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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. |
Answer the following : An artificial satellite orbitting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? |
| Answer» SOLUTION :K.E. INCREASES, but P.E. DECREASES, and the sum decreases due to DISSIPATION against FRICTION. | |
| 2. |
A body of mass 1 kg begins to move under the action of a time dependent force vec(F) =(2t hat(i)+3t^(2)hat(j)) N . Where hat(i) and hat(j) are unitvectors along X and Y axis . What power will be developed by the force at the time t . |
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Answer» `(2t^(2)+4t^(4))W` `vec(a) =(vec(F))/m = (2t hat(i)+3t^(2) hat(j))ms^(-2)` ` :. V = int_(0)^(t) "a dt"` ` = int_(0)^(t) (2t hat(i)+3t^(2)hat(j)) dt` ` = [ t^(2) hat(i) +t^(3) hat(j)]_(0)^(t)` `= t^(2) hat(i)+t^(3) hat(j)` ` :. ` Power `P = vec(F ) .vec(v)` ` = (2t hat(i) +3t^(2)hat(j)).(t^(2)hat(i)+t^(3)hat(j))` ` :. P = (2t^(3)+3t^(5))W` |
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| 3. |
Define and illustrate the following terms.Anti-parallel vectors, |
| Answer» SOLUTION :TWO vectors are SAID TOBE anti-parallel when they are in OPPOSITE directions . | |
| 4. |
If "1 g cm s"^(-1)=x newton- second, then the number x is equal to: |
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Answer» `10^(5)` `=(1gcms^(-1))/(10^(3)gxx10^(2)ms^(-2)xx1s)` `=(10gcms^(-1))/(10^(5)gcms^(-1))=10^(-5)` |
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| 5. |
Dimensional formula of the product of the two physical quantities P and Q is ML^2T^(-2), the dimensional formula of P/Q is MT^(-2). P and Q respectively are |
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Answer» FORCE, VELOCITY |
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| 6. |
It is ________ to open a door by pushing it near the outer edge than by pushing it near th e hinge.[ Fill in the blank ] |
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Answer» |
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| 7. |
Balu went to kuttraalam with his grandpa, when he saw the waterfalls falling down from the top of a mountain he could not believe his eyes. He loved admiring it. He adked grandpa how this is possible. Grandpa said even now the answer to this question from where the water is coming from the top of a mountain is a puzzle, but we got some scientific truths behind this flow. (ii) Give the formula for time of flight ofthe waterfall. |
| Answer» Solution :Actually the SCIENCE BEHIND this flow of water fall is it is an example for the projectile in horizontal projection water comes from the top of the MOUNTAIN with an initial horizontal velocity (U) | |
| 8. |
A body is thrown vertically up with certain velocity. If h is the maximum height reached by it, its position when its velocity reduces to 1/3 of its velocity of projection is at |
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Answer» `(8h)/(9)` from the ground |
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| 9. |
A light rays is travelling between two media as given below. The angle of incidence on the boundary in all the cases is 30^@. Identify the correct sequence of increasing order of angles of refraction. a) Air to water b) Water to glass c) Glass to water (refractive indices of glass and water are respectively3//2 and 4//3) |
| Answer» Answer :A | |
| 10. |
A uniform wire of mass m, length L, area of cross-section A and Young's modulus Y hange from the ceiling. Its elongation under its own weight will be |
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Answer» ZERO `:. Delta L= ((mg) ((L)/(2)))/(AY) = (mgL)/(2AY)` |
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| 11. |
If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes |
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Answer» Walking on ground would become more difficult. |
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| 12. |
Two bodies reIeased from different heights fall freeIy and reach the ground at the same time. The first body takes a time,t_(1) = 2s and second body takes a time, t_(2) = 1s. What was the height of the first body at the time of reIease of the second body? |
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Answer» SOLUTION :Let the height from which the first body is released be `h_(1)` above the ground. When it is at a height `h_(2)` the second body is released . So, the first body FALLS through a height `(h_(1) - h_(2))` in time `(t_(1)-t_(2))` before the release of thesecond body. According to the problem, `h_(1) = (1)/(2) "gt"_(1)^(2)" ""and"" "h_(1)-h_(2) = (1)/(2) G(t_(1)-t_(2))^(2)` or, `""(1)/(2)"gt"_(1)^(2)-h_(2)=(1)/(2)g (t_(1)-t_(2))^(2)` or, `"" h_(2) = (1)/(2)g(2t_(1)-t_(2))t_(2)` `= (1)/(2) xx 9.8 (2xx2-1)xx1 = 14 .7 `m. 
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| 13. |
At which position (point) pendulum of a clock have maximum speed ? |
| Answer» SOLUTION :At the LEAST POSITION of OSCILLATION of PENDULUM . | |
| 14. |
Can you find an example of decrease in internal energy at constant temperature? |
| Answer» SOLUTION :FREEZING WATER at `0^@ `C. | |
| 15. |
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (Mass of the sun = 2 xx 10^(30) kg). |
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Answer» Solution :`implies` Mass of SUN `= 2 xx 10^(30) kg` Mass of neutroní stars `M = 2.5 xx 2 xx 10^(30) kg` ` = 5 xx10^(30) kg` Radius of neutron stars `R = 12 km = 1.2 xx 10^(4 )m` Frequency revolution `F = 1.5 Hz ` Let the acceleration DUE to gravity on the surface of neutron stars is g, `g=(GM)/(R^2)` `= (6.67xx10^(-11) xx 5 xx 10^(30))/((1.2xx10^(4))^2)` `:. g = (GM)/(R^2) = 2.3 xx10^(12) ms^(-2)` Centrifugalacceleration `a_c` gets in body at equator `:. a_c = v^2/R` `:. a_c = Romega^2""[ :. v = 2Romega]` ` = R xx (2pif)^2` `= R pi^2Rf^2` `= 4xx(3.14)^2 xx1.2 xx10^(14) xx(1.5)^2` `= 1046.48 xx10^8` `:. a_c = omega^(2) R = 1.1 xx 10^(6) ms^(-2)` `g gt gt a_c` inward force at equator is more than centripetal force. So, the body will remain struck to the surface of the star due to gravitational force of star, |
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| 16. |
Statement - A : If a uniform metal disc is remoulded into a solid sphere, then the moment of inertia about the axis of symmetry increases than that before.Statement - B : For a given body for a given plane, the moment of inertia is minimum about an axis passing through the centre of mass. |
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Answer» Both A and B are WRONG. |
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| 17. |
A cannon and a target are 5.10 km apart and located at the same level. How soon will the shell launched with the initial velocity 240m/s reach the target in the absence of air drag? |
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Answer» Solution :Here, `v_(0)=240ms^(-1)`,R5.10km5100m, `g=9.8ms^(-2), alpha=?""R+(v_(0)^(2)sin2alpha)/(g)` `sin2alpha=(Rg)/(V_(0)^(2))""rArralpha=30^(@) or 60^(@)` using = `T =(2v_(0)sinalpha)/(g)` When, `alpha = 30^(@), T_(1)=(2xx240xx0.5)/(9.8)=24.5s` When,`alpha=60^(@), T_(2)=(2xx240xx0.867)/(9.8)=42.42s` |
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| 18. |
A glass rod radius 1. 0mm is inserted symmetrically into a vertical capillary tube of radius 2.0 mm such that their lower ends are at the same level. The arrangement is now dipped in water. The height to which the water rises in the capillary tube is (T=75xx10^(-3)N//m) |
| Answer» Answer :C | |
| 19. |
Quantity of heat lost in condensation of 10 gm of steam at 100^@C |
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Answer» `2.26 xx10^(5)J` |
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| 20. |
A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 30^(@). Find the work done by the force ? |
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Answer» ASSERTION is TRUE, REASON is true but it is not the correct EXPLANATION. |
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| 21. |
Two artificial satellites A and B of same mass are revolving round the earth at heights h_(1) and h_(2) in circular orbits. If h_(2) lt h_(1), satellite possessing greater KE is |
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Answer» A |
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| 22. |
Moment of inertia of a thin square plate about an axis passing through its diagonal is I. Its moment of inertia about an axis passing through its centre in the plane of the plate and making an angle theta with thediagonal is |
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Answer» I |
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| 23. |
The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length? |
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Answer» `Gh^(2)C^(3)` |
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| 24. |
If the amount of heat given to a system is 35 J and the amount of work done on the system is 15 J, then the change in internal energy of the system is |
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Answer» `-50 J` `DELTA Q = Delta U + Delta W` or `Delta U = Delta Q - Delta W` here , `Delta Q = 35 J, Delta W = -15 J :. Delta U = 35 J - (-15 J) = 50 J` NOTE :`Delta W` is negative because work done on the SYSTEM. |
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| 25. |
A ball is dropped from a height h on to a floor. If, in each collision, its speed becomes e times of its striking value (a) Find the total change in momentum of the ball (b) Find the average force exerted by the ball on the floor. |
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Answer» Solution :(a) Change in momentum in I collision `= mv_(1)-(-mv_(0))=m(v_(1)+v_(0))` Change in momentum in II collision `= m (v_(2)+v_(1))` Change in momentum in `n^(th)` collision `= m (v_(n)+v_(n-1))` Adding these all, total change in momentum of the ball is `DELTA p=m[v_(0)+2v_(1)+...+2v_(n-1)+v_(n)]` or `Delta p = mv_(0)[1+2e+2e^(2)+.....]` `["as "v_(1)=ev_(0), v_(2)=e^(2)v_(0)......]` `Delta = mv_(0)[1+2e((1)/(1-e))]=m sqrt(2gh)[(1+e)/(1-e)]` ....(1) (b) Now as `VEC(F)=(vec(dp))/(DT)` so, `F_(av)=(Delta p)/(Delta T)` We know `Delta T= sqrt((2H)/(g))((1+e)/(1-e))` from (1) `Delta p = m sqrt(2gh)[(1+e)/(1-e)]` `F_(av)=m sqrt(2gh)[(1+e)/(1-e)]xx sqrt((g)/(2h))[(1-e)/(1+e)]=mg` |
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| 26. |
In a rectangle ABCD, AB = 2l and BC = l. The centre of rectangle is at the origin of co-ordinate system. AB and CD are parallel to X-axis. The M.I. is least about |
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Answer» BD |
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| 27. |
What is a perfectly plastic body? |
| Answer» Solution :If a body has no TENDENCY at all to REGAIN its original shape and SIZE on the REMOVAL of deforming force it is called a PERFECTLY plastic body. | |
| 28. |
Two circular discs have their masses in the ratio 1:2 and radii in the ratio 2:1. The ratio of their inoments of inertia is |
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Answer» 0.084027777777778 |
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| 29. |
Under identical conditions of temperature, presure and volume, the ratio of mean free paths of the molecules of two gases having moleculaer diameter 1A^@ and 2A^@ is |
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Answer» `1:4` |
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| 30. |
A particle of mass 0.1 kg moving with an initial speed v collides with another particle of same mass kept at rest. If after collision total energy becomes 0.2 J, then the minimum and maximum values of .v. are |
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Answer» Solution :In case of perfectly INELASTIC COLLISION the velocity of combined mass will be v/2 `therefore KE=(1)/(2)(2m)((v^(2))/(4))i.e., (0.2)=(1)/(2)(0.2)((v^(2))/(4))` `RARR v=2sqrt(2)m//s`. In case of perfectly elastic collision, first particle STOPS and the second acquires the same velocity v `KE=(1)/(2)mv^(2)i.e., 0.2=(1)/(2)(0.1)v^(2) rArr v=2 m//s` Therefore, minimum value of v is 2 m/s in case of perfectly elastic collision and maximum value of v is `2sqrt(2)` m/s in case of perfectly inelastic collision. |
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| 31. |
If the gravitational force between two objects were proportional to 1/R(and not as 1/R^2 ) where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v_1proportional to ......... |
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Answer» `1/R^2` `:. V PROP R^@` |
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| 32. |
Half of the chain is lying on a table. The rest of the chain hangs in equilibrium from both of the ends of the table. Extra length of the hanging chain at one end is 1/8 th of the total length of the chain than that hanging at the other end. The coefficient of friction between table and chain is 25 xx 10^(-4) the value of x is |
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Answer» |
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| 33. |
If a spherical ball contract in volume by 1% under a normal uniform pressure of 200 atmophere, then the compressibility of the material of the ball is (1 atmosphere = 10^(5) Nm^(-2)) |
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Answer» `1 xx 10^(12)` |
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| 34. |
A disc of mass M and radius R moves in the x-y plane as shown in the figure. The angular momentum of the disc at tihe instant shows is |
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Answer» `(5)/(2)mR^(2) OMEGA` about O |
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| 35. |
If the momentum of a body increases by 50%,its K.E,increases by |
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Answer» 5 |
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| 36. |
Which of the following parameters does not characterise the thermodynamic state of matter? |
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Answer» Temperature |
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| 37. |
Explain why (a) there is no atmosphere on moon. (b) there is fall in temperature with altitude. |
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Answer» Solution :(a) GRAVITATIONAL FORCE of moon is roughly `1/6 th of the gravitational force of earth. Therefore, escape velocity of air molucules on moon is smaller than that on earth. Since moon is in proximity of earth, as seen from the sun, therefore, moon rececive almost same amoumt of heat per unit area as does the earth. A large number of molecules on moon have speeds greater than escape velocity on moon, and therefore, they escape. the rest of the molecules arrange the speed distribution for the equilibrium temerature. again, a significant number of molecules escape as their speeds exceed the escape speed. hence, over along PERIOD of time, moon has lost most of its atomsphere. (B) as molecules move higher, their potential ENERGY increases and hence KE decreases. The temperature falls. At greater height, more volume is available. the gas expands and hence cooling occurs. |
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| 38. |
Both ends of an aluminum rod of length 20 cm and cross-sectional area 4" cm"^(2) are kept at a temperature 0^(@)C. Thermal conductivity of aluminum is 205 W/(m-k). The initial distribution of temperature along the rod can be expressed as T=75sin(15.7x)""^(@)C. Here, length of the rod has been considered to be along the X direction and one of the ends as the origin so that .x. in the initial temperature distribution equation refers to the x position of any point on the rod. Also the angle in the sin term is in radian provided that x is expressed in the SI system Initially, the rate of heat flow at the centre of the rod will be |
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Answer» 0 |
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| 39. |
Find the speed of two objects if when they move uniformly towards each other, they get 4.0 metres closer each second and when they move uniformly in the same direction with original speeds, they get 4.0 metres closer each 10 sec. |
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Answer» SOLUTION :Let their SPEEDS be `v_(1)` and `v_(2)` and let `v_(1)gt v_(2)`. In First Case : relative VELOCITY, `v_(1) + v_(2) = 4` …..(1) In SECOND Case : Relative velocity, `v_(1) - v_(2) = (4)/(10) = 0.4` ….(2) Solving eqns. (1) and (2), we get `v_(1) = 2.2 MS^(-1), v_(2) = 1.8 ms^(-1)` |
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| 40. |
Both ends of an aluminum rod of length 20 cm and cross-sectional area 4" cm"^(2) are kept at a temperature 0^(@)C. Thermal conductivity of aluminum is 205 W/(m-k). The initial distribution of temperature along the rod can be expressed as T=75sin(15.7x)""^(@)C. Here, length of the rod has been considered to be along the X direction and one of the ends as the origin so that .x. in the initial temperature distribution equation refers to the x position of any point on the rod. Also the angle in the sin term is in radian provided that x is expressed in the SI system Initial temperature distribution along the length of rod is given above. Final temperature distribution after a long time, can be expressed as |
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Answer» `T=75cos(15.7x)""^(@)C` |
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| 41. |
Both ends of an aluminum rod of length 20 cm and cross-sectional area 4" cm"^(2) are kept at a temperature 0^(@)C. Thermal conductivity of aluminum is 205 W/(m-k). The initial distribution of temperature along the rod can be expressed as T=75sin(15.7x)""^(@)C. Here, length of the rod has been considered to be along the X direction and one of the ends as the origin so that .x. in the initial temperature distribution equation refers to the x position of any point on the rod. Also the angle in the sin term is in radian provided that x is expressed in the SI system Initially the rate of heat flow at any end will be |
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Answer» 0 |
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| 42. |
Define latent heat of fusion and latent heat of vaporisation. |
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Answer» Solution :Change of state of a substance from solid to liquid is called melting or fusion. The quantity of heat REQUIRED to change a solid at its melting points to liquid at the same TEMPERATURE is called L.H. of fusion. The quantity of heat required to change a liquid at its boiling POINT to its vapour at the same temperature is called the latent heat of vaporisation. Melting point and boiling point DEPENDS on pressure S.L. UNIT of latent heat is J kg. |
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| 43. |
A steel ball is dropped from the roof of a building . An obsever standing in front of a window 1.5m high notes that the ball takes 1//10s to fall from the top to the bottom of the window. The ball reappears at the bottom of the window 2s after passing it on the way down . If the collision between the ball and the ground is perfectly elastic, then find the height of the building ?Take g=10m//s^(2). |
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Answer» Solution :Since collision is PERFECTLY ELASTIC , the speed of the ball just before collision is equal to the speed of the ball just after collision . Hence time of descent is equal to the time of ascent. Therefore time taken by the ball to reach the ground from the bottom of th ewindow is 1sec. Let u be the speed of the ball when it is at the top of the window `rArr1.5=u1/(10)+1/2xx10xx1/(100).,(vecs=vecut+1/2vecat^(2))rArr u=14.5m//s` `:.`ball is DROPPED hence its initial speed is 0 It t be the time taken by the ball to acquire the speed of `14.5m//s` , then `14.5=0+10xxt,(vecv=vecu+vecat)rArr t =1.45sec` Hence total time of descent is GIVEN by `T=1.45+1/(10)+1=2.55s` If H be the height of the BUILDING , then `H=0+1/2"gt"^(2)` `rArrH=1/2xx10xx(2.55)^(2)rArrH=32.5m` |
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| 44. |
{:("List - 1","List - 2"),("a) Temperature","e) Light Year"),("b) Mass","f) Shake"),("c) Length","g) Fahrenheit"),("d) Time","h) Atomi Mass unit"):} |
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Answer» a-h B-f c-g d-e |
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| 45. |
What is meant by Doppler effect? Discuss the following cases. (1) Sources in motion and Observer at rest (a) Source moves towards observer (a) Source moves towards observer (b) Source moves away from the observer (2) Observer in motion and Source at rest. (a) Observer moves towards Source. (b) Observer recedes away from the Source. (3) Both are in motion (a) Source and observer approach each other (b) Source and Observer resides from each other ( c) Source chases Observer (d) Observer chases Source. |
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Answer» Solution :Doppler Effect: Often we have noticed that the siren sound coming from a police vehicle or ambulance increases when it comes away from us. When we stand near any passing train the train whistle initially increases and then it will decrease. This is known as Doppler Effect, named after Christian Doppler (1803-1853). Suppose a source PRODUCES sound with some frequency, we call it the as source frequency `f_(1)`. If the source and an observer are at a fixed distance then the observer observes the sound with frequency `f_(0)`. This is the same as the sound frequency produced by the source `f_(1)`, i.e., `f_(0)=f_(3)`, Hence, there is no difference in frequency, implying no Doppler effect is observed. (1) Source in motion and Observer at rest: (a) Source moves towards observer: A source S moves to the right (as shown in Figure 1) with a velocity `v_(s)` and let the frequency of the sound waves produced by the source be `f_(s)`. Velocity of sound in a medium is v. The components (sound wave front) produced by the source S at three successive instants of time are shown in the Figure 1. When S is a position `x_(1)` the compression is at `C_(2)` and similarly for `x_(3)` and `C_(3)`. Assume that if `C_(1)` reaches the observer's position A then at that INSTANT `C_(2)` reaches the point B and `C_(3)` reaches the point C as shown in the figure 1. The distance between compressions `C_(2)` and `C_(3)` is shorter than distance between `C_(1)` and `C_(2)`. This MEANS the wavelength decreases when the sources S moves towards the observer O (since sound travels longitudinally and wavelength is the distance between two CONSECUTIVE compression). But frequency is inversely related to wavelength and THEREFORE, frequency increases.  Let `lambda` be the wavelength of the source S as measured by the observer when S is at position `x_(1)` and `lambda`' be wavelength of the source observed by the observer when S moves to position `x_(2)`. Then the change in wavelength is `Delta lambda = lambda-lambda'=v_(3)t`, where t is the time taken by the source to travel between `x_(1)` and `x_(2)`. Therefore, `lambda'= lambda -v_(s)t` But `t= lambda/v` On substituting equation `t= lambda/v` in equation `lambda' = lambda - v_(s) t`, we get `lambda' = lambda(1- v_(r)/v)` Since frequency is inversely proportional to wavelength we have `f'= v_(s)/lambda'` and `f=v_(r)/lambda` Hence, `f' = 1/(1- v_(r)/v)` Since, `v_(s)/v lt 1`, we use the binomial expansion and retaining only first order in `v_(r)/v`, we get `f' = f(1+v_(s)/v)` (b) Source moves away from the observer: Since the velocity here of the source is opposite in direction when compared to case (a), therefore, changing the sign of the velocity of the source in the above case i.e., by substituting `(v_(s) to -v_(s))` in equation `lambda = lambda - v_(s) t`, we get `f' = 1/(1-v_(s)/v)` (2) Observer in motion and Source at rest: (a) Observer moves towards Source:  Assume that the observer O moves towards the source S with velocity `v_(0)`. The source S is at rest and the velocity of sound waves (with respect to the medium) produced by the sources is v. From the figure (2), we observe that both `v_(0)` and v moves in the same direction. Therefore, their relative velocity is `v_(r) =v-v_(0)` . The wavelength of the sound wave is `lambda =v/f`, which means the frequency observed by the observer O is `f'=v_(r)/lambda`. Then, `f'=v_(r)/lambda =((v+ v_(0)))/v f= f(1+ v_(0)/v)` (b) Observer recedes away from the Source: If the observer O is moving away (receding away) from the source S, then velocity `v_(0)` and v moves in the same direction. Therefore, their relative velocity is `v_(r) =v-v_(0)`. Hence, the frequency observed by the observer O is `f'=v_(r)/lambda =(v-v_(0))/v f =f(1-v_(0)/v)` (3) Both are in motion: (a) Source and Observer approach each other:  Let `v_(s)` and `v_(0)` be the respective velocities of source and observer approaching each other as shown in Figure 3. In order to calculate the apparent frequency observed by the observer, as a simple calculations, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, the dummy (observer) observes the apparent frequency equation `f' (f/(1-v_(s)/v))` as `f_(d) =(f/(1-v_(s)/v))`..........(1) `rArr (vf')/(v+v_(0)) =(vf)/(v-v_(s))` At that instant of time, the true observer approaches the dummy from the other side. Since the source (true source) comes in a direction opposite to true observer, the dummy (source) is treated as stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), from equation `f'=v_(r)/lambda = ((v+v_(0)))/vf =f(1+v_(0)/v)` is `f'=f_(d)(1+v_(0)/v) rArr f_(d) = (f')/(1+v_(0)/v)`.........(2) Since, this is true for any arbitrary time, therefore, comparing equation (1) and equation (2), we get `f/((1-v_(s)/v)) =(f')/(1+v_(0)/v)` `rArr (vf')/(v+v_(0)) =(vf)/(v-v_(s))` Hence, the apparent frequency as seen by the observer is `f'= ((v+v_(0))/((v-v_(s))))f`.......(3) (b) Source and observer receds from each other:  Here, we can derive the result as in the previous case. Instead of a detailed calculation, by inspection from Figure 4, we notice that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute `(v_(s) to v_(s))` and `(v_(0) to -v_(0))` in equation (3), and therefore, the apparent frequency observed by the observer when the source and observer recede from each other is `f'=((v-v_(0))/(v+v_(s)))f`.......(4) (c ) Source chases Observer  Only the observer's velocity is oppositely directed when compared to case (a). Therefore, substituting `(v_(0) to -v_(0))` in equation (3), we get `f'=((v-v_(0))/(v-v_(s)))f`.......(5) (d) Observer chases Source  Only the source velocity is oppositely directed when compared to case (a). Therefore, substituting `v_(s) to -v_(s)` in equation (3), we get `f'=((v+v_(0))/(v+v_(s)))f` |
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| 46. |
An object is held firmly in between the two studs of a screw gauge. By looking into the arrangement shown product the possible reading of the object. The main scale / pitch scale is graduated in 0.5 mm as can be seen in the diagram. The arrangement is past of the screw gauge apparatus with object in between the studs. (Consider head scale has 100 divisions).The column - I shows the possible situations of the screq gauge apparatus and column - II situations of the screw gauge apparatus and column - II shows corresponding reading of the object. Match the columns appropriately.{:(,"Column - I",,"Column - II"),((A),"Instument has zer error" +6,(P),3.72 mm),((B),"Instument hass zero error" +8,(Q),3.70 mm),((C ),"Instument has zero error"-6,(R ),3.076 mm),((D),"Instument has zero no error",(S),3.69 mm):} |
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| 47. |
A body has a time of 3 s under the action of one force and 4 s under the action of another force. Calculate the time period when both the forces act in the same direction simultaneously. Assume that the body undergoes the same displacement, under the action of the two different forces. |
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| 48. |
The ratio of the thermal conductivities of two different materials is 1 : 2. The thermal resistance of the rods of these materials having the same thickness are equal. Find the ratio of the length of rods. |
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| 49. |
Assertion : Surface tension decreases with increase in temperature .Reason : On increasing temperature kinetic energy increases and intermolecular forces decreases . |
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Answer» Assertion and REASON aretrue and reason is the CORRECT explaination of assertion. |
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| 50. |
A force Facting on a body depends on its displacement Sas F prop S^(-1//3). The power delivered by F will depend on displacement as |
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Answer» `S^(2//3)` `:.` ACCELERATION, `a prop S^(-1//3)` But `a = (DV)/(dt) = (dv)/(dS) (dS)/(dt) = (dv)/(dS) v :. V (dv)/(dS) prop S^(-1//3).` Intergrating both sides, we find `v^2 prop S^(2//3) or v prop S^(1//3)` As `P = Fv` `:. P prop (S^(-1//3))(S^(1//3)) = S^(0)`. i.e., power is independent of S. |
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