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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.
| 39801. |
A sphere made of a material of specific gravity 8 has a concentric spherical cavity and just sinks in water. Calculate the ratio of the radius of the cavity to that of the outer radius of the sphere. |
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Answer» SOLUTION :Let the radius of the SPHERE be R and that of the cavity be r. Then, since the sphere just sinks in water, `4/3 pi (R^3 - r^3) rho g = 4/3 pi R^3 rho_w g` `1 - (r/R)^3= (rho_w)/(rho) = 1/8` |
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| 39802. |
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net force at the lowest and highest points of the circle directed vertically downwards are: [ Choose the correct alternative] Lowest Point"" Highest point (a) mg +T_(1)""mg + T_(2) (b) mg +T_(1)""mg - T_(2) (c ) mg + T_(1)- (mv_(1)^(2))//R""mg - T_(2) + (mv_(1)^(2))//R (d) mg -T_(1) -(mv_(1)^(2))//R"" mg + T_(2) + (mv_(1)^(2))//R T_(1) and v_(1) denote the tension and speed at the lowest point. T_(2) and v_(2) denote corresponding values at the highest point. |
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Answer» SOLUTION :Alternative (a) is correct. Note: MG + `T_(2) = mv_(2)^(2)`/R, `T_(1) - mg = mv_(1)^(2)`/R in this EXAMPLE |
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| 39803. |
Show the nature of the following graph for a satellite orbiting the earth. (a) KE versus orbital radius R (b) PE versus orbital radius R (b) TE versus orbital radius R |
Answer» Solution :`implies` consider the diagram, where a satellite of mass m, moving around the earth in a circular ORBIT of radius R.  Orbital speed of the satellite orbiting the earth isgiven by `V_0=sqrt((GM)/R)` where, M and R are the mass and radius of the earth. (a) ` :.`KE of a satellite of mass m, `K=1/2 mv_0^2 =1/2 mxx(GM)/R` `:. K prop 1/R`  It means the KE decreases exponentially with radius. The graph for KE versus orbital radius R is shown in figure (b) Potential energy of a satellite. P.E. `=-(GMm)/R` `P.E. prop -1/R`  The graph for PE versus orbital radius R is shown in figure (c) Total energy of the satellite `E = K +U = (GMm)/(2R) - (GMm)/R` `=- (GMm)/(2R)`  The graph for total energy versus orbital radius R is shown in figure NOTE : We should keep in MIND that Potential energy (PE) and Kinetic Energy (KE) of the satellite-earth system is always NEGATIVE. |
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| 39804. |
The gravitational field in a region is given by vec(E) = (2hat(i) + 3hat(j))N//Kg. The workdone by the field when the particle is moved on the line 3y + 2x = 5 |
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Answer» SOLUTION :`vec(E) = 2hat(i) + 3 hat(j)` The field is represented as Slope `m_(1) = Tan theta_(1) = 3//2` The line `3Y + 2X = 5` can be represented as slope `m_(2) = Tan theta_(2) = -2//3` `m_(1) xx m_(2) = -1` Since, the direction of the field and DISPLACEMENT are PERPENDICULAR the workdone by the field on the particle is zero. 
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| 39805. |
A well insulated box has two compartments A and B with a conducting wall between them. 100 g of ice at 0^(@)C is kept in compartment A and 100 g of water at 100^(@)C is kept in B at time t = 0. The temperature of the two parts A and B is monitored and a graph is plotted for temperatures T_(A) and T_(B) versus time (t) [Fig. (b)]. Assume that temperature inside each compartment remains uniform. (a) Is it correct to assert that the conducting wall conducts heat at a uniform rate, irrespective of the temperature difference between A and B? (b) Find the value of time t_(1) and temperature T_(0) shown in the graph, if it is known that t_(0) = 200 s. Specific heat of ice = 0.5 cal g^(-1) .^(@)C^(-1) Specific heat of water = 1.0 cal g^(-1) .^(@)C^(-1) Latent heat of fusion of ice = 80 cal g^(-1) |
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| 39806. |
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion. |
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Answer» Solution :When suction pump connected through one end of U-tube, the mercury depressed by h in one limb and in other arm mercury raised by 2h. The difference in height of mercury in both the limbs = 2Y and pressure of `P = 2hpg` in this arms. Here p= density of mercury and g is the acceleration DUE to GRAVITY. If the cross sectional area of arms is A, then force produced due to height of 2h, `F= PA` `=2hpg xxA = (2PGA)h` `therefore F propto h ""therefore F = kh` Where `k= 2pgA` is constant. This force is opposite to displacement, `F= -kh` Hence, restoring force acts on oscillation of mercury column is directly proportional to displacement and opposite to displacement. So the motion of mercury in U-tube is simple harmonic motion. |
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| 39807. |
A capillary tube of radius 0.25mm is submerged vertically in water so that 25mm of its length is outside water. The radius of curvature of the meniscus will be ( T = 75 xx 10^(-3) "N/m") |
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Answer» 0.2 mm |
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| 39808. |
Obtain an expression for total energy of a satellite revolving in circular orbit around a planet . |
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Answer» SOLUTION :GravitationalP.E. at a point above the surface of the earth at a distance r = R + h is given by P.E. = `(-GM m)/(r)` where 'M' is mass of the earth and m is the mass of the satellite . kinetic ENERGY of satellite = `(1)/(2) (GMm)/(r)` Total energy of satellite = E = PE + KE Hence E = `(-GMm)/(r) + (1)/(2) (GMm)/(r)` i.e., `E = - (1)/(2) (GMm)/(r)` The `-ve` sign INDICATES that the satellite is bounded to the earth . |
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| 39809. |
At what points along the path of a simple pendulum is the tension in the string maximum and |
| Answer» SOLUTION :The TENSION is maximun at the MEAN POSITION (=MG) | |
| 39810. |
A lift is moving upward with an acceleration a. The apparent weight of a person standing in the lift will be |
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Answer» LESS than the ACTUAL weight |
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| 39811. |
A particle of mass m is moving yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is |
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Answer» MVA `bar(i)` |
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| 39812. |
Which physical quantity is conserved during the oscillation of a simple pendulum ? |
| Answer» SOLUTION :TOTAL MECHANICAL ENERGY of the BOB is conseved. | |
| 39813. |
An equilateral triangle ABCformed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and other along AC as shown. Neglecting frictional effects, the quantities that are conserved as the beads slide down, are: |
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Answer» ANGULAR velocity and total energy |
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| 39814. |
A solid cylinder of mass.m. rolls without slippling down an inclined plane making an angle theta with the horizontal. The fricitional force between the cylinder and the incline is: |
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Answer» `mg SIN theta` |
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| 39815. |
At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume gamma=1.41. Thework done on gas will be |
| Answer» Answer :C | |
| 39816. |
A horizontal stick of mass m has its right end attached to a pivot on a wall, while its left end rests on the top of a cylinder of mass m which in turn rests on an incline plane inclined at an angle theta. The stick remains horizontal. The coefficient of friction between the cylinder and both the plane and the stick is mu. Find the minimum value ofmuas function of theta for which the system stays in equilibrium. |
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| 39817. |
A ball of mass m strikes a bat with velocity v. If it retraces its path with the same speed, what is the impulse imparted by the bat ? |
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Answer» zero |
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| 39818. |
When a force of 10.0 N is exerted on the handle of a door, the door can be just opened. If the handle is a distance of 50 cm from the hinges find the torque applied on the door. |
| Answer» Solution :`tau = R F SIN theta = 0.50 xx 10.0 sin 90^(@)=5Nm` | |
| 39819. |
Two parallel plane sheets 1 and 2 carry uniform charge densities sigma_(1) and sigma_(2) as in fig. Find the electric field in the region marked prod (sigma_(1) gt sigma_(2)) |
| Answer» SOLUTION :`((sigma_1-sigma_2))/(2in_0)` | |
| 39820. |
Two solids spheres of diameter d_(1) and d_(2) having material densities rho_(1) and rho_(2) are placed in contact with each other on the horizontal plane. The height of the combined center of mass from plane |
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Answer» `h=((d_(1)^(2)rho_(1)+d_(2)^(2)rho_(2))/(d_(1)rho_(1)+d_(2)rho_(2)))` |
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| 39821. |
A projectile fired with velocity u at right angle to the slope which is inclined at an angle thetawith horizontal. The expression for R is |
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Answer» `(2u^2)/(G) TAN THETA` |
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| 39822. |
The velocity time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 sec are respectively. |
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Answer» 8 m, 16m |
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| 39823. |
Sometimes when the speed of vehicle increased, its body start to bounce why? |
| Answer» SOLUTION :Natural frequency of vehicle and frequency of vehicle DUE to rough road becomes EQUAL then rosonance occurs and HENCE the amplitude becomes large, So VEHICLES bounces. | |
| 39824. |
Four rods of equal length l and mass m each form a square as shown in figure. The moment of inertia of the system about the axis 1 and axis 2 is |
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Answer» `2/3ML^(2),5/3ml^(2)` |
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| 39825. |
(A) : In a measurement, two readings obtained are 10.001 and 10.0001. The second measurement is more precise. (R) : Measurement having more decimal places is more precise. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 39826. |
(A) : Power associated with torque is product of torque and angular speed of the body about the axis of rotation.(R ) : Torque in rotational motion is analogue to force in translatory motion. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 39827. |
A gas occupying a volume of 10^(-2) m ^(3).at a pressure of 5 atmospheres expands isothermally to a pressure of 1 atmosphere. Calculate the work done. |
| Answer» Solution :`W =2 . 303 RT log _(10) (V _(2) //V _(1)) = 2. 303 P _(1) V _(1) log _(10) (P_(1) //P_(2)) = 2. 303 xx 5 xx 101325 xx 10 ^(-2) log _(10) (5//1) = 8. 155 xx 10 ^(3) Nm` | |
| 39828. |
In the adiabatic compression, the decrease in volume is associated with |
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Answer» Increase in temperature and increase in PRESSURE |
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| 39829. |
Define wave vector and reciprocal vectors also write the relationship between velocity and reciprocal vectors. |
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Answer» SOLUTION :The wave number is the MAGNITUDE of a vector called wave vector. The points in space of wave VECTORS are called reciprocal vectors, `veck`. Dimensions of `vec k` is `L^(-1)` Velocity, `v= LAMBDA, f=(lambda)/(2pi) (2pi f)` `=(2pi lambda)/(2pi//lambda)` `=(omega)/(k)` |
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| 39830. |
If R is the range of a projectiel on a horizontal plane and h its maximum height, the maximum horizontal range with the same velocity ofprojection is: |
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Answer» 2h `=(u^(2))/(g) sin^(2)theta+(u^(2))/(g) cos^(2)theta` `=(u^(2))/(g)=R_("max")` |
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| 39831. |
A thin wire of mass.m. and length .L. is bent in the form of a ring. The M.I. of the ring about a tangential axis parallel to any of its diameters is: |
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Answer» `(mL^(2))/(2pi^(2))` |
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| 39832. |
A smooth sphere 'A' of mass 'm' is moving with a contant speed 'v' on the smooth horizontal surface collides elastically with an identical sphere at 'B' at rest. After elastic speed of sphere 'A' is V/2 then the speed of sphere of 'B' is |
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Answer» `(V)/(2)` |
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| 39833. |
Total energy of particle performing S.H.M. depends on |
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Answer» AMPLITUDE, TIME PERIOD |
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| 39834. |
A certain number of spherical drops of a liquid of a liquid of radius r coalesce to form a single drop of radius R and volume V. if T is the surface tension of the liquid, then |
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Answer» `"ENERGY"=4VT(1/R-1/R)"is released"` `A_i=nxx4pir^2=(V)/(4/3 PI r^3)4pir^2=(3V)/( r)` Hence, energy released `=(A_i-A_f)T=3VT(1/r-1/R)` |
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| 39835. |
In Q.4. work done by tangential force on the body is |
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Answer» ZERO |
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| 39836. |
What is average acceleration? |
| Answer» Solution :The AVERAGE acceleration is defined as the RATIO of change in velocity over the time INTERVAL `a_(avg)=(DeltavecV)/(Deltat)`. It is a VECTOR quantity. | |
| 39837. |
A flywheel of mass 100 kg and radius 1 m is rotating at the rate of 420 rev/min. Find the constant retarding torque to stop the wheel in 14 revolutions, the mass is concentrated at the rim. M.I. of the flywheel about its axis of rotation I = mr^(2). |
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Answer» Solution :Mass of flywheel = m = 100 kg Radius= R = 1m `I = mr^(2) = 100 xx 1^(2) = 100 kg m^(2)` Initial angular velocity `=omega_(0) = 2pir = 2 xx 3.14 xx 420/60 = 43.96 "rad"//s^(2)` Final angular velocity `=omega =0` Angular displacement in 14 REVOLUTION = `14 xx 2PI = 28 pi` radian. `alpha = (omega^(2) - omega_(0)^(2))/(20) = (0-43.96^(2))/(2 xx 28 pi) =-10.99 "rad"//s^(2)` Torque required to stop the flywheel `=tau = Ialpha = 100 xx 10.99 = 1099` Nm |
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| 39838. |
A couple produces ....... |
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Answer» PURE rotation |
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| 39839. |
In the given expression X=(abz^2)/(h^3), which physical quantity should be measured more accurately to give the most accurate answer? |
| Answer» SOLUTION :Since the power of h is highest, it should be MEASURED most accurately to GIVE accurate ANSWER. | |
| 39840. |
Given in figure. Are examples of some potential energy functions in one dimension (i) to (iv). The total energy E of the particle is indicated by the cross mark on energy axis. In each case, specify the refions, if any, in which the particle cannot be found for the given energy. Also, indicatethe minimum total energy the particle must have in each case. Think of simple physical contests for which these potential energy shapes are relevant. |
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Answer» Solution :We know that total ENERGY `E=K.E.+P.E.` or `K.E.=E-P.E.` and `K.E.` can never be negative. The object cannot exist in the region, where its K.E. would becomenegative. (i) For`X gt a`, P.E. `(V_(0) gt E` `:.` K.E. becomes negative. Hence, the objectconnot exist in the region `xgta`. (II)For `x lt a` and `x gt bP.E. (V_(0)) gt E.` `:.` K.E. becomes negative. Hencethe objectcannot be present in the region `x gt a` and `x gt b`. (iii) Object cannot exist in any region because P.E. ` (V_(0) gt E` in EVERY region. (iv) On the same basis, the object cannot exist in the region`-b//2 lt x lt -a//2` and `a//2 lt x lt b//2`. |
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| 39841. |
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, when the masses are released. |
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Answer» i.e.,a = 2m`cdot s^(-2)`, T = 96 N |
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| 39842. |
When a mass 'm' is attached to a vetical spring and slowly released, the extension in the spring is 'x'. When a mass '2m' is attached to the same spring and released suddenly, the extension in the spring is |
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Answer» x |
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| 39843. |
A bimetallic strip is made by rivetting two thin linear strips of copper ( alpha_C) and Iron ( alpha_1) at a temperature of theta_0. The thickness of each strip is 'd' and length of each is L_0. At some higher temperature theta, the bimetallic bends into an arc of angle theta as shown, so that the two electricalcontacts touch and electrical control system is activated. Now phi= (L_0)/( K alpha) ( alpha_( C) - alpha_(1) ) ( theta - theta_(0) ) , where K= |
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| 39844. |
If a body is raised from the surface of the earth upto height R, what is the change in potential energy? |
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Answer» MGR `h=(R )/(2)` `therefore P.E=(mgR)/(2)` |
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| 39845. |
A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is mu. The work done by the friction during the period the chain slips off the table is -(2muMgL)//x). Find x. |
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| 39846. |
10 gm of ice at- 10° C is mixed with 100 gm of water at 50° C contained in a calorimeter weighing 50gm. (Specific heat of water = cal, gm^(-1).°C^(-1), Latent heat of ice = 80 cal//gm,specific heat of ice =0.5 "cal/gm/°C" and specific heat of copper = 0.09 "cal/gm/"°C). The final temperature reached by the mixtures is |
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Answer» 25.5°C |
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| 39847. |
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ? (tan = 15° = 0.2679) |
Answer» Solution :Let O be the GROUND observation point and A, B, C be the POSITIONS of the aircarft at t = 0S, t = 5s and t = 10 s respectively,  Clearly, `tan15^(@) = (AB)/(OB)` or `AB= OB tan 15^(@)` THUS, distance travelled by the AIRCRAFT in 10s, i.e., `s = AC = 2AB = 2(OB tan 15^(@))` `= 2 xx 3400 xx 0.2679 = 1822 m` Speed of the aircraft `= (s)/(t) = (1822m)/(10s) = 182.2 m//s` |
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| 39848. |
State the law of equipartition of energy. |
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Answer» Solution :According to kinetic theory the average kinetic energy of system of molecules in thermnal equilibrium at tempreture T is uniformly distributed all degrees of freedom (x or y or z direction of motion ) so that each degree of freedom will GET `(1)/(2)kT` of energy. This is called LAW of EQUIPARTITION of energey. Average KINECTIC energy of MONATOMIC molecule `(with int=3)=3xx(1)/(2)kT=(3)/(2)kT` Average kinectic energy of diatomic molecule at low tempreture `(withint=5)=5xx(1)/(2)kT=(5)/(2)kT` Average kinectic energy of diamotic molecule at thehigh tempreture `(withint=7)=7xx(1)/(2)kT=(7)/(2)kT` Average kinectic energy of linear triatomicmolecule `(withint=7)=7xx(1)/(2)kT=(7)/(2)kT` Average kinectic energy of non linear triatomicmolecule `(withint=6)=6xx(1)/(2)kT=3kT`. |
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| 39849. |
The length of a pendulum is measured as 1.01m and time for 30 oscillations is measured as one minute 3 seconds. Error in length is 0.01m and error in time is 3 secs. The percentage error in the measurement of acceleration due to gravity is |
| Answer» Answer :C | |
| 39850. |
During an experiment, an idea/gas is found to obey an additional law VP^(2) = constant . The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2 V? |
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Answer» <P> Solution :According to the given problem `VP^(2)` =K. constant. So the gas equation PV = nRT in the light of above (eliminating P) YIELDS`((K)/(sqrt(V))) V = nRt , `i.e., `sqrt(V)= (nR)/(K ) T ` `therefore sqrt((V_(1))/(V_(2))) = ((T_(1))/(T_(2)))," i.e., " sqrt((V)/(2V)) = (T)/(T^(1))` or `T^(1) = (sqrt(2))T ` |
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