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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) : Smaller the orbit of the planet around the sun, shorter is the time it taken to complete one revolution. (R) : According to Kepler's third law of planetary motion, square of time period is proportional to cube of mean distance from sun. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 2. |
Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is.......... |
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Answer» `x (t) = B SIN ((2pi t)/(30))` Here `theta = omega t` In right triangle `triangleORQ` `sin theta = (OR)/(OQ)` `OR = OQ sin theta` `=r sin omega t"[Radius of circle OQ = r and " theta = omega t]` OR is the X-component `therefore x(t) = B sin ((2pi)/(T)t)""[therefore r= B, omega = (2pi)/(T)]` `therefore x(t) = B sin ((2pi)/(30)t)`. 
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| 3. |
An object is placed at A (OA gt f). Here f is the focal length of the lens. The image is formed at B. A perpendicular is erected at 0 and c is chosen such that angle BCA=90^@. Let OA =a, OB=b and OC=c. Then the value of f is |
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Answer» SOLUTION :From `1/v-1/u=1/F` we have `1/b+1/a=1/f or f=(AB)/(a+b)`…….(1) Further `AC^2+BC^2=AB^2 or (a^2+c^2)+(b^2+c^2)=(a+b)^2` or `a^2+b^2+2c^2=a^2+b^2+2ab, ab=c^2` Substituting this is Eq. (1) we get `f=c^2/(a+b)` 
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| 4. |
The resultant of two vectors vecA "and"vecB is perpendicular to vecA. Magnitude of Resultant vecR is equal to half magnitude of vecB . Find theangle between vecA"and"vecB? |
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Answer» SOLUTION :Since `VECR ` is perpendicular to `vecA `. Figure SHOWS the THREE vectors `vecA,vecB"and vecR`. Angle between `vecA`and `vecB` is `pi-theta` `sintheta =R/B=B/(2B)=1/2` `rArr theta rArr 30^(@)rArr "angle "vecA `between and `vecB` is `150^(@)` 
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| 5. |
The radius of mercury drop at 20^(@)C is 3mm. If the surface tesion of mercury at this temperature is 4.65xx10^(-1)Nm^(-1). Find the excess presure inside the liquid drop. |
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Answer» Solution :Excess PRESSURE `(P)=(2S)/R` Surface tension `(S)=4.6xx10^(-1)Nm^(-1)` radius of the MERCURY drop `r=3=mm=3xx10^(-3)m` `:.P=(2S)/r=(2xx4.65xx10^(-1))/(3xx10^(-3))=310Nm^(-2)` |
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| 6. |
A bullet of mass 5gm is fired horizontally into block of mass 1kg at rest on a horizontal surface and gets embedded in it. The combination moves 5m before coming to rest. If mu_(k)=0.40, what is the speed with which the bullet strikes the block ? |
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Answer» Solution :Mass of bullet `=m=5gm=0.005kg` Mass of block `=M=1kg` `mu_(k)=0.4` DISTANCE travelled `=s=5m` The KINETIC energy of the bullet is used to do work against the force of FRICTION between the block and the horizontal surface `(1)/(2)mv^(2)=Fs=mu_(s)Ns` But `N=(m+M)g` `(1)/(2)mv(2)=mu_(4)(m+M)g` `v=sqrt((2mu_(g)(m+M)g)/(m))=sqrt(((2xx0.4xx1.005xx9.8))/(0.005))=39.7m//s` |
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| 7. |
Discuss the factors on which amount of thermal energy passing perpendicularly between two nearby cross-section depends. Hence obtain expression for heat current. |
Answer» Solution :Figure SHOWS a rectangular block of solid of cross section .A..  Let us consider two cross sections ABCD and EFGH, at a distance x and `DELTAX` from one end of a block having temperature `T+DeltaT` and T respectively. Here, the distance between two cross section is `Deltax` and temperature difference is `DeltaT`. The ratio `DeltaT//Deltax` is called temperature GRADIENT Temperature gradient : The difference in temperature per unit distance between two parts of a solid is called distance between two parts of a solid is called as temperature gradient parts of a solid is called as temperature gradient. Its unit is Kelvin/meter and dimensional formula is `M^(0)L^(-1)T^(0)K^(1)`. Experiment shown that for a small change in temperature, the amount of heat passing between these two cross sectional planes in direction perpendicular to these planes in time `Deltat` depends on the following FACTORS : `DeltaQprop(DeltaT)/(Deltax)"(For given "Deltat" and A)"` `DeltaQpropDeltat" (For given "DeltaT//Deltax" and A)"` `DeltaQpropA" For given "Deltat " and "DeltaT//Deltax")"` `:.DeltaQ=-kA(DeltaT)/(Deltax)Deltat` `(DeltaQ)/(Deltat)=-kA(DeltaT)/(Deltax)` . . .(1) Here, k is constant of proportionality called thermal conductivity of the material of the block. Its value depends on the type of the material and to some extent on the temperature. The materials with higher value of thermal consuctivity are GOOD conductors of heat. Normalally, if the differnces in the temperatures of different parts of solid are not very large, k can be considered a constant. Negative sign in above equation indicates that with the increase in x the temperature decreases. Taking limits `Deltax to 0` and `Deltat to0` in equation (1) `(dQ)/(dt)=-kA(dT)/(dx)`. . . (2) `:.H=-ka(dQ)/(dt)` `(dQ)/(dt)=H` is called heat current. Heat current is the rate of flow of heat through any cross section. Its unit is cal/s or J/s, its dimensional formula is `M^(1)L^(2)T^(-3)`. |
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| 8. |
The tidal Waves in the sea are primarily due to |
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Answer» The gravitational EFFECT of the MOON of the earth |
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| 9. |
A simple pendulum consists of a bob suspended by inextensible and massless thread of length 1 m from a point O. When the bob is at extreme position of oscillation, the thread is suddenly caught by a bolt at a point A located at a distance 0.3 m from O. Now, the bob begins to oscillate in the new condition. Calculate the approximate change in time period of the pendulum. Take, g = 10 ms^(-2). |
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Answer» Solution :Time period , `T = 2pi sqrt((l)/(g))` As LENGTH,`l = 1 m, T = (2pi)/(sqrt(g))` When the thread is caught by the BOLT at A, the length of the pendulum becomes `l. = 1 - 0.3 = 0.7 m` New time period, `T. = 2pi sqrt((0.7)/(g)) = 2pi sqrt((7)/(10g))` CHANGE in time period, `T = T. = 2pi [(1)/(sqrt(g)) - sqrt((7)/(10g))] = (2pi)/(sqrt(g)) [1-sqrt(0.7)] = 3.162` |
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| 10. |
The direction of angular velocity vector is along: |
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Answer» the TANGENT to the circular path |
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| 11. |
While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a poistion where it can stay in equilibrium. In this condition, distance of the piston from the top is |
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Answer» `(2P_(0)PHI^(2))/(phir^(2)P_(0)+MG)(2L)` |
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| 12. |
In the circuit shown in figure |
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Answer» V = 10 V |
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| 13. |
In the above problem the time after which the bodies are chosest to each other is |
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Answer» `(VL)/(U^(2)+v^(2))` |
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| 14. |
In a machine, a horizontal arm of 2 m is connected with a vertical shaft and is rotated at an angular velocity of 600 rpm in a horizontal plane. A 10 kg block is attached to the free end of the arm. Find out the force on the block and the force on the vertical shaft due to rotation of the block. |
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Answer» |
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| 15. |
A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x * 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is |
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Answer» 45 J  Work done W = Area under F-X GRAPH with PROPER sign W = Area of triangle ABC + Area of rectangle CDEF +Area of rectangle FGHI + Area of rectangle IJKL `W = [1/2 xx 6 xx 10] + [4 xx (-5)] + [4 xx (5)] + [2 xx (-5)]` `= 30 - 20 + 20 - 10 = 20 J` According to work energy theorem, `K_f - K_i = W` or `(K_f)_(x = 16m) -(K_i)_(x = 0m) = W` or `(K_f)_(x = 16m) = (X_i)_(x = 0m) = W = 25J + 20 J = 45J` . |
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| 16. |
A Particle slides down a frictionless parabolic (y=x^(2)0 track(AtoBtoC) starting from rest at a point A as shown in fig. Point B is at the vertex of the parabola and point C is at a height less than that of point. A. After C, the particle moves freely in air as projectile. If the particle reaches heightest point at P, then which of the following are not correct. |
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Answer» K.E. at P=K.E. at B |
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| 17. |
The r.m.s speed of the molecules of a gas in a vessel is 400 ms""^(-1). If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be |
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Answer» `800ms^(-1)` |
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| 18. |
A ball ofmass M is hanging from the lower end of a string of length l. a bullet of mass m moving horizontally hits the ball and sticks to it. What should be the minimum velocity of the bullet so that the ball will be able to complete a full circle ? |
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Answer» |
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| 19. |
Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between road and tyres is 0.5, the shortest distance in which the car can be stopped is |
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Answer» 30 m Work done against FRICTION = `mu mg s` From work - energy theorem `mu m g s = 1/2 mv^2 " or " s = ((V^2)/(2 mu g))` Stopping distance , `s = ((v^2)/(2 mu g))` Given, `v = 72 km//h = 72 xx 5/18 = 20 m//s` `mu = 0.5 and g = 10 ms^(-2)` `:. s = (20 xx 20)/(2 xx 0.5 xx 10) = 40 m` . |
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| 20. |
Six identical particles each of mass m are arranged at the corners of a regular hexagon of side length a. If the mass of one of the particle is doubled, the shift in the centre of mass is |
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Answer» a |
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| 21. |
A cylinderical tank having cross sectional area ^^=0.5m^(2) is filled liquids of densities rho_(1)=900kgm^(3)&rho_(2)=600kgm^(3) to a height h=60cm as shown in the figure a small hole having area a=5cm^(2) is made in right vertical wall at a height y=20cm from the bottom calculate. (i). velocity of efflux (ii). horizontal force F to keep the cylinder in static equilibrium if it is placed on a smooth horizontal plane (iii). minimum and maximum value of F to keep the cylinder at rest. The coefficient of friction between cylinder and the plane is mu=0.1 (iv). velocity of the top most layer of the liquid column and also the velocity of the boundary separating the two liquids. |
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Answer» (II). `F=7.2N` (III). `F_(min)=0,F_(max)=52.2N` (IV). Both `4XX10^(-3)m//s` |
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| 22. |
One end of a uniform brass rod 20 cm long and 10cm^(2) cross-sectional area is kept at 100^(0)C. The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360 g of ice melts per hour. The thermal conductivity of the rod, given that the thermal conductivity of brass is 0.25 cal/s cm ""^(0)C and L = 80 cal/g, is 0.111xxK" ca"l//"cm-s-"^(0)C. Then K is equal to |
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Answer» |
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| 23. |
A cat is following a rat. The rat is running with a constant velocity u. The cat moves with constant speed v with her velocity always directed towards the rat. Consider time to be t = 0 at an instant when both are moving perpendicular to each other and separation between them is L. |
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Answer» Find acceleration of the cat at `t = 0` |
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| 24. |
Ten moles of a diatomic perfect gas are allowed to expand at constant pressure. The initial volume and temperature V_0and T_0respectively. If 7/2 = RT_0heat is transferred to the gas then the final volume and temperature are |
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Answer» `1.1V_0, 1.1 T_0` |
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| 25. |
In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures, t_(1) and t_(2). The liquid columns in two arms have heights I_(1) and I_(2) respectively. The coefficient of volume expansion of the liquid is equal to |
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Answer» `(I_1+I_2)/( (I_2 t_(1) - t_(1) I_(2) )` |
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| 26. |
The change in temperature of a body is 50^(@)C. The change on the kelvin scale is ……………. . |
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Answer» 50 K |
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| 27. |
A particle is moving in x-y plane withy = x/2 and v_(x) = 4-2t.The displacement versus time graphof the particle would be |
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Answer» |
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| 28. |
In a circus , the diameter of globe of death is 30 m. From what minimum height must a cylist start in order to roll down the inclined andgo round the globe successfully? |
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Answer» |
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| 29. |
("M")/("Vr") has the dimensions of (M = Magnetic moment, V = Velocity, r = radius) |
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Answer» POLE strength |
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| 30. |
The initial temperature of a gas is 100^(@)C. The gas is contained in closed vessel. If the pressure on the gas is increased by 5% calculate the increase in temperature of the gas |
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Answer» `1^(@)C` `T_(1)//T_(2) = P_(1)//P_(2)` `(100)/((100-DeltaT)) = (P)/(1.05P)` `105P =100P+P*Delta T` `(5P)/(P)=Delta T = 5^(@)C`. |
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| 31. |
What is the angle between velocity vector and acceleration vector in uniform circular motion? |
| Answer» SOLUTION :`90^(@)` | |
| 32. |
A shot putter with a mass of 80 kg pushes the iron ball of mass of 6 kg from a standing position accelerating it uniformly from rest at an angle of 45^(@) with the horizontal during a time interval of 0.1 seconds. The ball leaves his hand when it is 2m high above the level, ground and hits the ground 2 seconds later. The horizontal distance between the point of release and the point where the ball hits the ground : |
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Answer» 16 m |
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| 33. |
(A) The minimum distance between an object and its real image formed by a convex lens is 4l (R ) The distance between an object and its real image is minimum when its magnification is one. |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION of A |
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| 34. |
A shot putter with a mass of 80 kg pushes the iron ball of mass of 6 kg from a standing position accelerating it uniformly from rest at an angle of 45^(@) with the horizontal during a time interval of 0.1 seconds. The ball leaves his hand when it is 2m high above the level, ground and hits the ground 2 seconds later. The acceleration of the ball in shot putter's hand : |
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Answer» `11 SQRT(2) m//s^(2)` |
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| 35. |
A system consists of three particles located at the corners of a right triangle as shown in the figure. Fin the position vector of centre of mass of the system. |
Answer» Solution : Using the equation `X_(c )=(summ_(i)x_(i))/(M)=(2md+m(B+d)+3m(d+b))/(6M)=d+((2)/(3))b` `Y_(c )=(Summ_(i)y_(i))/(M)=(2m(0)+m(0)+3mh)/(6m)=(H)/(2)` `Z_(c )=0`, because the particles are in X-Y plane we can express the position of centre of mass from the origin using a position vector as `vecr_(c )=X_(c )hati+Y_(c )hatj+Z_(c )hatk,vecr_(c )=(d+(2)/(3)b)hati+(h)/(2)hatj` |
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| 36. |
A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is T_(i) K and the final temperature is alphaT_(f)K then determine the value of alpha. |
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Answer» |
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| 37. |
A shot putter with a mass of 80 kg pushes the iron ball of mass of 6 kg from a standing position accelerating it uniformly from rest at an angle of 45^(@) with the horizontal during a time interval of 0.1 seconds. The ball leaves his hand when it is 2m high above the level, ground and hits the ground 2 seconds later. The minimum value of the static coefficient of frictin if the shot putter does not slip during the shot is closet to : |
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Answer» 0.28 |
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| 38. |
An ideal fluid is flowing through four tube a, b,c, d of radii in the ratio 2:7:3:1 with velocities in the ratio 1:2:5:15 when maintained at different pressures. The ascending order of the amount of fluid flowing through the tubes per second is |
| Answer» Answer :A | |
| 39. |
A ball of mass m is just disturbed from the top of a fixed smooth circular tube of radius a in a vertical plane and falls impinging on a ball of mass 2m at the bottom. The coefficient of resitution is 1/2. Find the heights to whichthe balls rise after a second impact. |
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Answer» `(a)/(2), (a)/(8)` |
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| 40. |
The force required to move a body up a rough inclined plane is double the foce required to prevent the body from slinding down the plane. The coefficient of friction, when the angle of plane is 60^(@) is |
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Answer» `1/3` |
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| 41. |
If 10^(20) oxygen molecules per second strike 4cm^2 of wall at an angle of 30^@ with the normal when moving at a speed of 2xx10^3ms^(-1) , find the pressure exerted on the wall . (mass of 1 atom =1.67xx10^(-27)kg) |
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Answer» Solution :`{:("Mass of"),("ONE atom"):}} = 1.67xx10^(-27)kg` `{:("Mass of"),("oxygen atom "):}} = 1.67xx10^(-27)xx16` `=26.72xx10^(-27)kg` `{:("Momentum of "),("a molecule P"):}} = mv` Velocity = `2xx10^3ms^(-1)` `P = 26.72xx10^(-7) xx8xx2xx10^3` `=427.5xx10^(-4) kg MS^(-1)` Component of momentum normal to WALL, `P_C=P COS theta` Angle `theta=30^@` `P_C=427.5xx10^(-4)cos30^@` `=427.5xx10^(-4)xxsqrt3/2` `=370.2xx10^(-4)kg ms^(-1)` Pressure `=("Force")/("Area")` `= ("CHANGE in momentum")/("Area")` `=(370xx10^(-4))/(4xx10^(-4))` `=92.55Nm^(-2)` `:.` Pressure `=92.55Nm^(-2)` |
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| 42. |
Assertion : Areal velocity of a planet around of surface area and density is same for two planets, escape velocities will be same for both Reason : Areal velocity = (L)/(2m), Where L is angular momentum of planet about centre of sun. |
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Answer» If both Assertin and Reason are CORRECT and Reason is the correct explanation of ASSERTION i.e., `(dA)/(dt)` is independent of `m`. |
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| 43. |
(I ) Ina cyclieprocess changein internalenergyyiszero(ii) C_(V ) gtC_(p) whichone isincorrectstatement ? |
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Answer» 1 only |
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| 44. |
Discuss Newton’s formula for velocity of sound in air medium and apply Laplace’s correction. |
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Answer» Solution :Newton's assumed that when sound propagates in ari, the formation of compression and rarefaction takes PLACE in a very slow manner so that the PROCESS is isothermal in nature. It is found that, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (Pressure decreases, volume increases)occure over a period of time such in a wavy that the temperature of the medium remainsconstant. Hene, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle's law, Mathematically `PV="chonstant" ""..(1)` Differentiating equation (1) we get `PdV+VdP=0` (or) `P-V(dP)/(dV)=B_(T)""...(2)` Where `B_(T)` is an isothermal bulk modulus of air. `v= sqrt((B)/(rho))""...(3)` Substituting equatio (2) in equation (3), the speed of sound in air is `v_(T)= sqrt((B_(T))/(rho))= sqrt((P)/(rho))` Since P is the pressure of air whose value at NTP (Norma Temperature and Pressure) is 76 cm of mercury, we have `P=(0.76xx13.6xx10^(3)xx9.8)Nm^(-2)` `rho=1.239 kg m^(-3)`. Here `rho` is density of ari. Then the speed of sound in air at NORMAL temperature and pressure (NTP) is `v_(T) = sqrt(((0.76xx13.6xx10^(3)xx9.8))/(1.293))` `=270.80 ms^(-1)` `=280 ms^(-1)` (theoretical value)But the speed of sound in air at `0^(@)C` is experimentally observed as `332 ms^(-1)` that is close upto 16 % more than theoretical value. Lapace correction : Laplace satisfactorily CORRECTED this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occure very fast. Hene the exchange of heat produed due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a poor conductopr of heat. Since, temperature is no longer considered as a constant here, propagation of sound is an adiabatic process. By adiabatic considerations, the gas obeys Poisson's law (as Newton assumed), that is `PV^(gamma)= "Constant" ""....(4)` where `gamma=(C_(P))/(C_(v))`, that is the ration between specific heat at constant pressure and specific heat at constant volume. Differentiating equation (4) on both the sides, we get `V^(gamma)dP+P(gammaV^(gamma-1)+dV)=0` Or , `gamma P=-V(dp)/(dV)=-B_(A) ""...(5)` where `B_(A)` is the adiabatic bulk modulus of air. `v= sqrt((B)/(rho)) ""...(6)` Now, substituting equation (5) in equation (6), the speed of sound in air is `v_(A)= sqrt((B_(A))/(rho))= sqrt((gammaP)/(rho))` `= sqrt(gamma)v_(T)` Since air contains mainly, nitrogen, oxygen, hydrogenetc. ( diatomic gas ) , we take ` gamma = 1.47`. Hence, speed of sound in air is `v_(A)=(sqrt(1.4))(280 ms^(-1)=331.30 ms^(-1)` , which is very much closer to experimental data . |
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| 45. |
What is absolute zero ? |
| Answer» SOLUTION :The temperature at which the VOLUME of given mass of GAS REDUCES to zero at CONSTANT pressure (or) he temperature at which the pressure of given mass of gas reduces to zero at constant volume. | |
| 46. |
Find the power of an engine which can draw a train of 400 metric ton up the inclined plane of 1 in 98 at the rate of 10 ms^(-1). The resistance due to friction acting on the train is 10N per ton. |
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Answer» `350 kW` |
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| 47. |
Does the kinetic energy ofa\boay moving ipa vertical circular path also remain constant ? |
| Answer» SOLUTION :No. Because the SPEED is DIFFERENT at different POINTS of the CIRCLE. | |
| 48. |
A gas may expands adiabatically or Isothermally. If P,V graphs are drawn and if s_(1) and s_(2) are the slope of the process then a)s_(1)gts_(2) b) adiabatic curve and Isothermal curve may intersect c) s_(1)lts_(2) |
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Answer» both a and B are CORRECT |
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| 49. |
A piece of wood of density 0.88 gm cm ^(-3) floats in benzene of density 0.9 gmcm^(- 3) at 0^(@)C. At what temperature does the piece of wood sink in Benzene? (Coefficient of expansion of wood gamma _(w) = 8.4 xx 10 ^(-5) // ^(@)Cand Coefficient of expansion of benzene gamma _(B) = 1.2 xx 10 ^(-3) // ^(@)C) |
| Answer» SOLUTION :`20.4^(@)C` | |
| 50. |
If c is rms speed of molecules in a gas and v is the speed of sound waves in the gas, show that (c)/(v) is constant and independent of temperature for all diatomic gases. |
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Answer» <P> Solution :rms speed of MOLECULE of a gas,`c = sqrt((3 P )/( rho ))` (By fourmula) `= sqrt (( 3 PV )/( M ))` `= sqrt ((3 mu RT)/( mu M _(0)))` `c = sqrt (( 3 RT)/( M_(0))) ""...(1)` (Where `M _(0)=` molecular WEIGHT of a given gas, R = gas constant) Speed of sound in a gas, `v = sqrt (( gamma P )/( rho ))= sqrt (( gamma P V )/( M )) = sqrt (( gamma mu RT )/( mu M _(0))) = sqrt (( gamma RT)/( M _(0))) ""...(2)` Taking ratio of equation (1) and (2), `c/v = sqrt ((3)/( gamma )) = sqrt ((3)/( 7//5)) = sqrt ((15)/( 7)) = ` constant (Where `gamma = (C _(P))/( C _(V)) = 7/5 ` for diatomic GASES) `implies c/v` does not depend on temperature. |
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