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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 physical quantity of the unit Joule xx" second"^(2)? |
| Answer» SOLUTION :MOMENT of INERTIA. | |
| 2. |
A piece of metal floats on mercury. The volume coefficient of expansion of the metal and mercury are gamma_(1) and gamma_(2) respectively. If the temperature of both mercury and metal are increased by DeltaT, by what factor does the fraction of the voilume of the m.etal submerged in mercury change ? |
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Answer» Solution :Let the total VOLUME of metal in air and mercury be V and `V_(s)` RESPECTIVELY, `rho and sigma` be the densities of s . metal and mercury respectively. when the metal floats m equilibrium, Fraction of the volume submerged, `f_(s) = (V_(s))/(V) = (rho)/(sigma)""` .......(1) When the temperature changes, the fraction of volume submerged changes as densities change. `(Delta _(s))/(f_(s)) = (f_(s))/(f_(s))- 1 = (rho^(1))/(sigma^(1)) xx (sigma)/(rho) - 1 "" ` ...... (2) Densities of liquid and metal decrease as temperature INCREASES `rho = (rho)/(1 + gamma_(1) Delta T ) , sigma. = (sigma)/(1 + gamma_(2) Delta T)` Which on substitution in eqn. (2) yields `(Delta f_(s))/(f_(s)) = (1 + gamma_(2) Delta T)/(1 + gamma_(1) Delta T) - 1 = ((1 + gamma_(2) Delta T ) - (1 + gamma_(1) Delta T ))/((1 + gamma_(1) Delta T))` `((gamma_(2) - gamma_(1))Delta T )/((1 + gamma_(1) Delta T)) = (gamma_(2) - gamma_(1)) Delta T (1 - gamma_(1) Delta T)` = `(gamma_(2) - gamma_(1) ) Delta ` T `[ " As" (1)/(1 + gamma_(1) Delta T) = (1 + gamma_(1) Delta T)^(-1) = (1- gamma_(1) Delta T)` and `gamma_(1) gamma_(2)` small number, it has been neglected] When the solid is completely submerged, 1 = `f_(s) (1 + gamma Delta T ) " or "Delta T = (1 - f_(s))/(gamma f_(s))` (b) `f_(s_(1)) = f_(s) (1 + gamma Delta T_(1) ), f_(s_(2)) = f_(s) (1 + gamma Delta T_(2))` `(f_(s_(1)))/(f_(s_(2))) = ( (1 + gamma Delta T_(1))/(1 + gamma Delta T_(2)) ) . ` ON solving for `gamma` , we have ` gamma = (f_(s_(1)) - f_(s_(2)))/(f_(s_(2)) Delta T_(1) - f_(s_(1)) Delta T_(2))` (C) As `f_(s) = (rho)/(sigma) and f_(s) = (rho.)/(sigma.)` `f_(s). = ((rho)/(1 + gamma_(1) Delta T)) ((1 + gamma_(2) Delta T))/(sigma ) f_(s). = f_(s) ((1 + gamma_(2) Delta T )/(1 + gamma_(1) Delta T ))` If `gamma_(2)gt gamma_(1) , ` the solid sinks. If `gamma_(2) = gamma_(1)` , no effect on submergence. If ` gamma_(2) lt gamma_(1)`, the solid lifts up . |
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| 3. |
A smooth ring is fixed horizontally on a smooth horizontal surface. From a point A of the ring, a particle is projected at an angle 'theta' to the radius vector at A. If the coefficient of restitution between the find and the particle is 0.5 and the particle returns to the point of projection A after two reflection, then cot^(2) theta is equal to |
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Answer» 10 |
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| 4. |
Two blocks having masses 8 kg and 16 kg are connected to the two ends of a light spring. The system is placed on a smooth horizontal floor. An inextensible string also connects B with ceiling as shown in figure at the initial moment. Initially the spring has its natural length. A constant horizontal force F is applied to the heavier block as shown. What is the maximum possible value of F so that lighter block doesn.t lose contact with ground. |
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Answer» |
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| 5. |
A particle is thrown with a velocity u at an angle theta from the horizontal. Another particle is thrown with the same velocity at an angle theta from the vertical. The ratio of times of flight of the two particles will be |
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Answer» `Tan 2 THETA:1` |
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| 6. |
What is meant by Doppler effect? |
| Answer» Solution : When the SOURCE and the observer are in relative motion with respect to each other and to the MEDIUM in which sound propagates, the FREQUENCY of the sound wave observed is DIFFERENT from the frequency of the source. This phenomenon is called DOPPLER Effect. | |
| 7. |
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of the new unit, if light takes 8 min and 20 sec. to cover thedistance ? |
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Answer» 300 Speed of light in VACCUM `c=1` nw unit of LENGTH `s^(-1)` time taken by light of sun to reach the EARTH `t=8` min and 20 seconds `=(8xx60+20)s=500s` `therefore` Distance between the sun and the earth `S=ct=1` new unit of length `s^(-1)xx500s` `=500` new unit of length. |
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| 8. |
Breaking stress for steel is 8xx10^(6) Nm^(-2) . The density of steel is 8xx10^(3) kgm^(-3) and g=0ms^(-2). What will be the maximum length of steel wire which will hang without breaking under its load. |
| Answer» Answer :2 | |
| 9. |
A particle of mass m is subjected to a force vec(F) = F_0 [cos(t)hati + sin(t)hatj]. If initially (t=0) the particle was at rest, the kinetic energy particle as a function of time is given by: |
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Answer» `(F_0^2)/(m)[1 - COS (2T)]` |
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| 10. |
A steel ball is allowed to fall freely from a height of 36 cm onto a smooth floor. Find the height to which it rises after rebounding from the floor. Coefficient of restitution between the steel ball and the floor = 0.60. |
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Answer» Solution :Here `e=0.60, h=36 CM = 36xx10^(-2)m` The HEIGHT attained by the body after .n. NUMBER of rebounds `= e^(2n)h` Here n = 1 `THEREFORE` The height attained by the body after `1^(st)` rebound `= e^(2(1))h=(0.60)^(2)(36xx10^(-2))` `= 0.36xx0.36=0.13 m`. |
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| 11. |
The angular magnification of a simple microscope can be increased by increasing |
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Answer» FOCAL LENGTH of lens |
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| 12. |
If the linear momentum of the object is increased by 0.1%, then the kinetic energy is increased by |
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Answer» `0.1 %` |
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| 13. |
A horizontal jet of water coming out ofa pipe of area of cross - section 20 cm^(2) hits a vertical wall with a velocity of 10 ms^(-1) and rebounds with the same speed. The force exerted by water on the wall is, |
| Answer» ANSWER :C | |
| 14. |
A uniform meter stick AB of mass M is lying in state of rest on a rough horizontal plane. A small block of mass m is placed on it at its centre C. A variable force F is applied at the end B of the stick so as to rotate the stick slowly about A in vertical plane. The force F always remains perpendicular to the length of the stick. The stick is raised to theta = 60^(@)and it was observed that neither the end A slipped on the ground nor the block of mass m slipped on the stick. (a) F_(1) is force applied by the stick on the block. Plot the variation of F_(1) with theta (0 le theta le 60^(@)). (b) What must be the minimum coefficient of friction between the block and the stick. (c) f is the friction force acting at end A of the stick. Plot variation of f vs theta (0^(@) le theta le 60^(@)). |
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Answer» (B) `mu_("min") = SQRT3` (c) `(##IJA_PHY_V01_C06_E01_082_A02##)` |
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| 15. |
A stone projected vertically up from the ground reaches a height y in its path at t_(1) seconds and after further t_(2) seconds reaches the ground. The height y is equal to |
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Answer» `(1)/(2)G(t_(1)+t_(2))` |
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| 16. |
Consider a simulation with a car of mass 1000kg moving with a speed 18.0 km/h on a road and colliding with a horizontally mounted spring of spring constant 6.25 xx 10^(3)N m^(-1). Taking the coefficient of friction, mu to be 0.5 What si the maximum compression of the spring? |
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Answer» Solution :In presence of friction, both the SPRING force and the frictional force act so as to oppose the compression of the spring. We invoke the work-energy theorem, rather than the conservation of mechanical energy. The chenge in KINETIC energy is `Delta K= K_(f)-K_(i)= 0 -(1)/(2) m u^(2)` The work done by teh net force is `W= -(1)/(2) kx_(m)^(2)- mu mgx_(m)` Equating we have `(1)/(2) m u^(2) = (1)/(2) kx_(m)^(2)- mu mgx_(m)` Now `mu mg = 0.5 xx 10^(3) xx 10= 5 xx 10^(3)N` (TAKING `g= 10.0 ms^(-2)`). After rearranging the above equation we obtain the following quadratic equation in the unknown `x_(m)` `kx_(m)^(2) + 2mu mgx_(m)- m u^(2)= 0` `x_(m)= (-mu mg + [mu^(2) m^(2) g^(2) + mku^(2)]^(1//2))/(k)` where we take the positive square root since `x_(m)` is positive. Putting in numerical VALUES we obtainthe following quadratic equation in the unknown `x_(m)= 1.35m` |
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| 17. |
The amplitude of SHM is 5cm, period 4sec and initial phase is pi//6. Its displacement at t = 0 [ initial displacement] is |
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Answer» `2.5sqrt(2)cm` |
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| 18. |
Does the centre of mass of a body necessarily lie inside the body? |
| Answer» Solution :may LIE outside the BODY e.g., the CM of a CIRCULAR ring is at the centre of the circle which is outside the ring | |
| 19. |
In a smooth hemispherical shell of radius R, a rod of mass (sqrt3//2)kg is placed horizotally and is in equilibrium. The length of rod is R. Find the normal reaction at any end of the rod (in N). (Take g=10m//s^(2)) |
| Answer» Answer :A | |
| 20. |
Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are ....... |
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Answer» `5 m//s, 1m//s` If v is relative speed when they re moving in same direction, then `v = v _(1) + v _(2) ` `THEREFORE 6= v _(1) + v _(2) ""…(1)` If v. is relative speed when they are moving in same direction, then `v. = v _(1) -v _(2)` `therefore 4= v _(1) -v _(2) ""...(2)` By adding equation (1) & (2), `10 = 2v _(1) implies v _(1) = 5m//s` From equation (2), `v _(1) = 5m//s` |
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| 21. |
In the figure shon mass of both, the spherical body and blocks is m. Moment of inertia of the spherical body about centre of mass is 2mR^(2). The spherical body rolls on the horizontal surface. There is no sliiping between any two surfaces in contact. The ratio of kinetic energy of the spherical body to that of block is |
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Answer» `3/4` |
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| 22. |
Pressure is a scalar quanity because |
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Answer» It is the RATIO of FORCE to AREA and both force and area are vectors |
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| 23. |
If the angular momentum of a planet is given by vec(L)=5t^(2)hat(i)-6that(j)+3hat(k). What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum? |
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Answer» SOLUTION :Given:`vec(L)=5t^(2)HAT(i)-6that(J)+3hat(K)` To find:`vec(tau)="Torque"` Formula:`vec(tau)=(dvecL)/(dt)=(d)/(dt)(5t^(2)hat(i)-6that(j)+3hat(k))` `=10that(i)-6hat(j)` |
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| 24. |
If heat Q is added reversibly to a system at temperature T and heat Q. is taken away from it reversibly at temperature T., then which one of the following is correct ? |
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Answer» `Q/T- (Q.)/(T.) =0` |
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| 25. |
A hot air balloon is a sphere of radius 8m . The air inside is at a temperature of 60^(@)C. How large a mass can the balloon lift when the outside temperature is 20^(@)C ? Assume air is an ideal gas , R =8.314J" mole"^(-1)k^(-1),1atm=1.013xx10^(5)Pa, the membrane tension is =5N//m. |
Answer» Solution :The pressure inside the balloon be `P_(i) andP_(i)gtP_(o)`. Excess pressure `P_(i)-P_(o)=(2s)/(R )` where s= surface tension ,r =radius of balloon Considering the air to be ideal gas , `P_(i)V=n_(i)RT_(i)`...(1) Where V= is the volume of the air inside the balloon) `n_(i)` = the NUMBER of moles of inside air `T_(i)=` the temperature of inside air For air outside the balloon, `P_(o)V=n_(o)RT_(o)`....(2) where V= volume of the air displaced `P_(o)=` air pressure outside `n_(o)` = number of moles of air displaced `T_(o)=` outside temperature From equation (1), `n_(i)=(P_(i)V)/(RT_(i))=(M_(i))/(M_(A))`.....(3) where `M_(i)` mass of air inside and `M_(A)` is the molar mass of air .) From equation (2), `n_(o)=(P_(o)V)/(RT_(o))=(M_(o))/(M_(A))`....(4) (Where `M_(o)=` mass of air outside that has been displaced.) If W is the load it can raise , then `W+M_(i)g=M_(o)g` `W=M_(o)g-M_(i)g` ...(5)  There is `21%O_(2)and79%N_(2)`in air . `therefore` Molar mass of air `=(0.21xx32)+(0.79xx28)` `=28.84g` Weight carried by balloon , `W=(M_(o)-M_(i))g` From equation (3),(4) putting value of `M_(o)M_(i)` in above equation, `W=((M_(A)V)/(R)(P_(o))/(T_(o))-(M_(A)V)/(R)(P_(i))/(T_(i)))g` `W=(M_(A)V)/(R)[(P_(o))/(T_(o))-(P_(i))/(T_(i))]-g` (Volume of balloon `V=(4)/(3)pir^(3))` `W=(0.02884xx(4)/(3)pixx8^(3)xx9.8)/(8.314)` `[(1.013xx10^(5))/(293)-(1.013xx10^(5))/(333)]` `W=(0.0884xx(4)/(3)pixx8^(3))/(8.314)xx(1.013xx10^(5))` `[(1)/(293)-(1)/(333)]xx9.8` `W=3044.2N` `thereforemg=3044.2` `thereforem=(W)/(g)=(3044.2)/(10)=304.42kg` `m=305kg` |
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| 26. |
A body is thrown horizontally with a velocity of v m/s from the top of a tower of height 2h reaches the ground in .t. seconds. If another body double the mass is thrown horizontally with a velocity 5v m/s from the top of another tower of height 8h it reaches the ground in a time of |
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Answer» 2t |
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| 27. |
A horizontal pipe of no uniform cross section has water flow through it such that the velocity 1ms^(-1) at a point where the pressure is 50 Kpa The pressure at a point where the velocity flow is 2ms^(-1) is |
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Answer» 1500 Pa |
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| 28. |
(A) : In the uniform gravitational field centre of mass and centre of gravity coincide with each other. Where as in ther case of non - uniform gravitational field they will not coincide with each other.(R ) : Centre of mass is independent of acceleration due to gravity but centre of gravity depends on acceleration due to gravity. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 29. |
A gun is kept on an inclined plane. The maximum ranges up and down the plane are 250 m and 750m respectively what would be the maximum range in a direction at right angles to the line of greatest slope? |
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Answer» |
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| 30. |
(a) Give reason why transverse waves can be propagated in solids, (b) Give reason why longitudinal waves can be propagated through all the three types of media - solids, liquids and gases.(c) Describe in brief about two types of waves that can be propagated on the free surface of water. |
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Answer» Solution :(a) In order to have propagation of TRANSVERSE waves, we need to produce crests and troughs. TOR this we need to produce shearing strain WHEREBY we can change the shape of a medium. For this to happen, medium selected for the propagation of transverse waves, must be capable enough to sustain shearing stress i.e. They must have very high value of modulus of rigidity. Now we know that solids have very high value of modulus of rigidity. Hence transverse waves can be propagated in solids. (b) Longitudinal waves propagate in the form of condensations and rarefactions, causing changes in density and pressure. Accordingly, compressive and expansive stresses are produced. All the three media - solids, liquids and gases can sustain compressive and expansive stresses and so longitudinal waves can be propagated in all of these media. (c) Two types of waves which can propagate on free surface of water (i.e, on the uppermost layer of water which is in contact with atmosphere) are capillary waves and gravity waves. (i) Capillary waves : They are transverse waves of fairly short wavelengths (few centimeter only). Here necessary restoring force is produced due to surface tension property of water. (ii) Gravity waves : They are transverse as well as longitudinal in nature. They have large wave lengths ranging from several meters to several hundred meters. Here necessary restoring force is produced by the pull of gravity, which compels water surface to remain at lowest possible LEVEL. During the propagation of these waves on the free surface of water all the particles of water medium, from surface to the bottom oscillate with gradually decreasing amplitude Here particles of water medium oscillate up and down as well as back and forth. The waves in an ocean are the combination (superposition) of transverse and longitudinal waves. Transverse and longitudinal waves are FOUND to be travelling with different speeds even in the same solid medium (because for a given solid medium, its shear modulus and bulk modulus are different.) |
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| 31. |
Three measurements of the time for 20 oscillations of a pendulum give t_(1)= 39.6s, t_(2)= 39.9s and t_(3)= 39.5s The accuracy in the measurement is |
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Answer» 39.6 sec |
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| 32. |
A disc of radius r rolls without slipping on a rough horizontal floor. If veloocity of its centre of mass is v_(0), then velocity of point P, as shown in the figure (OP=r//2 and angleQOP=60^(@))is |
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Answer» `v_(0)` |
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| 33. |
In a faulty barometer some air is trapeed above the mercury column in the tube when the barometer reads 760mm and 750mm, a correct barometer reads 770mm and 750mm respectively.What is the length of the air column entrapped in the first case? When the faulty barometer reads 752mm,What is the reading in the correct barometer? Assume that the temperature remains constant throughout. |
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Answer» |
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| 34. |
If the kinetic energy of a body increases by 125%, the percentage increase in its momentum is, |
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Answer» 0.5 |
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| 35. |
A rope passess over a pulley which is sufficiently high. Two monkeys of equal mass climb the rope from opposite ends, one of them climbing quickly than the other relative to the rope. Assuming that the pulley and the rope are weightless and the rope is in extensible, which of monkeys will reach the top first? |
| Answer» Solution :Both will reach the top at the same time because the LENGTH of the TWO segments of the ROPE on the two SIDES of the pulley are always EQUAL. | |
| 36. |
A certain mass of an ideal gas having volume V_1 and pressure P_1 at absolute temperature T_1. When it undergoes adiabatic change , its pressure is increasedto P_2 and volume is increased to V_2 and temperature is increased to T_2. If gamma is the ratio of two specific heat capacities of the gas gamma=C_P/C_V then select the incorrect equation from the following. |
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Answer» `T_2=T_1(V_1/V_2)^(GAMMA-1)` |
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| 37. |
The pair of physical quantities having the same dimensional formula are |
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Answer» MOMENTUM, impulse |
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| 38. |
A number of small drops of mercury coalesce adiabatically to form a single drop. The temperature rature of drop |
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Answer» DECREASES |
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| 39. |
(A) : It is difficult for a man to walk on sand. But camel walks easily on sand. (R ) : Camels feet have larger area than feet of man hence pressure is less. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 40. |
The refrative index of a prism is 2 The prism can have a maximum refracting angle of |
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Answer» Solution :CRITICAL angle `theta_c= sin^-1 (1/mu)= sin^-1 (1/2)=30^@` if `A gt 2 theta_C` the RAY does not emerge from the PRISM , SO maximum refracting angle can be `60^@` |
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| 41. |
A rigid body is made of three identical thin rods, each with length fastened together in the form. of letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the arms of H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of H is vertical? |
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Answer» Solution :Using principle of conservation of mechanical enegy, we have `(mgl)/2+mgl=1/2Iomega^(2)` where `omega` is the angular speed of the rod when it becomes VERTICAL and `I=(ml^2/3)+ml^(2)`, assuming mass of EAC rod is `m`. SUBSTITUTING value of `I` in the above equation, we GET `omega=sqrt((9g)/(4l))` |
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| 42. |
Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero. |
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Answer» SOLUTION :Given:`m_(1)=m` `m_(2)=4m` distance = R To find:V = ??  `(Gm)/(x^2)=(G(m.4))/((r-x)^(2))` `(r-x)^(2)=4x^(2)rArr(r-x)=2X` `r=3x` `thereforex=(r)/(3)` `V=(Gm)/(r//3)-(G(4m))/(2r//3)` `=-(9Gm)/(r)` |
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| 43. |
The mass of 1.2 cm^3 of a certain substance is 5.74g. Calculate its density with due regard to significant figures. |
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Answer» SOLUTION :Mass, m= 5.74g, volume, V= `1.2 CM^(3)` density, `d= (m)/(v) = (5.74g)/(1.2 cm^(3))` 5.74 has significant digits and 1.2 has two significant digits. no need for rounding off, `therefore d = (5.74)/(1.2)= 4.783333`. This is to be rounded off to two significant digits. The density of the given SUBSTANCE is `4.783 = 4.8 g cm^(-3)` |
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| 44. |
A wooden box is placed on the back parth of a lorry moving with an acceleration of 6ms^(-2), if mu = 0.5, the acceleration of the box relative to lorry is |
| Answer» Answer :A | |
| 45. |
A larry and a car moving with same momentum are brough to rest by applying the same retarding force. Then |
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Answer» Lorry will come to REST in a SHORTER distance |
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| 46. |
Change in temperature of the medium changes. |
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Answer» frequency of sound waves. Also we have `v = F LAMDA implies lamda prop v ""…(2) (because f =` constant) From relations (1) and (2), we have `lamda prop sqrtT` `implies Wavelength of wave will change with the change in temperature. |
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| 47. |
In a simple pendulum experiment, length is measured as 31.4 cm with an accuracy of 1mm. The time for 100 oscillations of pendulum is 112.0s with an accuracy of 0.01s. The percentage accuracy in g is |
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Answer» 1 |
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| 48. |
A new unit of length is chosen such that the speed of light is vacuum is unity. What is the distance between the sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? |
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Answer» |
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| 49. |
Given reason why mechanical transverse waves are not possible in fluids and gases. |
| Answer» SOLUTION :MECHANICAL WAVES are not possible in fluids and GASES becauses of absenceof a strong cohesive force between the atoms `//` molecules in them. | |
| 50. |
Whenever a liquid is heated in a container, expansion of liquid as well as container lakes place. If gamma is the coefficient of volume expansion of the liquid and alpha is the coefficient of volume expansion of the container. {:("Column-I","Column-II"),("A)Liquid level raises with respect to container.","P)" gamma=2alpha),("B) Liquid level remains same with respect to container.","Q)" gamma lt 3 alpha),("C) Liquid level drops with respect to container.","R)" gamma= 3 alpha),("D)Liquid level remains same with respect to ground.","S)" gamma gt 3 alpha):} |
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Answer» |
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