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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. |
Derive an expression using the method of dimensions, the distance travelled by a body during an interval of time t, if its initial velocity is u and uniform acceleration is a. |
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Answer» Solution :Let the distance travelled by the body S be GIVEN by `S=ku^(x)d^(y)t^(z)` Substituting the dimensions on both sides `L=[LT^(-1)]^(x)[LT^(-2)]^(y)[T]^(z)` `M^(2)T^(0)L=L^(x+y)T^(-x-2y+z)` Equating the dimensions of L and T on both sides `x+y=1`..........i `-x-2y+z=0` `x+2y=z`....ii We have three variables x,y and z but to solve we have only TWO equations, so we split the problem into two parts `S=S_(1)+S_(2)` Where `S_(1)` is the distance covered with a=0 and `S_(2)` is the distance with U=0 Then `S_(1)=k_(1)u^(a)t^(B)`.......iii Substituting the dimensions `L=(LT^(-1))^(a)(T)^(b)=L^(a)T^(-a)T^(b)=L^(a)T^(-a+b)` Equating the dimensions of L and T on both sides. `a=1, -a+b=0, b=a=1` Substituting in equation (iii) `S_(1)=k_(1)ut`............iv For the second part `S_(2)=k_(2)a^(c)t^(d)`.........v Substituting the dimensionis `L=[LT^(-2)]^(c)[T]^(d)=L^(c)T^(-2c+d)` Equating the dimensions of L and T on both sides `c=1, -2c+d=0, d=2c=2` Substituting in equation (v) `S_(2)=k_(2)at^(2)` `S=S_(1)+S_(2)=k_(1)ut+k_(2)at^(2)`................vi |
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| 2. |
Derive the expression for mean free path of the gas. |
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Answer» Solution :Expression for mean free path We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves along straight path with uniform velocity. This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. Assume that only one molecule is in motion and all others are at rest. If a molecule moves with AVERAGE speed `nu` in a the distance travelled is nutIn this time t, consider the molecule to move in an imaginary cyinder of volume `pid^(2)nut` It collides with any molecule to whose center is within this cylinder. Therfore, the number of collisions is equal to the number of molecule in the volume of the imaginary cylinder. It is equal to ` pid^(2)nutn` The total path lengh divided by the number of collisions in time t is the mean free path. `"Mean free path ",lambda="Distance travelled"/"Number of collisions"` `lambda=(nut)/(npid^(2)nut)=1/(npid^(2))` .....(1) Though we have assumed that only one molecule is moving at a time and other molecule are at rest, in actual practice all the molecules are in random motion. So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path `therefore lambda =1/(sqrt2npid^(2))` .....(2) The equation (2) implies that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases so it decreases the distance travelled by the molecule before collisions. Casel: REARRANGING the equation (2) using .m. (mass of the molecule) But mn=mass per unit `=rho` (density of the gas) `therefore lambda =1/(sqrt2pid^(2)rho)` ......(3) Also we know that PV=NkT `P=N/VkT=nkT impliesn=P/(kT)` substituting `n=P/(kT)` in equation (2), we GET The equation (4) implies the following: (i) Mean free path increases with increasing temperature. As the temperature increases. the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several meter away than smell of eld food. (II)Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules. |
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| 3. |
Two bubbles A and B made from the same solution (AgtB) are joined through a narrow tube. Then |
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Answer» SIZE of A will INCREASES |
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| 4. |
A body is rolling down in inclined plane. Its translational and rotational kinetic energies are equal. The body is a |
| Answer» Answer :D | |
| 5. |
Two particles of equal mass move in a circle of radius r under the action of their mutual gravitational attraction. Find the speed of each particle if its mass is m. |
Answer» Solution : The particle will ALWAYS remain diametrically OPPOSITE, so that the force on each particle will be directed along the radius. When each particle is describing a CIRCULAR ORBIT, the gravitational force on one of the PARTICLES must be equal to the necessary centripetal force. `(mV^2)/r=(Gmm)/(2r)^2` i.e., `V=sqrt((Gm)/(4r))` |
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| 6. |
A uniform rod of mass m and length l is suspended by means of two light inextensible strings at the ends of rod. Tension in one string immediately after the other string is cut is |
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Answer» SOLUTION :`MG-T=ma`……….`(1)` `alpha=(TAU)/(I)=(mg.(l)/(2))/((ml^(2))/(3))=(3g)/(2I)`………..`(2)` `a=(l)/(2)alpha`………`(3)` Solving equations `(1)`, `(2)` and `(3)` we get, `T=(mg)/(4)` 
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| 7. |
If the radius of the earth were to shrink by one percent, keepig its mass, the acceleration due to gravity on the earth.s surface would |
| Answer» Answer :A | |
| 8. |
Water flows in a horizontal pipe whose one end is closed with a value and the pressure gauge falls to 1 xx 10^(5)N//m^(2) when the value is opend. Calculate the speed of water flowing in the pipe. |
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Answer» |
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| 9. |
In a SHM, when the displacement is one- half the amplitude, what fraction of the total energy is kinetic |
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Answer» ZERO |
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| 10. |
The set containing only vector quantities is |
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Answer» THERMAL capacity, Magnetic susceptibility and Electric CHARGE |
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| 11. |
A motor is used to lift water from a well of depth 10 m and to fill a water tank of volume 30 m^(3) in 10 minutes. The tank is at a height of 20 m above ground. If 20% of energy is wasted, the power of the motor is (in kW). |
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Answer» `18.75` |
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| 12. |
Angular momentum acts perpendicular to |
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Answer» `VECR` |
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| 13. |
A particles is moving on circular path of radius R with constant angular velocity omega. Its centripetal acceleration will be …... . |
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Answer» |
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| 14. |
A spherical imperfectly elastic ball strikes a plane with a velocity 8 m/s at an angle 30^(@)to the plane. Determine the magnitude and direction of the velocity after the impact, if e = 0.5 (Neglect gravity). |
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Answer» |
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| 15. |
If 'U' be the speed of projection and V_(1) be thevelocity of striking the inclined plane when projected such that range up the plane is maximum. And V_(2) be the velocity of striking the plane when projected such that range down the plane is maximum then |
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Answer» `U^(2)=V_(1)V_(2)` |
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| 16. |
A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height H. What is the speed of the centre of mas when the cylinder reaches the bottom ? |
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Answer» `SQRT(2gh)` |
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| 17. |
A uniform disc of mass m, radius R is placed on a smooth horizontal surface. If we apply a horizontal force F at P as shown in the figure. If F = 4 N, m= .1 kg, R = 1 m and r = 1/2 m then, find the: A rod of length l = 1 m leaning against a vertical wall is pulled at its lowest point A with a constant velocity v = 4 ms^(-1). In consequence, the rod rotates win the vertical plane. When the rod makes an angle theta=4ms^(-1) with vertical and find the angular velocity of the rod (in rad s^(-1)). |
Answer»  Since `v_(P)=0` we have `omega_(BP)=omega_(AP)(=omega)` substituting `PA=lcostheta` and `v_(A)=v` We have `omega=v/(costheta)=4/(1.4/5)=5rads^(-1)` |
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| 18. |
In damped oscillations, the amplitude of oscillations is reduced to one - third of its initial value a_(0) at the end of 100 oscillations. When the oscillator completes 200 oscillations, the amplitude must be |
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Answer» `(a_(0))/(2)` |
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| 19. |
Which of the following time measuring devices is most precise? |
| Answer» Answer :D | |
| 20. |
One of the rectangular components of a velocity of 20 ms^(-1)is 10 ms^(-1). Find the other component. |
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Answer» `10sqrt(3) ms^(-1)` |
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| 21. |
A bullet is fired with a speed of 1500 m/s in order to hit a target 100 m away. if g 10 m//s^(2), the gun should be aimed: |
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Answer» 15 cm above the TARGET  `OA=R=(u^(2) sin 2 theta)/(g)` `or sin 2 theta=(RG)/(u^(2))=(100xx10)/((1500)^(2))=(1)/(2550)` `:. 2 theta= sin^(-1)(1)/(2250)=0.025^(@)` `or theta= 0.0.125^(@)=((0.0125xxpi)/(80))^(C)` `or theta =(2.21xx10^(-4))^(c)` But `theta=(AB)/(OA)` Or `AB=theta x OA=(2.21xx10^(-4)xx100)` `=(2.21xx10^(-2))m` `=2.21xx10^(-2)xx10^(2) cm` `=2.21 cm =3 cm` |
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| 22. |
Answer carefully, with reasons : If the potential energy of two billard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). |
| Answer» SOLUTION :ELASTIC. | |
| 23. |
A body is projected with a velocity u at an angle of 60^(@) to the horizontal. The time interval after which it will be moving in a direction of 30^(@) to the horizontal is |
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Answer» `(U)/(SQRT(3)G)` |
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| 24. |
The situation of a wave pulse is given as y=(0.8)/((4x+5t)+4) the amplitude and velocity of pulse are |
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Answer» `0.2` units, `1.25` units along `-`ve x-axis `4x+5t+4=`cons`tantimplies4dx+5dt=0` `impliesv=(DX)/(DT)` |
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| 25. |
Describe the model and theory of planetary motions presented in earliest of times. |
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Answer» Solution : `implies` The earliest recorded model for planetary motions proposed by PTOLEMY about 2000 years ago was a .geocentric. model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. `implies`A more elegant model in which the sun was the CENTRE around which the planets revolved the "heliocentric. model was mentioned by Indian Rushi and scientist Shri Aryabhatta in his treatise. `implies`A thousand years later, Nicholas Copernicus proposed a model in which the planets moves in circles around the fixed central sun. His theory was discredited by the church but NOTABLE by Galileo. `implies`After Galileo, Tycho Brahe recorded observations of the planets with the naked eye. `implies`His compiled data were analyzed by Johannes Kepler. He gave three laws for planetary motion. `implies` These laws were known to Newton and enabled him to make a GREAT scientific leap in proposing his universal LAW of gravitation. |
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| 26. |
Assuming that shear stress at the base of a mountain is equal to the force per unit area due to its weight, calculate the maximum possible height of a mountain on the earth if breaking shear stress for a typical rock is 30xx10^(7)N//m^(2) and its density 3xx10^(3)kg//m^(3). |
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Answer» Solution :For a MOUNTAIN of heighth and base area A, WEIGHT `W= Ah rhog`. So PRESSURE at the base due to its own weight will be `p=(W)/(A)=h rhog` The mountain will exist if, `h rho G lt` Breaking SHEAR stress `i.e, h=(30xx10^(7))/(3xx10^(3)xx10) (or) (h)_(max)=10 km` which is nearly the height of Mount Everest. |
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| 27. |
When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium. |
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Answer» SOLUTION :For translation a EQUILIBRIUM `sum F_("ext")=0` For ROTATIONAL equilibrium `sum VEC tau_(ext)=0` |
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| 28. |
For a liquid to rise in capillary tube, the angle of contact should be .......... |
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Answer» ACUTE |
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| 29. |
The absolute temperature of a body A is four times that of another body B. For the two bodies, the difference in wavelengths, at which energy radiated is maximum is 3.0 μm. Then the wavelength at which the body B radiates maximum energy, in micrometers is |
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Answer» `2` |
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| 30. |
The property of metals whereby they could be drawn into thin wires beyond their elastic limit without breaking is |
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Answer» Ductility |
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| 31. |
A person travels along a straight road for half the distance with velocity V_1 and the remaining half distance with velocity V_2 the average velocity is given by |
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Answer» `v_(1)v_(2)` |
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| 32. |
Consider a circular leveled road of radius 10 m having coefficient of static friction 0.81. Three cars (A, B and C) are travelling with speed 7 ms^(-1), 8 ms^(-1) " and " 10 ms^(-1) respectively, which car will skid when it moves in the circular level road? (g=10ms^(-2)): |
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Answer» Solution :From the SAFE turn condition the SPEED of the VEHICLE (v) must be less than or equal to `sqrt(mu_srg)` `v lt = sqrt(mu_s rg)` `sqrt(mu_srg ) = sqrt(0.81 xx 10 xx 10) = 9ms^(-1)` For Car C, `sqrt(mu_srg)`is less than v. The speed of car A, B and C are` 7 ms^(-1),8 m s^(-1)`and `10 m s^(-1)`respectively. The cars A and B will have safe turns. But the car C has speed `10 m s^(-1)`while it turns which EXCEEDS the safe turning speed. Hence, the car C will skid. |
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| 33. |
Heat given to a system is 35 joules and workdone by the system is 15 joules. The change in the internal energy of the system will be |
| Answer» ANSWER :B | |
| 34. |
Write any five properties of vector product of two vectors. |
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Answer» Solution :(i) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors `vecA` and `vecB`, even though the vectors `vecA` and `vecB` may or may not be mutually orthogonal. (ii) The vector product of two vectors is not commutative, i.e., `vecA xx vecB ne vecB xx vecA`. But, `vecA xx vecB = - vecB xx vecA`. Here it is worthwhile to note that `|vecA xx vecB| = |vecB xx vecA|` = AB sin `theta` i.e., in the case of the product vectors `vecA xx vecB` and `vecB xx vecA`, the magnitudes are equal but directions are opposite to each other. (III) The vector product of two vectors will have maximum magnitude when sin `theta` = 1, i.e., `theta = 90^(@)` i.e., when the vectors `vecA` and `vecB` are orthogonal to each other. `(vecA xx vecB)_(max)` = AB n (iv) The vector product of two non-zero vectors will be minimum when sin `theta` = 0, i.e., `theta = 0^(@)` or `180^(@)` `(vecA xx vecB)_(min)` = 0 i.e., the vector product of two non-zero vectors VANISHES, if the vectors are either parallel or antiparallel. (v) The self-cross product, i.e., product of a vector with itself is the null vector `vecA xx vecA` = AA sin `0^(@)` n = `VEC0` In physics the null vector `vec0` is simply denoted as zero. |
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| 35. |
A quantity of dry air at 27 ^(@)C is compressed (a) slowly and (b) suddenly to 1/3 of its volume. Find the change in temperature in both case assuming y = 1.4 for dry air. |
| Answer» Solution :`(a) DT =0. (b) T _(2) = T_(1) (V _(1)//V_(2)) ^( gamma -1) = 300 xx 3 ^(0.4) = 465.5 K, dT = 165.5 K` | |
| 36. |
The positiion vectors of two point masses 10 kg and 5 kg are (3hati+2hatj+4hatk)m and (3hati+6hatj+5hatk)m respectively. Locate the position of center of mass. |
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Answer» Solution :`m_(1)=10 kg` `m_(2)= 5kg` `vecr_(1)=(-3hati+2hatj+4hatk)m` `vecr_(2)=(3hati+6hatj+5hatk)m` `vecr=(m_(1) vecr_(1)+m_(2) vecr_(2))/(m_(1)+m_(2))` `:. vecr=(10(-3hati+2hatj+4hatk)+5(3hati+6hatj+4hatk))/(10+5)` `=(-30 HATI+20 hatj+40 hatk+15hati+30 hatj+25 hatk)/(15)=(-15 hatk+50 hatj+65 hatk)/(15)` `vecr=(-hati+(10)/(3) hatj+(13)/(3)hatk)` The center of mass is located at position `vecr` |
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| 37. |
Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is |
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Answer» `1:2^((1)/(3))` |
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| 38. |
Water is flowing in a non viscous tube as shown in the diagram .The diameter of point A and point B are 0.5 m and 0.1 m respectively . The pressure difference between point A and B are DeltaP=0.8m, then the rate of flow is …..m^(3)s^(-1). |
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Answer» `0.0314` `P_(1)+(1)/(2)rhov_(1)^(2)=P_(2)+(1)/(2)rhov_(2)^(2)(because` TUBE is horizontal) `thereforeP_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))` but according to equation of continuity `A_(1)v_(1)=A_(2)v_(2)`=rate of flow Q `thereforeQ=A_(1)v_(1)rArrv_(1)=(Q)/(A_(1))` and `Q=A_(2)v_(2)rArrv_(2)=(Q)/(A_(2))` `THEREFORE` Equation (1), `P_(1)-P_(2)=(1)/(2)rho[(Q^(2))/(A_(2)^(2))-(Q^(2))/(A_(1)^(2))]` `therefore(2(P_(1)-P_(2)))/(rho)=Q^(2)[(A_(1)^(2)-A_(2)^(2))/(A_(1)^(2)A_(2)^(2))]` `thereforeQ^(2)=(A_(1)^(2)A_(2)^(2)[2(P_(1)-P_(2))])/(rhoA_(1)^(2)-A_(2)^(2))` `thereforeQ^(2)=A_(1)A_(2)sqrt((2(P_(1)-P_(1)))/(rho(A_(1)^(2)-A_(2)^(2))))` `A_(1)=pi(0.25)^(2)=0.19625m^(2)andA_(1)^(2)=0.0385m^(4)` `A_(2)=pi(0.05)^(2)=0.00785m^(2)andA_(2)^(2)=6.16xx10^(-5)m^(4)` `DeltaP=P_(1)-P_(2)=0.8m` `thereforeQ=0.19625xx0.00785xxsqrt((2xx0.8)/(10^(3)xx(0.03844)))` `thereforeQ=0.00154xxsqrt(0.0416)` `thereforeQ=0.00154xx0.0204` `thereforeQ=0.000314m^(3)s^(-1)` |
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| 39. |
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. the precise dependence is given by the so called law of atmosphere' : n_(2) = n_(1) exp [-mg (h_(2)-h_(1))//k_(B)T] where, n_(1). n_(2) refer to number density at height h_(1) and h_(2) respectively. use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid columns: n_(2) = n_(1) exp [-mg N_(A) (roh-roh^('))(h_(2)-h_(1))//(roh RT)] where, roh is the density of the suspended particle, and roh' that of surrounding medium. [N_(A) is Avogadro's number, and R is the universal gas constant]. |
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Answer» Solution :ACCORDING to the law of atmospheres, `n_(2) = n_(1)` exp. `[-(mg)/(k_(B)T)(h_(2)-h_(1))]` ...`(i)` Where `n_(2), n_(1)` refer to number density of particles at heights `h_(2)` and `h_(1)` respectively. If we consider the sedimention equilibriumof suspended particles in a liquid, then in place of mg, we will haveto take effective weight of the suspended particles. LET `V =` averagevolume of a suspended particle, `rho=` density of suspended particle `rho'=` density of liquid, `m=`mass of one suspendedparticle, `m' = `mass of one suspended particle, `m' =` equal volume of liquid displaced. According to Archimede'sprinciple, effective weight of one suspended particle = actual weight - weightof liquid displaced `= mg - m'G` `=mg-V rho' g = mg -(m/(rho)) rho' g = mg (1-(rho')/(rho))` Also, Boltzman constant, `k_(B)= (R)/(N_A)` where, R is gas constant and `N_(A)` is Avogadro's number. Putting , `mg (1-(rho')/(rho))` in place of mg and VALUE of `k_(B)` in `(i)` , we get `n_(2) = n_(1)` exp. `[-(mgN_(A))/(RT)(1-(rho')/(rho))(h_(2)-h_(1))]` which is the required relation. |
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| 40. |
Rain drops are falling with a constant speed, by the time they reach the ground because |
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Answer» rain drops are too light to be accelerated by gravity. |
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| 41. |
The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K |
| Answer» Answer :B | |
| 42. |
One end of a uniform wire of length l and weight W is attached to a point in the roof and weight W /2 is suspended from the lower end. If A is the area of cross section of the wire then stress in the wire at its mid point is |
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Answer» W/2A |
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| 43. |
At what temperatue will the rms speed of oxygen molecule will be sufficient for esca-ping from the earth ?[v_(e)=11.2 km/s,m=2.76xx10^(-26)Kg and k=1.38xx10^(-23)J/K] |
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Answer» SOLUTION :The MEAN square VELOCITY of GAS molecule given by `v_(av)=sqrt((2KT)/(m))` For the molecule to just escape from the earth `v_(av)=v_(e)` or `sqrt((2KT)/(m))=v_(e)` `therefore T=(mv_(e)^(2))/(2K)=(5.34xx10^(-26)xx(11.0xx10^(3))^(2))/(2xx1.38xx10^(-23))=2.34xx10^(5)K` |
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| 44. |
A hydraulic automobile lift is used to life a car of mass 3000kg. The cross-sectional area of the piston on which the car is supported is 425cm^(2). What pressure would the smaller piston have to bear if the bigger piston with the car is 3m above the smaller piston? The density of the oil filling the hydraulic machine is 800kg*m^(-3). Let the pistons be of equal mass. |
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Answer» Solution :The POSITIONS of the pistions (with a car) of a hydraulic automobile LIFT is Fig. Let the area of the LARGER PISTON = `A_(2)` and the area of the smaller piston = `A_(1)`. If the pressure applied on the smaller piston is `p_(1)`, then according to Pascal.s law,  `p_(1)` = pressure on the larger piston + pressure exerted by a 3m LIQUID column `W/A_(2)+` pressure exerted by a 3m liquid column[W = weight of the car] = `(3000xx9.8)/(425xx10^(-4))+3xx800xx9.8=7.155xx10^(5)N*m^(-2)`. |
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| 45. |
A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase angle is 0^@ and 90^@. |
| Answer» SOLUTION :(a) 7.0 CM, (B) 5.0 cm | |
| 46. |
The escape velocity of a body at the surface of planet P is (1)/(sqrt(5)) times than that at surface of earth. The radius of planet is (1)/(24) times the radius of earth. The acceleration due to gravity for planet is |
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Answer» <P>`2.4 g_(E)` `V_(e) = sqrt(2g_(e) R_(e))` `V_(P) = (1)/(sqrt(5)) V_(e)` `sqrt(2g_(P)R_(P)) = (1)/(sqrt(5)) sqrt(2g_(e)R_(e))` `R_(P) = (R_(e))/(24)` `RARR sqrt((2g_(P)R_(e))/(24)) = sqrt((2g_(e)R_(e))/(5))` `(2g_(P)R_(e))/(24) = (2g_(e)R_(e))/(5)` `g_(P) = (24)/(5)g_(e) = 4.8 g_(e)` |
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| 47. |
Find the stretch in the springs shown in figure. The respective data are given in the figure. The friction and masses in pulleys are negligible. Find the extension in spring? |
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Answer» Solution :a) As mass M is in equilibrium, the tension in the string with which it is hanging will be Mg. As pulley is MASSLESS the tension in the upper string must be half of the lower string, Mg/2. The stretch in the spring must be such that the restoring force in the string is equal to the tension in it. Thus we have `k_1 x =(Mg)/(2) or x_1=(Mg)/(2k_1)` (b) Here again mass M is in equilibrium, thus the tension in the string connected to it must be equal to Mg and hence the restoring force in the lower spring will also be Mg. If `x_1` be the extension in this spring, we have `k_1x_1= Mg or x_1=(Mg)/(k_1)`  As pulley is light, the tension in upper string must be twice that of the lower srting, 2Mg which is equal to restoring force in the upper spring. If `x_1` is the extension in the upper spring, we have `k_2x_2=Mg or x_2=(2Mg)/(k_2)` Wherever a body is in equilibrium, due to inertia it is maintained in a state of rest. If any of the supporting force a body is REMOVED, body start from rest with some acceleration in the direction opposite to the force applied by the support, as all other forces acting on body will have a resultant in that direction. We discuss the situation with an example shown in figure. Two masses 5 kg and 10 kg are hanging in equilibrium with the strings A and B. If tensions in the strings A and B are `T_1 and T_2` respectively, then we have `T_1 = 5 g= 50N, T_2=15g = 150N` If at some instant we break string B. tension `T_1` becomes zero instantane-ously and block will start falling from rest under gravity and 10 kg block, as it is still in equilibrium, `T_2` will instantly change to 10 N to balance its weight. If INSTEAD of string B, we break the string A, tension `T_2` becomes zero and both the blocks will start falling with acceleration g. Now `T_1` is just an internal force for the system of two blocks and as there is no interaction between the two blocks and both are falling under gravity `T_1` will also instantly become zero. Now to understand the concept in a better way we slightly modify the situation of previous example. Instead of strings, we use springs as shown in figure. Initially the blocks are in equilibrium . If spring B breaks, its tension will instantly become zero as we are using ideal springs (massless/ inertialess),but due to inertia initially 10kg mass will not move from its INITIAL position, so tension in upper spring will not change instantly hence it will accelerate in upward direction with acceleration given as ` a=(150-100)/(10)=5m//s^(2)` If spring A breaks tension in spring A will instantly become zero, but as due to inertia 5 kg and 10 kg blocks will start moving from rest. As initally just after breaking of spring A the two masses are at rest, tension in spring B will not change at this instant which is T = 5g =50 N, Thus at this instant (just after spring A is cut) acceleration of 10 kg mass is ` a_1 = (10 g + kx)/( 10) = (100 + 50)/(10) = 15 m//s^(2)` and that of 5 kg mass is ` a_2=(5g-kx)/(5) = (50-50)/(5) =0m//sec^(2)` 
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| 48. |
A capillary tube of radius is dipped vertically in a liquid of density d, surface tension T and angle of contact theta then the pressure difference just below the two surface, one in the beaker and the other in the capillary tube, is : |
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Answer» `(2T)/(R )` |
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| 49. |
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed. |
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Answer» Solution :Calorimetry: Calorimetry MEANS the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, the heat LOST by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be Anomalous expansion of water in lakes mathematically EXPRESSED as `Q_("gain") = -Q_("lost")` `Q_("gain") + Q_("lost") = 0`  Heat gained or lost is measured with a calorimeter. Usually the calorimeter is an insulated container of water A sample is heated at HIGH temperature `(T_(1))` and IMMERSED into water at room temperature `(T_(2))` in the calorimeter. After some time both sample and water reach a final equilibrium temperature `T_(f)`. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water. `Q_("gain") = -Q_("lost")` Note the sign convention. The heat lost is denoted by negative sign and heat gained is denoted as positive. From the definition of specific heat capacity  `Q_("gain")=m_(2)s_(2)(T_(f)-T_(2))` `Q_("lost")=m_(1)s_(1)(T_(f)-T_(1))` Here `s_(1) and s_(2)` specific heat capacity of hot sample and water respectively. So we can write `m_(2)s_(2)(T_(f)-T_(2))=-m_(1)s_(1)(T_(f)-T_(1))` `m_(2)s_(2)T_(f)-m_(2)s_(2)T_(2)=-m_(1)s_(1)T_(f)+m_(1)s_(1)T_(1)` `m_(2)s_(2)T_(f)+m_(1)s_(1)T_(f)=m_(2)s_(2)T_(2)+m_(1)s_(1)T_(1)` `"The final temperature "T_(f)=(m_(1)s_(1)T_(1)+m_(2)s_(2)T_(2))/(m_(1)s_(1)+m_(2)s_(2))` |
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