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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. |
To construct a barometer, tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. Thebarometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm and then closed by a cork. It is invertd in a mercury cup and the cork is removed. The height of mercury column in the tube over the surface in the cup will be |
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Answer» zero |
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| 2. |
A force F appllied to a body(A)of mas m_(1) produces an acceleratioin of 4m//s^(2). If the same force F is applied to another body (B) of mass m_(2), then an acceleratiionof 10m//s^(2), produced in the body. A and B are the tied together and the same force is applied to the combined body. What is the acceleration of the system? |
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Answer» `10/7m//s^(2)` and the COMBINEDMASS (M) `=m_(1)+m_(2)` `=(F)/(4)+(F)/(10)= (14F)/(40)` `therefore` Acceleration of the system `= (F)/(M)= (F)/(14F//40)` `=(40)/(14)= (20)/(7) m//s^(2)` |
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| 3. |
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water. [The surface tension of waterat temperature of the experiment is 7.30xx10^(-2)Nm^(-1). 1 atmospheric pressure = 1.01xx10^(5) Pa, density of water = 1000 kg m^(-3),g=9.80ms^(-2)]. |
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Answer» Solution :The excess pressure in a bubble of gas in a liquid is GIVEN by `2S//r`, where S is the surface TENSION of the liquid-gas interface. You should note there is only one liquid surface in thiscase. The radius of the bubble is r. Now the pressure OUTSIDE the bubble, `p_("out")` equals atmospheric pressure plus the pressure due to 8.00cm of WATER COLUMN. `P_("out")=(1.01xx10^(5)+0.08xx1000xx9.80)Pa` `=1.01784xx10^(5)Pa` Therefore, the pressure inside the bubble is `P_("in")=P_("out")+2S//r=1.01784xx10^(5)+(2xx7.3xx10^(-2)//10^(-3))` `=(1.01784+0.00146)xx10^(5)=1.02xx10^(5)Pa` |
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| 5. |
What is the largest average velocity of blood flow in an artery of radius 1x10 m if the flow must remain laminar ? (Given viscoisty of blood 2.084xx10^(-3) Pa s and density of blood 1060 kg m^(-3)). |
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Answer» `0.886 MS^(-1)` |
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| 6. |
A ball at rest is dropped freely from a height of 20 m. It loses 30% of its energy on striking the ground and bounces back. The height to which it bounces back is |
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Answer» 14 m |
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| 7. |
A particle strikes a horizontal frictionless floor with a speed u at an angle theta with the vertical, and rebounds with a speed v at an angle phi with vertical. The coefficient of restitution between the particle and floor is e. The magnitude of v is |
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Answer» `EU` i.e., `m u sin theta = mv sin phi" or " u sin theta = V sin phi "" ..(i)` And in vertical direction, `(v cos phi)/(u cos theta) = 3` or `v cos phi = eu cos theta "" ........(ii)` SQUARING and adding (i) and (ii), we get `v^2 (sin^2 phi+ cos^2 phi) = u^2 sin^2 phi + e^2 u^2 cos^2 theta` or `v = u sqrt(sin^2 phi + e^2 cos^2theta)`. |
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| 8. |
A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring constant 100 Nm^(-1) as shown in the figure.The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. The coefficient of friction between the block and the incline is (Take g = 10 ms^(-2) and assume that the pulley is frictionless) |
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Answer» `0.2` |
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| 9. |
A particle strikes a horizontal frictionless floor with a speed u at an angle theta with the vertical, and rebounds with a speed v at an angle phi with vertical. The coefficient of restitution between the particle and floor is e. The angle phi is equal to |
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Answer» `theta` i.e, `m U sin theta = mv sin phi " or " u sin theta = V sin theta "" ….(i)` And in VERTICAL direction, `(v cos phi)/(u cos theta) = e` or `v cos phi = eu cos theta "".....(ii)` Divide (i) by (ii), we get `tan phi = 1/e tan phi " or " phi = tan^(-1) [1/e tan phi]`. |
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| 10. |
Derive an expressionformaximumspeedthe carcan havesafeturnon a leveledcircular road . |
Answer» Solution :In aleveledcircularroad, skiddingmainlydependson thecoefficientof staticfriction`mu_s`thecoefficientofstaticfrictiondependson thenatureof thesurfacewhichhasamaximumlimiting value, toavoidthisis problem usuallythe outeredgeof theroadis slightlyraisedcomparedto inneredgethisis calledbankingof roadsor tracks. thisintroducesan inclination, andthe angleis calledbankingangle . Letthe SURFACEOF theroadmakeangle` theta`withhorizontalsurface, thenthe normalforcemakesthe sameangle `theta `with thevertical. whenthe cartakesa TURN, thereare twoforcesactingon the CAR : (a) GravitationalforceMg( downwards) (b )NormalforceN ( perpendicularto surface ) We canresolvethe normalforceintotwocomponentsN` cos thetaand Nsin theta `thecomponent balancesthe downwardgravitionalforce.mg.andcomponentwillprovidethe necessarycentripetalacceleration. by usingNewtonsecondlaw . Thebankingangle ` theta `and radiusof curvatureof theroador trackdeterminesthe safespeedof htecar attheturning. If thespeedofcarexceedsthissafespeedthenitstartstoskidoutward butfrictionalforcecomesintoeffect andprovidesand additionalcentripetalforceto preventthe outwardskidding.Atthe sametime, if thespeedof htecar islittlelesserthansafespeed, itstartsto skidinwardandfrictionalforcecomesintoeffect whcihreducescentripetalforce to preventinwardskidding .Howeverif htespeedof THEVEHICLE is sufficientlygreaterthanthe correctspeedthenfrictional forcecannotstopthecarfromskidding. 
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| 11. |
If vec(P) + vec(Q) = vec(R) and vec(P) - vec(Q) = vec(S), then R^(2) + S^(2) is equal to |
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Answer» <P>`P^(2) + Q^(2)` |
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| 12. |
Then two moles of oxygen is heated from 0^@C to 10^@Cat constant volume, its internal energychanges by 420 J. What is the molar specific heat of oxygen at constant volume ? |
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Answer» `5.75 JK^(-1) "MOL"^(-1)` |
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| 13. |
The displacement of a particle moving along x-axis with respect to times is given by x=at+bt^(2)-ct^(3). The dimensions of b are |
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Answer» `L^(0)T^(-3)` The dimensionally . `[L] = [LT^(-1)] + [T] + [ LT^(-2)] - [LT^(-3)] [T^3]` `therefore ` the dimension of .b. is `LT^(-2)` |
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| 14. |
The rotation of the earth about its axis speeds up such that a man on the equator becomes weightness. In such a situation, what would be the duration of one day ? |
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Answer» `2pi SQRT((R)/(g))` `:. omega=sqrt((g)/(R))rArr(2pi)/(T)=sqrt((g)/(R))rArrT=2pi sqrt((R)/(g))` |
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| 15. |
A transverse wave is represented by equation y= y_0 sin((2pi)/lambda)(vt-x).For what value of lambda is the maximum particle velocity to two times of the wave velocity? |
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Answer» `2piy_0` |
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| 16. |
Derive the expression for final speed of a particle moving in an inclined plane. |
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Answer» Solution :(i) Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in Figure. There are two forces acting on the object along the inclined plane. (ii) One is the component of gravitational force `(mgsintheta)` and the other is the static frictional force (f). The other component of gravitation force `(mgcostheta)` is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the thefree body diagram (FBP) of the object.  (iii) For translational motion, `mg sintheta` is the supporting force and f is the OPPOSING force, `""mgsintheta-f=ma` For rotational motion, let us take the TORQUE with respect to the center of the object. Then `mg sintheta` cannot cause torque as it passes through it but the frictionalforce f can set torqueof Rf. `""Rf=Ialpha` (iv) By using the relation, `a=ralpha`, and momentof inertia `I=mK^(2)`, we get, `Rf=mK^(2)a/R, f=ma(K^(2)/R^(2))` Now equation becomes, `mgsintheta-ma(K^(2)/R^(2))=ma` `mgsintheta=ma+ma(K^(2)/R^(2))` `a(1+K^(2)/R^(2))=gsintheta` After rewriting it for acceleration, we get, `a=(gsintheta)/((1+K^(2)/R^(2)))` (v) We can ALSO find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane `v^(2)=u^(2)+2as`. If the body starts rolling from rest, u=0. When h is the vertical height of the incline, the length of the incline s is, `s=h/sintheta` `v^(2)=2(gsintheta)/((1+K^(2)/R^(2)))(h/sintheta)=(2gh)/((1+K^(2)/R^(2)))` By taking square root, `v=sqrt((2gh)/((1+K^(2)/R^(2))))` (vi) The time taken for rolling down the incline could also be written from first equation of motion as, v=u+at. For the object which starts rolling from rest, u=0. Then, `t=v/a` `t=(sqrt((2gh)/((1+K^(2)/R^(2)))))(((1+K^(2)/R^(2)))/(gsintheta))` `t=sqrt((2h(1_+K^(2)/R^(2)))/(gsin^(2)theta))` (vii) The equation suggests that for a given incline, the object with the LEAST value of radius of gyration K will reach the bottom of the incline first. |
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| 17. |
State the principle and usage of Venturimeter. |
| Answer» Solution :This device is used to measure the rate of follows (or say FOLLOW SPEED) of the INCOMPRESSIBLE fluid flowing through a pipe. It works on the principle of Bernoulli.s theorem. | |
| 18. |
If an elevator is moving vertically upwards with an acceleration a, the force exerted on the floor by a passenger of mass M is |
| Answer» ANSWER :D | |
| 19. |
A compound microscope is used to enlarge an object kept at a distance 0.03m from its object which consists of several convex lenses in contact and has focal length 0.02m. IF a lens of focal length 0.1m is removed from the objective. Find out the distance by which the eyepiece of the microscope must be moved to refocus the image. |
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Answer» Solution :IF initially the objective forms the IMAGE at distance `v_1=1/v_1-1/(-3)=1/2 i.e, v_1=6cm` Now as in case of lenses in contact `1/F=1/f_1+1/f_2+…….or 1/F=1/f_1+1/F.` with `1/F.=1/f_2+1/f_3+….` So if ONE the lenses is REMOVED, the FOCAL length of the remaining lens system `1/F.=1/f-1/f_1=1/2 -1/10 i.e, F.=2.5cm` This lens will form the image of same OBJECT at a distance `v_2` such that `1/v_2-1/(-3)=1/2.5 i.e., v_2=15cm` So to refocus the image, eyepiece must be moved by the same distance through which the image formed by the objective has shifted i.e, 15-6=p cm away from the objective. |
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| 20. |
In the non-relativistic regime, if the momentum, is increased by 100%, the percentage increase in kinetic energy is |
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Answer» 100% Since m REMAINS costant , so `K PROP p^2` `:. (K_1)/(K_2) = ((p_1)/(p_2))^(2) = 1/4" or " K_2 = 4K_1` Percentage increase in kinetic energy = `(K_2 - K_1)/(K_1) xx 100%` `= (4K_1 - K_1)/(K_1) xx 100% = 300%`. |
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| 21. |
The acceleration of small block m with respect to ground is (all the surface are smooth): |
| Answer» ANSWER :a | |
| 22. |
Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself . |
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Answer» Solution :If `vec(A)"||"vec (B)` then `theta = 0^(@)` ` :. vec(A) .vec(B) = AB cos theta = AB ` and `vec(A). Vec(A) = |vec(A)"||"vec(A)| =A^(2)` ` :.|vec(A)|= sqrt(vec(A).vec(A))` hence the MAGNITUDE of a vector is equal to the SQUARE root of the scalar product of the vector with itself . MAGNITUDEOF vector is equal to the square root of the scalar product of the vector with itself . |
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| 23. |
A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon's orbit. The satellite completes one revolution in ........ lunar month. |
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Answer» Solution :`(T_(E))/(T_(m)) = ((a_(e))/(a_(m)))^(3/2)` `T_(e) = T_(m) XX (1/2)^(3/2) = 1XX (2)^((-3)/2)` `T_(e) = 2^((-3)/2 )`lunarmonth |
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| 24. |
. A sphere A moving with a speed .U. and rotating with an angualr velocity .o. makes head-on elastic collision with an identical stationary sphere. There is no friction between the surfacces A and B neglect gravity A will stop moving but continue to rotate with angular velocity.omega. A will come to rest and stop rotating B will move with speed .U. without rotating B will move with speed U and rotate with an angular velocity w. |
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Answer» A & C |
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| 25. |
Explain the importance of Reynold.s number. |
| Answer» Solution :It is unit less and DIMENSIONLESS quantity, R. is different for different liquids. For LAMINAR flow .R. LIES between 0 to 1000. For TURBULENT flow `R gt 1500`. For 1000 `lt R lt 1500` flow is unstable. | |
| 26. |
Calculate the refractive index of glass with respect to air if its critical angle is 37^@ |
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Answer» `(5//3)` |
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| 27. |
A ball falls under gravity from a height 10 m with an initial velocity v_0. It hits the ground, loses 50% of its energy in collision and it rises to the same height, what is the value of v_0? |
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Answer» 14 m/s then `v^2 = v_0^2 + 2gh = v_0^2 + 2 xx 9.8 xx 10` or `v^2 = v_0^2 + 196` Let `v.` be the velocity after impect and it RISE same height 10 m. `:. V.^(2) - 0 = 2 xx 9.8 xx 10 = 196 or v. = 14 m//s` Ratio of kinetic energy before impact and after impact `(1/2 mv^2)/(1/2 mv.^2) = (v^2)/(v.^2) = (v_0^2 + 196)/(196) = 2 " or " v_0 = 14 m//s` . |
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| 28. |
A sound detector D moves with constant speed on a circle or radius R and centre at O in xy plane. A pointsource of sound S lines in xy plane at a distance 2R from the point O and emits sound of a given frequnecy . The ratio of maximum frequencyrecorded by the detector is 11/9 and speed of sound is 340 m/s . the minimum time interval in seconds between reacording a maximum frequency and mimmim frequency is (takeR = 17) |
| Answer» ANSWER :A | |
| 29. |
When a force of 10.0 N is exerted on the handle of a door, the door can be just opened. If the handle is a distance of 50cm from the hinges find the torque applied on the door. |
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Answer» SOLUTION :`tau=rFsintheta=0.50xx10.0sin90^(@)` `=5NM` |
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| 30. |
If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now ? |
| Answer» Solution :The LOWER LAYERS of Earth.s atmosphere REFLECT infrared RADIATIONS from Earth BACK to the surface of Earth. Thus the heat radiation received by the earth from the Sun during the day are kept trapped by the atmosphere. If atmosphere of Earth were not there, its surface would become too cold to live. | |
| 31. |
A simple pendulum of length 0.2m has bob of mass 5gm, it is pulled aside through a angle 60^(@) from the vertical. A spherical body of mass 2.5 gm is placed at the lowest position of the bob. When the bob is released it strikes the spherical body and comes to rest. What is the velocity of the spherical body after collision ? (g = 9.8 ms^(-2)) (in m/s) |
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Answer» 1.4 |
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| 32. |
Marbles each of mass 'm' are dropped from height 'h' on pan of a balance at the rate 'R' per sec. The balance calibrated in units of mass reads zero initially. If marbles are dropped continuously, what will the balance read after 't' sec ? |
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Answer» Solution :Each marble hits the pan with VELOCITY `v=sqrt(2gh)`and in 1 sec no. of marbles hitting the pan = R `therefore` Force due to momentumchange (Impact), `F_(1)=MR sqrt(2gh)`, force due to weight `F_(2)=` mg. Rt `therefore` TOTAL reading, `F=mgRt + mR sqrt(2gh)=mgR(t+sqrt((2h)/(g)))` |
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| 33. |
sound is simultaneously produced at the ends of the two strings of the same length, one of rubber and the other of steel. in which string will the sound reach the order end eariler and why? |
| Answer» Solution :SPEED of sound in a STRING, `V= sqrt Y/P`. As the VALUE of Y/P is larger for steel than for rubber, sound will reach the other end earliar in the case of steel string. | |
| 34. |
Obtain the scalarproduct of unit vectors in Cartesian co - ordinatesystem . |
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Answer» Solution :Since `HAT(i) , hat(j) and hat(k)` are the unit vectors in the direction of X,Y Z in Cartesian co - ordinatesystem . (i) `hat(i) . hat(j) = (1) (1) cos 0^(@) "" [:. |hat(i)| = 1 , and hat(i) ||hat(i)] ` ` :. hat(i) .hat(i) = 1 "" [:. cos 0^(@) =1] ` similarly `hat(i).hat(j) = 1 and hat(k) .hat(k) = 1 ` (ii) `hat(i) . hat(j) = (1) (1) cos 90^(@)` `[ :. |hat(i)| = 1 , |hat(j)| = 1 and hat(i) bot hat(j)] ` ` :. hat(i) . hat(j) = hat(j) .hat(i) = 0 "" [:. cos 90^(@) = 0 ] ` similarly `hat(j).hat(k) = hat(k) .hat(j) = 0 ` `hat(k) .hat(i) = hat(i) .hat(k) = 0 ` |
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| 35. |
A lorry and a car moving with the same kinetic energy are brought to rest by the application of brakes, which provide equal retarding forces .Which will come to rest in less distance? |
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Answer» |
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| 36. |
A refrigerator whose coefficient of performance is 5 extract heat from the cooling compartment at the rate of 250J/cycle. How much electric energy is spent per cycle? How much heat per cycle is discharge to the room? |
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Answer» As `C.O.P= (Q_(2))/W` `:. W=(Q_(2))/(C.O.P)= (250)/5= 50J` Heat/ cycle discharge to the room `Q_(1)=Q_(2)+W= 250+50=300J` |
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| 37. |
Inside a satellite orbiting very close to the earth's surface, water does not fall out of a glass when it is inverted. Which of the following is the best explanation for this ? |
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Answer» The earth does not EXERT any force on the water |
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| 38. |
does the motion of a satellite obey Kepler's laws ? |
| Answer» Solution :YES, it OBEYS As eccentricityof the elliptical orbit is very SMALL, it can be treated as CIRCULAR. | |
| 39. |
The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward ans a car accelerating on a horizontal road. The above statement is corect in |
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Answer» The CAR only |
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| 40. |
Torricell 's barometer used mercury . Pascal duplicated it using French wine of density 984kgm^(-3) . Determine the height of the wine column for nirmal atmospheric pressure. |
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Answer» Solution :Normal atmosphetic pressure `P_(a)=1.03xx10^(5)` PA Suppose the height of French wine column is h and corresponding pressure P. `thereforeP=grhog` where `rho` = DENSITY of wine `=984kgm^(-3)` GIVE that `P.=P_(a)` `hrhog=P_(a)` `thereforeh=(P_(a))/(rhog)` `thereforeh=(P_(a))/(rhog)` `thereforeh=(1.013xx10^(5))/(984xx9.8)` `thereforeh=0.000105xx10^(5)` `thereforeh=10.5m` |
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| 41. |
A wheel of radius 0.4 m can rotated freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung, An angular acceleration of 8 rad s^(-2)is produced in it due to the torque. Then moment inergia of the wheel is (g = 10 ms^(-2)) |
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Answer» `2 kg-m^(2)` |
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| 42. |
The dimensions of impulse is |
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Answer» `[MLT^(-1)]` |
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| 43. |
If the ice on the polar caps of the earth melts and water collected about equator, how will that affect the duration of a day? |
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Answer» Day becomes short Since external torque on earth is not acting, its angular momentum remains constant `therefore omega_(1)I_(1)=omega_(2)I_(2),I_(2)ltI_(2)` `{:(omega_(1)gtomega_(2),|,"PERIODIC time for one"),(therefore (2pi)/(T_(1))GT(2pi)/(T_(2)),"revolution increases and",),(therefore T_(2)gtT_(1),"hence length of day increase",):}` |
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| 44. |
A boat moves relative to water with a velocity which is n times less than the river flow velocity . At what angle to the stream direction must the boat move to minimize drifting? |
Answer» Solution :In this problem, one thing should be carefully noted that the velocity of boat is less than the river flow velocity. In such a case, boat cannot reach the point DIRECTLY opposite to its starting point , i.e,. Drift can never be zero. Thus, to minimize the drift , boat starts at an angle `theta` from the normal directionup stream as shown. now, again if we find the components of velocity of boat along and perpendicular to the flow , these are, velocity along the river, `V_(x)=u-v sin theta` and velocity perpendicular to the river, `V_(y)=v cos theta`. time taken to cross the river is `t=d/v_(y)=d/(v cos theta)` In this time, drift `x=(v_(x))t=(u-vsintheta)d/(vcostheta)"(or)" x=(ud)/v SEC theta - d tan theta`. The drift x is minimum , when `(dx)/(dtheta)=0"(or)" ((ud)/v) (sec theta. tan theta) - d sec^(2) theta =0` (or)`u/vsin theta =1 "or" sin theta =v/u=1/n(as"" v=u/n)` So, for minimum drift , the boat must move at an angle `theta = sin ^(-1)(v/u)` from normal DIRECTION or an angle `pi/2+sin^(-1)(v/u)` from stream direction. |
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| 45. |
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both: (a)Motion of a kink in a longitudinal spring produced by displacing one end of the spring sideways. (b) Waves produced in a cylinder containing a liquid by moving its piston back and forth. (c ) Waves produced by a motorboat sailing in water. (d) Ultrasonic waves in air produced by a vibrating quartz crystal. |
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Answer» Solution :(a) When a long spring is given little push or pull, PERPENDICULAR to its length and then RELEASED, waves progagation on it are transverse. If a long spring is given little push or pull, parallel to its length and then released, waves propagating in it are longitudinal. (b) Here given substance is a fluid which cna sustain COMPRESSIVE stress nad can not sustain shearing stress. Hence longitudinal waves can propagate inside it but transverse waves can not propagate in it. (c ) (i) Free surface of water can sustain shearing stress due to its surface TENSION peoperty and so transverse waves can propagate on its free surface. (ii) Part of water MEDIUM, below the free surface can sustain compressive stress and so longitudinal waves can propagate in it. (d) Air is a fluid which can sustain compressive stress but can not sustain shearting stress. Hence longitudinal waves can propagate in air but transverse waves can not. |
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| 46. |
A load 'm' is attached to a spring of force constant 'k' and stretched it through 'x' and released. It makes oscillation in a vertical plane with a time period "T". It is further pulled down through another ‘x' and released. Now the time period of vertical oscillation will be |
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Answer» T/2 |
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| 47. |
Due to Doppler effect, the shift in wavelength observed is 0.1 Å, for a star producing a wavelength 6000 Å. The velocity of recession of the star will be |
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Answer» 20 km `cdot s^(-1)` `therefore v=c(Delta lambda)/ lambda = (3xx 10^(8))xx0.1/6000` `=5 XX 10^(3) m//s = 5 `km/s The OPTION (D) is correct. |
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| 48. |
SI unit and CGS unit of certain quantity vary by 10^(3) times. That quantity is |
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Answer» BOLTZMANN constant |
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| 49. |
Calculate the rms speed , average speed and the most probable speed of 1 mole of hydrogen molecules at 300K . Neglect the mass of electron. |
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Answer» SOLUTION :The hydrogen atom has ONE proton and one electron. The mass of electron is negligible compared to the mass of proton `=1.67xx10^(-27)` kg One hydrogen molecule =2 hydrogen ATOMS `=2xx167xx10^(-27)` kg The average speed `barnu=sqrt((8kT)/(PIM))=1.60sqrt((kT)/m)` `=1.60sqrt(((1.38xx10^(-13))xx(300))/(2(1.67xx10^(-27))))=178xx10^(3)ms^(-1)` (Boltzmann Constant `k=1.38xx10^(-32)JK^(-1)`) The rms speed `nu_(rms)=sqrt((3kT)/m)=1.73sqrt((kT)/m)` `=1.73sqrt(((1.38xx10^(-23))xx(300))/(2(1.67xx10^(-27))))=1.93xx10^(3)ms^(-1)` Most probable speed `nu_(mp)=sqrt((2kT)/m)=1.41sqrt((kT)/m)` `=1.41sqrt(((1.38xx10^(-23))xx(300))/(2(1.67xx10^(-27))))=1.57xx10^(3)ms^(-1)` Note that `nu_(rms) gt barnu gt nu_(mp)` |
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| 50. |
A trolley filled with sand moves on a smooth horizontal surface with a velocityV _(0). A smalll hole is made at the base of if from which sand is leaking out vertically down a constant rate. As the sand leaks out (a) the velocity of the trolley remains constant (b) the velocity of the trolley increases (c ) the velocit of trolley decreases (d) the momentum of trolley + leaked out sand is conserved |
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Answer» a & b are correct |
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