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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A train is moving due east and a car is moving due north with equal speed. A passenger in the train finds that the car is moving towards |
| Answer» Answer :B | |
| 2. |
The linear strain in x,y and z directions are e_(x), e_(y) and e_(z) respectively. Then the volumetric strain is given by |
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Answer» `e_(X) e_(y) e_(Z)` |
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| 3. |
A rocket with an initial mass of 1000 kg is launched vertically upwards from rest under gravity. The rocket burns the fuel at the rate of 10kg s^(-1). The burnt matter is ejected vertically downwards with a speed of 2000 ms^(-1) relative to the rocket. If burning is complete after 1min, find the maximum velocity of rocket. (GIven g = 10 ms^(-2) and In 2.5 = 0.916) |
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Answer» `1232 MS^(-1)` |
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| 4. |
A chain of 100 links is 1m long and has mass of 2kg. With the ends fastened together it is set rotating at 3000 rpm, in a horizontal plane. The cenripetal force each link is |
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Answer» 3.14 N |
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| 5. |
A 40 kg sphere suspended by a thread of legnth 1 is oscillating in a vertial plane, the angular amplitude being theta_(0). What is the tension in the thread when it makes an angle theta with the vertical during oscillations? If the thread can support a maximum tension of 80 kg-f, then what can be the maximum angular amplitude of oscillation of the sphere without breaking the rope? |
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Answer» `3mg COS theta-2mg cos theta_(0),theta_(0)=60^(@)` |
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| 6. |
A person is running at his maximum speed of 4 m/s to catch a train. When he is 6 m from the door of the compartment the train starts to leave the station at a constant acceleration of 1m//s^(2). Find how long it takes him to catch up the train |
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Answer» 2 sec |
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| 7. |
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation ? Explain. |
| Answer» Solution :No , it does not matter if one uses gauge instead of absolute pressure inapplying Bernoulli.s EQUATION provided the atmospheric pressure at the TWO points where Bernoulli.s equation is applied are SIGNIFICANTLY DIFFERENT. | |
| 8. |
The SI unit of energy is J=kg m^(2)s^(-2), that of speed v is ms^(-1) and of acceleration a is ms^(-2). What of the formulae for kinetic energy (k) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) : (a) K = m^(2)v^(2) (b) K = (1/2)mv^(2) ("c") K= ma (d) K = (3/16)mv^(2) (e) K=(1/2)mv^(2)+ma |
| Answer» SOLUTION :Every CORRECT formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with same PHYSICAL dimensions can be added or subtracted. The dimensions of the quantity on the right side are `[M^(2) L^(3) T^(-3)]` for (a), `[M L^(2) T^(-2)]` for (b) and (d), `[M L T^(-2)]` for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see chapter 6). The correct formula for kinetic energy is given by (b). | |
| 9. |
What is moment of inertia of a hollow sphere about an axis passing through its centre ? |
| Answer» Solution :`I = (2)/(3) MR^(2)`, where `M` is the MASS and `R` is RADIUS of the hollow SPHERE. | |
| 11. |
Using constraint method find the relation between acceleration of 1 and 2. |
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Answer» Solution :At any instant of time LET ` x_ and x_2 ` be the displacement of 1 and 2 from a FIXED line Thenm ` x_1 + x_2 `= constant ` or x_1 + x_2 = l` ( length of string ) Differentiating with respect to time , we have ` v_1 + v_2 = 0 or v_1= -v_2` Again differentiating with respect to time , we get ` a_1 + a_2 =0` ` or a_1 = -a_2` This is the required relation between `a_1` and `a_2` , i.e., ACCELERATION of 1 and 2 are equal but in opposite DIRECTIONS . , |
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| 12. |
Direction of frictional force between wheel of the car and road is |
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Answer» upward |
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| 13. |
Which nuclear process occur in hydrogen bomb ? Or write the principle of 'Hydrogen Bomb' ? |
| Answer» SOLUTION :NUCLEAR FUSION . | |
| 14. |
What are contact forces ? Give examples. |
| Answer» SOLUTION :FORCES which are EXERTED only when two bodies are in contact are CALLED contact forces .e.g. : a push or pull, the FORCE of sliding frictions. | |
| 15. |
The hot water pipes used to heat up the rooms are painted |
| Answer» Answer :B | |
| 16. |
A cylinderblockof length0.4m and area of cross - section 0.04m^(2) is placedcoaxiallyon a thin metaldisc of mass 0.4 kgand of the samecross - section. Theupperfaceof the block is maintainedat a constanttemperature of 400 Kand the initialtemperatureof the disc is 300 K . If the thermal conductivityof tehmaterialof the cylinderis 10 Watt/mK andthe specificheatof the material of the disc is 600J/kg - k, howlong with it take for the temperatureof the disc to increseto 350 K ? Assume , for purposeof calculation ,the thermal conducityof the disc to beveryhighand the systemto bethemallyinsulatedexceptfor theupperfaceof the cylinder . |
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Answer» Solution :Fundamental quationof heatconductionis steadystateis . `H = (Q)/(t) = (KA(T_(1) - T_(2)))/(l)………….(1)` In this cause the temperatureof upperfaceof cylinderis `T_(1)= 400 K`,while thetemperatureoflowerface ofcylinder is changing . Therefore ,we shall USE equation (1) in the form `(dQ)/(dT) = KA((400-T)/(l))` where Tis any intermediate temperatureof lowerfaceof cylinder, dQ is heatabsorbed by disc in timedtdQ = mc , dT , Here m = mass of disc 0.4 kg , c = specificheatof disc= 600 j/kg - K. `therefore(mc dT)/(dt) = KA ((400 - T)/(l)) (or) dt =(mc l)/(KA) ((dT)/(400-T))` INTEGRATING` t = (MCL)/(KA) int_(300)^(350) (dT)/((400 - T)) = (mcl)/(KA)[(log_(e) (400- T)/(-1))]_(300)^(350) = (mcl)/(KA) xx 2.3 log_(10) [(400-300)/(400-350)] = (mcl)/(KA) xx 2.3 log_(10) 2` substituting given values, we GET ` t = (0.4 xx 600 xx 0.4)/(10 xx (0.04)) = 2.3 xx 0.3010 = 166` sec =2 min . 46 sec . |
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| 17. |
What would be the pressure in side a small air bubble of 0.2 mm diameter situated just below the surface of water (surface tension of water = 0.072Nm^(-1)). |
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Answer» |
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| 18. |
The gravitational field at some point in space is g = 3hati + 4hatj N/kg. The force exerted on a 2kg mass placed at that point is |
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Answer» 5N, `53^(@)` with x-axis |
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| 19. |
A block of mass 100 g moved with,a speed of 5 m/s at the highest point in a closed circular tube of radius 10 ,cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it., The block makes several oscillations inside the tube and finally steps at the lowest point. The work done by t\Ie tube on the block during the process is: |
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Answer» 1.45 J |
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| 20. |
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brough below the cylinder and put in a positoon so that the water surface in the tank is at the same level as the top of the cylinder as shown in the fgure. The density of the water is rho.In equilibriium, the height H of the water column in the cylinder satisfies |
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Answer» <P>`rhog(L_0-H)^(2)+P_0(L_0-H)+(L_0P_0)=0` |
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| 21. |
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun. |
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Answer» 2 `T_(A) =8T_(B) rArr(8)^(2/3) a_(B) = a_(A) rArr a_(A) = 4a_(B)` |
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| 23. |
A refrigerator, whose coefficient of performance K is 5, extracts heat from the cooling compartment at the rate of 250 J per cycle. a0 How much work per cycle is required to operate the refrigerator cycle? ( b) How much heat per cycle is discharged to the room which acts as the high temperature reservoir? |
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Answer» SOLUTION :a) As coefficient of PERFORMANCE of a refrigerator is DEFINED as `K=Q_(L)//W," So "W=Q_(L)/K=250/5=50J` B)As `Q_(H)=Q_(L)+W" so "Q_(H)=250+50=300J` |
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| 24. |
Arrive at an expression for elastic collision in Dimension and discuss various case. |
Answer» Solution :Consider two elastic bodies of MASES `m_(1) and m_(2)` moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in FIGURE given below.  In order to have collision, we assume that the mass `m_(1)` move faster than mass `m_(2)` i.e, `u_(1) gt u_(2)`. For elastic collison, the total linear momentum and kinetic energy of the two bodies before and after collision must remain the same.  From the law of conservation of lienear momentum, Total momentum before collision `(p_(i))`= total momentum after collision `(p_(f)0` `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`...(1) or `m_(1)(u_(1)-v_(1))=m_(2)(v_(2)-u_(2))` ...(2) Further,   Total kinetic energy before collisioon `KE_(i)`=Total kinetic energy after collision `KE_(f)` `1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)` ....(3) After simplifying and rearranging the terms. `m_(1)(u_(1)^(2)-v_(1)^(2))=m_(2)(v_(2)^(2)-u_(2)^(2))` Using the formula `a^(2)-B^(2)=(a+b)(a-b)` , we can rewrite the above equation as `m_(1)(u_(1)+v_(1))(u_(1)-v_(1))=m_(2)(v_(2)+u_(2))(v_(2)-u_(2))` ...(4) Dividing equation (4) by (2) gives, `(m_(1)(u_(1)+v_(1))(u_(1)-v_(1)))/(m_(1)(u_(1)-v_(1)))=(m_(2)(v_(2)+u_(2))(v_(2)-u_(2)))/(m_(2)(v_(2)-u_(2))` `u_(10+v_(1)=v_(2)+u_(2)` `u_(1)-U_(2)=u_(2)=v_(2)-v_(1)` Rearranging, (5) Equation (5) canbe rewritten as `u_(1)-u_(2)=-(v_(1)-v_(2))` This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass. `v_(1)=v_(2)+u_(2)-u_(2)` ...(5) or`v_(2)=u_(1)+v_(1)-u_(2)`...(7) To find the final velocities `v_(1) and v_(2)` : Substituting equation (7) in equation (2) gives the velocity of `m_(1)` as `m_(u_(1)-v_(1)=m_(2)(u_(1)+v_(1)-u_(2)-u_(2))` `m_(1)(u_91)-v_(1))=m_(2)(u_(1)+v_(1)-2u_(2))` `m_(1)u_(1)-m_(1)v_(1)=m_(2)u_(1)+m_(2)v_(1)+2m_(2)u_(2)` `m_(1)u_(1)-m_(2)u_(1)+2m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(1)` `(m_(1)-m_(2))u_(1)+2m_(2)u_(2)=(m_(1)+m_(2))v_(1)` or `v_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)((2m_(2))/(m_(1)+m_(2)))u_(2)`...(8) Similarly by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of `m_(2)` as `v_(2)=((2m_(1))/(m_(1)+m_(2)))u_(1)+((m_(2)-m_(1))/(m_(1)+m_(2)))u_(2)`... (9) Case 1 : When bodies has the same mass i.e.,`m_(1)=m_(2)` equation (8) `rArr v_(1)=(0)u_(1)+((2m_(2))/(2m_(2)))u_(2)` `v_(1)=u_(2)` ... (10) equation (9) `rArr v_(2)=((2m_(1))/(2m_(1)))u_(1)+(0)u_(2)` `v_(2)=u_(1)` ...(11) The equations (10) and (11) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged. Case 2: When bodies have the same mass i.e, `m_(1)=m_(2)`and second body (usually called target) is at rest `(u_(2)=0)`, By substituting `m_(1)=m_(2)=andu_(2)` ,in equations (8) and equations (9) we get, from equation (8) `rArr v_(1)=0)`...(12) from equation (9) `rArr v_(2)=u_(1)`...(13) Equations (12) and (13) show that when the first body comes to rest the second body move with the initial velocity of the first body Case3 : The first body is very much lighter than the second body `(m_(1)ltltm_(2),m_(1)/m_(2)ltlt1)` than the ratio `m_(1)/m_(2)=0`and also if the traget is at rest `(u_(2)=0)` Dividing numerator and denomintor of equation (8) by `m_(2)`, we get `v-(1)=(((m_(1)/m_(2))-1)/(m_(1)/(m_(2))+1))u_(1)+((2)/(m_(1)/(m_(2))+1))(0)` `v_(1)=((0-1)/(0+1))u_(1)` `v_(1)=-u_(1)` ...(14) Similarly , dividing numerator and denominator of equation (9) by `m_(2)` ,we get `v_(2)=((2(m_(1))/(m_(2)))/(m_(1)/m_(2)+1))u_(1)+((1-m_(1)/m_(2))/(m_(1)/m_(2)+1))(0)` `v_(2)=(0) mu_(i)+ ((1-m_(1)/m_(2))/(m_(1)/m_(2)+1)) (0)` `v^(2)=0` ...(15) The equation (14) implies that the first body which is higher is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (15) implies that the second body whifch is heavier in mass continuous to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was throuw but in opposite direction. Case 4: The second body is very much lighter than the first body `(m_(2) ltlt m_(1), m_(2)/m_(1) lt lt 1)` then the ratio `m_(2)/m_(1)=0` and also if the target is at rest `(u_(2)=0)` Dividing numerator and denominator of equation (8) by `m_(1)`, we get `v_(1)=((1-m_(2)/m_(1))/(1+m_(2)/m_(1)))u_(1)+((2 m_(2)/m_(1))/(1+m_(2)/m_(1)))(0)` `v_(1)=((1-0)/(1+0))u_(1)+((0)/(1+0)) (0)` `v_(1)=u_(1)` Similary, Dividing numerator and denominator of equation (xiii) by `m_(1)", we get "v_(2)=((2)/(1+0)) u_(1)` `v_(2)=2u_(1)` The equation (16) implies that the first body which is heavier continuous to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will mass with TWICE the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision. |
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| 25. |
A uniform steel wire of density 7.8 g/cc weighs 8 g and is 125 cm long . It lengthens by 0.6 mm whwn stretched by a force of 8 kg-wt . Calculate young's modulus for steel . |
| Answer» SOLUTION :19.9*10^10N/m^2 | |
| 26. |
A particle at the end of the spring executes S.H.M with a period t_(1), while the corresponding period for another spring is t_(2). If the period of oscillation with two springs in series is .T., then |
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Answer» `T=t_(1) =t_(2)` |
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| 27. |
Threevectors vecA, vecB, vecB are shown in the figure. Find angle between (i)vecA and vecB (ii)vecB and vecC (iii)vecA and vecC. |
Answer» SOLUTION :To find the angle between two vectors we connect the tails of the two vectors. We can SHIFT the vectors PARALLEL to themselves such that tails of `vecA, VECB and VECC` are connected as shoown in figure.  Now we observe that angle between `vecA and vecB` is `60^(@),vecB and vecC` is `15^(@)` and between `vecA and vecC` is `75^(@)` |
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| 28. |
A metre stick is balanced on a knife edge at its centre. When two coint, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ? |
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Answer» 66.0 g |
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| 29. |
Name the unit in which size of a nucleus is measured. |
| Answer» SOLUTION :FERMI, 1Fermi `=10^(-15)`m. | |
| 30. |
A force of 25 N acts on a body at rest for 0.2s and a force of 70 N acts for the next 0.1 s in opposite direction. If the final velocity of the body is 5ms^(-1), the mass of the body is |
| Answer» ANSWER :D | |
| 31. |
A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment t=0 the speed of the particle equals V_(0)then The speed of the particle as a function of the distance coverd will be |
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Answer» `V= V_(0)E^(-s//R)` |
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| 32. |
Calculate the angle for which a cyclist bends when he turns a circular path of length 34.3 m in sqrt(22) s. |
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Answer» Solution :Given Data: `l= 34.3 m, t= SQRT(22), g = 9.8 MS^(-2), THETA = ?` If r is radius of CIRCULAR path, then length of path = `2pir = 34.3` m `r =(34.3)/(2pi)` m. and time taken t = `sqrt(22)` s As tan `theta = (v^(2))/(rg)` `:. tan theta = ((34.3)/(sqrt(22)))^(2) xx (2pi)/(34.3 xx 9.8)` tan `theta = (34.3xx34.3)/(22) xx (2xx22)/(7xx343xx9.8) = (34.3xx2)/(68.6) = 1`[`:. theta = 45^(@)`] |
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| 33. |
A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment t=0 the speed of the particle equals V_(0)then The total acceleration of the particle as function of velocity and distance covered |
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Answer» `a= SQRT(2)(V^(2))/(R)` |
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| 34. |
The critical velocity v for a liquid depends upon (a) coefficient of viscosity eta (b) density of the liquid rho and © radius of the pipe r. Using dimensions derive an expression for the critical velocity. |
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Answer» Solution :`V prop eta^(x)rho^(y)r^(Z)` `=k eta^(N) rho^(y)r^(z)` .i the dimensional formula of `eta` is `ML^(-1)T^(-1)` TAKING dimensions on both sides `M^(0)LT^(-1)=[ML^(-1)T^(-1)]^(x)[ML^(_3)]^(y)[L]^(z)` `M^(2)LT^(-1)=M^(x+y)L^(-x-3y+z)T^(-x)` Equating the dimensions of T, M nd L on both sides For T, `=1=-x,x=1` For M, `0=x+y,y=-x=-1` for L `1=-x-3y+z` `z=1+x+3y=1+1-3=-1,z=-1` Substiting in equation (i) `v=keta^(1)rho^(-1)r^(-1)=k(eta)/(rho r)` |
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| 35. |
A system absorbs 2 Kcals heat and does 500 J of work, then internal energy of the system will be ……. |
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Answer» 6400 J `THEREFORE` dQ=+2000 cal =2000 x 4.2 J =+8400 J System does work, `therefore DeltaW`=+500 J From first law of thermodynamics, `therefore DeltaW`=+500 J dQ=dU+dW `therefore` dU=dQ-dW =8400-500 `therefore` dU=7900 J It is POSITIVE so internal ENERGY INCREASES . |
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| 36. |
Does the energy stored in a spring changes when it stretched or compressed ? |
| Answer» Solution :Its POTENTIAL ENERGY (energy stored) increases when it is STRETCHED or COMPRESSED. | |
| 37. |
A sphere of mass M,, moving with a velocity V0 collides head on with a stationary sphere of mass M. The collision is elastic. V1 and V2 are respectively their velocities immediately after collision. |
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Answer» a-g, b-e, C-h, d-f |
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| 38. |
The planet mars has two moons. Phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 xx 10^(3) km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun. with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ? |
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Answer» Solution :(i) We know that `T^(2) = (4PI^(2))/(GM_(m)) R^(3)` `M_(m) = (4pi^(2))/(G)(R^(3))/(T^(2)) = (4 xx (3.14)^(2) xx (9.4)^(3) xx 10^(8))/(6.67 xx 10^(-11) xx (459 xx 60)^(2))` `M_(m) = (4 xx (3.14)^(2)(9.4)^(3) xx 10^(18))/(6.67 xx (4.59 xx 6)^(2) xx 10^(-5))` `= 6.48 xx 10^(23) kg`. (II) According to Kepler.s third law `(T_(M)^(2))/(T_(E)^(2)) = (R_(MS)^(3))/(R_(ES)^(3))` where `R_(MS)` is the mars-sun distance and `R_(ES)` is the earth-sun distance. `therefore T_(M) = (1.52)^(3//2) xx 365 = 684` DAYS |
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| 39. |
Assertion : At same level of same liquid pressure is always same. Reason : When any fluid travels from a region of higher pressure to lower pressure (At same levels) it gainssame speed. |
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Answer» If both Assertion and REASON are correct and Reason is the correct explanation of Assertion. |
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| 40. |
A spring of force constant 800 N/mhas an extension of 5 cm . The work done in extending it from 5 cm to 15 cm is |
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Answer» 16 J |
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| 41. |
Calculate the volume of one mole of any gas at STP and at room temperature (300K) with the same pressure 1 atm. |
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Answer» Solution :Here STP means standard temperature (T = 273 K or `0^(@)C`) and Pressure (P = 1 atm or 101.3 KPA) We can use ideal gas equation `V=(muRT)/(P)` Here `mu=1` mol and R = 8.314 J/mol.K By SUBSTITUTING the value `V=("(1 mol)"(8.314(J)/(mol)K)(273K))/(1.013xx10^(5)Nm^(-2))=22.4xx10^(-3)m^(3)` We know that 1 Litre `(L)=10^(-3)m^(3)`. So we can conclude that 1 MOLE of any ideal gas hasvolume 22.4 L By multiplying 22.4 by `(300K)/(273K)` we get the volume of one mole of gas at room temperature. It is 24.6 L |
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| 42. |
The difference between the apparent frequency of a source of sound as perceived by an observer during its spproach and recession is 2%of the naturalfrequency of the source . If the velocity of sound in air is 300 m .s^(-1), find the velocity( in m .s^(-1) ) of the source . |
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Answer» |
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| 43. |
When milk is chrned, cream gets seperated due to |
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Answer» centripetal FORCE |
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| 44. |
A body of mass 4 kg and density 2.5g*cm^(-3) is suspended with a metallic wire of length 1 m and diameter 2 mm. If the body is totally immersed in water, what will be the elongation of the wire? Young's modulus of the metal = 2xx10^(11)N*m^(-2)andg=9.8m*s^(-2) |
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Answer» |
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| 45. |
Twoforcesof magnitudeF havinga resultantof thesamemagnitudeof ,Fthe anglebetweenthe twoforcesis ______ |
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Answer» SOLUTION :Magnitudeof eachforceis F ` therefore` Theresultant force`= F = SQRT( F^2+F^2+ 2F .F costheta )` `F^2= 2F^2cos thetaimplies cos=- 1/2implies theta= 120^@` |
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| 46. |
Velocity is defined as the rate of change of displacement.A car travels from A to Bat 60 km./hr and returns to A at 90 km//hr. What is its average velocity and average speed? |
| Answer» SOLUTION :AVERAGE velocity=0 Average SPEED`(2v_1v_2)/(v_1+v_2)=(2xx60xx90)/(60+90)=72kmh^-1` | |
| 47. |
The behaviour of a body under zero resultant force is given by |
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Answer» FIRST LAW of motion |
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| 48. |
A body is performing SHM. Then its |
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Answer» AVERAGE total energy PER cycle is EQUAL to its MAXIMUM kinetic energy |
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| 49. |
A body projected vertically from the earth reaches height equal to earth.s radius before returning to the earth, The power exerted by the gravitational force is greatest |
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Answer» At the INSTANT just before the body hits the EARTH |
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| 50. |
A gas mixture consists of 2 moles of O_(2) and 4 moles of Ar at temperature T.Neglecting all vibrational modes ,the total internal energy of the system is: |
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Answer» 11RT |
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