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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.
| 12751. |
If a current of 2.33 A is passed through a resistance of 10.485 Omega, the potential is 24.43005, its value in proper significant figures would be |
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Answer» 24.43 V |
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| 12752. |
A block m_(1) strikes a stationary block m_(3) inelastically. Another block m_(2) is kept on m_(3). Neglecting the friction between all contacting surfaces, calculate the fractional decrease in KE of the system in collision. |
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Answer» Solution :Impact takes place between `m_(1)` and `m_(3)` horizontal. Since `m_(2)` is kept on `m_(3)` and all the surface in contact are SMOOTH, friction does not act between `m_(2)` and `m_(3)` during the displacement of `m_(3)` in the impact. Even though there is a friction between `m_(2)` and `m_(3)`, which is very less than the impace FORCE, the frictional force is assumed as non-impulsive force here. Since the impact between `m_(1)` and `m_(2)` is inelastic. `m_(1)` and `m_(3)` will move together toward RIGHT and `m_(2)` will not due to the absence of friction. The VELOCITY of the combined mass `=v^(')=(m_(1)v)/(m_(1)+m_(3))` `(|/_\KE|)/(KE)=(1/2m_(1)v^(2)-1/2(m_(1)+m_(3))v^('2))/(1/2m_(1)v^(2))` `=1-((m_(1)+m_(3))/(m_(1)))((v^('))/v)^(2)` `=1-((m_(1)+m_(3))/(m_(1)))((m_(1))/(m_(1)+m_(3)))^(2)` `=1-(m_(1))/(m_(1)+m_(3))=(m_(3))/(m_(1)+m_(3))` If `m_(2)` and ` m_(3)` are rigidly attached, both together behave as a single mass `(m_(2)+m_(3))` and the answer would have been `(m_(2)+m_(3))/(m_(1)+m_(3)+m_(3))` |
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| 12753. |
A mass M kg is suspended by a weightless string. The horizontal force required to hold the mass at 60^@with the vertical is |
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Answer» MG |
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| 12754. |
Two absolute scales A and B have triple points of water defined to be 200 A and 300 B.A temperature is measured on these scales as T_(A) and T_(B) What is the relation between T_(A) and T_(B)? Given, that triple point of water as 273.16K. |
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Answer» Solution :200 A on absolute scale A corresponds to 272.16 K on Kelvin scale. SIZE of ONE degree on absolute scale A in TERMS of the size of the degree on Kelvin scale = `(273.16)/(200)` . the value of temperature `T_(A) ` on Kelvin Scale `( (273.16)/(200)) T_(A)` Similarly, the value of temperature `T_(B)` on Kelvin scale `( (273.16)/(300)) T_(B)`. But , `T_(A) and T_(B)` represent the same temperature `((273.16)/(200))T_(A) = ((273.16)/(300)) T_(B) rArr T_(A) = (2)/(3) T_(B)` |
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| 12755. |
For a give perimeter in plane, following are some shpaes mentioned. Choose the shape for the loop such that which when placed inside magnetic field will have maximum torque acting on it. |
| Answer» ANSWER :C | |
| 12756. |
A car starting from position of rest, moves with constant acceleration x. Then it moves with constant deceleration y and become stationary. If the total time elapsed during this is t, then the total distance travelled by car in time t is |
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Answer» `((xy)/(x + y )) t ` `therefore In v = v _(0) + at, v _(0) =0 , a = x and ` PUTTING `t =t _(1)` `therefore v = xt _(1)` `therefore t _(1) = (v)/(x) ""...(1)` After gaining maximum velocity, it moves with constant deceleration y in times `t_(2)` and become stationary then, `therefore` Putting `v =v _(0) + at, v =0, v _(0) = v , a =- t and t = t _(2)` `therefore 0=v -yt ^(2) ` in it `therefore t _(2) = (v)/(y) ""...(2)` Total time `t =t _(1) + t _(2)` `therefore t = (v)/(x) + (v )/(y) = ((y+x)/( xy))v` `therefore v = ((xy)/(x + y ))t ` Now AVERAGE velocity `v.= (0+v)/(2) = 1/2 ((xy)/(x + y )) t` `therefore ` Distance covered `d =v. xx t = 1/2 ((xy)/(x + y )) t ^(2)` |
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| 12757. |
Explain gravitational force exerted by an extended object. |
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Answer» Solution :`implies`For the gravitational force between an extended object like earth and a point mass EQUATION `F= (Gm_1 m_2)/r^2` is not DIRECTLY applicable. `implies` Each point mass in the extended object will EXERT a force on the GIVEN point mass and these force will not all be in the same direction. To add up these forces vectorially for all the point masses in the extended object to GET the total force. `implies`This is easily done using Calculus. |
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| 12758. |
The moment of inertia of an uniform circular disc about its central axis is 'I'. Its M.I. about a tangent in its plane is equal to |
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Answer» 2I |
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| 12759. |
A film of water is formed between two straight parallel wires each 10 cm long and at separation 5 mm. To increase the distance between them to 6 mm, the work done is 144 xx10^(-7)J. The surface tension of water is |
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Answer» `7 xx 10^(-3) NM^(-1)` |
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| 12760. |
A boat flaoting in a water tank is carrying a number of large stones. If the stones are unloaded into the water, what will happen to the water level in the tank [ hint : Assume the initial level of the tank to be K and the length of the boat immersed in water to be h. Apply the law of floatition. find the volume of water in th tank. repeat the process after a plece of stone is unloaded. the stone will sinkto the bottom . assume the new height to be H' and h' respectively. make use of the fact that the volume of water remains constant . H-H=V/A((P_(s))/(P_(w))-1) where V = volume of stone , A = area of cross - section of tank , P_(s) = density of stone and P_(w) = desnsity of water .] |
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| 12761. |
SI is reational system of units while MKS system is not rational. Why ? |
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Answer» Solution :SI is a rational SYSTEM because it assigns only one unit to a PARTICULAR physical quantity. For example, SI unit of all types of energy is joule. On the contrary, on MKS system, mechanical energy is measured in joule, heat energyis measured is calorie, electric energy is measured in watt - HOUR. So MKS system in not rational. |
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| 12762. |
Find the maximum possible mass of a greased needle floating on water surface. T is surface tension of water, 1 is the length of the needle |
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Answer» `m_("MAX") = ( 2TI)/(G)` |
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| 12764. |
Aman standing on the edge of a roof of a 20m tall building projects a ball of mass 100g vertically up with a speed of 10 ms^(-1) and simulaneously throws another ball of same mass vertically down with the same speed. Find the kinetic energy of each ball when they reach the ground (g= 10 ms^(-2)) |
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Answer» Solution :Both the two BALLS reach the ground with the same velocity `h=20m, m= 100g= 10^(-1) kg, u= 10 ms^(-1)` `v^(2)- u^(2)= 2gh` `v^(2)= u^(2) + 2gh = 10^(2) + 2 xx 10 xx 20 = 500` K.E. of each ball `=(1)/(2) mv^(2) = (1)/(2) xx 10^(-1)xx 500 = 25J` `THEREFORE` K.E of balls = 25J |
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| 12765. |
If the significant figures are more |
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Answer» percentage ERROR is more and accuracy is LESS |
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| 12766. |
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. The initial length and the initial period of oscillation at a place where g = 9.8 ms^(-2) is. |
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Answer» 0.48 m, 1.39 SEC |
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| 12767. |
Statement I: The bridges are declared unsafe after a long use. Statement II: Elastic strength of bridges decreases with time. |
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Answer» STATEMENT I is TRUE, statement II is true, statement II is a CORRECT EXPLANATION for statement I. |
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| 12768. |
A gas is at temperature 80^(@)C and pressure 5xx10^(-10)Nm^(-2). What is the number of molecules per m^(3) if Boltzmann's constant is 1.38xx10^(-23)JK^(-1) |
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Answer» Solution :Temperature of a gas `(T)=80^(@)C+273=353K` Pressure of a gas (P) `=5XX10^(10)Nm^(-2)` Boltzmann.s CONSTANT `(K_(B)) =1.38xx10^(-23)JK^(-1)` `Vm^(3)` Number of molecules,`n=(PV)/(kT)=(5xx10^(-10)xx1)/1.38xx 10^(-23)xx353` `(5xx10^(-10))/(487.14xx 10^(-23))=0.01026xx10^(13)` `n=1.02xx10^(11)` |
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| 12769. |
What is mechanics ? Give definitions of its sub branches. |
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Answer» Solution :Mechanics : "The BRANCH of physics which STUDIES and deals with motion of object is CALLED mechanics." OR "Kinematics and dynamics are combindly known as Mechanics". The sub branches of mechanics are : (1) Kinematics : The branch which describes the motion without its causes is known as kinematics. (2) Dynamics : The branch which describes the motion with its causes is known as Dynamics |
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| 12770. |
Find the minimum value of F for which 5 kg being to slide if the coefficient of friction between the blocks is changed to 0.5 : |
Answer» Solution :  `f_(2)= mu N_(1) = F` |
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| 12771. |
An object is released from rest. The time it takes to final through a distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated a) the measured times are same b) the measured speed are same c) the actual time in the fall are equal d) the actual speeds are equal. |
| Answer» Answer :A | |
| 12772. |
The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon ? |
Answer» Solution :Angle subtended at distance r due to an arc of LENGTH 1 is  `theta=(l)/(r)` Given `l=R_(e), r=600 R_(e)` `theta=(R_(e))/(60R_(e))=(1)/(60) "rad" = (1)/(60)XX(180^(@))/(pi)` `:. theta=(3)/(pi)=1^(@)` Thus, angle subtended by diameter of the EARTH `=2 theta=2^(@)`. |
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| 12773. |
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping. |
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Answer» Solution :The moment of INERTIA of the hollow cylinder about it axis `I=MR^(2)` [`because` can considered as RING] `3xx(0.4)^(2)` `=0.48kgm^(2)` TORQUE on hollow cylinder `tau=RFsin90^(@)` [`because` Rope is as tangent to cylinder, `theta=90^(@)`] `=RF [because sin90^(@)=1]` `=0.4xx30` `=12NM` Now `tau=Ialphaimpliesalpha=(tau)/(I)` `therefore alpha=(12)/(0.48)=25" RAD s"^(-2)` and tangential ACCELERATION `a_(1)=Ralpha` and radial acceleration `a_(r )=0` `therefore a=sqrt(a_(T)^(2)+a_(r)^(2))` = `sqrt(((Ralpha)^(2)+(0)^(2))^(2))` `=sqrt((0.4xx25)^(2))` `=sqrt((10)^(2))` `therefore a=10ms^(-2)` |
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| 12774. |
A fly moving in a room has degrees of freedom of |
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Answer» one |
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| 12775. |
A bee of mass 0.000096kg sits on a flower of mass 0.0123kg . find the total mass in appropriate significant figures. |
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Answer» `7.202 xx10^(3)` cm |
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| 12776. |
A stone is projected vertically up from the ground with velocity 40 ms^(-1) . The interval of time between the two instants at which the stone is at a height of 60 m above the ground is (g = 10 "ms"^(-2)) |
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Answer» 4s |
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| 12777. |
When water boils at a constant temperature of 100^@C is there any change in the initial energy? |
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Answer» |
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| 12778. |
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m's to the heavier block in the direction of the lighter block. The velocity of the center of mass is |
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Answer» 30 m/s `MV=(M+M)V_(CM)` `:.V_(CM)=(MV)/((M+M))=(10xx14)/(10+4)=10 MS^(-1)` |
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| 12779. |
A wire is halved by cutting it. Would there by any change in the breaking load due to this? |
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Answer» |
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| 12780. |
A length -scale (l) depends on the permittivity (epsilon) of a dielectric material, Boltzmann constant (k_(B)), the absolute temperature (T), the number pr unit volume (n) of certain charged paticles, and the charge (q) carried by each of the particles. Which of the following expression for l is dimensionally correct? |
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Answer» `l = SQRT((nq^(2))/(epsilonk_(B)T))` |
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| 12781. |
The disc of radius r is confined to roll without slipping at A and B. If the plates have the velocities shown then : |
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Answer» ANGULAR VELOCITY of the DISC is 2V/r |
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| 12782. |
A uniform rod of mass M and length L lies radially on a disc rotating with angular speed omega in a horizontal plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Then the kinetic energy of the rod is : |
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Answer» `1/2momega^(2)(R^(2)+(L^(2))/(12))` |
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| 12783. |
An object reaches a maximum vertical heightof 23.0 m when thrown vertically upward on the earth. How highwould it travel on the moon where the acceleration due to gravity it about one sixth that on the earth ? Assume that initialvelocity is the same. |
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Answer» 138 m |
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| 12784. |
The dimensions of Planck's constant are same as |
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Answer» ENERGY |
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| 12785. |
The speed of a train is reduced from 60 km/h, to 15 km/h, while it travels a distance of 450 m. If the retardation is uniform, find how much further it will travel before coming to rest ? |
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Answer» Solution :Here, `u=60 xx (5)/(18)=(50)/(3)n` `v=15 xx (5)/(18)=(25)/(6)m//s` Using `v^(2)=u^(2)+2as`, we GET `((50)/(3))^(2)=((25)/(6))^(2)+2 xx a xx 450 or a=-(125)/(36 xx12)m//s^(2)` If `s^(1)` is the further distance travelled before coming to rest, then `s^(1)=(v^(2))/(2a)=(25)/(6) xx (25 xx 36 xx 12)/(6 xx 2 xx 125)=30m` |
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| 12786. |
The scale of a barometer is made of brass and is correct is 15^(@)C. The reading of the barometer is 0.754m at 5^(@)C. Correct this reading to 0^(@)C. (Linear expansivity of brass =18x10^(-6)K^(-1) and cubical expansivity of mercury =18xx10^(-5)K^(-1)) |
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Answer» |
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| 12787. |
A person of mass 60mg is in a lift The change in the apparent weight of the person when the lift moves up with acceleration of ms^(-2) and the down with an acceleration of 2ms^(-2) is (take g =10 m//sec^(2)) . |
| Answer» Solution :`w_(1) =m(G +a),w_(2) =MG(g -a)` | |
| 12788. |
Answer the following: (a) The casing of a rocket in flight burns up due to friction. At whose expense is the het energy required for burning obtained ? The rocked or the atmosphere ? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? (c)An aritificial satellite orbiting the earth in very thin atmosphere loses its energy graduallydue to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? (d) In figure, theman walks 2 m carrying a mass of 15 kg on his hands. In fig., he walks the same distance pulling the rope behind him. The rope goes over a pulley , and a mass of 15 kg hangs at its other end. In which case is the work done greater ? |
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Answer» SOLUTION :(a) The total energy of a rocket in flight depends on its mass i.e., `P.E. +K.E. =mgh +(1)/(2) mv^(2).` When the casing burns up, its mass decreases. The total energy of the rocket decreases. Henc, heat energy required for burning is obtained from the rocket itselfand not from the atmosphere. (b) This is because gravitational force is a CONSERVATIVE force. Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero. (c) Whenthe artificial SATELLITE orbiting the earth comes closer and closer to earth, its potential energy decreases. As sum of potential energy and kinetic energy is constant, therefore, K.E. of satellite and hence its velocity goes on increasing. However, total energy of the statellite decreases a little on account of dissipation against atmospheric resistance. (d) In figure, force is applied on the mass, by the man is vertically UPWARD direction but distance is moved along the horizontal. `:. theta=90^(@). W=F s cos90^(@)=Zero . ` In figure, force is applied along the horizontal and the distance moved is also along the horizontal. Therefore,, `theta=0^(@).` `W=Fs cos theta =mg xx s cos 0^(@).` `w=15xx9.8 xx2xx1=294 jou l e. :. ` Work done in 2nd case is greater. |
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| 12789. |
Match the following columns. {:(,"ColumnI",, "ColumnII"),((A),Inp=2/3E","Eis,(p),"isochoric"),((B),In U =3RT "for and monotomic gas"U is,(q),"Translational kinetic energy of unit volume"),((C),V/T= "constant is valid for",(r), "Internal energy of one mole"),((D), p/T= "constant is associated with",(s),"isobaric process"):} |
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Answer» `""U=n(f/2RT)""(f="DEGREE of FREEDOM")` For monotomic gas, f=3 `""U=(3nRT)/(2)` `or U=3RT ` for n = 2 is valid for any gas |
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| 12790. |
A swimmer crosses a flowing stream of width .d. to and fro in time t_(1). The time taken to cover the same distance up and down the stream is t_(2). If t_(3) is the time the swimmer would take to swim a distance 2d in still water, then n |
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Answer» SOLUTION :Let v be the RIVER VELOCITY and u the velocity of swimmer in STILL water. Then `t_(1)=2((d)/(SQRT(u^(2)-v^(2))))…….(i)` `t_(2)=(d)/(u+v)+(d)/(u-v)=(2ud)/(u^(2)-v^(2))…..(ii)` and `t_(3)=(2d)/(u)…….(iii)` from equation (i), (ii) and (iii) `t_(1)^(2)=t_(2)t_(3)=sqrt(t_(2)t_(3))` |
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| 12791. |
A wire of length L is fixed at one end. It is elongates by I when a load W is hanged from other end. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be ...... |
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Answer» `l/2` `therefore l = (WL )/(AY) ""[because F =W]` Now wire goes over a PULLEY and weight W is suspended then LENGTH in both PART increase suppose to l. `therefore . = (WL//2)/( AY) = 1/2 (WL)/(AY) = l/2` `therefore` Total extension in wire `= l ^(1) + l ^(1) = (l)/(2) + l/2=l` |
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| 12792. |
One litre of He gas at a pressure of 76cm of Hg and temperature 27°C is heated till its pressure and volume are doubled. The final temperature attained by the gas is |
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Answer» 900°C |
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| 12793. |
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 pi t + tan^(-1) 0.75) where x is in centimetre and t in second. The motion is started at t = 0When does the acceleration have its maximum magnitude for the first time? (b) When does the particle come to rest for the second time? |
| Answer» SOLUTION :(a)`1.6xx10^(-2) s , (B) 3.6xx10^(-2)s ` | |
| 12794. |
An equilateral triangle ABC is formed by joining three rods of equal length and D is the mid point of AB. The coefficient of linear expansion for AB is alpha_(1) and for AC and BC is alpha_(2), if distance DC remains constant for small changes in temperature. Find the relation between alpha_(1) and alpha_(2). |
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Answer» |
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| 12795. |
A given object takes n times more time to slide down a 45^(@) rough inclined plane as it takes to slide down a perfectly smooth 45^(@) incline. The coefficient of kinetic friction between the object and the incline is : |
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Answer» `(1)/(2-n^(2))` |
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| 12796. |
Statement I: When height of the tube is less than the rise in liquid in a capillary tube, the liquid does not overflow. Statement II: Product of radius of meniscus and height of liquid in the capillary tube always remains constant. |
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Answer» STATEMENT I is true, statement II is true , statement II is a CORRECT EXPLANATION for statement I. |
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| 12797. |
Kinetic energy can be negative. |
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Answer» |
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| 12798. |
A scientist performs an experiment and takes 100 readings. He repeats the same experiment and now takes 500 readings, By doing this what is the advantage? |
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Answer» |
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| 12799. |
A rigid horizontal smooth rod AB of mass 0.75 kg and length 40 cm can roate freely about a fixed vertical axis through its mid point O. Two rings each of mass 1 kg are initially at rest at a distance of 10 cm from O on either side of the rod. The rod is set in rotatiob with an angular velocity of 30 radians per second. Find the velocity of each ring along the length of the rod in m//s when they reach the ends of the rod. . |
Answer»  `I_(1) omega_(1) = I_(2) omega_(2)` `((ml^(2))/(12) + 2ma^(2)) omega_(1) = [((ml^(2))/(12) + 2M((l)/(2))^(2))]omega_(2)` `omega_(2) = 10 rad//sec` form conservation of energy `(1)/(2) I_(1) omega_(1)^(2) = (1)/(2) I_(2) omega_(2)^(2) + 2 xx (1)/(2) MV^(2)` `v = 3 m//s`. |
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| 12800. |
Coefficient of apparent expansions of a liquid in Gold vessel is G and when heated in a silver vessel is S. If coefficient of linear expansion of Gold is A, coefficient of linear expansion of Silver is |
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Answer» `(G+S- 3A)/( 3)` |
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