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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 38801. |
Explain why : (a) Two bodies at different temperatures T_1 and T_2 if brought in thermal contact do not necessarily settle to the mean temperature (T_1 + T_2 )//2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d)The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. |
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Answer» Solution :(a) It is due to that two bodies may have different thermal capacity. When they are brought in thermal contact, HEAT flows from the body at higher temperature to the body at lower temperature till equilibrium is achieved means, the temperature of both the bodies becomes equal. If both bodies have same thermal capacities their final mean temperature may have `(T_1+T_2)/2`. (b) This is because heat absorbed by a substance is directly proportional to be specific heat of the body so, for the same amount of increasing in temperature and body having high specific heat, higher is its heat absorbing capacity and hence this would prevent the PLANT from getting too hot. (c) When a car is in motion the air temperature inside the car increases because of the motion of the air molecules. According to Charles. law temperature is directly proportional to pressure ( P`prop`T). Hence, if the temperature inside a tyre increases, then the air pressure in it will ALSO increases. (d) In the atmosphere of harbour near the OCEAN side, the vapour of the ocean water mixes with the air near the ocean so, atmosphere stays warm. While in the atmosphere of jungle situated at same latitude, there is no water vapour present and so its atmosphere stays less warm. |
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| 38802. |
A black body radiates heat energy at the rate of 2 xx 10^(5)J//s m^(2) at the temperature of 127^(@)C. Temperature of the black body at which rate of heat radiation 32 xx 10^(5) J//s m^(2), is |
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Answer» 400K By Stefan.s LAW, the rate of emission of radiated energy per unit area per unit time is `E= sigma T^(4)` `:. (E_(2))/(E_(1)) = (T_(2)^(4))/(T_(1)^(4)) or (T_(2))/(T_(1)) = ((E_(2))/(E_(1)))^(1//4) = ((32 xx 10^(5))/(2 xx 10^(5)))^(1//4) = 2` or `T_(2) = 2 xxT_(1) =2 xx 400= 800K` |
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| 38803. |
A square hole is punched out of a circular lamena, the diagonal of the square coinciding with the radius of circle. If .a. be the diameter of the circle, the distance of the centre of mass of reminder from previous centre of mass. |
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Answer» `(4(2pi-1))/a` |
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| 38804. |
If A and B persons are moving with V_(A) and V_(B) velocities in opposite directions. Magnitude of relative velocity of B w.r.t. A is x and magni-tude of relative veloctiy of A w.r.t B is y. Then |
| Answer» ANSWER :B | |
| 38805. |
Two vibrating tuning forks produce waves whose equation is given by y_(1) = 5 sin (240 pi t) and y_(2) = 4 sin (244 pi t) . Compute the number of beats per second . |
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Answer» Solution :Given `y_(1) = 5 sin (240 pi t) ` and `y_(2) = 4 sin (244 pi t)` Comparing with `y = A sin (2 pi f_(1) t)` , we get `2pi f_(1) = 240 pi IMPLIES f_(1) = 120 Hz` `2pi f_(2) = 244 pi implies f_(2) = 122 Hz` The number of beats PRODUCED is `|f_(1) - f_(2)| = |120 - 122| = |-2| = 2 ` beats per sec |
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| 38806. |
Select the correct statement for a body in simple harmonic motion, loss of kinetic energy is proportional to : |
| Answer» Answer :B | |
| 38807. |
Consider ..n.. number of indentical cubes each of mass ..m.. and side ..a.. lying on a level ground. Find work done in piling them one above the other. |
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Answer» Solution :The initial total POTENTIAL energy of all the blocks lying on the ground is ` U_(1) = nxxmg""(a)/(2)` The potentia energy, when the blocks are piled to form a single COLUMN is `U_(2) = nmg""(na)/(2) = N ^(2) MG""(a)/(2)` `THEREFORE` Work done in arranging the blocks one above the other is `W = U_(2) - U_(1)` `W=n^(2)mg""(a)/(2)-nxxmg""(a)/(2)(n^(2)-n)` or `W = mg""(a)/(2)n (n-1)` |
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| 38808. |
Derive an expression for co-efficient of performance of refrigerator. |
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Answer» Solution :Coefficient of performance (COP)`(beta)`: COP is a MEASURE of the efficiency of a REFRIGERATOR. It is defined as the ratio of heat extracted from the cold BODY (sink) to the EXTERNAL work DONE by the compressor W. COP `= beta = (Q_(L))/W`...(1) From the equation `Q_(L)+W=Q_(H)` `beta = (Q_(L))/(Q_(H) - Q_(L))` `beta = (1)/((Q_(H))/(Q_(L)) - 1)`...(2) But we know that `(Q_(H))/(Q_(L)) = (T_(H))/(T_(L))` Substituting this equation into equation (1) we get `beta = (1)/((T_(H))/(T_(L)) - 1) = (T_(L))/(T_(H) - T_(L))` |
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| 38809. |
Answer carefully, with reasons : What are the answers to (a) and (b) for an inelastic collision? |
| Answer» SOLUTION :Linear MOMENTUM is conserved during an inelastic COLLISION, kinetic ENERGY is, of course, not conserved EVEN after the collision is over. | |
| 38810. |
A copper wire of 1mm diameter is stret- ched by applying a forcre of 10 N. Find the stress in the wire. |
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Answer» `1.273xx10^(7)N//m^(2)` |
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| 38811. |
The graph between the time period and the length of a simple pendulum is a: |
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Answer» Straight line `T PROP SQRT(L)` `T^(2) prop l` Graph between T and l will be a parabola symmetic about T-axis. |
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| 38812. |
A wire of length1 m and radius 1mm is subjected to a load. The extension is 'x'. The wire is melted and then drawn into a wire of square cross-section of side 1 mm. What is its extension under the same load? |
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Answer» `PI^(2) X` |
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| 38813. |
A brass disc at 0^(@) C has a diameter of 50Cm and a hole of diameter I 0cm. For Brass alpha = 18 xx 10^(-6)//^(@) C. When the disc is heated to 100^(@) C, the diameter of the hole becomes. |
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Answer» 10.018 CM |
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| 38814. |
A cannon and a target are 5.10 km apart and located at the same level. How soon will the shell launched with the initial velocity 240 m/s reach the target in the absence of air drag? |
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Answer» Solution :Hence `v_(0)=240ms^(-1),R=5.10ikm=5100m,,g=9.8ms^(-2), theta=?` From formula `R=(v_(0)^(2)sin 2 theta)/g` We have `sin 2 theta=(Rg)/(v_(0)^(2))`, Putting VALUES we get `sin 2 theta =(5100xx9.8)/(240xx240)=0.8677=(sqrt(3))/2=sin60^(@)` or `sin 120^(@)impliestheta=30^(@)` or `60^(@)` From formula `T=(2v_(0)sin theta)/g` When `theta =30^(@),T_(1)=(2xx240xx0.5)/9.8=24.5`sec When `theta=60^(@),T_(2)=(2xxd240xx0.867)/9.8=42.14` sec |
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| 38815. |
Mark the correct options |
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Answer» If the incident rays are converging we have a real object |
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| 38816. |
The gravitational potential energy of a body of mass m at the earth's surface is mgR_e . Its gravitational potential energy at a height R_efrom the earth's surface will be ..... (Here R_eis the radius of earth) |
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Answer» `-2 mg R_e` `:. U_2 - (-mg R_e) = (mgR_e)/2` `:. U_2 =-1/2 mg R_e` |
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| 38817. |
A car moving with a velocity of 72KMPH stops engine just before ascending an inclined road. It 25% of energy is wasted in overcoming friction, the car rises to a height of (g= 10 ms^(-2)) |
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Answer» 6M |
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| 38818. |
Is an oscillation a wave ? Give reason. |
| Answer» SOLUTION :No, an OSCILLATION is not a wave. The term wave implies the transfer of energy through successive vibrations of the particles of the medium. HENCE the oscillations of the bodydo not CONSTITUTE a wave. | |
| 38819. |
A body executes S H M . The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as function of displacement (x). Which of the following statement is true |
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Answer» K.E is MAXIMUM when X = 0 |
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| 38820. |
Tangential acceleration on a bob of a simple pendulum of mass 100gm when its amplitude is 1^(@) is |
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Answer» `0.017m//"SEC"^(2)` |
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| 38821. |
A satellite of mass 400 kg is in a circular orbit of raduis 2R about the earth where R is radius of the earth. How much energy is required to transfer it to a circular orbit of radius 4R? Find the changes in the kinetic and potential energies ? (R = 6.37 xx 10^(6)m) |
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Answer» Solution :Initial total energy is `-(GMm)/(2r_(2)) = -(GMm)/(8R) = E_(2)` Final total energy is `-(GMm)/(2r_(2)) = -(GMm)/(8R) = E_(2)` The CHANGE in the total energy is `DELTA E = E_(2) - E_(1) = (GMm)/(8R) rArr Delta R = ((GM)/(R^(2))) (mR)/(8)` `Delta R = (GMR)/(8) = (9.8 xx 400 xx 6.37 xx 10^(6))/(8) = 3.13 xx 10^(9)J` Change in kinetic energy = `K_(2) - K_(1) = -3.13 xx 10^(9)J` Change in potential energy `= U_(2) -U_(1) = -6.25xx 10^(9)J` |
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| 38822. |
Two bodies of masses m_(1) and m_(2) are separated by certain distance. If vec(F)_(12) is the force on m_(1) due to m_(2) and vec(F)_(21) is the force on m_(2) due to m_(1), then |
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Answer» `F_(12) = F_(21)` |
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| 38823. |
A 15Kg mass is accelerated from rest with a force of 100N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by F_(B) = A+ BV. whereA = 25 N and B= 0.5 N/(m//s). At what velocity m/sdoes the acceleration equal to one half of the initial acceleration ? |
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Answer» `25MS^(-1)` |
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| 38824. |
A staionary man observes that the rain is falling vertically downward. When he starts running with a velocity of 12 km/h observes that the rains is falling at an angle 60^(@) with the vertical. The actual velocity of rain is |
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Answer» `12SQRT(3)` km/h |
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| 38825. |
If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest |
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Answer» 1 cm |
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| 38826. |
A rocket is fired from the earth's surface to put the pay load in the required orbit/ The motion of the rocket is given by: |
| Answer» Answer :C | |
| 38827. |
In a specific heat experiment 100g of lead (sp. Heat=145Jkg^(-1)K^(-1)) at 100^(@)C is mixed with 200g of water at 20^(@). Find the change in entropy. |
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| 38828. |
What would be the angular speed of earth, so that body lying on equator may appear weightless? ( g = 10 m//s^2 , radius of earth = 6400 km) |
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Answer» `1.25 xx10^(-3)` rad/sec In weightless state `g.=0` Taking for equator `lamda=0` `0 = g - Romega^2` `:. Romega^2 =g` `omega=sqrt(g/R_e)=sqrt(10/(6400xx10^3))=1/800 (rad)/s` `omega=1.25 xx10^(-3) (rad)/s` |
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| 38829. |
A 100 m long train is travelling from North to South at a speed of 30ms^(-1). A bird is flying from South to North at a speed of 10ms^(-1). How long will the bird take to cross the train? |
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Answer» 3s BIRD and train are moving in a opposite direction `:.` Relative velocity `=30+10=40ms^(-1)` Time taken to cross the train (t) `=("distance")/("R.velocity")=(100)/(40)=2.5s`. |
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| 38830. |
When a spring is pulled with a force of 5xx10^(5) dyn, it elongates by 0.1 m. What mass should be attached to the end of the spring so that when pulled a little downwards and released, the mass can perform 2 complete oscillations per second? What will be the maximum velocity of the mass when the amplitude of vibration is 0.01 m? |
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| 38831. |
A block of mass 'M' is placed on the top of a wedge of mass '4M'. All the surfaces are frictionless. The system is released from rest. The distance moved by the wedge at the instant, the block reaches the bottom will be |
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Answer» 0.2 m |
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| 38832. |
Match the Columns |
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Answer» A - 3 , B - 5, C - 2 , D - 1 |
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| 38833. |
When a load is applied to stretch a wire of length 200 cm, it gets elongated by 2 mm. IF the diameter of its cross section is 1 mm, then calculate the decrease in its diameter due to stretching , Poisson's ratio for the material of the wire is 0.24 . |
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| 38834. |
Calculatethe totalnumberof degressof freedompossessedby themoleculesin onecm^3 ofH_2gasat NTP . |
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Answer» SOLUTION :22400`cm^2`of EVERY gascontains`6.02 XX 10^(23)`molecules ` therefore ` Numberof moleculesin `1 cm ^3 ` of ` H_2 ` gas`=( 6.02xx 10^(23))/( 22400 ) = 0.26875xx 10^(20)` Numberof degreesof freedomof a` H_2` gasmolecule = 5 ` therefore `Totalnumberof DEGRESS of freedomof `0.26875xx 10^(20)`molecules `= 0.26875xx 10^(20) xx 5 = 1.34 375xx 10^(20)` |
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| 38835. |
Mass is inertia and measure of inertia mean property to oppose change. |
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Answer» |
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| 38836. |
(A): During an adiabatic process, an ideal gas expands by decrease in its internal energy only (R ) : During an adiabatic process, heat cannot be exchanged between and its surroundings |
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Answer» Both (A) and(R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 38837. |
A body is projected at an angle 30^(@) to the horizontal with a speed of 30 ms^(-1). The angle made by the velocity vector with the horizontal after 1.5 s is ""(g=10ms^(2)) |
| Answer» ANSWER :A | |
| 38838. |
Consider Example 6.8 taking the coefficient of friction , mu to be and calculate the maximum compressionof 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 as shown in Fig. 6.9. We invoke the work-energy theorem, rather than the conservation of mechanical energy. The change in kinetic energy is `triangleK= K_(f)-K_(t)= 0-(1)/(2)mv^(2)` The work done by the net force is `W= -(1)/(2)kx_(m)^(2)- mu mg x_(m)` Equatting we have `(1)/(2)mv^(2)= (1)/(2)k x_(m)^(2)+ mu mg x_(m)` Now `mu mg= 0.5xx10^(3)xx10= 5xx 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)+ 2 mu mg x_(m)- mv^(2)= O` `x_(m)=(-mu mg+[mu^(2)m^(2)g^(2)+MKV^(2)]^(1/2))/(k)` where we take the positive square root SINCE `x_(m)` is positive. Putting in numerical VALUES we obtain `x_(m)= 1.35m` which, as expected, is less than the result in Example 6.8. If the two forces on the body consist of a conservative force `F_(c )` and a non-conservative force `F_(nc)`, the conservation of mechanical energy formula will have to be modified. By the WE theorem `(F_(c )+F_(nc)) trianglex= triangleK` But `F_(c ) triangle x= -triangle V` Hence, `triangle(K+V)= F_(nc) triangle x` `triangle E= F_(nc) trianglex` where E is the total mechanical energy. Over the path this assumes the form `E_(f)-E_(t)= W_(nc)` where `W_(nc)` is the total work done by the non-conservative forces over the path. Note that unlike the conservative force, `W_(nc)` depends on the particular path i to f. 
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| 38839. |
For the wave discribed in Exercise 15.8, plot the displacement (y) versus (t) graphs dor x = 0.2 and 4 cm. What are shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase? |
| Answer» Solution :All the GRAPHS are SINUSOIDAL. They have same amplitude and FREQUENCY, but DIFFERENT initial phases. | |
| 38840. |
Dimensional formulae are used (A) to convert one system of units into another (B) to find proportionality constants (C ) to check the correctness of an equation |
| Answer» Answer :C | |
| 38841. |
When a gas is compressed to half of its initial volume. If it is comparision between isothermal and adiabatic process, then a) Work done is less if the change is Isothermal b) Final pressure is more if the change is adiabatic c) Final pressure is less if the change is Isothermal |
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Answer» both a and B are CORRECT |
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| 38842. |
Assertion : Dynamic friction is always less than limiting friction Reason : Because once the actual motion starts inertia of rest has been overcome . |
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Answer» If both, ASSERTION and Reason are true and the Reason is the CORRECT explanation of the Assertion |
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| 38843. |
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass (m) hangs freely from the free end of the spring. The area of cross-section and young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released it will oscillate with a time period T equal to |
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Answer» `2pi SQRT((m)/( K))` |
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| 38844. |
A double star is a system of two stars moving around the centre of mass of the system due to gravitation. The distance between the components of the double star is d = ((GMT^(2))/(xpi^(2)))^(1//y) (given, the total mass is M and time periodis T ) Find the value x+ y |
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| 38845. |
The motion of a particle along a straight line is described by the function s=6+4t^(2)-t^(2) in SI units. Find the velocity, acceleration, at t=2s, and the average velocity during 3rd second. |
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Answer» SOLUTION :`s=6+4t^(2)-t^(4)` Velocity `=(dx)/(dt)=8t-4t^(3)` when t-2 Velocity `=8 xx 2-4 xx 2^(3)"Velocity =-16m/s"` Acceleration `a=(d^(2)s)/(dt^(2)) =8""12t^(2)` when t-2 acc `=8-12 xx 2^2=-40 ""acc=-40m//s^2` displacement in 3 seconds `s_(2)=6+4, 3^(2)-3^(4)=-39m` DISPLACMENT during 3rd second Average velocity during 3rd second `=(+45)/(1)=-45m//s` -ve sign indicates that the body is moving in OPPOSITE direction to the initial direction of motion. |
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| 38846. |
An open tank filled with water has a hole at its bottom. If the total pressure at the bottom of the tank is 2 atmospheres , the velocity of efflux is (nearly) ( 1 Atmosphere = 10^(5) "Nm"^(-2) ) |
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Answer» 30`"ms"^(-1)` |
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| 38847. |
The value of surface tension depends upon |
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Answer» NATURE of SOLID in CONTACT with LIQUID |
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| 38848. |
Beats are produced by two progressive waves. Maximum loudness at the waxing is x timesthe loudness of each wave. The value of x is ............ . |
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Answer» 1 Hence, maximum intensity is proportional to `4a^2` Loudness of ONE WAVE isgiven by ` x(4a^2)/(a^2) = 4` |
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| 38849. |
In an experiment, the values of refractive index of glass were found to be 1.54, 1.53 , 1.44 and 1.54 , 1.56 and 1.45 in successive measurements . Calculate (i) Mean value of refractive index of glass (ii) Absolute error in each measurement (iii) Mean absolute error (iv) relative error and (v) Percentage error. |
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Answer» 1.51, 0.04, 0.03,3% |
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| 38850. |
A uniform square sheet has a side length of 2R. If one of the quadrants is removed, the shift in the centre of mass: |
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Answer» `R/(3sqrt(2))` |
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