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.
| 13701. |
For (1)/(m) of the distance between two stations a train is uniformly accelerated and (1)/(n) of the distance it is uniformly retarded. It starts from rest at one station and comes to rest at the other. The ratio of the greatest velocity to the average velocity will be: |
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Answer» `(1 + (1)/(m) - (1)/(n))` Also `(v_("max"))^(2) = 2a_(2) xx (s)/(n)` AV. VELOCITY `= (s)/(t_(1) + t_(2))` Where `t_(1) = (v_("max"))/(a_(1))` and `t_(2) ((v_("max"))/(a_(2)))`
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| 13702. |
Assertion : Certain velocity of object freely falling in viscous liquid is know as terminal velocity .Reason : Terminal velocity and critical velocity are same . |
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Answer» Both ASSERTION and REASON are true and the reason is the correct EXPLANATION of the assertion. |
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| 13703. |
Explain resonance. Give an example . |
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Answer» Solution :The frequency of external periodic force (or driving force) matches with the natural frequency of the VIBRATING BODY (DRIVEN). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a PHENOMENON is known as resonance and the corresponding vibrations are known as resonance vibrations. EXAMPLE: The breaking of glass due to sound |
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| 13704. |
The wavelength of ligth is vaccum is 5000A^@ when it travels normally through diamond of thickness 1.0mm find the number of waves of light in 1.0mm of diamond. |
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Answer» 4834 waves |
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| 13705. |
Two bodies are thrown with the same initial velocity at anglesalpha and(90-alpha)to the horizon. What is the ration of the maximum heights reached by the bodies. |
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Answer» `COT^(2) ALPHA` |
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| 13706. |
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. The kinetic energy associated with the rotation of the cylinder is |
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Answer» 3025 J `= (1)/(2) xx ((1)/(2) MR^(2)) omega^(2)= (1)/(4) xx 20 xx (0.25)^(2) xx 100 xx 100 = 3125 J` |
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| 13707. |
A particle P is moving uniformly along a straight line XX with a speed of 4m/s. The particle crosses the shown position at t=0, what would be the angular velocity of P about O at t=1 s? |
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Answer» `4/3` rad/s |
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| 13708. |
A vertical U-tube contains a liquid. The total length of the liquid column inside the tube is l. When the liquid is in equilibrium, the liquid surface in one of the arms of the U-tube is pushed down slightly and released. The entire liquid column will undergo a periodic motion. a) The motion is not simple harmoic motion. b) The motion is simple harmonic motion. c) If it undergoes simple harmonic motion, the time period will be 2pisqrt(l//g). d) If it undergoes simple harmonic motion, the time period will be 2pisqrt(l//2g) |
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Answer» a and B are correct |
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| 13709. |
In a certain region of space there are only 5 molecules per cm^(3) on an average. The temperature is 3K. What is the average pressure of this very dilute gas ? (K=1.38 times 10^(-23)J//mol" "K) |
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Answer» <P> Solution :According to ideal GAS equation PV = `MU` RT.`P=(mu" "N_(A)kT//V)["as "R=N_(A)k]" (or) "P=(N//V)kT["as "mu" "N_(A)=N]" (or) "P=nkT" "["as "n=N//V]` here n = 5 molecules/`cm^(3)=5" molecules"//(10^(-2)m)^(3)=5 times 10^(6)"molecules"//m^(3)` So `P=(5 times 10^(6))(1.38 times 10^(-23))(3)=2.07 times 10^(-16)Pa` |
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| 13710. |
An ideal gas is compressed to half of the volume. How much work is done if the process of compression isAdiabatic ? In which of the two cases will the work done be more ? How do you account for the difference? |
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Answer» Solution :In case of adiabatic change : ` W=( nR[T_o- T_1 ])/((1- gamma) )` with ` TV^(-1) ` = costant `i.e.,w = (nRT_1)/(( gamma -1))[1- (T_p)/(T_1 )]with(T_f)/(T_l)]` with ` (T_F )/( T_l) = ((V_1)/( v_F))^)( gamma-1)` so ` W= (nRT)/([ (5//3 )-1]) [1-2^(2//3)][ asgamma= 5/3,T_f =T]` ` (or)W_ A=- 3/2nRTxx 0.5874=- 0.88nRT....(2)` Negative sign means that work is done on the gas. Comparing EGNS. (1) and (2) we find that for same compression `W_A gtW_1`. i.e., more work is required in adiabatic compression than in ISOTHERMAL. In adiabatic compression temperature and hence internal ENERGY of the gas also increases and so more work will be required in comparison to isothermal compression in which temperature and hence internal energy remains constant. |
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| 13711. |
Get an expression for stopping distance of a vehicle in terms of intial velocity v_(a) and deceleration "a"… |
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Answer» Solution :Let s be the distance travelled by a VEHICLE before it stops USING `v^(2)-u^(2)=2as,` `0_(0)^(2)=-2as` `s=(v_(0)^(2))/(2a)` The stopping distance is directily proportional to `v_(0)^(2)`. By DOUBLING intial velocity it increase stopping distance by 4 TIMES. Provided deceleration is kept as constant. |
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| 13712. |
An ideal gas is compressed to half of the volume. How much work is done if the process of compression isIsothermal |
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Answer» Solution :In case of ISOTHERMAL change `W = NRT log. (V_s //V_t)` So `W_t= nRT log, (1//2) = -0.6930nRT .`... (1) |
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| 13713. |
How will you .weigh the Sun., that is estimate its mass? You will need to know the period of one of its planets and the radius of the planetary orbit. The mean orbital radius of the Earth around the Sun is 1.5 xx 10^8km. Estimate the mass of the Sun. |
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Answer» SOLUTION :To estimate the mass of the Sun, we require, the time period of revolution TONE of its planets (say the Earth). Let `M_s, M_e`be the masses of Sun and Earth respectively and r be the MEAN orbital radius of the Earth around the Sun. The gravitationalforce acting on Earth due to Sun is `F = (GM_s.M_e)/(r^2)` Let the Earth be moving in circular orbit around the Sun with a uniform angular velocity `omega`. The centripetal force acting on Earth is `F. = M_e r omega^2 = M_e r (4pi^2)/(T^2)`. As this centripetal force is provided by the gravitational pull of Sun on Earth, so `(GM_sM_e)/(r^2) = M_e r (4pi^2)/(T^2) " or " M_s = (4pi^2 r^3)/(GT^2)` Knowing r and T, mass `M_s` of the Sun can be estimated. In this equation. we are GIVEN, `r = 1.5 xx 10^8 KM = 1.5 xx 10^11 m , T = 365 `days = `365 xx 24 xx 60 xx 60s` `M_s = (4 xx (22//7)^2 xx (1.5 xx 10^11)^3)/((6.67 xx 10^(-11)) xx (365 xx 24 xx 60 xx 60)^2) = 2 xx 10^30 kg ` |
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| 13714. |
A uniform rod of mass 2kg. And length 1m is lying on a horizontal surface.If the work done in raising oneend of the rod through an angle 45^(@) is .W., the work done in raising it further through 45^(@) is, |
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Answer» W |
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| 13715. |
Simple harmonic motion is represented by the equation x = 4 sin(2pi t+pi//4) metre. What is the initial displacement? |
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Answer» Solution :Given x = 4 `SIN (2PI t + pi/t)` to CALCULATE initial displacement put t=0 in the given equation. `therefore x = 4 sin (0 + pi/4) = 4 sin 45^@ = 4 1/(sqrt(2)) = 2SQRT2 ` metre. |
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| 13716. |
The teeter toy consists of two identical weights hanging from a peg on dropping arms as shown. The arrangement is surprisingly stable. Let us consider only oscillatory motion in the vertical plane. Consider the peg and rods (connecting the weights to the peg) to be very light. The length of each rod is l and length of the peg is L. In the position shown the peg is vertical and the two weights are in a position lower than the support point of the peg. Angle alpha that the rods make with the peg remains fixed.(a) Assuming the zero of gravitational potential energy at the support point of the pegevaluate the potential energy (U) when the pegis tilted to anangle theta to the vertical. The tip of the peg does not move.(b) Knowing that U shall be minimum in stable equilibrium position prove that theta = 0 is the stable equilibrium position for the toy if the two weights are in a position lower than the support point of the peg |
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| 13717. |
Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure of 8N.m^-2. The radii of bubbles A and B are 2 cm and 4 cm respectively. Surface tension of the soap water. Used to make bubbles is 0.04N.m^-1. Find the ratio n_B//n_A, where n_A and n_B are the number of moles of air in bubbles A and B respectively.(Neglect the effect of gravity), |
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| 13718. |
The average depth of Indian Ocean is about 5 km. Calculate the fractional compression,(Delta V)/(V) of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 xx 10 ^(9) Nm ^(-2)(Take g=10 ms ^(-2)) |
| Answer» Solution :FRACTIONAL compression `0.45 xx 10 ^(-2)` | |
| 13719. |
Adiabatic expansion produces cooling, why? |
| Answer» SOLUTION :Adiabatic PROCESS is a physical change taking place suddenly. Pressure, volume and temperature change. No heat is absorbed from SURROUNDINGS. When the gas expands it does work. For doing work the necessary energy is absorbed from its internal energy. THUS internal energy decreases. So temperature falls. | |
| 13720. |
What is the excess of pressure inside a spherical drop of water of radius 1mm. Surface tension of water is 70xx10^(-3)N//m. |
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Answer» Solution :Radius of WATER drop, R = 1mm = `1xx10^(-3)m`, Surface tension of water, `T=70xx10^(-3)Nm^(-1)` Excess PRESSURE inside the drop, `P=(2T)/r=(2xx70xx10^(-3))/(1xx10^(-3))=140Nm^(-2)`. |
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| 13721. |
An iron ball and a feather are both falling from a height of 10 m. (a) What are the time taken by the iron ball and feather to reach the ground? (b) What are the velocities of iron ball and feather when they reach the ground? (Ignore air resistance and take g=10ms^(-2) |
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Answer» Solution :SINCE kinematic equations are independent of mass of the object, according to equation the time TAKEN by both iron ball and feather to REACH the ground are the same. This is given by `T=sqrt((2h)/(g))=sqrt((2xx10)/(10))=sqrt2s =1.414s` THUS, both feather and iron ball reach ground at the same time. By FOLLOWING equation both iron ball and feather reach the Earth with the same speed. It is given by `v=sqrt(2gh)=sqrt(2xx10xx10)=sqrt200ms^(-1)~~14.14ms^(-1)` |
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| 13722. |
Which of the following is not a state function ? |
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Answer» Temperature |
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| 13723. |
For cyclic process which of the following quantity is zero ? |
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Answer» Solution :As internal energy of a state variable and first state coincide with the initial state in CYCLIC process. `THEREFORE DeltaU=0` |
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| 13725. |
The efficiency of a camot's engine is 2/5. When the temperature of the source is increased by 20 ^(@)C then its efficiency is found to increase to 9/20. Calculate the temperature of source and sink. |
| Answer» SOLUTION :`T _(2) //T_(1) =1-0.4 = 0.6 , 0.45 = 1 - T _(2)// (T _(1) + 20),` solving,` T _(1) = 220 K, T _(2) = 132 K` | |
| 13726. |
The work done in lifting a body of mass 20kg and specific gravity 3.2 to a height of 8m in water is, (g= 10 ms^(-2)) |
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Answer» 110J |
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| 13727. |
When any object is said to have retardation ? |
| Answer» SOLUTION :When velocity and acceleration of object are in opposite, then it is SAID to have RETARDATION | |
| 13728. |
A car and truck have some momentum. Which will have more kinetic energy? |
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Answer» Solution :From the EXPRESSION K`= (p^(2))/(2m)` If .p. is CONSTANT .K. is inversely proportional to the MASS of the body. Hence, the KINETIC energy of the CAR is more than that of the truck. |
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| 13729. |
A vehicle is moving on a banked road theta is angle of banking and mu is coefcient of friction between tyre of vehicle and road. The vehicle is moving along horizontal circular path of radius (r) velocity of vehicle for which it does not slip is 'v'. Choose the correct alternative |
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Answer» V should between 'o' and `SQRT(mugr)` |
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| 13730. |
A rectangular plank of mass m_(1) and height a is on a horizontal surface. On the top of it another rectangular plank of mass m_(2) and height b is placed. The potential energy of the system is |
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Answer» `(m_(1)+m_(2))((a+b))/2G` |
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| 13731. |
A copper block of mass 5 kg slides downalong a rough inclined plane of inclination 30^(@)with a constant speed. The increase in the temperature of the block as it slides down through 100cm assuming that the loss of mech anical energy goes into copper block as thermal energy. (specific heat of copper420 J kg^(-1)k^(-1) g=10ms^(-2)) |
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Answer» `1.19xx10^(-3)""^(@)C` |
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| 13732. |
Two SHMs are represented by the equations y_(1)=10sin(3pit+pi//4) and y_(2)=5[sin3pit+sqrt(3)cos pit]. Their amplitudes are in the ratio |
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Answer» `1:2` |
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| 13733. |
If a capillary tube of radius r is immersed in waster of surface tension T and density rho, how much heat will be evolved when capillary rise takes place completely ? |
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| 13734. |
For a mono - atomic gas, the molar specific heat at constant pressure divided by the molar gas constant R, is equal to …………… . |
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Answer» 2.5 |
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| 13735. |
"It is hotter at the same distance over the top of a fire than in front of it"Why ? |
| Answer» Solution :In front of the fire heat REACHES only by RADIATION. But above the fire heat reaches DUE to CONVECTION and radiation. | |
| 13736. |
A wire of length L_0 is supplied heat to raise its temperature by T. gamma is the coefficient of volume expansion of the wire and Y is Young's modulus of the wire then the energy density stored in the wire is |
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Answer» `(1)/(2) gamma^(2) T^(2) Y` |
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| 13737. |
A solid sphere of mass M and radius R spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is |
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Answer» `(2)/(5) pi^2 MR` Here , `I= (2)/(5) MR^(2) , omega = (600 xx 2pi)/(60) = 20 pi` rad/s `therefore K_(R) = (1)/(2) xx (2)/(5) MR^(2) (20 pi)^(2) = 80 pi^(2) MR^(2)` |
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| 13738. |
Calculate the energy radiated per minute from a filament of an incandescent lamp at 3,000 K if the surface area is 10^(-4) m^(2) and its relative emitted is 0.425. |
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| 13739. |
A 30 kg block is to be moved up an inclined plane at an anglue 30^(@) to the horizontal with a velocity of 5ms^(-1). If the frictional force retarding the motion is 150 N find the horizontal force required to move the block up the plane. (g=10 ms^(-2)). |
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Answer» Solution :The FORCE required to a body up an inclined plane is `F = MG sin THETA +` frictional force `= 30(10)sin 30^(@)+150=300 N`. If P is the horizontal force, `F = P cos theta` `P=(F)/(cos theta)=(300)/(cos theta)=(300xx2)/(sqrt(3))` `= 200 sqrt(3)=346 N`. |
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| 13740. |
In giving a patient blood transfusion the bottle is set up so that the level of blood is 1.3 m above the needle which has an internal diameter of 0-36 mm, and is 0-03 m in length. If 4-5 ce of blood passes through the needle in one minute calculate the viscosity of blood if density is 1020 kg m^(-3) |
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Answer» ` eta = pi Pr^(4) //8vl = pi xx 1.3 xx 1020 xx 9.8xx (1.8 xx 10^(-4))^(4) //8xx 7/5 xx 10 ^(-8)xx 0.03 ` ` = 0.0024 ` DECAPOISE |
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| 13741. |
(A) : When an algebraic equation has been derived, it is advisable to check it fordimensional consistency. (R) : Dimensional correctness of an equation guarantees that it is correct. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 13742. |
(A): The root mean square and most probable speeds of the molecules in a gas are the same.(R): The Maxwell distribution for the speed of the molecules in a gas is a symmetrical. |
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Answer» Both (A) and ( R) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 13743. |
How much work is done on a body of mass m in moving it around a horizontal circle of radius R? |
| Answer» Solution :CENTRIPETAL FORCE is always directed along the perpendiculer to the DIRECTION of motion.Hence no WORK is doen by this force `W=overlineF.overlineX=overlineFXcos(pi/2)=0` | |
| 13744. |
A wire of length L and area of cross section A is made of a material of Young's modulus Y. If it is stretched by an amount x, the work done is given by |
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Answer» Solution :RESTORING FORCE in extension x = F =`(AY x)/(L)` Work done in stretching it by dx = dw = F.dx Work done in stretching it from ZERO to x = W = `INT dw = int_(0)^(x)Fdx` `W = int_(0)^(x)(AY )/(L)dx=(1)/(2)(AY x^(2))/(L)` |
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| 13745. |
(A) There is no dispersion of light refracted through a rectangular glass slab. ( R) Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent colours. |
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Answer» Both A and R are true and R is the correct EXPLANATION of A |
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| 13746. |
A wire kepton two bridges 25 cm apart is elongated by o . 0 4 cm by application of a tension. The density and the Young's modulus of the metrial of the wire are 10 g *cm^(-3) and 9 xx 10^(11) dyn * cm ^(-2) , respectively . Find the frequency of the fundamental tone of such a stretchedstring . |
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| 13747. |
Assertion : The time period of revolution of a statellite close to surface of earth is smaller than that revolving away from surface of earth. Reason : The square of time period of revolution of a satellite is directely proportional to cube to its orbital radius. |
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Answer» Both Assertion and REASON are true and Reason is the CORRECT EXPLANATION of Assertion |
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| 13748. |
Determine the number of light years in one metre. |
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Answer» Solution :1 light YEAR = `9.46 XX 10^(15)` m 1 m `= (1)/ (9.46 xx 10^(15)) = 1.057 xx 10^(-16)` light year |
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| 13749. |
An electric kettle has two coils. When one of these is switched on, thje water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes. If the two coils are connected in series, the time taken to boil the water in the kettle is |
| Answer» SOLUTION :The WEIGHT PROVIDES the CENTRIPETAL FORCE. | |
| 13750. |
A person may stand on the lower steps or at the top of a ladder. In which of these cases does the possibility of sliding of the ladder become maximum? |
| Answer» Solution :The possibility of sliding BECOMES maximum when the person STANDS at the top of the ladder, because the COMBINED CENTRE of gravity of the system is very high and the system becomes UNSTABLE. | |