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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. |
Statement A: When no external force acts on a body its centre of mass will be at rest or under uniform motion. Statement B: When a force acts on a body at its centre of mass then will have only translation motion |
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Answer» only A is CORRECT |
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| 2. |
A mass m is suspended from a spring of force constant k. The period of oscillation is T_0The spring is cut into 4 equal parts: The same mass m is suspended from one of the parts. What is the new period? |
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Answer» |
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| 3. |
Is there any difference between an error and a mistake ? Explain. |
| Answer» Solution :Yes. ERROR is the difference between the CAREFULLY MEASURED value and true value. The errors arising from the fact that the observer has not taken all the precautions NECESSARY to avoid errors are CALLED mistakes. | |
| 4. |
A block of wood mass 5kg is placed on a plane making an angle 30^(@) with the horizontal. If the co-efficint of friction between the surface of contact of the body and plane is 0.5. What force is required to keep the body sliding down with uniform velocity |
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Answer» 1.6 N |
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| 5. |
Imagine a light planet is revolving round a very massive star in a circular orbit of radius R with a time period of revolution T. If the gravitational force of attraction between the star and planetis propostional to R^(-n), then T^2 is propostional to |
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Answer» `R^(n+1)` |
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| 6. |
If the Radius of earth were shrink by 10% its mass remaining same, the acceleration due to gravity on earth.s surface would |
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Answer» DECREASE |
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| 7. |
What is meant by mean by force constant of a spring ? |
| Answer» SOLUTION :The FORCE constant (or) spring FACTOR is defined as the restoring force produced PER unit DISPLACEMENT. | |
| 8. |
A man of mass 80 kg is riding on a small cart of mass 40 kg which is rolling along a level floor at a speed of 2 m/s. He is running on the cart so that his velocity relative to the cart is 3 m/s in the direction opposite to the motion of cart. What is the speed of the centre of mass of the system ? |
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Answer» |
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| 9. |
A sphere rolls without slipping on an incline of inclination theta. The minimum coefficient of static friciton to support pure rolling is to be |
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Answer» `2/3Tantheta` |
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| 10. |
Parallel rays of light are falling on convex sphere surface of radius of curvature R = 20 cm as show. Refractive index of the medium is mu = 1.5. A refraction from the spherical surface parallel rays |
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Answer» actually meet at some point |
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| 11. |
Derive expression for maximum height, time of flight and range of a projectile. |
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Answer» Solution :The maximum vertical distance travelled by the projectile during its journey is called maximum height. (II) For the vertical part of the motion, `v_(y)^(2)=u_(y)^(2)+2a_(y)s_(y)` (III) Here, `V_(y)=0, S=_(y)=h_("max"), u_(y)=u and a_(y)=-g` Therefore. `0=u^(2)sin^(2)theta-2gh_("max"),""h_("max")=(u^(2)sin^(2)theta)/(2g)` Time of flight : `(T_(f))` (i) The time of flight `(T_(f))` is the time taken by the projectile to hit the ground after thrown. (ii) The downward distance travelled by the projectile at a time t can be written as, `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)` (iii) Here substituting the values `S_(y)=0, t=T_(f).u_(y)=u sin theta, and a_(y)=-g ` we get, `0=u sin theta-(1)/(2)gT_(f)^(2)` Therefore, `T_(f)=(2u sin theta)/(g)` Horizontal range : (R) (i) The horizontal range (R) is the maximum horizontal distance distance between the point of projection and the point where the projectile HITS the ground. (ii) The horizontal distance travelled by the projectile at a time t can be written as, `s_(X)=u_(x)t+(1)/(2)a_(x)t^(2)` (iii) Here, `S_(x)=R, u_(x)=u cos theta, a_(x)=0 and t=T_(f)` `R=u cos theta.T_(f)` `R=u cos theta.(2u sintheta)/(g)=(2u^(2)sin theta cos theta)/(g)` `""[because T_(f)=(2usin theta)/(g)]` (iv) Therefore, `""F=(u^(2)sin 2theta)/(g)` `""[because sin 2theta=2 sin theta. cos theta]` (v) For maximum range is, `R-(u^(2))/(g)` |
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| 12. |
A stone of m ass 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 d irected vertically dow nw ards are : [Choose the correct alternative] Lowest Point (a) mg - T_(1)(b) mg + T_(1) (c) mg + T_(1) - (mv_(1)^(2))//R (d) mg - T_(1) - (mv_(1)^(2))//R Highest Point mg + T_(2) mg - T_(2) mg - T_(2) + (mv_(1)^(2))//R mg + T_(2) + (mv_(1)^(2))/R T_(1) and 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 :Whenstoneis tieldat endof stringand whirledin CIRCULARPATH inverticalplaneletH behighestpointand L belowest POINT AtpointL : Tension`T_(1)`towardcentreand weigth mg isin downwarddirection  Resultantforceactingon STONEIS indownwarddirection Resultantforce=mg= `T_(1)` At pointH : Heretensiion `T_(2) `and weightmg Henceoption( A)iscorrect. |
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| 13. |
How many 2.5 kg bricks can man carry up a staircase 3.6 m high in one hour if he works at an average rate of 9.8 watt? |
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Answer» SOLUTION :Work DONE by man in one hour= power `XX` time `= 9.8 xx 1 xx 60 xx 60 J` Work done by man in raising one brick = `mgh = 2.5 xx 9.8 xx 3.6 J` . Number of bricks , `N = (9.8 xx 1 xx 60 xx 60)/(2.5 xx 9.8 xx 3.6) = 400`. |
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| 14. |
When two ice blocks are rubbed against each other some ice melts in between the two. Which law can explain the process ? |
| Answer» SOLUTION :Joule.s LAW, | |
| 15. |
A block of mass 0.5 kg rests on a wedge of mass 2 kg as in Fig. 7.87 . The wedge is acted on by a horizontal force F and slides ona frictionless surface . If the coefficient of static friction between the wedge and the block is mu_(s) = 0.8 , and the angle of inclination is 30^(@) , find the maximum and minimum values of F for which the block doesnot slip . Take (g = 10 m//s^(2)) |
Answer»  `F + N "sin" theta - f"cos" theta = ma "" ……. (i)` N cos` theta +f "sin" theta =mg "" …… (ii) ` `a= (F)/(m + M) "" ....... (iii) ` and `"" f = mu_(s) N "" ........ (iv)` Solving, `"" F = (m + M) G [(mu "cos"theta - "sin" theta )/(mu "sin" theta + "cos" theta)]` = `3. 82 N ` (min) If we increase F , the direction of friction will GET reversed then again WRITTING force equation in horizontaland vertical directions and solving, we get `F = (m + M) g [(mu "cos"theta + "sin" theta )/( "cos" theta - mu"sin" theta)] = 64.6 N` (MAX) |
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| 16. |
A vertical hollow cylinder of height 1.52 m is fitted with a movable piston of neglible mass and thickness. The lower half portion of the cylinder contains an ideal gas and the upper half is filled with mercury. The cylinder is initially at 300 K. When the temperature is raised, half of the mercury comes out of the cylinder. Find this temperature, assuming the thermal expansion of mercury to be negligible. |
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Answer» SOLUTION :INITIALLY, Pressure = 76 + 76 = 152 CM of Hg From `PV=nRT rArr (152 times 76 timesA)/(300)=nR"…….(1)"` Final pressure = 76 + 38 = 114 cm of Hg `therefore (114 times (76+38)A)/T=nR"……(2)"` From (1) and (2) `rArr (152 times 76)/(300)=(114 times 114)/T, T=(300 times 114 times 114)/(152 times 76)=(300 times3)/4 times 3/2=3375K=337.5^(@)C` 
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| 17. |
A uniform circular disc of radius r . 1placed on a rough horizontal plane has initial velocity v_(0) and an angular velocity omega_(0) as shown. The disc comes to rest after moving some distance in the direction of motion. Then |
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Answer» the friction force acts in the backwards direction |
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| 18. |
If the temperature of a wire is doubled then Young's modulus of elasticity ...... |
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Answer» will also double. `L _(t) = L _(0) (1 + alpha Delta T)` where `Delta T` is the change in temperature, `L _(0)=` original length, `L_(t)` is length at temperature `t, alpha=`COEFFICIENT of linear expansion. `therefore L _(T) - L _(0) = L _(0) prop Delta T` `therefore Delta L - L _(v) ~~ Delta T""...(1)` Now Young.s modulus `Y = ("Stress")/("Strain") = (F//A)/(Delta L //L _(0)) = (FL _(0))/( A Delta L )` `= (FL _(0))/( AL _(0) prop Delta T) ` [ From eq. (1)] `therefore Y prop (1)/(Delta T)` (ALl other terms are constant) Hence, Y decrease with increasing temperature. |
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| 19. |
A massive box is dragged along a horizontal floor by a rope. The rope makes an angle of 60^(@) with the horizontal. Find the work done if the tension in the rope is 200N and the box is moved through a distance of 20m. |
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Answer» Solution :Tension T= 200N, distance `S= 20m, THETA= 60^(@)` Word done `W= vec(F).vec(S) = FS cos theta = (F cos theta)s` `=200 xx cos 60^(@) xx 20= 2000J` |
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| 20. |
A bus between Vijayawada and Hyderabad passed the 100km, 160-km and 220-km points at 10.30a.m, 11.30a.m and 1.30 p.m. Find the average speed of the bus during each of thefollowing intervals: 10.30 a. m, to11.30 a.m. 11.30 a.m . to1.30 p.m 10.30 a.m to 1.30 p.m |
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Answer» SOLUTION :The DISTANCE covered between `10.30 a.m "and " 11.30 a.m =160km=60KM ` The time interval =`11.30-10.30`=1 hour The AVERAGE speed during this time interval `=v_(1)=(60km)/(1h)=60km//h`. The distance covered between `11.30a.m "and"1.30p.m =220km-160km=60km` The time interval =1.30 p.m .-11.30=2 hours the average speed during this time interval =`v_(2)=(60km)/(2h)=30km//h.` The distance covered between 10.30 a.m to 1.30p.m.=220km-100km=120km. The time interval `=1.30p.m.-10.30 a.m=3` hours The average speed during this time interval `=v_(3)=(120km)/(3h)=40km//h`. |
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| 21. |
A particle starts rotating from rest accordingto the formulatheta = (3 t^(3)//20) - (t^(2)//3) .Find the angular velocity and the acceleration at the end of 5s. |
| Answer» Solution :`OMEGA = d theta//dt = 9T^(2)//20-2t//3`, at `t=5s, omega = 7.92 "rad"s^(-1), alpha = domega//dt = 18t//20 - 2//3`, at t=5s, `alpha = 3.83 "RADS"^(-1)` | |
| 22. |
The x - t graph of a particle undergoing simple harmonic motion is shown . The acceleration of the particle at t=4/3 is |
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Answer» `(sqrt(3))/(32)picm//s^(2)` |
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| 23. |
Two linear simple harmonic motions of equal amplitude and frequancy are imposed on a particle along x and y axis respectively. The initial phase difference between them is (pi)/2.The resultant path followed by the particle is: |
| Answer» Answer :A | |
| 24. |
To what temperature should the hydrogen at room temperature (27^(@)C) be heahted at constant perssure so that the RMS velocity of iths molecules becomes double its previous value? |
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Answer» `1200^(@)C` |
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| 26. |
Assertion : For a simple pendulum performing oscillation , the tension in its strings is constant for all position of the job . Reasion : The speed of job is different at different position which gives different values of tension in the string at different positions |
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Answer» If the assertion and reason are CORRECT and reason is a correct EXPLANATION of the assertion `T = MG cos theta = (MV^(2))/(l)` where v is the period when sring MAKES an angle `theta` with the vertical. |
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| 27. |
Ideal gas undergoes an adiabatic change in its state form (P_1 , V_1 , T_1) to (P_2, V_2, T_2) . The work doneW in the process is : |
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Answer» `W = MU (T_1 +T_2) C_v` |
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| 28. |
If the ratio of lengths, radii and Young.s modulus of the steel and brass rods in the figure are a, b,c respectively, find the ratio of increase in their lengths ? |
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Answer» Solution :`Y=(F)/(A),(L)/(e) rArr e=(F,L)/(AY),e=(F.L)/(pi r^(2)Y)` `rArr (e_(S))/(e_(B))=(3M)/(2M),((a)/(1)) ((1)/(b^(2)))((1)/(c)), (e_(S))/(e_(B))=(3a)/(2b^(2)c)` |
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| 29. |
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, ((dr)/(r)),is..... |
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Answer» `(Ka)/(mg)` `= (("Force")/("Area "))/( ( "change in volume")/("volume"))` `therefore (Delta V )/( V) = (mg )/(ka)` but, vlume of sphere `V = 4/3 PI R ^(3),` `therefore (Delta V )/(V) =3 (Delta r )/(r)` `therefore (Delta r )/( r ) = 1/3 xx (Delta V )/(V) = (mg )/(3Ka )` |
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| 30. |
An object A is moving with 5m//s and B is moving with 20m//sin the same direction . (positive x-axis ) (i)Find velocity of B with respect to A. (ii) Find velocity of A with respect to B. |
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Answer» Solution :(i)`V_(B)=20hati m//s , V_(A)=5hatim//s, V_(B)-V_(A)=15HATI m//s` (ii) `V_(B)=20hati m//s, V_(A)=5hati m//s, V_(AB)=V_(A)-V_(B)=-15hati m//s` `V_(BA)=-V_(AB)` |
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| 31. |
The arithmetic mean of several measurements is called |
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Answer» PRACTICAL value |
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| 32. |
The essence of kirchoff's law is that |
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Answer» a good ABSORBER must be a BAD radiator |
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| 33. |
The nearest star to our solar system is 4.29 light year away. How much is this distance in terms of parsecs/? |
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Answer» `1.32` `=4.29xx9.46xx10^(15)m` [ `:'` 1 light year `=9.46xx10^(15)m`] `=(4.29xx9.46xx10^(15))/(3.08xx10^(16))` PARSEC [ `:'` parsec `=3.08xx10^(16)m`] `=1.32` parsec |
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| 34. |
Two blocks of masses 3kg and 2kg are placed beside each other in contact with each other on a rough horizontal surface. A horizontal force of 20N is applied on 3kg. The coefficient of friction between blocks and the surface is 0.1 and g = 10ms^(-2). The force of contact between the two blocks is |
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Answer» 6N |
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| 35. |
A stone of mas 1 kg is tied to one end of a string of length 0.5 m. It is whirled in a vertical circle. If the maximum tension in the string is 58.8 N, the velocity at the top is |
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Answer» `1.82ms^(-1)` |
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| 36. |
A bus accelerates from rest at a constant rate alpha for some time, after which it decelerates at a constant rate beta to come to rest. If the total time elapsed is t seconds then, evaluate. (a) the maximum velocity achieve and (b) the total distance travelled graphically. |
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Answer» Solution :Let `t_(1)` be the time of acceleration and `t_2` that of DECELERATION of the bus. The total time is `t=t_(1)+t_(2)`. Let `v_("MAX")` be the maximum velocity. As the acceleration and decleratioin are CONSTANTS the velocity time graph is a straight line as shown in the FIGURE with +ve slope for acceleration and -ve slope for decleration. From the graph, the slope of the line OA gives the acceleration `alpha`. `alpha ="slope of the line "OA=(v_("max"))/(t_(1)) rArr t_(1)=(V_("max"))/(alpha)` the slope of AB gives the decleration `beta` `beta="slope of AB"=(v_("max"))/(t_(2)) rArr t_(2)=(v_("max"))/(beta)` `t=t_(1)+t_(2)=(v_("max"))/(alpha)+(v_("max")).(beta)` `t=v_(max) ((alpha+beta)/(alpha beta))`  `v_("max")=((alpha beta)/(alpha+beta))t` (b) Displacement =area under the v-t graph =area of `triangle OAB` `=1/2 ("base") ("height") =1/2 t v_("max") =1/2 t ((alpha beta t)/(alpha +beta))=1/2 ((alpha beta t^(2))/(alpha +beta))` |
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| 37. |
Does the work done in raising a box onto a platform depend upon how fast it is raised? Justify your answer. |
| Answer» SOLUTION :No. WORK done W = Fscos `THETA` which is independent of how FAST it is RAISED. | |
| 38. |
A motor car is going due north at a speed of 50 km/h. It makes a 90^(@) left turn without changing the speed. The change in the velocity of the car is about. |
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Answer» 50 km/h TOWARDS west |
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| 39. |
The relation between Pand T for monoatomic gas during adiabatic process is P prop T^(C ). The value of C is |
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Answer» `5//2` |
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| 40. |
Explain Wien's law and why our eyes are sensitive only to visible rays? |
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Answer» Solution :Wien.s law and Vision: The Sun is approximately taken as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation, Its surface temperature is about 5700 K. By SUBSTITUTING this value in the equation of Wien.s law. `lambda_(m)=(b)/(T)=(2.898xx10^(-8))/(5700)~~"508 nm"` It is the wavelength at which maximum intensity is 508 nm. Since the Sun.s temperature is around 5700 K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum, It is shown in Figure.  The humans evolved under the Sun by receiving its radiations. The HUMAN eye is sensitive only in the visible not in infrared or X-ray ranges in the spectrum. Suppose if humans had evolved in a planet near the star Sirius (9940K), then they would have had the ABILITY to SEE the ULTRAVIOLET ray! |
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| 41. |
The dimensional formula of pressure is |
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Answer» `[MLT^(-2)]` |
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| 42. |
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas ineach cylinder. If the rise in temperature of the gas in A to 30 K, then the rise in temperature of the gas in B is |
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Answer» Solution :Heat is given to the gas in cylinder A at constant pressure while the same amount of heat is given to the gas in cylinder B at constant VOLUME. Heat given to the gas in A is `Delta Q_(A) = nC_(P)Delta T_(A)` and that given to the gas in B is `Delta Q_(B) - nC_(V)delta T_(B)` Since `Delta Q_(A) = Delta Q_(B) :. nC_(P)Delta T_(A) = nC_(V)Delta T_(B)` or `Delta T_(B) = (C_(P))/(C_(V)) Delta T_(A) = (7)/(5) xx 30 K = 42 K`(`because` for a diatomic gas , `C_(V)//C_(P) = 5//7`) |
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| 43. |
Which of the following statements are correct ?(a) centre of mass of a body always coincides with the centre of gravity of the body.(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero.(c ) A couple on a body produce both translational and rotational motion in a body(d) Mechanical advantage greater than one means that smell effort can be used to lift a large load. |
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Answer» (C ) and (d) |
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| 44. |
Assertion: A solid sphere placed in the fuid under high pressure is compressed uniformly on all sides. |
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Answer» |
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| 45. |
One of the following graphs that represents the variation of gravitational field intensity E with the distance r from the centre due to the uniform solid sphere is |
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Answer»
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| 46. |
On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle theta to its initial direction and has a speed v/3. The second block.s speed after the collision is |
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Answer» `3/(sqrt2) v`  Let v. be speed of SECOND block after the COLLISION. As the collision is elastic, so kinetic ENERGY is conserved. According to conservation of kinetic energy. `1/2 Mv^2 + 0 = 1/2 M(v/3)^2 + 1/2 Mv.^(2)` `v^2 = (v^2)/9 + v.^(2) " or " v.^(2)= v^2 - (v^2)/9= (9v^2 - v^2)/(9) = 8/9 v^2` `v. = SQRT(8/9 v^2) = (sqrt8)/(3) v = (2sqrt(2))/(3) v`. |
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| 47. |
Three particles A, B & C start from the origin at the same time, A with a velocity 'a' along X-axis, B with a velocity 'b' along y-axis and C with velocity 'c' in XY plane along the line x = y. The magnitude of 'c' so that the three always remain collinear is: |
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Answer» `(a+b)/(2)` |
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| 48. |
(A) : A man is on the northern bank of a river. To cross the river in a shortest time he should swim due south. ( R) : Man should move opposite to the river flow to cross in shortest time. |
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Answer» Both (A) and ( R) are TURE and ( R) is the correct EXPLANATION of (A) |
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| 49. |
In the experiment to determine the Young's modulus of the material of a wire under tension used in the arrangement as shown the percentage error in the measurement of length is a, in the measurement of the radius of the wire is b and in the measurement of the change in length of the wireis C. Percentage error in the measurement of Young's modulus for a given load is |
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Answer» `a-2 b+c` TAKING log of both the sides `log Y = log F + log l - log pi - 2 log r - log Delta l` Differentiating it, we have `(Delta Y)/(Y)=0 +(Delta l)/(l) - 0 -(2Delta r)/(r) - (Delta l)/(Delta l)` % ERROR, `(Delta Y)/(Y)xx100= ((Delta l)/(l) + (2Delta r)/(r) + (Delta(Delta l))/(Delta l))xx100` `=a + 2b + c` |
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