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.
| 14651. |
The rate of flow of a liquid through a capillary tube of radius r is under a pressure difference of P. Calculate the rate of flow when the diameter is reduced to half and the pressure difference is made 4 P? |
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Answer» |
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| 14652. |
A 1kg ball moving at 12ms^(-1) collides headon with a 2kg ball moving in the opposite direction at 24 m/s. If the coefficient of restitution is 2/3,then the energy lost in the collision is |
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Answer» 60J |
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| 14653. |
The time period of a seconds pendululm is measured repeatedly for three times by two stop watches A, B . If the readings are as follows {:("S.NO.", A, B)(1, 2.01 "sec", 2.56 "sec"),(2, 2.10 "sec", 2.55 "sec"),(3, 1.98 "sec", 2.57"sec"):} |
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Answer» A is more ACCURATE but B is more PRECISE |
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| 14654. |
Ratio of adiabatic bulk modulus to isothermal bulk modulus is |
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Answer» <P>`C_(P)-C_(V)` |
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| 14655. |
The limiting friction between two bodies in conact is independent of |
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Answer» NATURE of SURFACES in CONTACT |
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| 14656. |
A body is projected with a velocity 60m/s at an angle 30^(@) with horizontal . It's displacement after 2sec is |
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Answer» `30 sqrt(3) hat(i) + 30 hat(j)` |
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| 14657. |
The excess pressure in an bubble just below the water surface is (surface tension of liquid is 72 dyne /cm and atmosphericl pressure =1.013xx10^(5)N//m^(2), Radius of bubble =0.1 mm) |
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Answer» `1.4xx10^(-3)N//m^(2)` |
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| 14658. |
A space-ship is launched into a circular orbit close to the earth's surface. What additional velocity should now be impartedto the space-ship in the orbit to overcome the gravitational pull. [Radius of earth 6400 km and g = 9.8 m//s^(2)] |
| Answer» SOLUTION :Additional VELOCITY to be IMPARTED to the ORBITING satellite for escapping `= V_(e) - V_(0) = 11.2 - 8 = 3.2 km//s` | |
| 14659. |
A constant power .P. is applied to a particle of mass .m.. What is the distance travelled by the particle when its velocity increases from v_(1) to v_(2). |
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Answer» |
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| 14660. |
State aristotelian of motion what is the flaw in this law ? |
| Answer» SOLUTION :Aristotelian LAWOF motionstates that Forcecausesmotion. THISIS wrong. Itisbased oncommonsense. Inpractiseforceisrequiredtocountertheopposingforce of FRICTION . | |
| 14661. |
In the previous question, a) Work is done by the gas b) Work is done on the gas c) Heat is absorbed by the gas d) Heat is given out by the gas |
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Answer» Only a, B are CORRECT |
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| 14662. |
A steel ball of mass 0.1 kg falls freely from a height of 10m and bounces to a height of 5.4m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 JKg^(-1)K^(-1)) (g=10ms^(-2)) |
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Answer» `0.01^@C` |
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| 14663. |
A block of mass 3kg which is on a smooth inclined plane making an angie of 30^(@) to the horizontal is connected by cord passing over light frictionless pulley to second block of mass 2kg hanging vertically. What is the acceleration of each block and what is the tension of the cord ? |
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Answer» `0.98m//s^(2), 17.6N` |
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| 14664. |
A metal wire is held at the two ends of rigid supports at 20^(@)C, the wire is just taut. The speed of transverse wave in this wire at 25^(@)C will be (alpha=16xx10^(-6)//C,Y=9xx10^(11)N//m^(2), density of metal =5gm//c.c) |
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Answer» `120ms^(-1)` |
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| 14665. |
The temperature inside a refrigerator is t_(2)^(@)C . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be |
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Answer» `(t_(1))/(t_(1)-t_(2))` Room temperature = `t_(1)^(@)C` For refrigerator, `(" HEAT given to high temperature" (Q_(1)))/("Heat taken from lower temperature" (Q_(2)))=(T_(1))/(T_(2))` `(Q_(1))/(Q_(2))=(t_(1)+273)/(t_(2)+273)` `RARR (Q_(1)/(Q_(1)-W)=(t_(1)+273)/(t_(2)+273)` or , `1-(W)/(Q_(1))=(t_(2)+273)/(t_(1)+273)` or `(W)/(Q_(1))=(t_(1)-t_(2))/(t_(1)+273)` The amount of HET delivered to the room for each joule of electrical energy (W=1 J) `Q_(1)=(t_(1)+273)/(t_(1)-t_(2))` |
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| 14666. |
Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1, 2, 3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1, 3, -2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system? |
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Answer» (0 , 0 , 0) Let the third particle of mass 5 kg PUT at `(x_(3) , y_(3) , z_(3))` i.e, `m_(1) = 6 kg , (x_(1) , y_(1) , z_(1)) = (1 , 2 , 3)` `m_(2) = 5 kg, (x_(2) , y_(2) , z_(2)) = (-1 , 3 , -2)` `m_(3) = 5 kg, (x_(3) , y_(3) , z_(3)) = ?` Given , `(X_(CM) , Y_(CM) , Z_(CM) ) = (1 , 2 , 3)` `X_(CM) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3)) or 1 = (6 xx 1 + 5 xx (-1) xx 5x _(3))/(6 + 5 + 5)` `5x_(3) =16 - 1 = 15 or x_(3) = 3` Similarly `, y_(3) = 1` and `z_(3) = 8` |
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| 14667. |
(A) : A sphere cannot roll on a smooth inclined surface.(R ) : For a smooth inclined surface force of friction is equal to zero. |
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Answer» Both 'A' and 'R' and TRUE and 'R' is the correct EXPLANTATION of 'A' |
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| 14668. |
Volume of the bulb of a mercury thermometer at 0^(0)Cis V_(0)and area of cross section of the capillary tube is A_(0)coefficient of linear expansion of glass is alpha_(g)per ""^(0)C , and the cubical expansion of mercury is gamma_(m) per ""^(0)C . If the mercury fills the bulb at 0^(0) C find the length of mercury column in capillary tube at T^(0)C |
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Answer» SOLUTION :EXPANSION of mercury `= V_(0) gamma_(m) T ` Expansion in GLASS BULB = ` V_(0) 3 alpha_(g) T ` Apparent expansion in mercury ` = V_(0) gamma_(m) T- V_(0) 3alpha_(g) T ` i.e.,`A_(t) xx Delta l = v_(0) T (gamma_(m) - 3 alpha_(g))` `Delta l = (V_(0) T (gamma_(m) - 3 alpha_(g) ) )/(A_(t)) = (V_(0) T (gamma_(m) - 3 alpha_(g)))/(A_(0) (1 + 2 alpha_(g) T))` |
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| 14669. |
Three identical spheres , each of mass M , are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 2 m . Taking their point of intersection as the origin , find the position vector of centre of mass |
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Answer» SOLUTION :`x = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3))` `therefore= ((M XX 0) + (M xx 0) + (M xx 2))/(M + M + M) = (2M)/(3 M) = (2)/(3) m` ![]() `therefore`x-coordination = `2//3 hati` Again `y = ((M xx 2) + ( M xx 0) + ( M xx 0))/(M + M + M)` or `y = (2M)/(3M) = (2)/(3)` m `therefore` y-coordinate = `2//3 hatj` `therefore` Position vector of CENTRE of mass= `(2)/(3) hati + (2)/(3) hatj = (2)/(3) (hati + hatj)` |
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| 14670. |
When a tensile or compressive load 'P' is applied to rod or cable, its length changes. The change length x which, for an elastic material is proportional to the force Hook' law). P alpha x or P = kx The above equation is similar to the equation of spring. For a rod of length L, area A and young modulue Y. the extension xcan be expressed as. x = (PL)/(AY) or P = (AY)/(x), hence K = (AY)/(L) Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to (1)/(2)Px. The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation. Four identical rods of geometry as described in problem (2) are attached with lift. If weight of the lift cage is 1000N, and elastic limit of each rod is taken as 9 xx 10^(4)N//m^(2) then the number of persons it can carry safety is equal to. (g = 10m//sec^(2)), assume average mass of a person as 50 kg and lift moves with constant speed) |
| Answer» SOLUTION :N//A | |
| 14671. |
The rotation of earth about its axis is……….. |
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Answer» periodic motion Obviously, the rotation of earth is not a to and fro TYPE of motion about a fixed point, HENCE it is not a simple harmonic motion. |
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| 14672. |
''Technological advancements owe a great deal to the developments in physics'' - Explain. |
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Answer» Solution :The application of knowledge for practical purposes in various fields to invent and produce useful PRODUCTS or to SOLVE PROBLEMS is known as technology. Technology is the application of the principles of physics. Thus, physics and technology can both TOGETHER impact our society directly or indirectly. Eg: Basic laws of electricity and magnetism led to the DISCOVERY of wireless communication technology which has shrunk the world with effective communication over large distances. |
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| 14673. |
A ray of light is incident at an angle of 60^@ on asqrt3cm thick plate (mu=sqrt3) The shift in the path of the ray as it emerges out from the plate |
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Answer» 1 |
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| 14674. |
A calorimeter of water equivalent 100 grams contains 200 grams of water at 10^@C . A solid of mass 500 grams at 45^@Cis added to the Calorimeter. If equilibrium temperature is 25^@Cthen, the specific heat of the solid is |
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Answer» `0.45 "CAL G"^(-1)""^@C^(-1)` |
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| 14675. |
Two physicists, both of mass 50 Kg climb up identical ropes suspended from the ceiling of a gymnassium. The ropes are 15 m long. Physicist 1 reaches the top twice as quickly as physicist 2 does. After physicist 2 also reaches the top, they argu about who did more work against gravity. Physicist 1: I did more work against gravity, because I was over coming gravity more quickly. your climb was lazier, and therefore, you did less work. Physicist 2: No was, I did more work against gravity because I spend more time climbing the rope. Answer the following questions: When physicist -2 has falled on third of the way from the calling to the floor, his kinetic energy is approximately |
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Answer» 10J |
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| 14676. |
Two physicists, both of mass 50 Kg climb up identical ropes suspended from the ceiling of a gymnassium. The ropes are 15 m long. Physicist 1 reaches the top twice as quickly as physicist 2 does. After physicist 2 also reaches the top, they argu about who did more work against gravity. Physicist 1: I did more work against gravity, because I was over coming gravity more quickly. your climb was lazier, and therefore, you did less work. Physicist 2: No was, I did more work against gravity because I spend more time climbing the rope. Answer the following questions: Physicist-1 climbed his rope in 30 sec. What average power did he exert against gravity (g=10m//s^(2)) |
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Answer» 25W |
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| 14677. |
Two physicists, both of mass 50 Kg climb up identical ropes suspended from the ceiling of a gymnassium. The ropes are 15 m long. Physicist 1 reaches the top twice as quickly as physicist 2 does. After physicist 2 also reaches the top, they argu about who did more work against gravity. Physicist 1: I did more work against gravity, because I was over coming gravity more quickly. your climb was lazier, and therefore, you did less work. Physicist 2: No was, I did more work against gravity because I spend more time climbing the rope. Answer the following questions: Which physicist if either did more work against gravity |
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Answer» PHYSICIST-1 |
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| 14678. |
A man standing still finds rain pouring at an angle 30^(@) with vertical , when he starts walking forward at a speed of 4 kmph it appears to him that rainpours vertically down . When he runs at riple the speed, what is the angle at which rain appears to pour? |
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Answer» Solution :Case (i) : When the man is still `V_(R)`=actual (absolute ) velocity of rain. ![]() Case (ii) : When man WALK at 4kph, rain falls vertically means horizontal component of rain.s velocity=4 `:.V_(R)sin 30^(@)=4rArr V_(R)=8` ![]() Case (iii) : The man runs at triple the SPEED `V_(M)=12kph` We have superimposed `-V_(M)"to " V_(R)`to get the resultant. from figure , we see `tanphi=(O.B)/(OO)=(12-4)/sqrt(8^(2)-4^(2))=8/sqrt(48)=2/sqrt3:. phi=49` and raindrop makes an angle of `49.6^(@)` in opposite DIRECTION .
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| 14679. |
The displacement of particle is represented by the equation y = sin^(3)omegat. The motion is |
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Answer» non periodic |
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| 14680. |
A needle of 12 cm long is a on the surface of water. The extra force required, over its weight, to take it out from the surface of water is (surface tension = 72 dyne cm^(-1)) |
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Answer» 72N |
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| 14681. |
A body of mass 10gm moving with velocity 5ms^(-1) collides with another body of mass 15gm moving with 3ms^(-1) in the same direction. After collision if they stick and move together, the loss of kinetic energy of the system is |
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Answer» 12J |
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| 14682. |
Let V and E represent the gravitational potential and field at a distance r from the centre of a uniform solidx sphere. Consider the two statements. A) The plot of V against r is discontinuous B) The plot of E against r is discontinuous |
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Answer» Both A and B are CORRECT |
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| 14683. |
A 670 kg meteorite happens to be composed of aluminium. When it is far from the earth, its temperature is -15^@C and it moves with a speed of 14 kms^(-1)relative to the earth. As, it crashes into earth, let total energy is equally shared between planet and meteorite. Specific heat of liquid and gaseous aluminium be 1170 J kg^(-1)C^(-1).The final temperature of meteorite is (x + 0.87)104^(4@)""^@C . Find .x. |
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| 14684. |
Define momentum ? Give its unit . |
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Answer» SOLUTION :MOMENTUM is defined as the PRODUCT of mass with velocity . The UNITOF the momentum is `kg ms^(-1)` |
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| 14685. |
A bullet strikes a wooden block resting on a frictionless horizontal surface and stops in it. Later the combined system moves. In this case (i) Only KE is conserved (ii) Only momentum is conserved (iii) Both KE and momentum are conserved |
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Answer» (i) is CORRECT |
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| 14686. |
Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis. Because the water-ice interface must have a temperature of 0^(@)C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions . It can lose energy and freeze only by sending energy up(through distance L) to the top of the icicle, where the temperature T_(r) can be below 0^(@)C. Take L=0.10m and T_(r)=-5^(@)C.Assume that the central tube and the upward conduction path both have crosse-sectional area A=0.5m^(2). The thermal conductivity of ice is 0.40 W//mcdotK, latent heat of fusion is L_(F)=4.0xx10^(5) J//K and the density of liquid water is 1000 kg//m^(3). The rate at which mass converted from liquid to ice at the top of the central tube is |
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Answer» `7.5XX10^(-3) kg//s` |
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| 14687. |
Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis. Because the water-ice interface must have a temperature of 0^(@)C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions . It can lose energy and freeze only by sending energy up(through distance L) to the top of the icicle, where the temperature T_(r) can be below 0^(@)C. Take L=0.10m and T_(r)=-5^(@)C.Assume that the central tube and the upward conduction path both have crosse-sectional area A=0.5m^(2). The thermal conductivity of ice is 0.40 W//mcdotK, latent heat of fusion is L_(F)=4.0xx10^(5) J//K and the density of liquid water is 1000 kg//m^(3). The rate at which the energy conducted upward is |
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Answer» 20 W |
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| 14688. |
Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis. Because the water-ice interface must have a temperature of 0^(@)C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions . It can lose energy and freeze only by sending energy up(through distance L) to the top of the icicle, where the temperature T_(r) can be below 0^(@)C. Take L=0.10m and T_(r)=-5^(@)C.Assume that the central tube and the upward conduction path both have crosse-sectional area A=0.5m^(2). The thermal conductivity of ice is 0.40 W//mcdotK, latent heat of fusion is L_(F)=4.0xx10^(5) J//K and the density of liquid water is 1000 kg//m^(3). At what rate does the top of the tube move downward because of water freezing there? |
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Answer» `5xx10^(-8) m//s` |
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| 14689. |
The value of acceleration due to gravity at earth's surface is 9.8 ms^(-2). The altitude above its surface at which the acceleration due to gravity decreases to , is close to: (radius of earth = 6.4 xx 10^(6) m) |
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Answer» `2.6 XX 10^(6)m` |
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| 14690. |
A homogeneous cylinder of mass M and radius R is pulled on a horizontal plane by horizontal force F acting through its mass centre. Assuming rolling without slipping what is the angular acceleration of the cylinder ? |
| Answer» SOLUTION :`ALPHA = (tau)/(I)=(F.R)/((3)/(2)MR^(2))=(2F)/(3MR)` | |
| 14691. |
If the linear density of rod of length 3 m varies as lambda=2+x then the distance of centre of gravity of the rod is…………. |
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Answer» `(7)/(3)m` `therefore dm=lambdadx` `therefore` Position of centre of mass `x_(CM)=(intdmxxx)/(intdm)=(UNDERSET(0)overset(3)int(lamdadx)x)/(underset(0)overset(3)intlambdadx)` `=([x^(2)+(x^(3))/(3)]_(0)^(3))/([2x+(x^(2))/(2)]_(0)^(3))=(36)/(21)=(12)/(7)m` |
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| 14692. |
For ideal gas, absolute zero temperature is 273.15^(@)C.true or false |
| Answer» SOLUTION :`-273.15^(@)C` | |
| 14693. |
Consider a thin uniform circular ring rolling down in an inclined plane without slipping. Compute the linear acceleration along the inclined plane if the angle of inclination is 45' |
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Answer» Solution : The linear ACCELERATION along the INCLINED plane can be computed by `a=(g sin theta)/(1+(K^(2))/(R^(2)))` For a thin UNIFORM circular ring, axis passing through its center is `I=MR^(2)` `:.K^(2)=R^(2) rArr (K^(2))/(R^(2))=1` And the angle of inclination `theta=45^(@) rArr sin 45^(@)=(1)/(sqrt(2))` Hence, `a=((g)/(sqrt(2)))/(1+1), a=(g)/(2 sqrt(2)) ms^(-2)` |
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| 14694. |
The velocity of a motor boat with respect to still water is 7ms^(-1) and velocity of water in a river is 3ms^(-1) . When the boat starts moving upstream a floatwas dropped from it . The boat travels 4.2 km and returns and reaches the float. The time taken by the goat to cath the float is |
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Answer» 25 min |
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| 14695. |
Two SHM are represented by the equations y_(1) = 10 sin(3pi t + pi//4) and y_(2) = 5[sin 3pi t + sqrt(3) cos 3pi t]. Their amplitudes are in the ratio |
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Answer» `1:2` |
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| 14696. |
(I) Two vecotrs which are perpendicular to each other are called Orthogonal vectors. (II) To find resultant of two of addition. Which statement is is correct? |
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Answer» I only |
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| 14697. |
An organ pipe of cross-sectional area 100 cm^(2) resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of waterto be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m s^(-1)) |
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Answer» `800 cm^(3)` `1000=(V)/(4l_(1))=(3v)/(4l_(2))` Using `v= 320" m "s^(-1)`, we get, `l_(1)`= 8 cmand `l_(2)` = 24 cm `therefore""` Minimum volume `=16xx100=1600" cm"^(3)`. |
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| 14698. |
On a temperature scale Y, water freezes at -160 ""^(@)Y and boils at -50 ""^(@)Y. On this Y scale, a temperature of 340K is |
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Answer» `-106.3 ""^(@)Y` |
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| 14699. |
(A) : Dimensionless quantities have no units. (R) : Unitless quantities have no dimensions. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 14700. |
The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency 2.2 Hz . The fundamental frequency of the closed organ pipe is 110 Hz, find the lengths of the pipes . Take velocity of sound= 330 m//s. |
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Answer» SOLUTION :Let `l_(1) and l_(2)` be lengths of open organ pipe and closed organ pipe respectively . FIRST overtone of open organ pipe ` = 2n_(1) = 2 xx (v)/( 2 l_(1)) = (v)/( l_(1))` First overtone of closed organ pipe ` = 3 n_(2) = 3 xx (v)/( 4 l_(2))` According to question , `(v)/(l_(1)) - ( 3v)/( 4 l_(2) = +- 2.2`(i) As `n_(2)` is the FUNDAMENTAL frequency of closed organ pipe `n_(2) = (v)/( 4 l_(2))` `l_(2) = (v)/( 4 n_(2)) = (330)/( 4 xx 110) = 0.75 m` From Eq.(i). `(v)/(l_(1)) = ( 3v)/( 4 l_(2)) +- 2.2` `(v)/(l_(1)) = 3 n_(2) +- 2.2` Taking positive SIGN `(v)/(l_(1)) = 3 xx 110 + 2.2 = 332.2` `:. l_(1) = (v)/( 332.2) = (330)/(332.2) m = 0.993 m = 99.3 cm` Taking negative sign `(v)/( l_(1)) = 3 n_(2) - 2.2 = 3 xx 110 - 2.2 = 327.8` `:. l_(1) = (v)/( 327.8) = (330)/(327.8) = 1.006 m = 100.6 cm` |
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