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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.
| 37651. |
A bimetallic strip made of brass and steel strips shown in the figure given below is heated. What shape does it acquire? (alpha of brass =19xx10^(-6)//""^(0)C and alpha of steel =11xx10^(-6)//""^(0)C.) ("Brass")/("Steel") |
| Answer» Solution :It bends with brass on convex side as `ALPHA` of brass is GREATER than `alpha` of STEEL. | |
| 37652. |
State stokes'law .By using it deduce the expression for : (i) Initial acceleration of smooth sphere and (ii)Equation of terminal velocity of sphere falling freely through the viscousmedium. (iii) Explain : Upward motion of bubbles produced in fluid. |
Answer» Solution :Scientist Stokes. said that , viscous force `F_(V)` on small spherical solid body of radius r and moving with velocity v through a viscous medium of large dimensions having coefficient of viscosity `eta` is `6pietarv`. As shown in FIGURE a small spherical body of radius r, density `rho` falling in viscous medium of density `sigma`. Following forces acted on it . (i) Weight `F_(1)=mg=("volume"XX"density")g` `=((4)/(3)pir^(3)rho))g`...(1) (In downward) Where m = mass of spehre (ii) Buoyant force `F_(2)=m_(o)g` `=((4)/(3)pir^(3)sigma)g=`(2) (in upward) where `m_(0)` = mass of liquid having volume of sphere (iii) According to stokes.law `F_(v)=6pietarv` ...(3) RESULTANT force acting on the sphere, `F_(R)=F_(1)-F_(2)-F_((v))` `F_(R)=((4)/(3)pir^(3)rho)g-((4)/(3)pir^(3)sigma))g-6pietarv` `F_(R)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv` ...(4) (i)Equation of initial acceleration If mass of sphere is m and initial acceleration `a_(0)` then `F_(R)=ma_(0)=(4)/(3)pi^(3)rhoa_(0)` From equation (4), `((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv` Taking initial velocity of freely falling sphere v=0 `((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)` `thereforerho(a_(0))=g(rho-sigma)` `thereforea_(0)=((rho-sigma)g)/(rho)` ...(5) (ii)Terminal velocity : As the time passes velocity of sphere will increase hence in equation (3), the force `6pietarv` wil increase and resultant force `F_(R)` decreases. For any one velocity of spehere `F_(R)`becomes zero and hence according to NEWTON first law , spehre moves with constant velocity . This constant velocity is known as final or terminal velocity `v_(t)` of sphere. In equation (4) `v=v_(t)andF_(R)=0`, `therefore(6pietar)v_(t)=(4)/(3)pir^(3)g(rho-sigma)` `thereforev_(t)=(2)/(9)(r^(2)g(rho-sigma))/(eta)` ...(6) Hence terminal velocity of small sphere in viscous medium `v_(t)` (i) Proportional to the square radius of sphere `v_(t)PROPR^(2)` (ii) Proportional to the difference of densities of sphere and fluid. `v_(t)prop(rho-sigma)` (iii)Inversely proportional to the coefficients of viscosity of fluid `v_(t)prop(1)/(eta)`. (iv)Upward motion of bubbles produced in fluid : The denisty `(rho)` of air bubble produced in the fluid is less then the density of fluid `(sigma)` .Hence `(rholtsigma)`, so terminal velocity is negative and hence bubble would travel in upward direction. |
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| 37653. |
A perfect gas goes from state A to state B by absorbing 8xx10^(5)J of heat and doing 6.5xx10^(5)J of external work. It is now transferred between the same two states in another process in which it absorbs 10^(5)J of heat. In second process. Find the work done in the second process. |
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Answer» SOLUTION :According to the FIRST LAW of thermodynamics `DeltaU=DeltaQ-DeltaW` In the first process `DeltaU=8xx10^(5)-6.5xx10^(5)=1.5xx10^(5)J` Now `DeltaU`, being a state function, remains the same in the second process, `DeltaW=DeltaQ-DeltaU=1 xx10^(5)-1.5xx10^(5)` `DeltaW=-0.5xx10^(5)J` The negative sign shows that work is doen on the GAS. |
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| 37654. |
A particle is moving along a circular path. The angular velocity, linear velocity,. Angular acceleration, and centripetal acceleration of the particle at any instant. Respectively are vec omega, vec v, vec prop, and vec a_c. Which of the following relations is not correct ? |
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Answer» `vec omega _|_ vec v` |
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| 37655. |
Consider two cylinders with same radius and same mass. Let one of the cylinders be solid and another one be hollow. When subjected to some torque, which one among them gets more angular acceleration than the other? |
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Answer» Solution :Moment of inertia of a solid cylinder about its axis `I_(s)=(1)/(2)MR^(2)` Moment of inertia of a hollow cylinder about its axis `I_(K)=MR^(2)` `I_(s)=(1)/(2)I_(h) or I_(h)=2I_(s)` torque `TAU =I alpha=(tau)/(1)` `alpha_(s)=(tau)/(I_(s)) and a_(h)=(tau)/(I_(b))` SINCE `I_(h) gt I_(s) rArr (I_(h))/(I_(s)) gt I` `:. alpha_(s) gt alpha_(h)` For the same torque, a solid cylinder gets more ACCELERATION than a hollow cylinder. |
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| 37656. |
The pressure P for a gas is plotted against its absolute temperature T for two different volumes V_1 and V_2 where V_1 gt V_2. If P is plotted on y-axis and T on x-axis, then: |
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Answer» The curve for `V_1` has greater slope than that for `V_2` |
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| 37657. |
Statements :(a) Motion of centre of mass depends on external forces applied on it. (b) The motion of the centre of mass of the body is called the translational motion of the body. |
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Answer» both a and B are TRUE |
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| 37658. |
When the road is dry and coefficient of friction is mu. the maximum speed of a cur in a circular path is 10m//sec, If the road becomes wet and mu^(1) = mu//2. What is the maximum speed permitted |
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Answer» `5m//SEC` |
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| 37659. |
A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ? |
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Answer» `eta=(T_1-T_2)/T_1` , where `T_1`=TEMPERATUREOF the SOURCE `T_2`=temperature of the sink `THEREFORE 0.40=(T_1-300)/T_1` `therefore T_1-300=0.40T_1` `therefore 0.6T_1=300 rArr T_1=300/0.6=3000/6`=500 K Now, 50% efficiency increases , `therefore 0.60=(T_1-300)/T_1 rArr T_1-300=0.6T_1` `therefore 0.4T_1=300 rArr T_1 =300/0.4=(300xx10)/4` =750 Increase in temperature of source =750-500=250 K |
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| 37660. |
Two clocks are being tested against a standard clock located in a national laboratory . At 12:00:00 noon by the standard clock, the regarding of the two clocks are : If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer ? |
| Answer» Solution :The range of variation over the seven DAYS of observations is 162 s for clock 1, and 31s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The IMPORTANT point is that a clock.s zero error is not as SIGNIFICANT for precision work as its variation, because a .zero-error. can always be easily corrected . HENCE clock 2 is to be preferred to clock 1. | |
| 37661. |
As we move from poles to equator of earth, the magnitude of acceleration due to gravity increases/decreases correct the sentence. |
| Answer» Solution :The magnitude of ACCELERATION DUE to gravity DECREASES. | |
| 37662. |
(A) : In the given Fig . 7 . 184 , a force20 N pulls a block of mass 10 kg on a rough floor . If coefficient of friction is mu = 0.4 , force of friction acting on the block will be 20 N. (R) : Force of kinetic friction is slightly less than the limiting value of force of friction . |
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Answer» If both A and R are true and R is the correct explanation of A . |
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| 37663. |
The temperature of a solid object is observed to be constant during a period. In this period. |
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Answer» heat may have SUPPLIED to the body |
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| 37664. |
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container as shown in Fig a) Find the work done when the gas is taken from state 1 to state 2. b) What is the ratio of temperature T_(1)//T_(2), if V_(2)=2V_(1) ? c) Given the internal energy for one mole of gas at temperature T is (3//2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V_(2)=2V_(1) . |
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Answer» SOLUTION :a) Work done by the GAS (Let` PV^(1//2)=A`) `DeltaW=int_(V_(1))^(V_(2))pdv=A int_(V_(1))^(V_(2))(dV)/sqrtV=A[(sqrtV)/(1//2)]_(V_(1))^(V_(2))=2A(sqrtV_(2)-sqrtV_(1))=2P_(1)V_(1)^(1//2)[V_(2)^(1//2)-V_(1)^(1//2)]` b) SINCE `T =pV//nR=A/(nR).sqrtV` Thus `T_(2)/T_(1)=sqrt(V_(2)/V_(1))=sqrt2` c) Then, the change in internal energy `DeltaU=U_(2)-U_(1)=3/2R(T_(2)_T_(1))=3/2RT_(1)(sqrt2-1)` `DeltaW=2AsqrtV_(1)(sqrt2-1)=2RT_(1)(sqrt2-1)` `DeltaQ=(7//2)RT_(1)(sqrt2-1)` |
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| 37665. |
In the figure. The time taken by the projectile to reach from A to B is t. Then the distance AB is equal to |
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Answer» `( U t)/(SQRT(3))` |
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| 37666. |
A relay satellite transmits the television programme from one part of the world to another part continuously because its period |
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Answer» is GREATER than the period of the earth about its axis |
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| 37667. |
An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (C_P) and at constant volume (C_v) is : |
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Answer» `7/2` |
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| 37668. |
A rocket is moving at a speed of 200 m s^(-1) towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. |
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Answer» Solution :(1) The observer is at rest and the source is moving with a speed of `200 m s^(-1)`. SINCE this is COMPARABLE with the VELOCITY of sound, `300 m s^(-1)`, we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, `v_(o) = 0`, and `v_(s)` must be replaced by `- v_(s)`. thus , we have `v = v_(o) (1 - (v_(s))/(v))^(-1)` `v = 1000 Hz xx [1 - 200 m s^(-1)//330 m s^(-1)]^(-1)` `~= 2540 Hz` (2) The target is now the source (because it is the source of echo) and the rocket.s DETECTOR id now the detector or observer (because it detects echo). thus , `v_(s) = 0` and `v_(o)` has a positive value. The drequency of the sound emitted by the source (the target) is v. Therefore , the frequency as registered by the rocket is `v. = v ((v + v_(o))/(v))` `= 2540 Hz xx ((200 m s^(-1) + 330 m s^(-1))/(330 m s^(-1)))` `~= 4080 Hz` |
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| 37669. |
A point mass is orbiting a significant mass M lying at the focus of the elleptical orbit having major and minor axes given by 2a and 2b respectively. Let r be the distance between the mass M and the point of major axis. The velocity of the particle can be given as |
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Answer» `(ab)/(2r) sqrt((GM)/(a^(3)))` time of fall `(T')/2=T/(2sqrt(8))=(sqrt(2)T)/8` |
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| 37670. |
In the following question statement 1 represent Assertion and statement 2 represents Reason .Statement 1 :The velocity of a particle at a point on its trajectory is equal to the slope at that point .Statement 2 : The velocity of the particle acts along the tangent to the trajectory atthe point .Select the correct statement from the following statements . |
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Answer» Statement 1 is true, statement 2 is true and it explains statement 1 CORRECTLY. |
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| 37671. |
Two particles initially at rest are approaching each other due to their mutual force of attraction. What is the velocity of the centre of mass of the system when the particles have a relative velocity v ? |
| Answer» Solution :For a two-particle system, the MUTUAL attraction is an INTERNAL FORCE and has no EFFECT on the motion of the centre of mass. Hence the velocity of the centre of mass is zero. | |
| 37672. |
(i) In the arrangement shown in the figure the coefficient of friction between the 2 kg block and the vertical wall is mu= 0.5. A constant horizontal force of 40 N keeps the block pressed against the wall. The spring has a natural length of 1.0 m and its force constant is k = 400 Nm^(–1). What should be the height h of the block above the horizontal floor for it to be in equilibrium. The spring is not tied to the block. (ii) A block of mass M is pressed against a rough vertical wall by applying a force F making anangle of theta with horizontal (as shown in figure). Coefficient of friction between the wall and the block is mu = 0.75. (a) If F = 2 Mg, find the range of values of theta so that the block does not slide [Take "tan" 37^(@) = 0.75, "sin" 24^(@) = 0.4] (b) Find the maximum value of theta above which equilibrium is not possible for any magnitude of force F. |
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Answer» `37^(@)` |
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| 37673. |
These questions have statement I and statement II. Of the four choices given below, choose the one that best describes the two statements. Statement I: The centre of mass of the system will not alter in any direction if the external force is not exerted on it. Statement II: If net external force is zero then the linear momentum of the system remains constant. |
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Answer» Statement I is true, statement II is true, statement II is a CORRECT EXPLANATION for statement I. |
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| 37674. |
Two rotating bodies A and B of masses mand 2m with moments of inertia I_(A)and I_(B)(I_(B)>I_(A)) have equal kinetic energy of rotation. If L_(A) and L_(B) be their angular momenta respectively, then |
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Answer» `L_(B) GT L_(A)` |
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| 37675. |
A ring is cut from a platinum tube of 8.5cm internal and 8.7cm external diameter. It is supported horizontally from a pan of a balance so that it comes in contact with the water in a glass vessel. If an extra 3.97g weight is required to pull it away from water, then surface tension of water is nearly (18.03) X dyne / cm. Find x (g=980cm/s^2) |
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Answer» |
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| 37676. |
The pressure p of a gas is plotted against itsabsolute temperature T for two different constant volumes V_(1) and V_(2),where V_(1) gt V_(2).pis plotted on the y-axis and T on the x-axis |
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Answer» The curve for `V_(1)`has GREATER slope than THECURVE for `V_(2)`. |
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| 37677. |
The ring R in the arrangement shown can slide along a smooth, fixed, horizontal rod XY. It is attached to the block B by a light string. The block is released from rest, with the string horizontal |
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Answer» One point in the STRING will have only vertical MOTION |
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| 37678. |
A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water at 35^(@)C. How much mass of ice at 0^(@)C should be added to decrease the temperature of the calorimeter to 20^(@)C. |
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Answer» Solution :Let .m. be the mass of ICE added, Heat gained by ice during melting = ML = `mxx0.335xx10^(6)J` Heat gainerd by molten ice (water) during rise in temperature = `mxx4186xx(20-0)=mxx83720J` `therefore` Heat gained by the ice = Heat LOST by the CALORIMETER and water `mxx[0.335xx10^(6)+83720]=1255.8+30139.2=31395J` `m[335000+83720]=31395` `m=31395/418720=0.07498kg`. |
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| 37679. |
A cylinder wooden float whose base area S=4000cm^(2) & the altitude H=50cm drifts on the water surface specific weight of wood d=0.8gf//cm^(3) (i). What work must be performed to take the float out of the water? (ii). Compoute the work to be performed to submerged completely the float into the water. |
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Answer» (II). `(1)/(2)SH^(2)(1-d^(2))=2Kgf-m` |
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| 37680. |
On what factors the value of "g" at any place on the Earth depends? |
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Answer» Solution :The FACTORS on which the value of "g" on the EARTH DEPENDS are (i) ALTITUDE (ii) Rotation (iii) DEPTH (iv) Shape |
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| 37681. |
Two capillary tubes of same radiusr but of lengths, l_(1) and l_(2) are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before |
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Answer» `l_(1)+l_(2)` |
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| 37682. |
If a simple pendulum of length 'l’ is about to start to make SHM, when it is brought a position so that its length has maximum angular displacement 'q'. If the mass of the bob is 'm", then the maximum K.E at its mean position is |
| Answer» Answer :D | |
| 37683. |
(A) : The size and shape of the rigid body remains unaffected under the effect of external forces.(R ) : The distance between two particles remains constant in a rigid body. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 37684. |
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume Vis given by PV^(n)= constant, then n is given by (Here C_(P) and C_(V) are molar specific heat at constant pressure and constant volume, respectively): |
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Answer» `N -(C-C_V)/( C-C_P)` |
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| 37685. |
The internal radius of one arm of a glass capillary U tube is r_(1) and for the second arm it is r_(2)(gt r_(1)). The tube is filled with some mercury having surface tension T and contact angle with glass equal to 90^(@) + theta. (a) It is proposed to connect one arm of the U tube to a vacuum pump- so that the mercury level in both arms can be equalized. To which arm the pump shall be connected?(b) When mercury level in both arms is same, how much below the atmospheric pressure is the pressure of air in the arm connected to the pump? |
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Answer» |
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| 37686. |
The menual of a car instructs the owner to inflate the tyres to pressure of 200 kPa. (a) What is the recommended gauge pressure ? (b) What is the recommended absolute pressure ? (c ) If after the required inflation of the tyres, the car is driven to a mountain peak. where the atmospheric pressrre is 10 % below that at sea level, what will be the tyre gauge read ? Atmospehric pressure=1.01 xx 10^5Pa. |
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Answer» Solution :(a) Pressure Instructed by manual = `P_(g) = 200 K P_(a)` (b) Absolute Pressure = `101 k P_(a) + 200 k P_(a) = 301 k P_(a)` At MOUNTAIN PEAK Pa′ is 10% less `P_(a)=90 kP_(a)` If we ASSUME absolute pressure in tyre does not change during driving then `P_(g) = P – P_(a)′ = 301 – 30 = 211 k P_(a)` So the tyre will read 211 k `P_(a)`, pressure. |
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| 37687. |
Which of the following statements is /are true |
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Answer» The direction of the excess pressure in the meniscus of a liquid of angle of CONTACT 2`PI`/3 is UPWARD |
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| 37688. |
Four particles, each of mass M and equidistant "from each other, move along a circle of radius R under the action of the mutual gravitational attraction. The speed of each particle is ......... (G = universal gravitational constant) |
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Answer» `sqrt((GM)/R (1+2sqrt2))` `F_1 = F_2 = (GM^2)/(2R^2)`  and force attraction exerted by particle at C on particle at A `F_3 = (GM^2)/(4R^2)` `:.` The TOTAL force on particle A `F= F_1 + F_2 + F_3` but resultant force of `F_1 +F_2` `F.=sqrt(F_1^2 +F_2^2+2F_1^2 cos 90^@) ` `F. = sqrt(F_1^2 +F_1^2) = sqrt(2F_1^2)` `:. F. = sqrt2F_1` `:. F. = sqrt2 F_1 +F_3` `=sqrt2 (GM^2)/(2R^2) + (GM^2)/(4R^2)` `:. (Mv^2)/R = (GM^2)/(2R^2) [sqrt2 +1/2]` `v^2 =(GM)/(2R) [(2sqrt2 +1)/2]` `:. v = 1/2 sqrt((GM)/R (1+ 2sqrt2)` |
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| 37689. |
A satellite moves around the earth in a circular orbit with speed V. If 'm' is the mass of the satellites, its potential energy is |
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Answer» `mv^2` |
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| 37690. |
Which of the following examples represent periodic motion? |
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Answer» A swimmer completing ONE (return) TRIP from one bank of a river to the other and back. |
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| 37691. |
What is the need for banking of tracks? Draw free body diagram of a car negotiating a banked track [ Take friction into account]. Write the expression for maximum velocity to negotiate the curve safely. |
| Answer» Solution :A vehicle moving along a curved path needs some centripetal force. If the path is banked. i.e., inclined with the horizontal , then the horizontal components of friction and normal REACTION help to provide this centripetal force. The vehicle may ATTAIN a velocity greater than that if the path WOULD have been horizontal along the CURVE. | |
| 37692. |
A force F is given by F = at + bt^(2), where t is time. The dimensions of a and b are |
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Answer» `[MLT^(-3)]` and `[MLT^(-4)]` `[a]= ([F])/([t])= ([MLT^(-2)])/([T])= [MLT^(-3)]` `[b]=([F])/([t^(2)])= ([MLT^(-2)])/([T^(2)])= [MLT^(-4)]` |
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| 37693. |
A rod of length 'l' and mass m, pivoted at one end, is held by a spring at its mid point and a spring at far end, both pulling in opposite directions. The springs have spring constant K, and equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position |
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Answer» Let `x_1 and x_2` be the changes in the lengths of the springs . Coresponding restoring forces will ACT as shown. Here `x_2 = L theta and x_1 =(L theta)/(2)` Net restoning torque on the rod will be `tau =(K x_2)L + (Kx_1) L/2 =-(KL^2 + K (L^2)/(4))theta` So, `(ML^2)/(3) ALPHA = K ((5L^2)/(4)) theta` `alpha = - ((15L^2)/(4mL^2)K)theta , alpha =- ((15K)/(4m)) theta` frequency `=1/(2PI) sqrt((alpha)/(theta)) = 1/(2pi) sqrt((15K)/(4m))` 
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| 37694. |
If angle between vec(a) and vec(b) is (pi)/(3), then angle between 2bar(a) and -3vec(b) is |
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Answer» `(PI)/(3)` |
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| 37695. |
Explain characteristics of streamlines. |
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Answer» Solution :Characteristics of streamlines : (i) In a steady flow streamline can never intersect each other . (ii)The tangent drawn at point on the streamline INDICATES the direction of velocity of fluid particle at that point. (iii) The velocity of each fluid particles passing at a point on streamline is same but it is different at different point. (iv)Widely SPACED streamlinesindicates REGION of low speed and CLOSELY streamlines indicates region of high speed . |
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| 37696. |
When 1 g of water 100^(@) C is converted into steam at100^(@)C,. It occupies a volumn of 1671 cc at normal atmospheric pressure . Find the increase in internal energy of the molecules of steam. |
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Answer» Solution :`1" atmosphere" = 1.013xx10^(5) Nm^(-2)` , volume of 1 gm of water, `V_(1) = 1cc = 10^(-6)m^(3)`, volume of steams = `1671 cc =1671xx10^(-6) m^(3)` External WORK done dW = `P(V_(2)-V_(1)) = 1.013xx10^(5)(1671xx10^(-6) -1xx10^(-6))=1.013xx167=169.2 J`. Specific latent heat of vaporisation of steam=540cal/g. So, heat supplied to CONVERT 1g of water into steam. `Delta Q = 540xx4.2J=2268J` By first LAW of THERMODYNAMICS `Delta U = Delta Q-Delta W = 2268-169.2=2098.8J` |
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| 37697. |
A ball strikes against the floor and returns with double the velocity. What type of collision is it? |
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Answer» PERFECTLY ELASTIC |
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| 37698. |
A solid sphere of mass 1.5 kg rolls on a frictionless horizontal surface, its centre of mass moving at a speed of 5 ms^(-1). Then it rolls up an incline of 30° to horizontal. The height attained by the sphere before it stops is (g = 10 sm^(-2)) |
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Answer» 5 m |
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| 37699. |
A ship'A' steams down to North at 16 kmph, and ship 'B' due west at 12 kmph. Relative velocity of B of w.r.t. A is |
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Answer» 10 kmph |
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| 37700. |
Two particle are projected at the same instant from the same point at inclinations alpha and beta to the horizontal. If they simultaneously hit the top and bottom of a vertical pole subtending angle theta at the pointof projection, find (tan alpha-tan beta). |
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Answer» |
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