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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 38501. |
(A) : When a body is dropped from a height explodes in mid air, but its centre of mass keeps moving in vertically downward direction.(R ) : Explosion occur due to internal forces only. |
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Answer» Both 'A' and 'R' and true and 'R' is the CORRECT EXPLANTATION of 'A' |
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| 38502. |
Condition for no deviation when light travels from one medium to another is a) angle of incidence is zero (b) refractive index of two media must be same |
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Answer» a is only true |
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| 38503. |
A cyclinderical tank is open at the top and has cross sectional area a_1. Water is filled in it up to a height h. There is a hole of cross sectional area a_2 at its bottom. Given a_1=3a_2. iii. The time taken to empty the tank is |
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Answer» `SQRT((2H)/(G))` |
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| 38504. |
Two metal spheres each of radius 'r' and made of same material are separated by a certain fixed distance or (d gt 2r) in a medium. The gravitational force between those spheres is proportional to |
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Answer» `(1)/(R^(2))` |
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| 38505. |
A cyclinderical tank is open at the top and has cross sectional area a_1. Water is filled in it up to a height h. There is a hole of cross sectional area a_2 at its bottom. Given a_1=3a_2. ii. The initially velocity with which the water emerges from the hole is |
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Answer» `1/2sqrt(GH)` |
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| 38506. |
The resultant of two vectors is maximum when the angle between them is |
| Answer» ANSWER :A | |
| 38507. |
A particle moves in the X-Y plane under the action of a force vec(F) such that the value of its linear momentum (vec(P)) at any time .t., and its components are P_(X) = 2 cos t, P_(Y) = 2 sin t. According to defination of force vec(F) = (d vec(P))/(dt) The angle .theta. between vec(F) and vec(P) at a given time .t. will be |
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Answer» `THETA = 0^(@)` |
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| 38508. |
A cyclinderical tank is open at the top and has cross sectional area a_1. Water is filled in it up to a height h. There is a hole of cross sectional area a_2 at its bottom. Given a_1=3a_2. i. The initially velocity with which the water falls in the tank is |
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Answer» `SQRT(2gh)` |
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| 38509. |
A particle moves in the X-Y plane under the action of a force vec(F) such that the value of its linear momentum (vec(P)) at any time .t., and its components are P_(X) = 2 cos t, P_(Y) = 2 sin t. According to defination of force vec(F) = (d vec(P))/(dt) The vector form vec(F) is |
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Answer» `-2 sin t hat(i) + 2 cos t hat(J)` |
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| 38510. |
The escape velocity of projectile on the surface of earth is 11.2 Kms^(-1). If a body is projected out with thrice of this speed , find the speed of the body far away from the earth . Ignore the presence of other planets and sun. |
| Answer» Solution :`V=SQRT(v_p^2-v_e^2)=sqrt((3V_e)^(2)-v_(e)^(2))=sqrt8xx11.2 = 31.68kms^(-1)`. | |
| 38511. |
Young.s modulus of a wire is 2 xx 10^(11)N//m^(2). If a stress of 2 x 108 N/m2 is applied at the free end of a wire, find the strain in the wire. |
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Answer» |
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| 38512. |
The significant figure in 0.050 is |
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Answer» 1 |
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| 38513. |
To get the best possible true value of the quantity ……...…. has to be taken. |
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Answer» RMS VALUE |
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| 38514. |
A solid ball of mass m and radius r spinning with angular velocity omega falls on a horizontal slab of mass M with rough upper surface (coefficient of friction mu) and smooth lower surface. Immediately after collision the normal component of velocity of the ball remains half of its value just before collision and it stops spinning. Find the velocity of the sphere in horizontal direction immediately after the impact (given: Romega= 5). |
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Answer» `muJR=intmu(N dt)R=(2/5mR^(2)omega-0)=2/5mR^(2)omega`……..II  From eqn i and ii we get `3/2mv rmu=2/5mR^(2)omega`……….iii let `V` and `V_(1)` be the speeds of the plank and the sphere, respectively in the horizontal direction. `muJ=intmj Ndt=Mv=mV_1`........iv FORM eqn i and iv `mu(3/2)mv=MV` `V=3/2(mumv)/M=3/2 4/15 (mRomega)/M=2/5Romega` and `V_(1)=2/5Romega=2m//s` |
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| 38515. |
Select the incorrect statement from the following |
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Answer» Rolling MOTION is the COMBINATION of translational and rotation motion. |
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| 38516. |
A cubical block of density rho is floating on the surface of water .Out of its height L, fraction x is submerged in water . The vessel is in an elevator accelerating upward with acceleration a . What is the fraction immersed ? |
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Answer» Solution :Let density of water is `rho_(w)`. A BLOCK of height L float on it .x be the height of block submerged in water . Volume of block `V=L^(3)` Mass of block `m=Vrho=L^(3)rhog` WEIGHT of the block `=mg=L^(3)rhog`  First Case : Volume ofpart of cube submerged in water `=xL^(2)` `therefore` Weight of water displaced by block `=xL^(2)rho_(w)G` Weight of block =weight of water displaced by block . `L^(3)rhog=xL^(2)rho_(w)g` `therefore(x)/(L)=(rho)/(rho_(w))` `thereforex=(rho)/(rho_(w))L`....(1) Second Case : When vessel is PLACED in an elevator moving upward with acceleration a , then effective acceleration =g=(g+a) More acceleration a is due to Pseudo force `therefore` Weight of block =mg `=m(g+a)` Suppose , and elevator is moving upward .Let new fraction of block submerged in water is `x_(1)` For floating of block , Weight of block =Weight of displaced water , `L^(3)rho(g+a)=x_(1)L^(2)rho_(w)(g+a)` `therefore(x_(1))/(L)=(rho)/(rho_(w))`....(2) `x_(1)=(rho)/(rho_(w))*L` From equation (1)and (2) , `x=x_(1)` Hence , the fraction of the block submerged is independent of acceleration. |
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| 38517. |
Odd one out………. |
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Answer» strain |
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| 38518. |
Givethe signconvention forQ and W. |
| Answer» SOLUTION :`{:("System gains HEAT","Q is positive"),("System LOSES heat","Q is negative"),("WORK DONE on the system","W is negative"),("Work done by the system","W is positive"):}` | |
| 38519. |
The velocity of river grows in proportion to the distance from it's bank and reaches maximumvalue V_(0) in the middle. Near the banks the velocity is zero. A boat is driven perpendicular to river flow, speed of boat in still water is u ( width of river is w) |
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Answer» time TAKEN to CROSS river `W/u` |
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| 38520. |
Whena stone tied to the end ofa stringwhireled in a circular path the centripetal force is providedby the |
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Answer» weight of the STONE |
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| 38521. |
A body is acted upon by a number of external forces can it remain at rest |
| Answer» SOLUTION :If thevectorsumof all theexternalforces ISZERO, THENTHE bodywillremainat REST . | |
| 38522. |
A ship is moving towards east at 10km*h^(-1). A boat ismoving north of east making an angle of30^@ with the north. What should be the velocity of the boat so that the boat always appears, from the ship , to move towards north ? |
Answer» Solution :Let the velocity of the SHIP be `vecu=vec(OA)` and velocity of THEBOAT be `vecv=vec(OB)` [Fig.2.43]  Hence, the velocityof the boat with RESPECT to the ship, `vecw=vecv-vecu=vec(AB)` From thefigure, `sin 30^@=u/v or, 1/2 =u/v` or, `v=2u=2xx10km*H^(-1) =20km*h^(-1)` [Given , `u=10km*h^(-1)`] |
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| 38523. |
Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ? |
| Answer» SOLUTION :`T=2pisqrt(m/k)` for (a) and `2pisqrt(m/(2K))` for (B) | |
| 38524. |
Zero errors of measuring instruments are called |
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Answer» INDETERMINATE errors |
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| 38525. |
Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F. What is the maximum extension of the spring in the two cases ? |
| Answer» SOLUTION :F/k for both (a) and (B). | |
| 38526. |
Given 50^(@)F , 50^(@)Cand 50K arrange them in increasing order of temperature |
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Answer» `50^(@)F, 50^(0) C " &" 50`K |
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| 38527. |
A seconds pendulum is taken from the surface of the earth to that of the moon. In order to maintain the period constant |
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Answer» LENGTH of the PENDULUM has to be DECREASED |
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| 38528. |
A balloon from rest accelerates uniformly upward with 'a' ms^(-2), for t seconds of time. A stone is released from the balloon. Now, read the following statements to pick the right ones. a) The stone's initial velocity is zero, relative to balloon b) The stone's initial velocity is nonzero, relative to earth c) The time taken to reach the ground from the balloon's frame of reference is inversely proportional to sqrt((a+g)). d) The time take to reach the ground from earth's frame of reference is directly proportional to sqrt((a+g)). |
| Answer» Answer :A | |
| 38529. |
A tube 2m long and closed at on e end is half filled with mercury and is then inverted with its open end just dipping into a mercury trogy. Find the height of mercury which will stand in the tube if the reading of the barometer is o.75 m. [hint: P_(1) = 0.765,V_(1)= 1 propm^(3), P_(2)=(3//4-x),V_(2)= (2-x)prop m^(3),use p_(1)V_(2)=P_(2)V_(2).] |
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Answer» |
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| 38530. |
A closed copper vessel contains water equal to half of its volume. When the temperature of the vessel is raised to 447°C, the pressure of the steam in the vessel is (treat stream as an ideal gas, R=8310 "JK"^(-1)//kg mole, density of water = 1000 "kgm"^(-5) molecular weight of water= 18 ) |
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Answer» `33.24 XX 10^(7)` PA |
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| 38531. |
A: The stretching of a coil is determined by the shear modulus. R: Share modulus change only shape of a body keeping its volume unchanged. |
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Answer» Both are TRUE and the REASON is the CORRECT EXPLANATION of the ASSERTION. |
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| 38532. |
Given |vec(A)_(1)| =2, |vec(A)_(2)|=3 and |vec(A)_(1) + vec(A)_(2)| =3. If the value of |(vec(A)_(1) + 2 vec(A)_(2)). (3vec(A)_(1) -4 vec(A)_(2))| " is " 4^(n) find n |
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Answer» |
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| 38533. |
A light rod of length 2 m is suspended from the ceiling horizontally by mean.s of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 10^(-3)m^(2) and the other is of brass of cross -section 2xx10^(-3)m^(2). Find out the position along of corss-section 2xx10^(-3)m^(2). Find out the position along the rod at which a weight may be hung to produce,(i) equal stress in bothwires (ii) equal strains in both wires Young.s modulus of brass =1xx10^(11)N//m^(2) Young.s modulus of steel =2xx10^(11)N//m^(2) |
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Answer» Solution :Suppose `a_(1) and a_(2)` are the corss - sectional areas, and `Y_(1) and Y_(2)` are the Young.s MODULI of steel and brass wirerespectively. Let `T_(1) and T_(2)` are tensions in the steel and brass wires respectively. Let x is distance of the position of the hanging weight from the steel wire. (i) First case : For EQUAL stress in both wires, wehave `(T_(1))/(a_(1))=(T_(2))/(a_(2)) (or) (T_(1))/(10^(-3))=(T_(2))/(2xx10^(-3))(or) T_(2)=2T_(1)""......(i)` As the whole system is in equilibrium, so `Sigma bartau=0`. Taking moment of all the forces acting on the rod about C, we have `T_(1)x-T_(2)(2-x)=0""...( ii)` Solving equations (i) and (ii), we get `x=(4)/(3)m` (ii) Second case: For equal STRAINS in both the wires `e_(1)=e_(2)` `(T_(1)l)/(a_(1)Y_(1))=(T_(2)l)/(a_(2)Y_(2))(or) (T_(1))/(10^(-3) xx 2xx10^(11))=(T_(2))/(2xx10^(-3)xx10^(11))` (or) `T_(1)=T_(2)""...(iii)` From equations (ii) and (iii), we get x = 1M |
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| 38534. |
Two satellites of masses m_(1) and m_(2) (m_(1) gt m_(2)) are revolving the earth in circular orbits of radius r_(1) and r_(2)(r_(1) gt r_(2)) respectively. Which of the following statement is true regarding their velocities v_(1) and v_(2). |
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Answer» `v_(1) = v_(2)` |
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| 38535. |
The escape velocity of projectile on the surface of earth is 11.2 Kms^(-1). If a body is projected out with thrice of this speed, find the speed of the body far away from the earth. Ignore the presence of other planets and sun. |
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Answer» Solution :`V = sqrt(v_(p)^(2) - v_(e)^(2)) = sqrt((3V_(e))^(2) - v_(G)^(2))` `= sqrt(8 XX 11.2) = 31.68 kms^(-1)`. |
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| 38536. |
The equation of motion of a particle is x= a cos (alpha t)^(2). The motion is……….. |
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Answer» PERIODIC but not OSCILLATORY Now putting `t+T` INSTEAD of t, `x(t+T)= a cos [alpha (t+T)^(2)]""[therefore x(t) = a cos (alpha t)^(2)]` `=a cos [alpha t^(2) + alphaT^(2) +2 alpha t T] ne x(t)` where, T is a period of the function `omega (t)` Hence given motion is oscillatory but not periodic. |
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| 38537. |
A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z |
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Answer» MUST be in THERMAL equilibrium |
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| 38538. |
A vertical cylinder of height 100cm contains air at a constant temperature and its top is closed by a frictionless piston at atmospheric pressure (76 cm of Hg) as shown figure - a. If mercury is slowly poured on the piston, due to its weight air is compressed. Find the maximum height of the mercury column which can be put on the piston. |
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Answer» SOLUTION :When MERCURY is poured on the top of the piston, due to increase in pressure, the volume of air will DECREASE. If final mercury column of height x is poured on the piston then gas pressure in EQUILIBRIUM can be given as `P_(f)=(76+x)` cm of Hg As atmospheric pressure is equivalent to the pressure due to a mercury column of height 76 cm. If A be the area of cross section of cylinder then we have according to Boyle.s LAW `P_(1)V_(1)=P_(2)V_(2)" or "(76 cm)(100A)=(76+x)(100-x)A" (or) "7600=7600+24x-x^(2)" (or) "x=24cm` |
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| 38539. |
What is the Boltzmann's constant? Give its value. |
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Answer» Solution :Boltzmann.s constant is DEFINED as the gas constant per molecules. `k_(B)=(R)/(N_(A))=1.38 xx 10^(-23) JK^(-1)` |
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| 38540. |
Can you cite a situation where no force acts on a body. |
| Answer» SOLUTION :No. Atlest ONE FORCE- theforceof gravitymustacton THEBODY. Onecancitemanyexamplewhereforceon thebody ISZERO . | |
| 38541. |
Calculate the angle of (a) 1^(@) (degree) (b) 1' (minute of arc or arcmin) and ("c") 1" (second of arc or arc second ) in radians. Use 360^(@) = 2pi rad, 1^(@)=60' and 1'=60". |
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Answer» SOLUTION :(a)We know that, `180^(@)= PI` rad then `1^(@)=(?)` `:.1^(@)=(pi)/(180)=(3.412)/(180)` `:.1^(@)=0.01745` `:.1^(@)~~1.745xx10^(-2)` rad (b) We know that, `60.(min)=1^(@)` and `1^(@)=1.745xx10^(-2)` rad `:.60.=1.745xx10^(-2)` rad `:. 60.=1.745xx10^(-2)` and then `1.=(?)` `1=(1.745xx10^(-2))/(60)=0.02908xx10^(-2)` `:.1=2.91xx10^(-4)` rad We know that, `60..(sec)=1.(min)` `and 1.(min)=2.908xx10^(-4)` rad `:. 60..=2.908xx10^(-4)` rad then `1,,=(?)` `I..=(2.908xx10^(-4))/(60)=0.048466xx10^(-4)` `:.1..~~4.85xx10^(-6)` rad |
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| 38542. |
Which one of the following physical quantities cannot be represented by a scalar? |
| Answer» Solution :momentum | |
| 38543. |
In deriving Bernoulli's equation , we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery ofdiameter 2xx10^(-3)m ifthe flow must remain laminar ? (b)Do thedissipative forcebecome more important as the fluid velocity increases ?Discuss quanlitatively . |
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Answer» Solution :Diameter of artery `d=2XX10^(-3)m` The coefficient of viscosity of blood `eta=2.084xx10^(-3)`PaS Density of blood `rho=1.06xx10^(3)kg//m^(3)` SUPPOSE , Reynold.s numberfor LINEAR flow , `R_(e)=2000` Maximum average velocityof blood, `v_("avg")=(R_(e)eta)/(rhoD)=(2000xx2.084)/(1.06xx10^(3)xx4xx10^(-3))` `=0.98m//s` (b)Volume of fluid passing per second, `Q=av_("arg")` `=(pi^(r2))v_("avg")` `=(22)/(7)xx(2xx10^(-3))^(2)xx0.98` `=1.23xx10^(-5)m^(3)s^(-1)` As per fluidvelocity increases , the dissipative forces become more important . This is because of the rise of turbulence . Turbulent flow causes dissipative loss in a fluid. |
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| 38544. |
(A) Dispersion of light occurs, because velocity of light in a material depends upon its colour. ( R) the dispersion power depends only upon the material of the prism, but not upon the refracting angle of the prism. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 38545. |
(A): When two bodies of equal masses collide, their velocities are always interchanged. (R ): Momentum is conserved in any collision |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A's |
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| 38546. |
The internal energy of a system remains constant when it undergoes a) Cyclic process b) an Isothermal process c) Any process in which heat given out by the system is equal to work done on the system |
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Answer» both a and B are CORRECT |
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| 38547. |
Rain, pouring down at an angle alphawith the vertical has a speed of 10 m/s. A girl runs against the rain with a speed of 8 m/s and sees that the rain makes an angle betawith the vertical, then relation between alpha and beta |
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Answer» `tan beta = (8+10 sin ALPHA)/(10 cos alpha)` |
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| 38548. |
A vehicle of mass is moving on a rough horizontal road with kinetic energy 'E'. If the coefficient of friction between the tyres and the road be mu, find the stopping distance |
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Answer» SOLUTION :`E/(MUMG)` |
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| 38549. |
A square plate of side 15 cm. is floating on the surface of water. It the surface tension of water is 60 dyne/cm., the excess force applied to separate this plate from water will be |
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Answer» 600 dyne |
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| 38550. |
Pressure P, volume V and temperature T of a certain material are related byP = (alpha T^2)/(V).Here, alphais aconstant. Work done by the material when temperature changes from T_0 to 2T_0while pressure remains constant is |
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Answer» `6 ALPHA T_0^3` |
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