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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.
| 13851. |
What is the frequency of oscillation of apendulum mounted in a lift that is freely falling under gravity ? |
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Answer» Solution :`implies T = 2PI SQRT(l/G) and g = 0 :. T= oo` `:. F = 1/oo=0` |
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| 13852. |
A particle is released from height h on a smooth track terminating in a circular path of radius R. A and C are points at top and at horizontal level, respectively of the circular path. Column I represents the different values of height of inclined plane and column Ii gives the conditions during the motion of the particle. |
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| 13853. |
(a) Algebraic sum of moments of masses about centre of mass is zero (b) For small bodies centre of mass coincides with centre of gravity (c) Position of centre of mass depends on co-ordinate system (d) Position of centre of mass is independent of mass distribution |
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Answer» a & b are CORRECT |
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| 13854. |
What is potential energy ? Explain gravitational potential energy . |
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Answer» Solution :Potential energy is the stored energy by virtue of the position or CONFIGURATION of the body .If configuration of the system changes , then its potential energy changes . Gravitational potential energy : MULTIPLICATION of weightof the body and its height from the surface of earth is called the gravitational potential energy relative to the surface of earth . Suppose a ball of mass is lie down of earth surface the gravitational force acting on it is mg . If this ball lifted upto height h from the surface ofearth , then externalforce of mg exerted on it in upward direction . Work done by external force `W = Fd COS theta ` but F = mg, d = h and `theta =0^(0)`(External force and displacement are in upward direction hence `theta = 0^(@)`) ` :. W = mgh cos 0^(@)` ` :. W = mgh ` This type of amount of work is stored as gravitational potential energy . Work doneby gravitational force `W = - mgh` ` :. ` Magnitude of gravitational potential energy is equal to the negative value of work done by gravitational force . ` :. V(h) = 0 mgh` where V(h) = gravitational potential energy ` :, (dV(h))/(dh) = - mg ` `= - F ` where F is the gravitational force and it is in downward direction and so it is negative. When a ball fall freely then its speed increases . Its speed is obtained when it touches the earth by equation of MOTION , `v^(2) - u^(2) = 2gh ` For free fall , u = 0 ` : v^(2) =2gh` MULTIPLYING `m/2` on both sides . ` 1/2 mv^(2) = mgh` It indicates that gravitational potential energy of free fall body converted into kinetic energy . Dimensional formula of potential energy `[M^(1)L^(2)T^(-2)]` and its SI unit is Joule . |
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| 13855. |
A heavy stone is thrown from a cliff of height h in a given direction. The speed with which it hits the ground |
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Answer» is same as the SPEED of projection |
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| 13856. |
Bob of a simple pendulum of length L is projected horizontally with a speed of u=sqrt(4gh), , from the lowest position. Find the distance of the bob from vertical line AB, at the moment its tangential acceleration becomes zero. |
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| 13857. |
For the circuit shown in figure, the equivalent capacitance of the combination is |
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Answer» `13 MUF` |
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| 13858. |
Two lens of focal length -20cm and +10cm are put in combination, find the power of the combination |
| Answer» ANSWER :C | |
| 13859. |
The opposite face of a cubic block of iron of cross-section 8 sq.cm are kept in touch with steam and melting ice.Determine the quantity of ice melted at the end of 5 minutes (K=0.2 CGS units). |
| Answer» SOLUTION :212.13 G | |
| 13860. |
Calculate the average velocity of the particle whose position vector changes from vec(r_(1))=5hati+6hatj" to "vec(r_(2))=2hati+3hatj in a tine 5 second. |
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Answer» `vecV_(avg)=(-3hati-3hatj)/(5)=(-3)/(5)(hati+hatj)` |
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| 13861. |
Match the following :{:("Column I","Column II"),(1."Centripetal force","(i)"(u^(2)sin^(2) theta)/(2g)),(2."Relative velocity "vec(V)_(AB),"(ii)"u t + (1)/(2) at^(2)),(3."Distance in linear motion s","(iii)"(mr)/(v^(2))),(4.h_("Mas")" of a" ,"(iv)"grad_(A)-grad_(B) ),(,"(v)" (mv^(2))/(r)):} |
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Answer» 1 - (IV), 2 - (i), 3 - (III), 4 - (V) |
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| 13862. |
The radius of gyrationof a body is independent of : |
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Answer» Mass of the BODY |
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| 13863. |
A body covers20 m in the 2nd second and 40 m in the 4th second of its motion . If its moiton is uniformaly accelaerated, hwo much distance will it cover in the 7 th second of its motion ? |
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| 13864. |
A cannon of mass 1000 kg located at the base of an inclined plane fires a shell of mass 50 kg in horizontal direction with velocity 180 km//h. The angle of inclination of the inclined plane with the horizontal is 45^(@). The coefficient of friction between the cannon and inclined plane is 0.5. The maximum height, in metre, to which the cannon can ascend the inclined plane as a result of recoil is |
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Answer» `5/6` Let `V` be the velocity of the cannon after firing, then form the conservation of linear mometum,  `0=50x50-1000VimpliesV=2.5m//s` Let the cannon go up by height h. Net work doen by external forces `=/_\KE` `implies Mgh-"mu"Mgcosthetaxxh/(sintheta)=0-1/2MV^(2)` where `M` is the mass of cannon. SOLVING we GET `h=5/24m` |
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| 13865. |
If a particle executes uniform circular motion, choose the correct statement |
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Answer» The VELOCITY and SPEED are constant |
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| 13866. |
Equal number of hydrogen and helium molecules are kept in two identical gas jars at the same temperature .What will be the ratio of the pressures of the gases in the two jars? |
| Answer» SOLUTION :Pressure is the same in both the JARS. As per Avagadro.s law all gases of equal volume contain the same number of molecules under indentical VALUES of temperature and pressure. In this example ,volume temperature and number of molecular are the same. SO the pressure will be equal in the TWO jars are the RATIO is `1:1`. | |
| 13867. |
Calculate the speed of a communication satellite that appears stationary from a place on the equator. |
| Answer» SOLUTION :`3.076 KMS^(-1)` | |
| 13868. |
Identify the increasing order of angular velocities of following (a) Earth rotating about its own axis (b) Hour.s hand of clock (C) Seconds hand of clock (d) Fly wheel of radius 2m making 300 rp.m |
| Answer» Answer :A | |
| 13869. |
Choose the graph showing the correct relationship between the stopping potential V_0 and the frequency v of light for potassium and tungsten. |
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Answer»
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| 13870. |
The temperatures T_(1)and T_(2)of two heat reservoirs in the ideal Carnot engine are 1500^(@)C and500^(@)C respectively. Which of these, increasing T_(1)by 100^(@)C or decreasing T_(2)by 100^(@)C, would result in a greater improvement in the efficiency of the engine? |
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Answer» Solution :As the efficiency of a heat ENGINE is given by `ETA= 1-(T_(2))/(T_(1))= (T_(1)-T_(2))/(T_(1))` so when `T_(1)` is increased by `100^(@)C`, `eta_(1)=((1600-500))/((1600+273)) = (1100)/(1873)=59% ……(1)` and when `T_(2)` is DECREASED by `100^(@)C`, `eta_(2)=((1500-400))/((1500+273))=(1100)/(1773)=62%........(2)` From Eqns. (1) and (2), it is clear that decreasing `T_(2)` results in greater improvement in the efficiency. |
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| 13871. |
A spectrometer gives the following reading, when used to measure the angle of a prism. Main scale reading = 58.5^(@) , Vernier scale reading = 09 divisionGiven that 1 division on main scale corresponding to 0.5^(@). Total division on the vernier scale is 30 and match with 29 divisions on the main scale. The angle of the prism, from the above data is |
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Answer» `58.59^(@)C` |
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| 13872. |
A piece of ice (heat capacity = 2100 J//kg//^@Cand latent heat = 3.30xx10^(5) J/kg) of mass m gram is at -5^@Cat atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is : |
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| 13873. |
Find the change in internal energy in joule,when 10g of air is heated from 30^(@)C to 40^(@)C. (C_(v)=0.172 kcal//kg" K", J= 4200 J//kcal) |
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Answer» 72.24 J |
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| 13874. |
Calculate the velocity of sound in a mixture of oxygen, nitrogen and argon at 0^(@)C when their masses are in the ratio 2 : 7 : 1. The molecular weights of gases are 32, 28 and 40 respectively |
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| 13875. |
A sample of ideal gas (gamma = 1.4) is heated at constant pressure. If 100 J of heat is supplied to the gas the work done by the gas is |
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Answer» 28.57 J `Delta Q = nC_(P) Delta T` and `Delta U = nC_(V) Delta T` From the FIRST law of THERMODYNAMICS, `Delta W = Delta Q - Delta U = n(C_(P) - C_(V)) Delta T` `:. (Delta W)/(Delta Q) = (n(C_(P) - C_(V))Delta T)/(nC_(P)Delta T) = 1 - (C_(V))/(C_(P)) = 1-(1)/(GAMMA)` `= 1-(1)/(1.4) = (0.4)/(1.4) = (4)/(14) = (2)/(7)` or `Delta W = (2)/(7) Delta Q = (2)/(7) xx 100 J = 28.57 J` |
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| 13876. |
A body of mass 2kg has an initial speed 5 m/s. A force acts as it for some time in the direction of motion. The force - time graph is shown in figure. Find the final speed of the body. |
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Answer» SOLUTION :Area of OAF `= (1)/(2)xx2xx4=4` Area of BGHC `= (1)/(2)(4+2.5)xx0.5=1.625` Area of ABGF `= 2xx4=8` Area of CDEH `= 2xx2.5=5` Total area = Change in mementum = 18.625. `m(v-u)=18.625 V =(18.625)/(2)+5` `V=14.25 m//s` |
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| 13877. |
The potential energy of a body of mass 'm' is given by U = px + qy + rz. The magnitude of the acceleration of the body will be |
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Answer» `(p+q+R)/(m)` |
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| 13878. |
To determine the real and apparent expansion of a liquid experimentally, we have some experimental observations and/or theoretical concepts. Some list of experiments are mentioned in Column-I and relevant information is mentioned in Column-II. Match the two columns. {:("Column-I","Column-II"),("A) Dulong & Petit's Method","P) Coefficient of real expansion"),("B) Specific Gravity Bottle Method","Q) Concept involved is pressure"),("C) Weight - thermometer Method","R) Coefficient of apparent expansion"),("D) Sinker Method","S) Concept involved is specific volume of the liquid"),(,"T) Volume is not involved in the measurement process"):} |
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| 13879. |
Steam burns are more painful than those caused by boiling water at the same temperature because, |
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Answer» steam contains more heat than the same amount of water at the same temperature. When steam condenses it gives out this extra latent heat |
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| 13880. |
The physical quantity which has dimensional formula as that of ("Energy")/( "mass" xx "length") is |
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Answer» Force |
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| 13881. |
The specific heat of air at constant pressure is 1.005 kJ/kg K and the specific heat of air at constantvolume is 0.718 kJ/kgK. Find the specific gas constant. |
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Answer» 0.287 kJ/kg K |
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| 13882. |
A solid sphere is rotationg in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected ? |
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Answer» MOMENT of inertia |
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| 13883. |
A container of volume 1m^3is divided into two equal compartments, one of which contains an idealgas at 300 K. The other compartment is vaccum. The whole system is thermally isolated from its . surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be |
| Answer» ANSWER :A | |
| 13884. |
A particle of mass 2 is projected at an angle 45^(@) with horizontal with a velocity of 20sqrt2m//s. After 1 sec, explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. The maximum height attained by the other part from the ground is (g=10m//s^(2)) |
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Answer» Solution :`M=2m, theta=45^(@), U=20sqrt2ms^(-1)` `u_(x)=u cos theta =20sqrt2xx(1)/(sqrt2)=20ms^(-1)` `u_(y)=u sin theta =20sqrt2 xx(1)/(sqrt2)=20ms^(-1)` But height attained before explosion, `H_(1)=UT-(1)/(2) g t^(2)=20xx1-(1)/(2)xx10xx1^(2)=15m` After 1 sec, `v_(x)=20ms^(-1)` `v_(y)=u_(y)-g t =20-10=10ms^(-1)` Due to explosion ONE part COMES to rest, `m_(1)=m_(2)=m, v_(1)=0` `M(v_(x)i+v_(y)j)=v_(1)bar(v_(1))+m_(2)bar(v_(2))` `2m(20i+10j)=m(0)+mbar(v_(2))` `v_(2)=40i+20j` `v_(y)^(1)=20ms^(-1)` Height attained after explosion, `H_(2)=((v_(y)^(1))^(2))/(2g)=(20xx20)/(2XX10)=20m` `H_("TOT")=H_(1)+H_(2)=15+20=35m` |
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| 13885. |
A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from this hole. a) The water level will rise up in the vessel b) The pressure at the surface of the water will decrease c) The force by the water on the bottom of the vessel will decrease d) The density of the liquid will decrease |
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Answer» B and C are CORRECT |
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| 13886. |
A uniform metal disc of radius R is taken and out of it a disc of diameter R is cut off from the end. The centre of mass of the remaining part will be |
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Answer» `(R )/(4)` from the centre |
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| 13887. |
Can a quantity have units but still be dimensionless? |
| Answer» Solution :YES. For EXAMPLE, ANGLE is dimensionless but has UNIT radian. | |
| 13888. |
A wooden plank of mass M and length L is floating in still water. A persons of mass m starts at one end of the plank and reachesto other end other end in time t_(0), moving with a constant speed. Choose the correct option. (i) The speed of the person as seen from the ground is smaller than (L)/(t_(0)). (ii) The speed of the plank as seen from the ground is ((m)/(m + M)) (L)/(t_(0)). (iii) The speed of the plank as seen from the ground is ((M)/(m + M))(L)/(t_(0)). (iv) The total K.E. of the system is (1)/(2) (m + M)((L)/(t_(0)))^(2). |
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Answer» `(i) ,(ii)` `u = (L)/(t_(0))` When the man moves to the left , the plank moves to the right. Velocity of the plank w.r.t. the ground `= v` Velocity of the man w.r.t. the ground ` = u - v` By the momentum conservation `m(u -v) = Mv rArr v = (MU)/(M + m)` Velocity of the man w.r.t. the ground `= u -v` `(Mu)/(M + m) = ((M)/(M + m)) ((L)/(t_(0))) LT ((L)/(t_(0)))` (i) is `O.K.` `v = ((m u)/(m + M)) = ((m)/(m + M)) ((L)/(t_(0)))` (ii) is `O.K.` `K.E`. of the system `= (1)/(2) Mv^(2) + (1)/(2) m(u - v)^(2)` `= (1)/(2) ((m M)/(m + M))((L)/(t_(0)))^(2)` (iv) is wrong. 
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| 13889. |
Why a brass tumbler feels much colder than a wooden tray on a chilly day? |
| Answer» SOLUTION :Brass is good conductor of heat than WOOD and its thermal CONDUCTIVITY is HIGH. | |
| 13890. |
What will happen of a rod is tied with fixed supports rigidly at both ends and temperature is increased ? |
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Answer» Solution :If a rod is tied with FIXED supports rigidly at both ends, then it will have expansion with increase with increase in TEMPERATURE. At both ends, forces are applied due to rigid supports hence compressive strain is produced. And corresponding thermal stress is produced in rod. As result rod can be banded. E.g. length of track of steel is 5 m and cross sectional area is `40" cm"^(2)`. Thermal expansion is prevented by decrease of `10^(@)C` in temp. Coefficient of linear expansion is `alpha_(l"(Steel)")=1.2xx10^(-5)K^(-1)` There will be slight change in length due to temperature change (it is compressive strain). `:.` Compressive strain `=(DELTAL)/(l)` `:.` Compressive strain `=(alpha_(l)lDeltaT)/(l)` `:.(Deltal)/(l)=alpha_(l"(Steel)")Delta T` `=1.2xx10^(-5)xx10` `=1.2xx10^(-4)` Now, Young.s modulus `"Y = "("Compressive stress")/("Compressive strain")` `:.` Compressive stress = Y `xx` Compressive strain `(DeltaF)/(A)=Yxx(Deltal)/(l)` `:.DeltaF=AYxx1.2xx10^(-4)` `=40xx10^(-4)xx2xx10^(11)xx1.2xx10^(-4)` `=96xx10^(3)N` `~~10^(5)N` Thus, by this value of force with two rigid supports, tracks can be banded. |
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| 13891. |
An inventor claims to have developed an engine that during a certain time interval takes in 110 MJ of heat at 415 K, rejects 50 MJ of heat at 212 K while manages to do 16.7 kWh of work. Do you agree with the inventor's claim? |
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Answer» Solution :The claimed EFFICIENCY `eta_(C)=W/Q_(H)=(16.7kWh)/(110MJ)` But as `1 kWh=10^(3)xx(J//s)xx(60xx60s)=3.6MJ` So `eta_(C)=(16.7xx3.6)/110=0.55=55%"………."(1)` while MAXIMUM possible theoretical efficiency, `eta_(MAX)=1-T_(L)/T_(H)=(T_(H)-T_(L))/T_(H)=(415-212)/415=0.49%"……….."(2)` From Eqns. (1) and (2) it is clear that claimed efficiency is greter than maximum possible theoretical efficiency, so inventor.s claim does not APPEAR to be correct. |
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| 13892. |
A glass rod of diameter d_1=1.5 mm is inserted symmetrically into a glass capillary with inside diameter d_2=2.0 mm. Then the whole arrangement is vertically oriented and brought in contact with the surface of water. To what height (in cm) will the liquid rise in the capillary ? Surface tension of water =73xx10^-3 N.m^-1. |
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| 13893. |
Which of the following statement is incorrect regarding the polar satellite ? |
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Answer» A polar satellite GOES AROUND the earth's pole in north-south direction. Hence,option ( c) is an INCORRECT STATEMENT . While all the other statements are correct. |
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| 13894. |
At 0^(@)C, Pressure measured by barometer is 760 mm. What will be the pressure at 100^(@)C? |
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Answer» 760 |
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| 13895. |
Many great rivers flow towards the equator, what effect does the sediment they carry to sea have on the rotation of the earth ? |
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Answer» The rotation of the earth SLOWS down |
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| 13896. |
A lead shot of 1 mm diameter falls through a long column of glycerin. The variation of its velocity v with distance covered is represented by |
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| 13897. |
Two needles are floating on the surface of water. A hot needle when touches water surface between theneedles, then they move |
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Answer» closer |
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| 13898. |
When a hole is drilled along the diameter of the earth and if a body is dropped into it, it executes SHM with a period T equal to (R&D are radius and density of the earth) a) 2pisqrt((R)/(g)) b) sqrt((3pi)/(GD)) c) 2pisqrt((pi)/(GD)) d)2pisqrt((R)/(2g)) |
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Answer» a and B are TRUE |
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| 13899. |
Swimming is possible on account of |
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Answer» FIRST law of mtion |
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| 13900. |
Eye of an experimenter detects the green light (lambda =4000 Å) at 5xx10^4 photons per square meter per second and ear can detect 10^-13 watt per square metre. The eye is sensitive as a power detector in comparison to the ear by a factor of |
| Answer» Answer :D | |
