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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.
| 13951. |
A ball is suspended by a thread from the ceiling of a tram car. The brakes are applied and the speed of the car changes uniformly from 36 kmh^(-1) to zero in 5 s. The angle by which the ball deviates from the vertical is (g=10 ms^(-2)) |
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Answer» `TAN^(-1)((1)/(3))` |
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| 13952. |
An engine water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the kinetic energy is imparted to water ? |
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Answer» `3/2mv^(3)` |
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| 13953. |
Explain damped oscillation . Give an example. |
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Answer» Solution :The oscillations in which the amplitude decreases gradually with the passage of time are CALLED DAMPED Oscillations. EXAMPLE: 1. The oscillations of a pendulum or pendulum oscillating inside an oil filled container. 2. Electromagnetic oscillations in a tank circuit 3. Oscillations in a dead beat and ballistic GALVANOMETERS. |
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| 13954. |
What is the maximum speed with which a car can move over a convex bridge? |
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Answer» SOLUTION :To move over a convex bridge, the car requires a centripetal FORCE. This centripetal force is provided by the weight of the car. Let us consider that when the car moves with a certain velocity v, the weight of the car is just sufficient to provide that necessary centripetal force, but if the car moves with a velocity greater than v, then car loses its contact with the bridge. Let the mass of the car be m and the radius of the circular path be R. So, the CONDITION for the car tonot lose contact with the surface of the bridge is `(mv^(2))/(r) le "mg" " "or, " " v le sqrt(GR)` So, a car can move with a maximum speed of `sqrt(gr)` over a convex bridge of radius of curvature r such that it does not lose contact with the bridge. |
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| 13955. |
The arrangement shown in figure consists of two identical uniform, solid cylinder, each of mass m, and radius R on which two light threads are wound symmetrically. Find the tension of each thread in the process of motion. The friction in the axel of the upper cylinder is assumed to be absent mf/2x where 'x' is |
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| 13956. |
A water pipe has an internal diameter of 10 cm. Water flows through it at the rate of 20 m/sec. The water jet strikes normally on a wall and falls dead. Find the force on the wall. |
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| 13957. |
What should be the properities of a liquid to satisfy Bernoulli's theorem? |
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Answer» SOLUTION :The liquid should be (i) non-viscous (II) INCOMPRESSIBLE (III) streamlined and (iv) irrelational |
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| 13958. |
Give the binding energy of body lying on the earth at distance r from the centre of earth |
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| 13959. |
Water is flowing down from a water tap of diameter 1.12 cm at the rate of 2 L/min. Discuss the type of flow - laminar or turbulent. The coefficient of viscosity of water is 10^(-3) Pas. |
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Answer» Solution :`D=1.12 xx 10^(-2)m` `V=2L"/MIN"=2/(60) xx 10^(-3) m^(3) s^(-1)` As `V=a v_(c)` `v_(c)=V/a =(V)/(pi D^(2)//4)` As REYNOLD Number `=N_(R)=(pDv_(c))/(eta)` `=(pDV)/(eta n D^(2)//4)` `=(4pV)/(eta pi D)` `=(4 xx 1000 xx 2 xx 10^(-3))/(60 xx 10^(-3) xx 3.14 xx 1.12 xx 10^(-2))` `=3791 gt 3000` The flow is turbulent in nature. |
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| 13960. |
What is the value of Reynold's number for turbulent flow ? |
| Answer» SOLUTION :GREATER than 2000. | |
| 13961. |
A flask with a capillary tube is filled with water at room temperature. When it is kept in boiling water for sufficiently long time water overflows. Does the amount of overflow of water depend on atmospheric pressure? |
| Answer» Solution :Yes, since the boiling point changes with ATMOSPHERIC PRESSURE and the EXPANSION depends on the DIFFERENCE between the boiling point and the INITIAL temperature. | |
| 13962. |
If vec(P) = I + 2j + 6k, its direction cosines are |
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Answer» `(1)/(41), (2)/(41) and (6)/(41)` |
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| 13963. |
{:(,"List - I",,"List - II"),((a),"changes only in direction",(e ),"external torque"),((b),"zero magnitude",(f),"angular momentum"),((c ),"acts along axis of rotation",(g),"kinetic energy"),((d),"none zero and constant (having no direction)",(h),"linear momentum"):} |
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Answer» a-e, b-h, c-f, d-g |
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| 13964. |
In case of stationary sound waves in air the correct statements(s) is a/are (A) each air particle vibrates with the same amplitude (B) amplitude is maximum for some particles and minimum for some other particles (C ) the particles do not execute periodic motion (D) phase of particles in a loop is same |
| Answer» SOLUTION :CONCEPTUAL | |
| 13965. |
When a metal rod attains same temperature through out its length, then its conductivity is |
| Answer» ANSWER :B | |
| 13966. |
Drops of water fall at regular intervals from the roof of a building of height H = 16m, the first drop striking the ground at the same moment as the fifth drop detaches from the roaf. Find the distance between the successive drops. |
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Answer» Solution :Step -I : Time taken by the first drop to touch the ground `=t=sqrt((2h)/g) ` For h=16m, `t=sqrt((2 XX 16)/g) =4sqrt(2/g)` Time interval between two drops is `t_("interval")=((1)/(n-1))t=(1/4)t=sqrt(2/g)` where n=number of drops. Step -ii : DISTANCE between first and SECOND drops `=S_(1)-S_(2)=1/2 gt_("interval")^(2) [4^(2)-3^(2)]=7M` Distance between second and third drops `=S_(2)-S_(3)=1/2 "gt"_("internal")^(2) [3^(2)-2^(2)]=5m` Distance between third and fourth drops `=S_(3)-S_(4)=1/2 "gt"_("internal")^(2) [2^(2)-1^(2)]=3m` Distance between fourth and fifth drops `=S_(4)-S_(5)=1/2 "gt"_("interval")^(2) [1^(2)-0]=1m`. |
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| 13967. |
A thief's car moving on a straight road with a speed of 80km/h it is followed by a police car moving at a constant speed of 120km/h, crossing a T-point 20s later than the thief's car. At what distance from the T-point the police car will catch the thief's car? |
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| 13968. |
Which of the followig inctruemnts has minium least count? |
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Answer» A screw gauge of PITCH 1 mm and 100 divisions n the CIRCULAR scale. `=1MSD-1VSD=1MSD-(19)/(20)MSD` `=(1)/(20)MSD=(1)/(20)mm=(1)/(200)cm=0.005cm` (b) Least count of screw gauge `=("Pitch")/("No. of divisions ono circular scale")` `=(1)/(100)mm=(1)/(1000)cm=0.001m` (c) Least count of spherometer `=("Pitch")/("No. of divions on circular scale")` `=(0.1mm)/(100)=(1)/(1000)mm=0.0001cm` (d) Wavelength of light `lamda~~10^(-5)cm=0.00001cm` CLEARLY the optical instrument is the most precise. |
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| 13969. |
A vec(F) = (5hat(i) +3hat(j) +2hat(k))Nis appliedover a particle which displaces it from its origin to the point vec(r ) = (2 hat(i) - hat(j))m. The work done on the particle in joule is ………. |
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Answer» Solution :Work DONE = `VEC(F) . vec(r )` ` = ( 5hat(i) +3 hat(j) + 2hat(k)) . (2hat(i) - hat(j))` ` = 10 - 3 = 7` J |
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| 13970. |
A rigid uniform rod of mass M and length .L. is resting on a smooth horizontal table. Two marbles each of mass.m. and travelling with uniform speed .V collide with two ends of the rod simultaneously and inelastically as shown. The marbles get struck to the rod after the collision and continue to move with the M rod. If m=M/6and V = L mts/sec, then the time 6 taken by the rod to rotate through pi/2 is: |
| Answer» ANSWER :D | |
| 13971. |
For an ideal gas a) The change in internal energy in a constant-pressure process from temperature T_(1) to T_(2) is equal to nC_(V)(T_(2)-T_(1)), where C_(V) is the molar heat capacity at constant volume and n is the number of moles of the gas b) The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process c) The internal energy does not change in an isothermal process d) No heat is added or removed in an adiabatic process |
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Answer» Only a, B are CORRECT |
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| 13972. |
(a) A steel wire of mass u per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform an lateral strains lt lt longitudinal strains find the extension in the length of the wire. The density of steel is 7860 kgm ^(-3) and Young's modulus =2 xx 10 ^(11) Nm ^(-2) (b) If the yield strength of steel is 2.5 xx 10 ^(8) Nm ^(-2), what is the maximum weight that can be hung at the lower end of the wire ? |
Answer» Solution :(a) Consider an element dx at a distance x from the load `(x =0) . If T (x) and T ( x + dx)` are tensions on the two cross sections a distance dx apart then,  `therefore T(x+dx) -T(x) = gdm = mu g dx ` (where DM = mass of wire dx and `mu =` is the mass per unit length `= (dm )/(dx))` `therefore dT = mu g dx [ because T (x + dx) - T (x) = dT]` Integration on both side, `therefore T(x) = mu g x + C` but at `x =0,` tension `T(0)=0 +C` `therefore Mg = C` where M is suspended mass `therefore T(x) = mu gx + Mg ""...(1)` If element dx, increases by length dr then strain `= (dr)/(dx)` YOUNG modulus `Y= ((T (x))/( A))/( (dr)/( dx)) (dr)/(dx) = (T (x))/( YA)` Integrating on both the sides, `int _(0) ^(r) (dr)/(dx) = (1)/(YA) int T` `therefore r=(1)/(YA) int _(0) ^(1) (mu g x + Mg ) dx` `= (1)/(YA) [ ( mu g x ^(2))/( 2 ) + Mgx ] _(0 ) ^(L) ` `= (1)/(YA) [ (mu g L ^(2))/( 2) + MgL]` Now `mu =(m)/(L) implies mu L = m`( let) `= (1)/(YA) [ (mgL )/(2) + MgL]` but m = area ` xx` length `xx` density `= AL rho` `r = (1)/(YA) [ ( A L rho xx gL )/(2) + MgL ]` `= (1)/(YA) [ (A ro g L ^(2))/( 2 ) + MgL ] ` `=(1)/( 2 xx 10 ^(11) xx pi (0.1 xx 10 ^(-2)) ^(2))` `[ ((pi xx (0.1) xx 10 ^(-2)) xx 7860 xx 10 xx (10) ^(2))/( 2)+ 25 xx 10 xx 10 ]` `therefore r = 15.92 xx10 ^(-7)` `((3.14 xx 786 xx 10 ^(-6) xx 10 ^(3))/( 2 ) + 2500)` `= 15.72 xx 10 ^(-7) (1.234+ 2500)` `=15.72 xx 10 ^(-2) xx 2501.234` `= 3.98196xx 10 ^(-3)` `therefore r ~~4 xx 10 ^(-3) m` (b) The maximum tension would be at `x = L` `T = mu g L + Mg [ because `Putting `x = L ` in eq. (1) ] `= mg + Mg [ because mu = (m)/( L ) implies m=mu L]` `T= (m +M) g""...(2)` and maximum tension force = stress `xx` area `=250 xx 10 ^(6) xx pi xx 10 ^(-6)` `= 250 pi N ""...(3)` `(m + M) g = 250 pi N` here, `m lt lt M , m `is neglected compare to M `Mg = 250 pi` `therefore M = (250 pi)/(g) = ( 250 xx 3.14 )/(10)` ` therefore M =78.5 kg` |
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| 13973. |
A unifrom rod weighs 10 kg and with loqd 5kg attached to one end, it balances on a knife edge at 2m from that end. The length of rod is |
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Answer» 10 m |
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| 13974. |
A particle moves in the xy-plane according to the law x = a sin(omegat) "and" y = a(1-cos omegat) where 'a' and 'omega' are constants. Then the particle. follows : |
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Answer» a parabolic path |
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| 13976. |
Match Column-I with Column-II properly : {:("Column-I","Column-II"),("(1) Distance between the earth and stars","(a) kilmeter"),("(2) Wavelength of infrared wave","(b) Light year"),(,"(c) Angstrom"):} |
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| 13977. |
A soap bubble of radius (1)/( sqrt(pi) cm is expanded to double its radius. If the surface tension is 30 dynes/cm, the work done is |
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Answer» 700 ERGS |
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| 13978. |
A mark on the surface of a glass sphere is viewed froma diametrically opposite position. It appears to be at a distance 10cm from its actual position Find the radius of sphere. |
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Answer» Solution :As the mark is viewed from the diametrically opposite position, refraction takes place at side II of the surface (the mark being on side I as shown) Here `mu_1=1.5, mu_2=1, u=-2R` USING `mu_2/v-mu_1/u=(mu_2-mu_1)/R` `1/v-(1.5)/((-2R))=(1-1.5)/(-R)` `therefore 1/v=0.5/R-1.5/(2R)=-0.5/(2R) or v=-4R` negative sign indicates that the image is FORMED to the left of refracting surface as shown in figure. Further it is GIVEN that, the image of mark is at a DISTANCE 10CM from the object. Hence: `4R=2R+10 therefore R=5cm` 
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| 13979. |
When the temperature of a gas is increased at constant pressure, the number of molecules per c.c. |
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Answer» DECREASES |
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| 13980. |
A hollow sphere of radius 2R is charged to V volts and another smaller sphere of radius R is charged to V//2 volts. Then the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres would becomes V//n. find value of n |
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| 13981. |
A ball A, moving with kinetic energy E, makes a head on elastic collision with a stationary ball with mass n times that of A. What is the maximum potential energy stored in the system during the collision ? |
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| 13982. |
Atmospheric pressure at aheight of about 6km decreases to nearly half of its value at the sealevel , though the height of the atmosphere is more than 100 km . |
| Answer» Solution :The variation of air density with height is non - linear. So, pressure also DEOS not REDUCE linearly h isgiven by `P=P_(0)e^(-alphah)` where `P_(0)` REPRESENT thepressureof air at sea level and `alpha` is a constant. | |
| 13983. |
One - fourth chain is hanging down from a table. Work done to bring the hanging part of the chain on tothe table is (mass of chain = M and length = L) |
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Answer» `(MGL)/(32)` |
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| 13984. |
An annular ring with inner and outer radii R_(1) and R_(2) is rolling without slipping with a uniform angular speed. The ratio of force experienced by the two particles situated on the inner and outer parts of the ring. (F_(1))/(F_(2)) = ………… |
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Answer» `(R_(1))/(R_(2))` `(mv^(2))/(R )=F` `therefore (MR^(2)OMEGA^(2))/(r)=F` `therefore F=mromega^(2)` `therefore F_(1)=mR_(1)omega^(2)` `therefore F_(2)=mR_(2)omega^(2)` `therefore (F_(1))/(F_(2))=(R_(1))/(R_(2))` |
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| 13985. |
In pure rolling velocity of center of mass is equal to |
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Answer» ZERO |
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| 13986. |
A boy of mass M is applying a horizontal force to slide a box of mass M' on a rough horizontal force to slide coefficient of friction between the shoes of the boy and the floor is mu and that between the box and the floor is mu. In which of the following cases it is certainly not possible to slide the box ? |
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Answer» `MU LT mu', M lt M'` |
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| 13987. |
Among solids, liquids and gases, which can have all the three modulus of elasticity? |
| Answer» Solution :SOLIDS have young's modulus. LIQUIDS and gases have BULK modulus. | |
| 13988. |
Which of the following statements is an incorrect statement ? |
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Answer» Kinematics deals with the motion of OBJECTS . |
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| 13989. |
Find the acceleration of the block of mass M in the situation of figure. The coefficient of friction between the two blocks is mu_1 and that between the bigger block and the ground is mu_2. |
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Answer» M it will also have acceleration a towards right. So it will EXPERIENCE two nertial forces as shown the free body diagram. From free body digram........1  `R_1-ma=0` `rarr R_1=ma`..........i Again, `2ma+T-Mg+mu_1R_1=0` `rarr T=Mg-(2+mu_1)ma`.........II From free body diagram -2, `T+mu_1R_1+Mg-R_2=0` Putting the VALUE of `R_1` from i `R_2=T+mu_1 ma+mg` Putting teh value of T from ii, `R_2=(Mg-2ma-mu_1ma)=mu_1ma+Mg+mg` R_2=Mg+mg-2ma`.....iii Again fromn the free body diagram 2, `T+T-R-Ma-mu_2R_2=0` `rarr 2T-Ma-ma-mu_2(Mg+mg-2ma)=0` Putting the values of `R_1 and R_2` from i and iii `2T=(M+m)a+mu_2(Mg+mg-2ma)`....iv From EQUATION ii and iv we have `2T=2mg-2(2+mu_1)ma` `=(M+m)a+mu_2(Mg+mg-2ma)` `rarr 2mg-mu_2(M+m)g` `=a[M+m-2mu_2m+4m+2mu_1m]` `rarr a=([2m-m_2(M+m)]g)/(M+m[5+2(mu_1-mu_2)])` |
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| 13990. |
Amplitude of an SHO is A. When it is at a distance y from the mean position of the path of its oscillation, the SHO receives blow in the direction of its motion, which doubles its velocity instantaneously. Find the new amplitude of its oscillations. |
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Answer» Solution :Velocity of SHO at displacement y. `v= omega_(0) sqrt(A^(2) - y^(2))"""……."(1)` Now instantaneous velocity after giving blow in the direction of motion is `v_(1)`, `v= omega_(0) sqrt(A^(2) - y^(2))` where `A_(1)` is new AMPLITUDE `2v= omega_(0) sqrt(A^(2) - y^(2))""[therefore v_(1)= 2v]` `therefore 2 omega_(0) sqrt(A^(2) - y^(2)) = omega_(0) sqrt(A_(1)^(2) - y^(2))""[=therefore " From EQ. (1) "]` `therefore 4 (A^(2)- y^(2))= A_(1)^(2) -y^(2)""[therefore " Squaring "]` `therefore 4A^(2) - 4Y^(2) = A_(1)^(2) -y^(2)` `therefore 4A^(2)- 3y^(2) = A_(1)^(2)` `therefore A_(1) = sqrt(4A^(2)- 3y^(2))`. |
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| 13991. |
A certain mass of saturated water vapour is contained in a cylindrical vessel under a weightless piston at one atmosphric pressure. The piston is slowly lowered till 0.7g of vapour gets condensed. The volume under thepiston is now only 1/5 th of the original volume. Find the original mass of the vapour. Neglect volume of the liquid formed. What is the temperature of the vapour? |
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| 13992. |
A uniform rod AB of mass m is hinged at one A. The rod is kept in horizontal position by a massless string tied to point B. The reaction of the hinge on the end A of the rod at the instant, when string is cut is mg/x. Where 'x' is |
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| 13993. |
Learned Indian classical vocalists do not like the accompaniment of a harmonium because |
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Answer» INTENSITY of the notes of the HARMONIUM is too LARGE |
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| 13994. |
Of the following situations which are possible in practice |
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Answer» ZERO velocity and non-zero ACCELERATION |
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| 13995. |
Which of the following is not a unit of time |
| Answer» Answer :D | |
| 13996. |
The maximum error in the measurement of mass and density of the cube are 3% and 9% respectively. The maximum error in the measurement of length will be |
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Answer» 0.02 `:. (Deltarho)/(rho)XX 100= (Deltam)/(m)xx 100+ 3((DeltaL)/(L)) xx 100` or ` 3((DeltaL)/(L)) xx 100=((Deltarho)/(rho)) xx 100-((Deltam)/(m)) xx 100= 9%-3%= 6%` or `((DeltaL)/(L))xx 100= 2%` |
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| 13997. |
A clock with a metallic pendulum is 5 seconds fast each day at a temperature of 15^(0)C and 10 seconds slow each day at a temperature of 30^(0)C. Find coefficient of linear expansion Jor the metal. |
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Answer» Solution :Loss (or) gain of time per day `= (1)/(2) prop Delta t xx 86 , 400 ` s At `15^(0)C` clock is 5s fast. At ` 30^(0)C ` it is 10S slow. Between `15^(0)`C & `30^(0)`C at temperature `t^(0)C` it will show correct time. `Delta T prop Delta rArr (5)/(10) = ((t - 15))/((30 - T)) rArr t = 20^(0)C ` Loss of time per day = `(1)/(2) prop Delta t xx 86 , 400` 10 `= (1)/(2) xx prop xx (30 - 20) xx 86, 400` `rArr prop = 2.31 xx 10^(-5) //^(0)` C |
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| 13998. |
A bullet of mass m moving with velocity u passes through a wooden block of mass M=nm as shown in figure. The block is resting on a smooth horizontal floor. After passing through the block, velocity of the bullet becomes v. Its velocity relative to the block is |
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Answer» Solution :`m u =MV + mn v ^(-1) (or) v ^(1) = ((u -v)/( N)) THEREFORE` velocity of bullet relative to block will be `v _(R) = v - v ^(i) = v - ((u-v)/( n)) = ((1+n)v -u)/( n )` |
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| 13999. |
A body falls for 5s from rest. If the accleration due to gravity of earth ceases to act, the distance it travels in the next 3s is |
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Answer» 73.5m |
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| 14000. |
A vector vecA points vertically upward and vecB points towards north. The vector product vecAxxvecB is |
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Answer» `ABcostheta` |
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