without leaving the ground at the highest point ?. |
| Answer» Solution :`V = sqrt(rg) = sqrt(20 xx 9.8) = 14 MS^(-1)` | |
| 4. |
Dimensional formula of capacitance is |
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Answer» `M^(-1) L^(-2) T^(-4) I^(2)` |
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| 5. |
A cable is cut into half its original length.What will be effect on the increase in its length under a given load? |
| Answer» Solution :Y=FL//A delta L, deltaL=FL//AY, since F,A and Y re FIXED,delta L PROP L.so elongation will become half. | |
| 6. |
The relation between acceleration and displacement of four particle are given below, which one of the particle is exempting simple harmonic motion? |
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Answer» `a_(x)= +2x` `therefore a propto -x` `therefore a= -KX` In given option D, `k= -2` and displacement MUST be LINEAR. |
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| 7. |
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100^@ C, while the other one is at 0^@ C. If the two bodies are brought into contact, then assuming no heat loss the final common temperature is _____ |
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Answer» less than `50^@ C` but greater than `0^@C` According to principle of calorimetry, Heat loss = heat gain `S_h[100^@C-theta]=S_c theta` `therefore S_h xx 100^@ -S_h xx theta =S_c theta` `therefore S_h xx 100^@ =(S_h +S_c)theta` `therefore theta =(S_h/(S_h+S_c))xx100^@` `therefore theta = 1/(1+S_c/S_h)xx100^@` ...(1) but `S_c lt S_h rArr S_c/S_h lt 1` `rArr 1+ S_c/S_h lt 1+1` `rArr 1+ S_c/S_h lt 2` `therefore` From equation (1), `theta gt 100^@/2` `therefore theta gt 50^@C` |
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| 8. |
If the other factors remain unaltered, does the change in the volume of a solid , with the change in temperature, depend on the presence of a hole in the solid? Answer with reason. |
| Answer» Solution :The change in volume of a SOLID with the change in TEMPERATURE, does not depend on the presence of a HOLE in the solid. The hole SIZE will change by the same amount as the solid that would have exactly fitted in the hole. | |
| 9. |
One mole of an ideal monoatomic gas undergoes the process P = alpha T^(1//2)where alphais a constant. |
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| 11. |
A block of mass sqrt(3) kg is kept on a frictional surface with mu=(1)/(2sqrt(3)). The minimum force to be applied as shown to move the block is |
| Answer» Solution :`N=mg+F sin THETA, F COS theta=f, f=muN` | |
| 12. |
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = (kx^(2))/2where k is the force constantof the oscillator . For k = 0.5 "Nm"^(-1) , the graphof V(x)versusx is shown in figure . Show that a particleof total energy 1 J movng under this potential must 'turn back ' when it reaches x = pm 2 m. |
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Answer» Solution :Total ENERGY of a simple HARMONIC oscillator , `E = K+V` ` 1= 1/2 MV^(2)+1/2 kx^(2) ""…(1)` Whensimple harmonic oscillator TURN back , at this time its speed or velocity will be zero . v = 0 ` :.1 = 0 + 1/2 (0.5)x^(2) "" [ :. "From equation (1)"]` |
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| 13. |
Show dimensionally that the pressure is given by (i) energy per unit volume (ii) force per unit area. |
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Answer» Solution :DIMENSIONS of PRESSURE `=ML^(-1)T^(-2)` Dimensions of energy per unit volume `=(ML^(2)T^(-2))/(L^(3))=ML^(-1)T^(-2)` Dimensionless of FORCE per unit area `=(MLT^(-2))/(L^(2))` So all the three are the same. |
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| 14. |
A stone is projected vertically upward from the top of a tower with a velocity u andstrikes the bottom of the tower with a velocity 3u. The distance travelledby the stone is |
| Answer» Answer :B | |
| 15. |
A given mass of a gas is compressed isothermally untill its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure . The ratio of the specific heats of the gas is approximately . |
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Answer» `1.20` |
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| 16. |
A small ball strikes a stationary uniform rod, which is free to rotate, in gravity-free space. The ball does not stick to the rod The rod will rotate about |
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Answer» its CENTRE of MASS |
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| 17. |
The adjacent sides of a parallelogram are vec(P) = 2 I - 3j + k and vec(Q) = -2i + 4j - k.The are aof the parallelogram is sqrtn units. Find the value of n |
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| 19. |
A body of mass 2kg is projected from the ground with a velocity 20ms^(-1)at an angle 30^@with the vertical. If t_1 is the time in seconds at which the body is projected and t_2is 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 | |
| 20. |
A body of mass 'm' slides down a smooth inclined plane having an inclination of 45° with the horizontal. It takes 2s to reach the bottom. If the body is placed on a similar plane having coefficient friction 0.5, what is the time taken for it to reach the bottom? |
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Answer» SOLUTION :2.28 SEC |
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| 21. |
A piston of cross-sectional area 100 cm is used in a hydraulic press to exert a force of 10.dyne on water. The cross-sectional area of the other piston which supports a truck of mass 200 kg is, |
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Answer» `1.96cm^(2)` |
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| 22. |
(A): Different gases at the same conditions of temperature and pressure have different root mean square velocities. (R ):Average kinetic energy of gas is directly proportional to absolute temperature. |
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Answer» Both (A) and ( R) are true and (R ) is the CORRECT explanation of (A) |
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| 23. |
Which of the following is represented by the area under a-t curve ? |
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Answer» CHANGE in acceleration |
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| 24. |
(A) : The surface tension of cold solution is more than the surface tension of hot solution. (R ) : It is better to wash the clothes in cold soap solution. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 25. |
Derive the expression for mean free path the of the gas. |
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Answer» Solution :Let us CONSIDER a system of molecules each with diameter d. Let n be the number of molecules per unit volume . It is assumed that only one molecule is in motion and all other are at rest as shown in the figure.  `implies` Mean free path If a molecule moves with average speed v in a timet , the distance travelled is vt. In this time t, let us consider the molecule to move in an imaginary cylinder of volume `pid^2vt`. It collides with any molecule whose centre is within this cylinder . Hence , the number ofcollisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equation of `pid^2vtn.` The total path length divided by the number of COLLISION in time t is the mean free path . Mean free path, `lamda = ("distance travelled")/("Number of collisions")` `lamda=(vt)/(npid^2vt) =1/(npid^2) "" ....(1)` It is assumed that only one molecule is moving at a time and other molecules are at rest. But in actual practice all the molecules are in random motion . Hence the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct. expression for mean free path, `lamda=1/(sqrt2npid^2)""..(2)` It is implied from the equation that the mean free path is inversely proportional to number density. When the number density INCREASES the molecular collisions increases. Hence it decreases the distance travelled bu the molecule before collisions. REARRANGING the equation (2) using 'm' (mass of the molecule) `:. lamda=m/(sqrt2pid^2mn)` But mn= mass per unit volume = `rho` (density of the gas) `lamda=m/(sqrt2pid^2rho) ""...(3)` Also we know that PV = NkT `P = N/V KT = nkT "" [:.n=N/v]` `:. n=P/(kT)` Substituting `n= P/(kT)` in equation (2), we get `lamda(kT)/(sqrt2pid^2P)` |
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| 26. |
Define volume stress. |
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Answer» Solution :Force PER unit AREA is called VOLUME STRESS. `sigma_(V)=(F)/(A)` which is the same as the pressure. |
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| 27. |
A square plate is rotating with angular velocity omega about its diogonal. Suddenly the rotation axis is shifted parallel to itself passing throuhj B. The new angular velocity is omega//b where b is |
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| 28. |
A voltmeter and an ammeter are connected in the circuit as shown. Resistance of ammeter is say (R)/(10) and that of voltmeter is 10R. Then |
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Answer» PERCENTAGE ERROR in the reading of ammeter (compared to that MEASURED, if both ammeter and voltmeter were ideal) is - 1.0% |
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| 29. |
A string AB of length 21 is fixed at A to a point on a smooth horizontal table. A particle - of mass m attached to B is initially at a distance I from A as shown in figure. The particle is projected horizontally with speed u at right angles to AB. Find the impulsive tension in the string when it becomes taut and the velocity of the particle immediately afterwards. |
Answer» Solution :When the string b eomes thut `AB. = 2l and angle B. AB = 60^(@)`  Just before the string JERKS, the particles is VELOCITY components parallel and perpendicular to AB. are u `sin 60^(@)` and u `cos 60^(@)` respectively. When the string becomes thut the length of AB is fixed and particle cannot travel in the direction AB.. After the jerk, the velocity of the particle is, therefore, perpendicular to AB.. Using impulse = change in momentum (a) Along `B.A , J =0- (- m u sin 60^(@)) (or) J = (SQRT3)/(2) m u` (b) Perpendicular to B.A (no impulsive component) `0= MV - m u cos 60^(@) (or) v = u cos 60^(@) = (u)/(2)` Therefore, the velocity of the particle just after the string becomes thut is `(u)/(2) ` perpendicular to the string. |
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| 30. |
The volume of one mole of an ideal gas changes from V to 2V at temperature 300 K. If R is universal gas constant, then work done in this process is |
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Answer» 300 Rln2 `W = nRT ln ((V_(2))/(V_(1)))` Here, `n = 1 mol, T = 300 K, V_(1)V` and `V_(2) = 2V` `:. W = 1 xx R xx 300 ln ((2V)/(V)) = 300 R ln 2` |
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| 31. |
A stone of mass 2 kg is tied to one end of a string of length 0.5 m. It is whirled in a vertical circle. If the velocity of stone at the top is circle is 5 ms^(-1), the tension in the string at the bottom of circle is |
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Answer» 125 N |
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| 32. |
Two rods of different metals, having the same area of cross-section A, are placed end to end between two massive walls as shown in fig. If the temperature of both the rods are now raised by Deltat^(0)C then (a) Find the force with whichthe rods acts on each other at higher temperature. (b) Find the lengths of the rods at the higher temperature. Assume that there is so change in the corss-sectional area of the rodsdo not bend. There is no deformationof walls. |
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Answer» Solution :(a) Due to HEATING the increases in length of the composite rod will be `(Deltal)_(I)=l_(1) alpha_(1) Deltat +l_(2) alpha_(2)DELTA t=(l_(1)alpha_(1)+l_(2)alpha_(2)) Deltat …(1)` due to compressiveforce .F. from the walls, due to elasticity, the decreasein length will be `(Deltal)_(D) = (Fl_(1))/(AY_(1)) +(Fl_(2))/(AY_(2)) =(F)/(A) ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2)))....(2)` As the lengthof the composite rod remainsunchangedthe increase in length due to heatingdue to heatingmustbe equalto decreasein lengthdue to compression. `(l_(1) alpha_(1) +l_(2)alpha_(2)) Deltat=(F)/(A) ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2)))` `RARRF=(A(l_(1)alpha_(1) +l_(2) alpha_(2)) Deltat)/( ((l_(1))/(Y_(1)) +(l_(2))/(Y_(2))))` (b)As initiallythe lengthof one rod in `L_(1)` and due to heatingits length increases by `(Delta l_(1))_(H)=alpha_(1) l_(1)Deltat`, whiledue tocompression itslengthdecreases by `(Deltal_(1))_(C )=(Fl_(1))/(AY_(1))` so its final length`l_(1)^(1) =l_(1) +(Delta l_(1))_(H) -(Deltal_(1))_(C )` `=l_(1) +l_(1)alpha_(1)Deltat-(Fl_(1))/(AY_(1))` for the other rod`l_(2)^(1) =l_(2)+(Deltal_(2))_(H) -(Deltal_(2))_(C )` `l_(2)^(1) =l_(2) +l_(2) alpha_(2)Deltat-(Fl_(2))/(AY_(2))` |
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| 33. |
As shown in fig a bob of mass m is attached by one end of string of length 1 metre and makes an angle 60^(@) with vertical. When it is released it strikes perfectly elastically with block of mass 3m (which is rest on smooth horizontal table). The height of table is also 1 metre. The range obtained by 3m block from vertical side CD of table will be ....... |
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Answer» `SQRT(2)` metre |
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| 34. |
A cylinderof radius Rmadeof materialof thermal conductivityK_(1) is surrondeb bycylindricalshellof innerradius R andouterradius 2Rmade ofmaterialof thermal conductivityK_(2) .The two ends of combinedsystemare maintainedat twodifferent temperatures. There is no lossof heatacross the cylindrical surfaceand thesystemis in steadystate. Findthe effetivethermalconductivityof the system . |
| Answer» SOLUTION :`(K_(1) + 3K_(2))/(4)` | |
| 35. |
Gravitational potential versus distance r graph is represented in fig. The magnitude of gravitational field intensity is equal to (##AKS_TRG_AO_PHY_XI_V01_B_C04_E02_109_Q01.png" width="80%"> |
| Answer» ANSWER :A | |
| 36. |
The temperature of the system decrease in the process of |
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Answer» free EXPANSION |
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| 37. |
A smooth inclined plane of length L, having an inclination theta with horizontal is inside a lift which is moving down with retardation a. The time taken by a block to slide down the inclined plane from rest will be |
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Answer» `sqrt((2L)/(sqrt(a sin theta)))` `a_(R)=(g+a) sin theta`(down the plane) From , `"" L=(1)/(2)a_(r)t^(2)"" (s=(1)/(2)at^(2))` `t=sqrt((2L)/(a_(r)))=sqrt((2L)/((g+a)sin theta))` |
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| 38. |
Two parallel rays are travelling in a medium of refractive index mu_1=4/3 One of the rays passes through a parallel glass slab of thickness t and refractive index mu_2=3/2. The path difference between the two rays due to the glass slab will be |
| Answer» SOLUTION :`DELTA X= (mu_2/mu_1-1)t=((3//2)/(4//3)-1)t=1/8` | |
| 39. |
Which of the following forces is called a conservative force ? |
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Answer» FRICTIONAL force |
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| 40. |
When a thermometer is taken from the melting ice to a warm liquids, the mercury level rises to 2//5^(th) of distance between the lower and the upper fixed points. Find the temperature of liquid in ""^@C and K |
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| 41. |
What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces? |
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Answer» Solution :Critical angle at `A=C_1` `therefore sin C_1=mu_1/mu_2=sqrt2/2=1/sqrt2, C_1=45^@` Critical angle at `B=C_2` `therefore sinC_2=mu_3/mu_2=sqrt3/2 therefore C_2=60^@` `therefore` Minimum angle of incidence for TOTAL internal REFLECTION to occur on both the slabs should be `60^@` `therefore i_min=60^@` 
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| 42. |
The modulus of elasticity is dimensionally equivalent to |
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Answer» Stress |
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| 43. |
A stone is just dropped from the window of a train moving along a horizontal straight track with uniform speed. The path of the stone is |
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Answer» a parabola for an observer STANDING by the SIDE of the track |
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| 44. |
An iron sphere of diameter 10 mm falls through a column of oil of density 940 kg//m^(3) . The density of iron is. 7.8 xx 10^(3)kg//m^(3). The coefficient of viscosity of oll is 4.48 Ns//m^(2). Caleulate the terminal velocity attained by the ball. |
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Answer» SOLUTION :Radius of the BAIL = ` a = 5 xx 10 ^(-3) m ` Density of oil `= sigma = 940 kg//m^(3) ` Density of iron sphere = ` rho = 7.8 xx 10 ^(3) kg//m^(3)` Coefficient of viscosity of oil =` eta = 4.78 Ns//m^(2)` Terminal velocity `= V= ? ` ` "" v= (2(rho - sigma)a^(2) g)/( 9eta )= ( 2(7.8 xx 10 ^(3) -940)xx 25xx 10 ^(-6) xx 9.8)/( 9xx 4. 78) ` ` "" = 0.078 m//s ` |
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| 46. |
The pendulum of a certain clock hs time period 2.04s. How fast or slow does the clock run during 24 hour? |
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Answer» 28.8 MINUTES slow |
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| 47. |
Why colours and lubricating oil have less surface tension ? |
| Answer» SOLUTION :Colours and LUBRICATING oil have LESS surface TENSION so they spread EASILY on the surface . | |
| 48. |
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 9.5). The combined mass of all the persons performing the act, and the table, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by ehich each thighbone gets compressed under the extra load. |
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Answer» Solution :Total mass of all performers, tables PLAQUES etc. =280kg Mass of the performer = 60 kg Mass supported by the legs of the performer at the BOTTOM of the pyramid `=280-60=220` kg Weight of this supported mass `=220" kg wt."=220xx9.8" N = 2156 N"`. Weight supported by each thighbone of th performer `=1//2(2156)N=1078N`. From Table 9.1, the Young.s MODULUS for bone is GIVEN by `Y""=9.4xx10^(9)"N m"^(-2)`. Length of each thighbone `L=0.5` m the radius of thighbone `=2.0cm` Thus the cross - sectional area of the thighbone `A=pixx(2xx10^(-2))^(2)m^(2)=1.26xx10^(-3)m^(2)`. Using Eq. (9.8), the compression in each thighbone `(DeltaL)` can be computed as `DeltaL=[(FxxL)//(YxxA)]` `=[(1078xx0.5)//(9.4xx10^(9)xx1.26xx10^(-3))]` `=4.55xx10^(5)" m or "4.55xx10^(-3)" cm"`. This is very small change! The fractional decrease in the thighbone is `DeltaL//L=0.000091` or `0.0091%` |
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| 49. |
A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. |
| Answer» SOLUTION :(a) `50PI rads^(-2)`, (b)200 rev | |
| 50. |
The radii of a planet and its satellites are 2r and r and their densities are rho and 2 rho respectively. Their centres are separated by a diatance d. The minimum speedwith which a body should be projected from the mid point of the line joining their centres so that the body escapes to infinity is (G - universal gravitational constant) |
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Answer» `4[sqrt((10G PI r^(3) rho)/(3D))]` |
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