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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. |
What is the difference between sliding and slipping ? |
Answer» SOLUTION :
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| 2. |
On a two-lane road, car A is travelling with a speed of 36 km h^(-1). Two cars B and C approach car A in opposite directions with a speed of 54 km h^(-1)each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. The minimum acceleration of car B is required to avoid an accident is |
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Answer» `1m//s^2` |
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| 3. |
A particle of mass m is moving in a circular path of radius r with a uniform speed v. The centripetal force on the body is (m omega^(2))/( r). When the particle is displace half of the distance, then what is the work done by the centripetal force? |
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| 4. |
Apply Newton's second law to a mango hanging from a tree. (Mass of the mango is 400 gm) |
Answer» Solution : Note: Before applying Newton.s laws, the following STEPS have to be followed: 1. Choose a suitable inertial COORDINATE system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system. 2. Identify the system to which Newton.s laws need to be applied. The system can be a single object or more than one object. 3. Draw the free body diagram. 4. Once the forces acting on the system are identified, and the free body diagram is drawn, APPLY Newton.s second law. In the left hand side of the equation, write the forces acting on the system in vector notation and equate it to the right hand side of equation which is the product of mass and acceleration. Here, acceleration should also be in vector notation. 5. If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated. By following the above steps: We fix the inertial coordinate system on the ground as shown in the figure.  The forces acting on the mango are (i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis. (ii) Tension (in the cord attached to the mango) acts upward alongpositive y-axis. The free body diagram for the mango is shown in the figure ` vecF_g = mg(-hatj)=- mg hatj ` Here, mg is the magnitude of the gravitational force and (`-hatj` ) represents the unit vector in negative y-direction `VECT = T hatj` Here T is the magnitude of the tension force and (`hatj`) represents the unit vector in positive y direction. `vecF_("net") = vecF_(g) + vecT = - mg hatj + T hatj = (T = mg) hatj` From Newton.s second law`vecF_("net") = m veca ` Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero `(veca = 0)` . So,`F_("net")= mvec a = 0` `(T - mg) hatj = 0` By comparing the components on both sides of the above equation, we get T - mg = 0. So the tension force acting on the mango is given by T = mg. Mass of the mango m= 400 g and `g = 9.8 m s^(-2)` Tension acting on the mango is `T=0.4 xx 9.8 = 3.92 N` |
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| 5. |
A 2 m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radians at the other end. The shear strain developed will be |
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Answer» `0.002` |
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| 6. |
The moment of inertia of a rod about an axis through its centre and perpendicular to it is (1)/(12)ML^(2) (where M is the mass and L, the length of the rod). The rod is bent in the middle so that the two halves make an angle 60^(@). The moment of inertia of the bent rod about the same axis would be.......... |
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Answer» `(1)/(48)ML^(2)`  Moment of inertia of UNIFORM rod about ONE end = `(ML^(2))/(3)` Moment of inertia of a rod about an AXIS passing through O and perpendicular to the plane of the rod will be `I.=2((M)/(2))(((L)/(3))^(2))/(3)` `=(ML^(2))/(4xx3)` `therefore I.=(ML^(2))/(12)` |
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| 8. |
At what height above the earth's surface is the value of g the same as in a mine 80km deep? |
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Answer» `9.8 ms^-2` |
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| 9. |
A mass .m. is released, with a horizontal speed v from the top of a smooth and fixed, hemispherical bowl, of radius r. The angle theta w.r.t. the vertical where it leaves constant with the bowl is |
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Answer» `SIN^(-1)((V^(2))/(3rg)+(2)/(3))` |
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| 10. |
Resultant of two vectors of magnitudes P and Q is of magnitude 'Q'. If the magnitude of vecQis doubled now the angle made by new resultant with vecPis |
| Answer» ANSWER :B | |
| 11. |
A ray of light travels in the way as shown in the figure. After passing through water, the ray grazes along the water air surface. The value of mu_g interms of I is |
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Answer» `1/(SIN i)` |
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| 12. |
Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus ,calculate the longitudinal strain and the volume strain on the unt cube. |
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Answer» Solution :Each side of a unit cube=1. `therefore` Volume of the cube =1. surface area of each of the SIX faces =1. Let x,y and z axes be chosen along three mutually perpendicular SIDES of the cube. Now at first, we choose the two opposite faces perpendicular to the x-axis. Due to the FORCES (F,F) acting on them, the side along the x-direction will have elongation=l(suppose). `therefore` LONGITUDINAL strain along the x-direction `=(elongati on)/( i nitial l eng th)=l/1=l` Also longitudinal stress `=(applied forces)/(surface area)=F/1=F` `therefore` Young modulus , `Y=(Stress)/(strai n)=F/lso,l=F/Y` Simultaneously the side along the x-direction will also suffer lateral strains due to the forces acting along y and z directions we know, Poisson.s ratio `sigma=("lateral strai n")/("longitudi nal strai n")` `thereforeLateral strai n=sigmatimeslongitu d i nal strai n=sigmal=(sigmaF)/Y` Now, an elongation is always associated with a lateral contraction. Therefore, considering the strain along all the three axes, the effective longitudinal strain along the x-axis is , `F/Y-(sigmaF)/Y-(sigmaF)/Y=F/Y(1-2sigma)` From symmetry, the effective longitudinal strain along each of y and z axes `=F/Y(1-2 sigma)` As a result the final length of each of the three sides of the cube `=1+F/Y(1-2sigma)` `therefore` Final volume of a unit cube `=[1+F/Y(1-2sigma)]^3=1+(3F)/Y(1-2sigma)` [neglecting higher order terms] `therefore` Volume EXPANSION `=1+(3F)/Y(1-2sigma)-1` `=(3F)/Y(1-2sigma)` `therefore` Volume strain `=((3F)/Y(1-2sigma))/1=(3F)/Y(1-2sigma)` |
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| 13. |
A body is suspended at the end of a spring hanging vertically and it oscillates with a time period of 1.414 s. The spring is then cut into two equal parts and the same body attached to one of the pieces is made to oscillate. What be the time period of oscillation ? |
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Answer» Solution :The time period of oscillation of the SPRING mass system is given by `T= 2pisqrt((m)/(K))` ,……….(1) When the spring is CUT into two pieces the force constant of each piece is given by `k^(1) = 2k`………(2) `:.` Period of oscillation `T^(1)= 2pisqrt((m)/(k))` ..........(3) From (1) & (3) we get `(T)/(T^(1)= sqrt((k^(1)/(k))` but `T= 1.414s,T^(1)=?` `:. (1.414)/(T^(1))= sqrt((2k)/(k))= sqrt(2):. T^(1)= 1s` |
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| 14. |
Two satellites of masses 50 kg and 100 kg revolve around the earth in circular orbits of radii 9R and 16R respectively, where R is the radius of the earth. The speeds of the two satellites will be in the ratio |
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Answer» `(3)/(4)` |
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| 15. |
A system consists of two stars of equal masses thatrevolve in a circular orbit about a centre of mass midways between them. Orbital speed of each star is v & period ofrevolution isT. Find the mass M ofeach star. (G is gravitational constant ) |
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Answer» `(2Gv^(3))/(PIT)` |
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| 16. |
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 x× 10^(8) km away from the sun ? |
| Answer» SOLUTION :`1.43 XX 10^(12)` m | |
| 17. |
A coin is dropped in a lift. It takes time t_(1) to reach the floor when lift is stationary. It takes time t_(2) when lift is moving up with constant acceleration. Then |
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Answer» `t_(1) GT t_(2)` |
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| 18. |
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0^(@)C ? (R= 8.31Jmol^(-1)K^(-1)) |
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Answer» Solution :We know that 1 mol of any (ideal) gas at STP occupies a VOLUME of 22.3 litres. Thus the cylinder in this example contains 2 mol of helium. SINCE helium is monatomic ,molar specific heat at constant volume ,`C_(v)=(3//2)R` , and molar specific heat at constant pressure ,`C_(P)=(5//2)R` `THEREFORE` The Heat required to raise the temperature by `15^(@)`C DQ= `ne_(V)dt=2xx1.5Rxx15.0=45R` `=45xx8.31`=374J (approximately) |
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| 19. |
(A) Work done by frictional force is always negative (B) A body at rest can have mechanical energy (C ) Mechanical energy of freely falling body decrease gradually |
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Answer» Only A is true |
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| 20. |
When a body is suspended from spring its time period 'T'. If the spring is cut into two equal parts and the body is suspended from one part, its time period will be |
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Answer» `T"/"SQRT(2)` |
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| 21. |
A quantity X is given by epsi_0 L(Delta V)/(Delta t)where epsi_0is the permittivity of free space, L is a lengthDelta Vis a potential difference and Delta tis a time interval. The dimensional formula for X is the same as that of |
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Answer» resistance |
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| 22. |
If man is walking, direction of friction is |
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Answer» OPPOSITE to direction of motion |
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| 23. |
A person travels along a straight road for half the distance with velocity V_(1) and the remaining half distance with velocity V_(2) the average velocity is given by |
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Answer» `v_(1)v_(2)` |
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| 24. |
A spring, suspended vertically, elongates by 0.02m when a mass of 0.08 kg is attached to it. An object of mass 0.6 kg is attached to its free end and it is displaced by 8 cm from its equilibrium position. Determine the energy of the system at this position. From the viewpoint of energy, determine the velocity of the body when it is at a distance of 0.04 m from the equilibrium position. |
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| 25. |
A steel tape measures the length of a copper rod as 80.00cm when at 15°C, which is the calibration temperature for the tape. The reading of the tape when both are at a temperature of 65°C will be (linear expansivities of steel and copper are 12 x10^(-6)//°C and 17 xx 10^(-6)//°C respectively ) |
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Answer» 80.01 cm |
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| 26. |
The period of oscillation of abob suspended by a string oflength L is 2pisqrt(l/g). This expression is valid only for |
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Answer» A light bob and no friction |
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| 27. |
Starting with the same initial conditions an ideal gas expands from volume V_(1)" to "V_(2) in three different ways. The work done by the gas is W_(1) if the process is purely isothermal, W_(2) if purely isobaric and W_(3) if purely adiabatic. then |
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Answer» `W_(2)gtW_(1)gtW_(3)` |
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| 28. |
A small hole is made at a height of (1//sqrt(2))m from the bottom of a cylindrical water tank. The length of the water column is sqrt(2)m. Find the distance where the water emerging from the hole strikes the ground. |
| Answer» ANSWER :D | |
| 29. |
The pulley of Atwood's machine has a moment of inertia 'I' about its axis and its radius is R. Find the magnitude of acceleration of the two blocks assuming the string is light and does not slip on the pulley. |
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Answer» Solution :Suppose the block of mass M goes down with an acceleration a. The ANGULAR acceleration of the pulley is THEREFORE `alpha=(a)/(R )` `Mg-T_(1)=Ma` `T_(2)-mg=ma` `T_(1)R-T_(2)R=Ialpha=I(a)/(R )` Solving the EQUATIONS `a=((M-m)GR^(2))/(I+(M+m)R^(2))` 
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| 30. |
A block is on horizontal slab which is moving horizontally. If block is not separated from slab then determine angular frequency of oscillation. |
Answer» Solution :If block is not separated from slab then restoring force due to SHM should be less than FRICTIONAL force between slab and block.  `F_("restoring") LE F_("FRICTION") rArr ma_("max") le mu mg` `:. A_("max") le mg rArr m OMEGA^(2)A le m mug` `omega^(2) le (m mug)/(mA)` `omega le sqrt((m_(g))/(A))` `:. le sqrt((mu_(g))/(A))` |
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| 31. |
A ball of mass 5kg hangs from a spring of force constant K N/m and oscillates tain time period. If the ball is removed, the spring shortens by |
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Answer» `(K/G)m` |
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| 32. |
What is latent heat? Is it constant? |
| Answer» Solution :The AMOUNT of heat energy absorbed or lost by UNIT mass of substance during its CHANGE of STATE at constant temperature. Its VALUE changes slightly with pressure. | |
| 33. |
When lift is moving in upward direction with constant acceleration pseudo weight of object will …... . |
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Answer» |
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| 34. |
Which of the following quantities are always positive In a SHM |
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Answer» `BAR F. BARA` |
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| 35. |
If the radius of the earth decreases by 10% with out change in its mass what happens to the acceleration due to gravity |
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Answer» DECREASES by 19% |
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| 36. |
A petrol engine consumes 5 kg of petrolin one hour, the calorific value of petrol being 10^(8) calorie kg^(-1) . Assuming that only 12% of heat is converted into mechanical energy, find the average power developed in kilowatt. |
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| 37. |
What will be shape of wave propagationg on sea when observed from sea shore ? Why ? |
| Answer» Solution :WAVE PROPAGATING on SEA surface is the resultant of capillary waves (transverse waves on the surface of sea WATER) and gravity waves (longitudinal waves below the surface of sea water) which assumes elliptical shape. | |
| 39. |
Consider two trains A and B moving along parallel tracks with the same velocity in the same direction. Let the velocity of each train be 50kmh^(-1) due east. Calculate the relative velocities of the trains. |
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Answer» Solution :Relative VELOCITY of B with respect to A, `v_(BA)=v_(B)-v_(A)=50kmh^(-1)+(-50)KMH^(-1)=0kmh^(-1)` Similarly, relative velocity of A with respect to Bie. V is also zero. Thus each TRAIN will appear to be at rest with respect to the other. |
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| 40. |
If iron sphere is replaced by wooden sphere of same mass, time period of simple pendulum |
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Answer» Increases |
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| 41. |
Densities of a liquid at 30^(@) C and 80^(@) C are 0.6gm/c.c and 0.54 gm/c.c. The coefficient of real expansion of liquid is |
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Answer» `0.0002//^(@)C ` |
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| 42. |
Show that the frequency of sound vibrations emitted by a source will appear to be doubledto a listener fixed in space when the source approaches the listener with a speed equalling half the speed of sound. |
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| 44. |
An orbiting satellite will escape if a) its speed increases by 41.4% b) its speed in the orbit is made sqrt(1.414)times that of its initial value c) its kinetic energy increases by 100% d) its kinetic energy is made 12 times that of its initial value |
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Answer» only a & B are CORRECT |
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| 45. |
When we try to close a water tap with our fingers , fast jets of water gush through the oopenings between our fingerss |
| Answer» Solution :ACCORDING to equation of contiunity av = constant as the the gap between fingers DECREASES , area of outlet of WATER jet is reduced , so velocity of water INCREASES. | |
| 46. |
A particle is moved from (0,0) to (a, a) under a force vecF = (3hati +4hatj) from two paths. Path 1 is OP and path 2 is OQP. Let W and W be the work done by this force in these two paths.Then |
| Answer» ANSWER :A | |
| 47. |
A gas is enclosed in a container which is then placed on a fast moving train . The temperature of the gas .............. . |
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Answer» rises |
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| 48. |
A particle moves along a straight line to follow the equation ax^2 + bv^2 = k, where a,b and k are constants and x and u are x-coordinate and velocity of the particle respectively. Find the amplitude: |
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Answer» `SQRT(k/b)` |
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| 49. |
Name some physical quantities that have same dimension. |
| Answer» SOLUTION :WORK ENERGY and TORQUE. | |
| 50. |
Subtract according to significant digit. 3.9xx10^(5)-2.5xx10^(4) |
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Answer» Solution :`3.9xx10^(15)=39.0xx10^(4)` `2.5xx10^(4)=(2.5xx10^(4))/(36.5xx10^(4))=3.65xx10^(5)` Here, LEAST SIGNIFICANT digit is 2 hence it will be REPRESENTED as after ROUNDING off `3.9xx10^(5)-2.5xx10^(4)=3.6xx10^(5)` |
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