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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The period of a satellite in a circular orbit around a planet is independent of |
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Answer» the mass of the PLANET |
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| 2. |
If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature (27^(@)C), find the temperature inside the refrigerator. |
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Answer» Solution :`(Q_(2))/(W)=5, Q_(2)=5W`, `Q_(1)-Q_(2)=W` `:.Q_(1)=6W` `(Q_(2))/(Q_(1))=(T_(2))/(T_(1))=(5)/(6)=(T)/(300)`, `T_(2)=250K=-23^(@)C` |
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| 3. |
An insulated container monoatomic gas of molar mass m is moving with a velocity nu_(0). If the container is suddenly stopped. Find the change in temperature. |
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Answer» SOLUTION :Suppose the container has N moles of the monoatomic gas. Then the loss in K.E. of the gas `DeltaE=1/2(mn)nu_(O)^(2)` If the TEMPERATURE of the gas change by `DeltaT,` then heat gained by the gas, `DeltaQ=nC_(V)DeltaT` `DeltaQ=3/2 nRDeltaT` Now, DeltaQ=DeltaE `3/2nDeltaT=1/2(mn)nu_(O)^(2)` `DeltaT=(mnu_(O)^(2))/(3R)` |
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| 4. |
The time dependence of a physical quantity p is given by p=p_(0)e^(-alpha ^(2)) where alpha is constant and t is time. The constant alpha |
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Answer» <P>is DIMENSIONLESS |
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| 5. |
The K.E acquired by a mass m travelling a certain distance d starting from rest under the action of a constant force is directluy prportional to |
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Answer» m |
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| 6. |
The length and breadth of a rectangle are (5.7 pm 0.1) cm and (3.4 pm 0.2) cm respectively. Calculate the area of the rectangle with error limits. |
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Answer» Solution :Lenght`L=(5.7 pm 0.1)` cm Breadth`B = (3.4 pm 0.2)` cm Area A with error LIMIT = `A pm DELTAA=?` Area A = `lxxb = 5.7 xx 3.4 = 19.38 = 19.4 "cm"^(2)` `(DeltaA)/(A) = (DELTAL)/(l) + (Deltab)/(b) , DeltaA = ((Deltal)/(l) + (Deltab)/(b))A` `DeltaA = ((0.1)/(5.7) + (0.2)/(3.4)) 19.4 = (0.0175 + 0.0588) xx 19.4 = 1.48 = 1.5` Area with error limitA = `(19.4 pm 1.5) "cm"^(2)` |
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| 7. |
A stone is thrown vertically up from a bridge with velocity 3 ms^(-1). If it strikes the water under the bridge after 2 s, the bridge is at a height of (g = 10 "ms"^(-2) ) |
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Answer» 26m |
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| 8. |
Give an example of irreversible reaction |
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Answer» Solution :All naturally occurring processes are irreversible. Here we give some interesting examples. (a) When we open a gas bottle, the gas MOLECULES slowly spread into the entire room. These gas molecules can never get back in to the bottle. (b) Suppose one DROP of an ink is dropped in water, the ink droplet slowly spreads in the water. It is impossible to get the ink droplet back. (c) When an object falls from some height, as soon as it hits the earth it COMES to rest. All the KINETIC energy of the object is CONVERTED to kinetic energy of molecules of the earth surface, molecules of the object and small amount goes as sound energy. The spreaded kinetic energy to the molecules never collected back and object never goes up by itself. 
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| 9. |
Two moles of helium gas undergo cyclic processes shown as assuming the gas to be ideal.calculate the following quantities in this process. (a) The net change in the heat energy (b) The network done (C) The net change in internal energy. (T_A = 300K, T_B = 400K) |
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Answer» Solution :We shall first calculate here work done in the cyclic process. As AB is isobaric process, HENCE work done during this process from A to B is `W_(AB)= P_A(V_B -V_A)=P_BV_B-P_AV_A(:.P_A =P_B )` According to perfect gas EQUATION,` P_A V_A= nRT _Band P_BV_B= nRT_B ` ` thereforeW_(AB )= mu RT(T_B -T_A) 2 xx 8.32 xx ( 400-300= 2 xx 832 = 1664` joule workdoneduringisothermalprocessfromB to C ` W_(BC )= nRTlog_e((V_f)/(V_i))= mu RT_Clog_e((V_C)/(V_B))=muRT _C log((P_B)/(P_C))[ :.P_BV_B=P_C V_C ]` `= 2 xx 8.32 xx 400 xx log _e21[ :.P_B= 2AND P_C= 1 atm]= 4068` joule Work done during isobaric process from C to D `W_(CD)= nR(T_D -T_C ) 2 xx 8.32 xx (300 -400 ) =- 1664` joule Work done during isothrmal process from D to A ` W_(DA ) = nRTlog_e((V_f)/(V_i))= nRT _Dlog_e((V_A)/(V_D)) = nRT_D log_e (P_d //P_A) [:.P_AV_A=P_DV_D ]` `=- 2 xx 8.32 xx 300 xxlog_e 2 =- 3456` joule ` therefore` Networkdone ` =W_(AB) +W_(BC) +W_(CD) +W_(DA) = 1664+ 4608- 1664- 3456= 1152` joule (a)INA cyclicprocess `DeltaU=0therefore `Fromfirstlawof thermodynamics Netchangein heatenergy ` Delta Q = Deltaw= 1152` joule (b )Networkdone ` DeltaW= 11.52` joule `(c )` Netchange ininternalenergy ` Delta U=0`. |
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| 10. |
If an object is thrown horizontally with an initial speed "10 ms"^(-1) from the top of a building of height 100 m. What is the horizontal distance covered by the particle. |
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Answer» R=45m |
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| 11. |
Structure of an equilateral triangle ABC is made using three thin rods. Another rod AD connects the vertex A with D, the mid-point of BC. Coefficient of linear expansion for AB and AC is alpha and that for the base BC is beta. Show that, if the coefficient of linear expansion for the rod AD is 1/3(4alpha-beta), the arms of the system will not show any tendency to bend, for a small rise in temperature. |
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Answer» Solution :LET the initial length of each side of the triangle ABC be 2l [Fig. 5.11]. Hence, BD = CD = L `therefore "" AD=sqrt(AB^(2)-BD^(2))` `""=sqrt(4l^(2)-l^(2))=sqrt(3)l` Let `gamma` be the coefficient of linear expansion for rod AD. For a rise in temperature t, `""` changed length of `AB=2l(1+alphat)`, `""` changed length of `BD=l(1+betat)` and changed length of`AD=sqrt(3)l(1+gammat)` If the sides do not bend, the triangle ABD continues to be a right angled triangle, i.e., `""[2l(1+alphat)]^(2)=[l(1+betat)]^(2)+[sqrt(3)l(1+gammat)]^(2)` or, `"" 4l^(2)(1+2alphat+alpha^(2)t^(2))=l^(2)(1+2betat+beta^(2)t^(2))+3l^(2)(1+2gammat+gamma^(2)t^(2))` or, `" " 4(l+2alphat)=1+2betat+3(1+2gammat)` (Neglecting the terms CONTAINING `alpha^(2), beta^(2) " and " gamma^(2)` for their very small values) or, `""4+8alphat=1+2betat+3+6gammat " or, " 8alphat=2betat+6gammat` or, `" " 4alpha=beta+3gamma " or, " gamma=1/3(4alpha-beta). 
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| 12. |
A microscope consists of two convex lenses of focal lengths 2.0cm and 6.25cm placed 15.0cm apart. Where must the object be placed so that the final virtual image is at a distance of 25cm from the eye? |
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Answer» 2.5cm |
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| 13. |
y=-kx^(2) is represented by |
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Answer»
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| 14. |
When a girl sitting on a swing stands up, the periodic time of the swing will decrease. In standing position of girl, the length of the swing will increase. |
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Answer» Both 'A' and 'R' are ture and 'R' is the correct explanation of 'A' |
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| 15. |
Three identical uniform thin metal rods from the three sides of an equilateral triangle. It the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular to the plane of the triangle is n times the moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length, the value of 'n' is |
| Answer» Answer :B | |
| 16. |
A ring has a mass of 0.5 kg and radius 1 m. The moment of inertia of the ring about its diameter is |
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Answer» `0.75 KGM^(2)` |
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| 17. |
A physical quantity P is related to four quantities a,b , c and d as follows . P = (a^3 b^2)/(sqrt(c) d) The percentage errors of measurement in a,b , c and d are 1% , 3% , 4% and 2% respectively . What is the percentage error in the quantity P ? |
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Answer» |
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| 19. |
A chain is kept on a smooth horizontal table. 1/5th of the length of the chain is hanging off the table. If l is the length of the chain and m is its mass, how much workis to be done to full the hanging partof the chain onto the table ? |
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Answer» |
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| 20. |
A solid cylinder rolls up an inclined plane of angle of inclination 30^(@) . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. (a) How far will the cylinder go up the plane ? (b) How long will it take to return to the bottom ? |
| Answer» SOLUTION :(a) 1.913m, (B)1.53s | |
| 21. |
Let V,T,L,K and r. denote the speed, time period, angular momentum, kinetic energy and radius of satellite in circular orbit, then a) V prop r^(-1) b) L prop r^(1//2) c) T prop r^(3//2) d) K prop r^(-2) |
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Answer» only a & B are TRUE |
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| 22. |
There is a hole at the bottom of a large opien vessel. If water is filled upto a height h, it flows out in time t. If water is filled to a height 4th, it will flow out in time |
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Answer» 4t |
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| 23. |
Select the odd man out from the following parameters. |
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Answer» Velocity GRADIENT |
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| 24. |
An alpha-particle and a proton having same kineti energy enetrs in uniform magnetic field perpendicularly. Let x be the ratio of their magnitude of accelation any y the ratio of their time periods. Then |
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Answer» `x=(1)/((2)^(3//2))` |
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| 25. |
The displacement x of the particle moving in one direction, under the action of a constant force is related to the time t given by the equation t=sqrt(x)+3 where x is in metre and t in second. Then find the workdone by the force in the first 6sec |
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Answer» |
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| 26. |
An external device, e.g., an electric motor, supplies constant power to a rotating system, e.g., a flywheel, through a torque tau. The angular velocity of the system is omega. Both tau and omega are variable |
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Answer» `omega prop TAU` |
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| 27. |
The lower end of a capallary tube of diameter 2.50 mm is dipped 8.00 cm below the surface of water in a beaker .What is the pressure required in the tube in order to blow a hemispherical bubble at the end in water ? The surface tension of water at temperature of the experiments is 7.30xx10^(-2)Nm^(-1). 1 atmospheric pressure 1.01xx10^(5)Pa. Density of water =1000kg//m^(3),g=9.80ms^(-2).Also calculate the excess pressure . |
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Answer» <P> SOLUTION :`P_(i)-P_(o)=116.8Pa` |
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| 28. |
The smallest division on main scale of a vernier callipers is 1 mm and 10 vernier divisions coincide with 9 scale divisions. While measuring the length of a line, the zero mark of the vernier scale lies between 10.2 cm and 10.3 cm and the third division of vernier scale coincide with a main scale division. (a) Determine the least count of the callipers (b) Find the length of the line |
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Answer» Solution :(a) Least count (L.C) `= ("SMALLEST division on main scale")/("Number of divisions on vernier scale")= (1)/(10)mm= 0.1mm= 0.01cm` (B) Length of the line = `(10.2+ 3 XX 0.01) cm = (10.23)cm` |
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| 29. |
All surfaces are smooth. The acceleration of mass m relative to wedge is |
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Answer» `G sin theta` |
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| 30. |
Assertion : The air pressure in a tyre of a bus increases during driving. Reason : The pressure of a given mass of a gas is inversely proportional to its volume. Choose the correct choice from the following. |
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Answer» Both ASSERTION and REASON are true and reason is the correct explanation of the assertion. When temperature increases, definitely pressure will be increased as PER Charles law. Hence pressure is directly proportional to temperature . So reason is false. |
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| 31. |
A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms^(-1) and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms^(-1), respectively are |
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Answer» 120 Hz and 5 m |
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| 32. |
A book is lying on the table what is the angle between the action of the book on the table and the reaction of the table on the book |
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Answer» `0^(@)` `:.` Anglebetweenactionand REACTIONIS `180` |
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| 33. |
Four lead spheres a, b, c & d whose radii 1cm, 2cm, 3cm and 4cm respectively are dropped into different liquids of co-efficients of viscosity 1xx10^(-3)Pa-s, 2xx10^(-5)Pa-s, 6.5xx10^(-3)Pa-s, 15.4xx10^(-3)Pa-s. Arrange the spheres with their terminal velocity in descending order |
| Answer» Answer :B | |
| 34. |
An infinite number of particles each of mass 1kg are placed on the postive x-axis at 1m, 2m, 4m, 8m… from the origin. The magnitude of the resultant gravitational force on 1kg mass kept at the origin is |
| Answer» ANSWER :D | |
| 35. |
Two identical balls 'A' and 'B' are moving with same velocity. If velocity of 'A' is reduced to half and of 'B' to zero, then the rise in temperature of 'A' to that of 'B' is |
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Answer» `3:4` |
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| 36. |
A block of mass 10 kg lying on a smooth horizontal surface is being pulled by means of a rope of mass 2 kg. If a force of 36 N is applied at the end of the rope, the tension at the mid point of the rope is, |
| Answer» Answer :A | |
| 37. |
Two wires are made of the same material and have the same volume. The area of cross sections of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by Deltal on applying a force F, how much force is needed to stretch the second wire by the same amount? |
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Answer» 2 |
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| 38. |
The rate of cooling of water in a calorimeter is one-tenth of a degree per second when its temperature is 40^(@)C above that of the sorroundings. What is the rate of cooling and heat per second when the excess temperature over the surroundings is 30^(@)C ? Thermal capacity of water and calorimater is 4600J//""^(@)C. |
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Answer» Solution :RATE of COOLING, `(d theta)/dtprop` excess temperature over the surroundings. In the first case `1/10prop40`, In the second case `(d theta)/(dt)prop30` divifing `(d theta)/(dt)//1/10=30/40` `rArr(d theta)/(dt)=3/(4XX10)=3/40""^(@)C//sec` Rate of heat radiation = `MS(d theta)/(dt)` = Thermal capacity `XX(d theta)/dt=4600xx3/40=345J//sec.` |
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| 39. |
A disc of radius .r. is removed from the disc of radius .R. then(a) The minimum shift in centre of mass is zero. (b) The maximum shift in centre of mass cannot be greater than (r^(2))/((R+r))(c ) Centre of mass must lie where mass exists(d) The shift in centre of mass is (r^(2))/((R+r)) |
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Answer» only a and B are CORRECT |
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| 40. |
The trajectory of a particle is as shown here and its trajectory follows the eqution y=(x-1)^(3)+1.Find co-ording of the point A on the curve such that direction of instantaneous velocity at A is same as direction of average velocity for the motion O ti A: |
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Answer» `(3//2,9//8)` |
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| 41. |
What shoud be the radius of a capillary tube if the water has to rise to a height of 6 cm in it. Surface of water =7.2xx10^(-2)Nm^(-1) (g=10ms^(-2)) |
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Answer» SOLUTION :Height `(h)=6cm=6XX10^(-2)m` Surface tension of WATER (S)`=7.2xx10^(-2)NM^(-1)` for water `theta=0, cos theta =1` Density of water `(rho)=1000kgm^(-3)` Surface tension `(S)=(HR rho g)/(2 cos theta)` Substituting the values `7.2xx10^(-2)=((6xx10^(-2))(r)(1000)(10))/(2xx1)` `r=0.24xx10^(-3)m` |
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| 42. |
An ideal gas is expanded so that amount of heat given is equal to the decrease in internal energy. The gas undergoes the process TV^(1//5)= constant. The adiabatic compressibility of gas when pressure is P, is - |
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Answer» <P>`(7)/(5P)` `C= -C_(V) = (-R)/(gamma-1)=(+R)/(gamma-1)+(P)/(n) (dV)/(dT)` `-(P)/(n) (dV)/(dT)= (2R)/(gamma-1)` `T^(5)V=` CONST. `V=("const.")/(T^(5))` `(dV)/(dT)= -5("const")/(T^(6))` `PV=nRT` `P//n=RT//V` `+ (RT)/("const.")T^(5) xx (-5 ("const")/(T^(6)))=(2R)/(gamma-1)` `(5)/(2)=(1)/(gamma-1) rArr gamma-1 =2//5` `gamma=7//5` adiabatic compressibility `beta=(1)/(gammaP)=(5)/(7P)` |
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| 43. |
Two rods of different materials but having same length L and cross-sectioinal areas A are arranged end-to-end between fixed supports. The temperature is T and there is no initial stress. The rods are heated so that their temperature increases by DeltaT. Find : (a) the displacement of the interface (b) the stress at the interface. The coefficinets of linear expansioin of the materials are alpha_(1) and alpha_(2) and their Young's modulii are E_(1) and E_(2) respectively. |
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Answer» |
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| 44. |
A man of mass "m" is standing on the floor of the lift. Find his apparent weight when:(i) elevator is rest(ii) elevator is accelerating upwards(iii) elevator is accelerating downwards. |
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Answer» Solution :(i) When the elevator is at rest:The acceleration of the man is zero. Hence the net force ACTING on the man is zero. When Newton's SECONS law is applied with respect to inertial frame on the man, `vec(F)_(G)+vec(N)=0` `-mghat(j)+Nhat(j)=0` By comparing the components, we can write `N-mg=0(or)N=mg` Since weight, W = N, the apparent weight of the man is equal to his actual weight. (ii) When the elevator is accelerating upwards:If an elevator is moving with upward acceleration then with respect to inertial frame when Newton's second law is applied on the man, `vec(F)_(G)+vec(N)=mvec(a)` The above equation can be written in terms of unit vector in the vertical direction, `-mghat(j)+Nhat(j)=mahat(j)` By comparing components, `N=m(g+a)` Therefore, apparent weight of the man is GREATER than his actual weight.  (iii) When the elevator is accelerating downwards:If the elevator is moving with downward acceleration `(vec(a)=-ahat(j))`, by APPLYING Newton's second law on the man, we can write `vec(F)_(G)+vec(N)=mvec(a)` In terms of unit vector in the vertical direction, we get the above equation as `-mghat(j)+Nhat(j)=mahat(j)` By comparing the components, `N=m(g-a)` `therefore` Apparent weight `[W=N=m(g-a)]` of the man is lesser than his actual weight. |
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| 45. |
A ladder is kept at rest with is upper end against a wall and the lower end on the ground. The ladder is more like to slip when a mass stands on it at the top than at the bottom. Why ? |
| Answer» Solution :When the man is at the TOP the torque due to his WEIGHT will have a MAXIMUM VALUE so | |
| 46. |
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv_(e), where ve is its escape velocity from the surface of the earth. The value of f is |
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Answer» `SQRT(2)` |
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| 47. |
An external pressure P is applied on a cube atm 0^(@)C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and alpha is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by ...... |
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Answer» <P>`(3 ALPHA )/(PK)` `therefore P =- B xx (Delta V )/(V)` But `Delta V = 3 alpha V Delta T` `therefore Delta T` `therefore P =-B xx (3 alpha V Delta T )/(-V)` `therefore P = + B xx 3 alpha Delta T` `therefore Delta T = (P)/(3 alpha B)` But here B is k. `therefore Delta T = (P)/(3 alpha K)` |
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| 48. |
Kinematics is the branch of mechanics which delas with the motion of objects without taking _________ into account |
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Answer» kinetics |
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| 49. |
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^(-2). The crew and the passengers weigh 300 kg. Give the magnitude and direction of the (a) force on the floor by the crew and passengers, (b) action of the rotor of the helicopter on the surrounding air, (c) force on the helicopter due to the surrounding air. |
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Answer» Solution :(a) ‘Free body’:CREW and passengers Force on the system by the floor = F upwards, WEIGHT of system = mg DOWNWARDS, ` thereforeF – mg = ma` ` F- 300xx 10 = 300 xx 15` ` F= 7.5 xx 10^3 N ` upward By the Third Law, force on the floor by the crew and passengers = `7.5 xx 10^3 N`downwards. (b) ‘Free body’ :helicopter plus the crew and passengers Force by air on the system = R upwards, weight of system = mg downwards ` THEREFORE R – mg = ma` `R - 1300 xx 10=1300 xx 15` ` R= 3.25 xx 10^4 N `upwards By the Third Law, force (action) on the air by the helicopter = `3.25 xx 10^4 N ` downwards. (c ) `3.25 xx 10^4 N ` upwards |
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