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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. |
Force constants of two wires A and B of the same material are k and 2k respectively. If the two wires are stretched equally, then the ratio of workdone in stretching (( W_A)/( W_B) ) is |
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Answer» `1/4` |
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| 2. |
The following graphs represent velocity- time graph. Identify what kind of motion a particle undergoes in each graph |
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Answer» SOLUTION :(a)`vec(a) `- constant (B) `vec(V) -`constant (C)`vec(a)` - constant but greater than first graph (d) `vec(a)` - variable |
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| 3. |
If eta =(A)/(B)log (Bx + C) is dimensionally true, then (here h is the coefficient of viscosity and x is the distance) |
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Answer» C is dimensionless constant |
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| 4. |
The angular frequency of a fan of moment of inertia 0.1kg mo is increased from 30 rpm to 60 rpm when a torque of 0.03 Nm acts on it. Find the number revolutions made by the fan while the angular frequency is increased from 30 rpm to 60 rpm. |
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Answer» Solution :WORK done `=(1)/(2)I(omega_(F)^(2)-omega_(i)^(2))tautheta=(1)/(2)I(omega_(f)^(2)-omega_(i)^(2))` `theta=(1)/(2)((I)/(tau))(omega_(f)^(2)-omega_(i)^(2))=(0.1)/(2xx0.03)((2pi)^(2)-PI^(2))=5pi^(2)` The number of revolutions, `(theta)/(2pi)=(5pi^(2))/(2pi)=(5pi)/(2)=7.855rev` |
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| 5. |
Explain how geocentric theory is replaced by heliocentric theory using the idea of retrograde motion of planets. |
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Answer» Solution :(i) During the observation of motion of the planets it is in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called "retrograde motion" of planets. (ii) Ptolemy introduced the concept of "epicycle" in his geocentric model. To explain this retrograde motion, according to this theory, while the planet orbited the Earth, it also underwent another circular motion TERMED as "epicycle". A combination of epicycle and circular motion AROUND the Earth gave rise to retrograde motion of the planets with respect to Earth. (iii) But Ptolemy's model became more and more complex as every planet was found to undergo retrograde motion. (IV) According to heliocentric model, the Sun is at the centre of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is due to the relative motion of the planet with respect to Earth. The retrograde motion from the heliocentric point of view.  The Earth orbits around the Sun faster than Mars. Due to the relative motion between Mars and Earth, Mars APPEARS to move backwards from July to October. In the same way the retrograde motion of all other planet was explained successfully by the Copernicus model. The heliocentric model slowly REPLACED the geocentric model, because of its simplicity. |
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| 6. |
How is surface tension related to surface energy? |
| Answer» SOLUTION :The surface energy per unit area of a surface is numerically EQUAL to the surface TENSION. | |
| 7. |
The energy would be …. If 1 large drop of water drop splitting into 8 small drop of water. |
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Answer» `therefore` From `R_(E)=(rhovD)/(eta),R_(e)` = infinite Hence , such type of FLOW can never be a streamline. |
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| 8. |
A cricket ball of mass 100 g moving with a speedof 20 ms^(-1) is brought to restby a player . Find the changein momentum of ball. |
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Answer» `0.5 Ns ` intial VELOCITY `mu = 20 ms^(-1)` Final velocity v = 0 change in momentum in impulse= m(v - u) = 0.1 (0-20)= - 2NS |
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| 9. |
Two small balls of the same size and of masses m_(1) and m_(2) (m_(1) gt m_(2)) are tied by a thin weightless thread and dropped from a balloon. The tension T of the thread during the flight towards the ground after the motion of the balls has become steady state is (steady state means that the balls are coming down with uniform velocity due to the forces weight, air friction and upthrust). Consider that air is still without any wind below, during the motion of the balls through air: |
| Answer» Solution :`2T=(m_(1)-m_(2))g` | |
| 10. |
Motion of a coin on a carrom board is an example of |
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Answer» ONE DIMENSIONAL motion |
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| 11. |
What is water equivalent? |
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Answer» SOLUTION :WATER equivalent is numerically equal to thermal capacity of substance in CGS units and EXPRESSED in grams of water. Eg: If heal capacity of a substance is 60 CAL `""^(0)C^(+)`, then its water equivalent is 60 GRAM. |
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| 12. |
There is a small hole in a table. A string of length 1m passes through it. Two bodies of masses 70g and 100g are attached at its ends. The 100g mass hangs freely at a depth of 60 cm from the table. If this mass is to be in equilibrium, the other mass should rotate in a circle with a frequency equal to |
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Answer» `4 pi// 140 HZ` |
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| 13. |
When a fluid passes through the constricted part of a pipe, its………….. . |
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Answer» VELOCITY and PRESSURE DECREASES |
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| 14. |
The densities of two substances are in the ratio 5 : 6 and the specific heats are in the ratio 3:5 respectively. The ratio of their thermal capacities per unit volume is |
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Answer» `2:1` |
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| 15. |
A particle slides down from rest on an inclined planet of angle theta with horizontal and the distances are as shown. The particle slides down from top to the position A, where it's velocity is v. Then below correct options are |
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Answer» `(V^(2)-2gH)` will remain constant |
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| 16. |
Using the following data, calculate refractive index of the material of convex lens. Radii of curvature of both the surfacea are 0.2 m.Data recorded in the displacement//shift method to determine focal length of convex lens is shown here. |
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Answer» Solution :`f_1 = (D_1^2 - x_1^2)/(4 D_1) = (0.82^2 - 0.128^2)/(4 XX 0.82) = 0.2 m` `f_2 = (D_2^2 - x_2^2)/(4 D_2) = (0.9^2 - 0.3^2)/(4 xx 0.9) = 0.2 m` `:.` AVERAGE FOCAL length, `f - (f_1 + f_2)/(2) = (0.2 + 0.2)/(2) = 0.2 m` Now, `R_1 = 0.2 m, R_2 = -0.2 m, mu = ?` From `(1)/(f) = (mu - 1)((1)/(R_1) - (1)/(R_2))` `(1)/(0.2)=(mu - 1) ((1)/(0.2) + (1)/(0.2)) = (mu - 1)/(0.1)` `mu - 1 = (1)/(2), mu = (1)/(2) + 1 = 1.5`. |
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| 17. |
When a body is projected vertically up from ground, the ratio of its potential to kinetic energies at a height .h. above the ground is 3:4. The height above ground at which the ratio of potential to kinetic energies becomes 4:3 is |
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Answer» `3:25` |
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| 18. |
What is the angular displacement made by a particle after 5s, when it starts from rest with an angular acceleration 0.2 and s^(-2) ? |
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Answer» 4rad ` theta= omega t + 1/2 prop t^2` ` theta = 1/2 prop t^2 = 1/2 XX 0.2(5)^2 = 2.5 rad ` |
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| 19. |
Passage - II : Liquids do not have definite shape and size, they are taken in a vessel. Their expansion (or) contraction takes place with respect to the container in which they arc considered. The volume coefficient of expansion relative to the vessel is termed as coefficient of apparent expansion ( gamma_(a) ) and the volume coefficient of expansion of the liquid is termed as absolute ( gamma_(R) "or" gamma_(L ) ). In some of the arrangements mentioned as per the instructions. solve the problems. A metallic scale reads correct values at 0°C. It registers the height of liquid in the vessel (cylindrical shape) at 0°C as h_0. If gamma_( C), gamma_( L) are the coefficients of cubical expansion of metallic scale and liquid respectively and alpha_(S) is the coefficient of linear expansion of the material of the cylindrical vessel made of material different to that of the scale, then the scale reading. Registering the height of the liquid in the vessel as function of temperature ( theta ) of the surroundings is |
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Answer» `h_( theta ) = h_( 0) [ 1 - ( gamma_( C ) - gamma_( L ) - 2alpha_(S)) theta ]` |
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| 20. |
A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip istowards right hand. The bimetal strip is then heated. The bimetal strip |
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Answer» REMAINS straight |
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| 21. |
Passage - II : Liquids do not have definite shape and size, they are taken in a vessel. Their expansion (or) contraction takes place with respect to the container in which they arc considered. The volume coefficient of expansion relative to the vessel is termed as coefficient of apparent expansion ( gamma_(a) ) and the volume coefficient of expansion of the liquid is termed as absolute ( gamma_(R) "or" gamma_(L ) ). In some of the arrangements mentioned as per the instructions. solve the problems. A thread of liquid in a uniform capillary tube of length L, as measured by a ruler. The temperature of the tube and the thread of the liquid is raised by 'Delta theta'. If gamma is the coefficient of volume expansion of the liquid and alpha is the coefficient of linear expansion of the material of the tube, then the increase in length 'Delta L' of the thread as measured by the ruler will be |
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Answer» `Delta L = L ( gamma - alpha) Delta THETA` |
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| 22. |
The sum of magnitudes of two forces acting at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N. Then the forces are |
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Answer» SOLUTION :LET `vec(F)` be the resultant of two forces `vec(F)_(1)` and `vec(F)_(2)` as shown in figure with `F_(2) gt F_(1)`. `F_(2) sin theta = F_(1)` ….(i) `F_(2) cos theta = F = 8` …..(ii) Squaring and adding EQS (i) and (ii), we get `F_(2)^(2) = F_(1)^(2) + 64`.....(iii) Given `F_(1) + F_(2) = 16`.....(iv) SolvingEqs. (iii) and (iv), we get `F_(1) = 6N` and `F_(2) = 10 N` |
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| 23. |
A particle is projected from the ground with an initial speed v at an angle theta with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is |
Answer» SOLUTION : R/2 `vecv_(avg) = (vecv+vecu)/(2) = (u COS theta HAT(i) +(u cos theta hat(i)+u sin theta hat(J)))/(2)` ` v_(av) = (v)/(2)SQRT(1+3cos^(2)theta)` |
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| 24. |
Passage - II : Liquids do not have definite shape and size, they are taken in a vessel. Their expansion (or) contraction takes place with respect to the container in which they arc considered. The volume coefficient of expansion relative to the vessel is termed as coefficient of apparent expansion ( gamma_(a) ) and the volume coefficient of expansion of the liquid is termed as absolute ( gamma_(R) "or" gamma_(L ) ). In some of the arrangements mentioned as per the instructions. solve the problems. The height of the liquid column in the cylindrical container is h_0 at 0^(@) C. If the temperature of the liquid raises to 0^(@) C, assuming the coefficient of volume expansion of liquid and solid as gamma_(L ) and gamma_(S) respectively, the height of the liquid in the vessel ( h_( theta )) as a function of temperature ( theta) is |
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Answer» `h_( theta ) = h _(0) [ 1+ ( gamma_(L)- gamma_(s) ) theta ]` |
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| 25. |
The component of a vector is |
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Answer» ALWAYS LESS than its MAGNITUDE |
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| 26. |
What would a person hear, if he moves away from a source of sound with the speed of sound? |
| Answer» SOLUTION :Nothing. The RELATIVE velocity of sound WAVES with respect to PERSON is zero. | |
| 27. |
A moving particle of mass.m. makes head-on elastic collision with a particle of mass .2m. which is initialy at rest. The fraction of K.E. lost by colliding particle is: |
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Answer» `1//3` |
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| 28. |
Five vectors vecA, vecB, vecC,vecD and vecE have magnitude10, 12 sqrt(2), 20, 20 and10 unitrespectively, they aredireacted as shown in thefig. 3.58 Answer the following questions : |
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Answer» `vecC = - sqrt(3) 10hati + 10 hatj" ", "" vecp= 10hati-10sqrt(3) hatj` `VECD = -10hatj` `(vecA + vecB +vecC +vecD + VECE)=(11-10sqrt(3)) hati + (12-10sqrt(3))hatj` X COMPONENT = `12-10sqrt(3)` y component = `12-10sqrt(3)` |
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| 29. |
A cyclist is riding with a speed of 27 km /h . As he approaches a circular turn on the road of radius 80 m , he applies brakes and reduces his speed at the constant rate of 0.50m//s every second . Whatis the magnitude and direction of thenet acceleration of the cyclist on the circular turn ? |
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Answer» |
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| 30. |
Five vectors vecA, vecB, vecC,vecD and vecE have magnitude10, 12 sqrt(2), 20, 20 and10 unitrespectively, they aredireacted as shown in thefig. 3.58 Answer the following questions : |
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Answer» `VECC = - SQRT(3) 10hati + 10 hatj" ", "" vecp= 10hati-10sqrt(3) hatj` `VECD = -10hatj` `(vecA + vecB +vecC ) = (22-10sqrt(3)) hati + 22hatj` x COMPONENT = `22-10sqrt(3)` y component = `22` |
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| 31. |
The cycle is shown in figure is made of one mole of perfect gas in a cylinder with a piston . The processes A to B and C to D are isochoric wheras process B to C and D to A are adiabatic, the work done in one cycle is (V_(A)=V_(B)=V,V_(C)=V_(D) =2V "and " gamma=5//3) |
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Answer» `[1-4^(3)/(2)](P_(B)-P_(A))V` |
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| 32. |
Five vectors vecA, vecB, vecC,vecD and vecE have magnitude10, 12 sqrt(2), 20, 20 and10 unitrespectively, they aredireacted as shown in thefig. 3.58 Answer the following questions : |
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Answer» `VECC = - sqrt(3) 10hati + 10 hatj" ", "" vecp= 10hati-10sqrt(3) hatj` `VECD = -10hatj` `(vecA + vecC+ vecD) = - 10sqrt(3)hati + (10-10sqrt(3))hatj` x component = `-10sqrt(3)` y component = `10-10sqrt(3)` |
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| 33. |
An iron needle slowly placed on the surface of water floats on its because |
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Answer» when inside in water it will displaced water more than its weigths |
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| 34. |
What is the differences between accuracy and precision? |
| Answer» Solution :ACCURACY means the extent to which a MEASURED value agress with the TRUE value. But precision means the extent to which a given set of measurements of the same QUANTITY agree with their MEAN value. Precise measurement need not be accurate. | |
| 35. |
A ray of light is incident is 60^@ on a prism of refracting angle 30^@. The emerging ray is at an angle 30^@ with the incident ray. The value of refractive index of the prism is: |
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Answer» `SQRT3/4` |
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| 36. |
In which of the following cases the net force is not zero ? |
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Answer» A kite skillfully held STATIONARY in the sky |
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| 37. |
Internal energy, volume and mass are intensive variable while pressure, temperature and density are extensive variables. |
| Answer» SOLUTION :FALSE | |
| 38. |
A body freely falling from a height h describes 7h/16 in the last second of its fall. The height h is (g=10ms^(-2)) |
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Answer» 80 m |
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| 39. |
The unit of the quantity g/G in SI will be |
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Answer» `KGM^(-2)` |
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| 40. |
A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in fron of the tower C and spots a very distant objct O in line with AC. He then walks pependicular to AC up to B, a distance of 100 m and look at O and C again. Since O is very distant the direction BO is practivally the same as AO, but he finds the line of slight by of C shifted fro the original line of slight by an angle theta=30^(@) (theta is known as parallax) estimate the distance of the tower C from his original position A. |
Answer» SOLUTION : Parallax ANGLE `theta=40^(@)` `"tan" theta=(AB)/(AC)` `:.AC=(AB)/(tan theta)` `=(100)/(tan 40^(@))` `=(100)/(0.8391)` `=119.175` `:.AC~~119 m` |
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| 41. |
ABC is a right angled triangular plate of uniform thickness. The sides are such that AB gt BC as shown in figure. I_(1),I_(2) and I_(3) are moments of inertia about AB,BC andAC respectively. Then, which of the following relation is correct ? |
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Answer» `I_(1)=I_(2)=I_(3)` The mass is farthest from axis BC, so `I_(2)` is maximum. Mass is nearest to axis AC, so `I_(3)` is minimum. Hence, the CORRECT sequence will be `I_(2) gt I_(1) gt I_(3)` |
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| 42. |
Two closed organ pipes of length 100 cm and 101 cm produce 16 beats in 20 sec. When each pipe is sounded in its fundamental mode calculate the velocity of sound in m/s. |
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Answer» 303 `f _(1) = (v)/(4L) = (v)/( 4 xx 100) = (v)/(400)` For second closed, pipe, `f._(1) = (v)/(4L), = (v)/(4 xx 101) =(v)/(404)` `IMPLIES f _(1) gt f _(1)` Here, `f _(1) -f ._(1)= (16)/(20)= ("TOTLA no. of BEATS")/(total time")` `therefore (v)/(400)-(v)/(404) = (16)/(20)` `therefore v ((404 - 400)/(400 xx 404))= (16)/(20)` `therefore v = 32320 cm //s =323.2 m//s` |
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| 43. |
In an experiment to measure density of a cube percentage error in measurement of mass and length of sphere is 0.26% and 0.38% respectively, then find % error in measurement of density. |
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Answer» 14% VOLUME of cube `=l^(3)` `:.` % error in measurement of density, `(Deltarho)/(rho)xx100=(DeltaM)/(M)xx100+3(Deltal)/(l)xx100` `:.(Deltarho)/(rho)xx100=0.26+3xx0.38` `=0.26+1.14=1.40%` |
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| 44. |
A force vecF = alpha hati + 3hatj + 6hatk is acting at a point vecr = 2hati - 6hatj - 12hatk . The value of alpha for which angular momentum about origin is conserved is |
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Answer» zero By definition, `vectau = vecrxx vecF` Here `vecr = 2hati- 6hatj - 12 hatk` and `vecF = alpha hati + 3hatj + 6hatk` `therefore vectau = |{:(hati , HATJ , hatk) , (2 , -6 , -12) , (alpha , 3 , 6):}|` `= hati (-36 + 36) - hatj (12 + 12 alpha) + hatk (6 + 6 alpha)` `=- hatj (12 + 12 alpha) + hatk (6 + 6 alpha)` But `vectau = 0` `therefore 12 + 12 alpha = 0 or alpha = -1` and `6 + 6 alpha = 0` and `alpha = -1` |
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| 45. |
For an ideal gas the molar heat capacity varies as C = C_v +3a T^2.Find the equation of the process inthe variables (T,V) where a is a constant. |
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Answer» |
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| 46. |
(A) When a ball hits a wall (stationary) it returns back with the same velocity, if collision is elastic. (B) A ball hits a bus moving towards it. If the collision is inelastic, they move together. (C ) Inelastic collision arises due to loss in K.E during collision |
| Answer» Answer :C | |
| 47. |
If the sum of two unit vectors is also a unit vector. Find the magnitude of their difference? |
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Answer» Solution :Let`HATA "and" HATB` are the given UNIT vectors and `hatR` is their RESULTANT then `|hatR|=|hatA+hatB|` `1=sqrt((hatA)^(2)+(hatB)^(2)+2|hatA||hatB| COS theta), 1=1 +1+2 cos theta rArr cos theta =-1/2` `|vecA-vecB|=sqrt((hatA)^(2)+(hatB)^(2)-2|hatA||hatB|cos theta)=sqrt(1+1-2xx1xx1(-1/2))=sqrt3` |
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| 48. |
Circular disc of mass 2 kg and radius 1 metre is rotating about an axis perpendicular to its plane and passing through its centre of mass with a rotational kinetic energy of 8Joules. The angular momentum in(J-sec) is |
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Answer» 8 |
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| 49. |
(i) For the wave on a string described in Exercise 15.11. do all the points on the string oscillate with the same (a) frequency , (b) phase , (c ) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end? |
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Answer» Solution :Here given stationary wave is `y = 0.06 SIN ((2pi)/(3)x) cos (120 pi t ) ""…(1)` Comparing above equation with, `y = 2a sin (kx) cos (omegat)""…(2)` we get ,` omega 2pi f = 120 pi impliesf = 60 Hz` which is same for all the particles of a string. In case of stationary wave, PHASES of oscillation for particles in any given loop are same because they oscillate simulataneously PERPENDICULAR to string. (In other words, phase differentce between any two particles in any oop is ZERO). In case of stationary wave AMPLITUDES of all the particles in a given string are not same because amplitude of oscillation of a given particle at distnace x from one end of string is, `A =2a sin (kx)` `implies` Amplitude of oscillation of a particle depends on its position. Here,` A = 3 a sin (kx)` `= 0.06 sin ((2pi)/(3) xx 0.375) = 0.067 sin (0.25pi)` `=0.06 sin ((pi)/(4))` `=0.06 xx 0.7071` `therefore A = 0.04343 m` |
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