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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.
| 12401. |
The coefficient of linear expansion alpha of the material of a rod of length 1.5m varies with absolute temperature T as alpha= a T- b T^(2) where a = 2.4 xx 10^(-8) //K^(2) and b= 4.2 xx 10^(-12) //K^(3), Determine the total linear expansion of the rod when heated from room temperature (27^(@) C) to ( 127^(@) C) |
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Answer» 2.206 mm |
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| 12402. |
What will be the change in phase of wave due to reflection from rigid support? |
| Answer» SOLUTION :Its PHASE INCREASES by `PI` RAD. | |
| 12403. |
A carnot engine works between 200^(@)C and 0^(@)C first and then between 0^(@)C and -200^(@)C. Compare the values of efficiencies in the two cases. |
| Answer» Solution :`ETA _(1) = ( 473- 273) 473 = 200 // 473 ) , eta _(2)= (273 - 73) //273 = 200 // 273 , eta _(1) // eta _(2) = 0.577` | |
| 12404. |
Three particles of masses 1g, 2gand 3g are at distances of 1cm, 2cm and 3cm from the axis of rotation. Findi) the moment of inertia of the system and ii) the radius of gyration of the system. |
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Answer» Solution :(i) The MI of the system , `I=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)+m_(3)r_(3)^(2)` `=1xx1^(2)+2XX2^(2)+3xx3^(2)=1+8+27=36gcm^(2) ` (ii) The radius of GYRATION `k=SQRT((I)/("TOTAL mass"))=sqrt((36)/(1+2+3))=sqrt(6)=2.449cm` |
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| 12405. |
A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n_(1) . Then (if the sound velocity in air is 300 m/s)........ . |
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Answer» `n_(1) = 10`n |
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| 12406. |
Solidcopper cubeof edge1 cmis suspended in an evacuatedenclosure . Itstemperatureis foundto fall from100^(@)Cto 99^(@)Cin 100 seconds . Anothersolid copper cubeof edge2cm with similar surface natureis suspendedin similar manner . Findthe timerequiredfor this cubeto cool from 100^(@)C to 99^(@)C |
| Answer» SOLUTION :200 SEC | |
| 12407. |
Ball A of mass m moving with velocity V collides head on with a stationary ball B of mass m. If e be the coefficient of restitution, then which of the following statements are correct ? |
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Answer» The ratio of VELOCITIES of ball A and B after the collision is `((1+e)/(1-e))` |
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| 12408. |
A man is spinning in the gravity free-space changes the shape of the body by spreading his arms. By doing this he can change his (a) moment of inertia (b) angular momentum (c) angular velocity (d) rotational kinetic energy Which one of the following are correct? |
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Answer» a,B and c |
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| 12409. |
A body constrained to move in y directions is sunjected to a force F=(-2hati+15hatj+6hatk)N. What is the work doen by the force in moving through a distance of 10 m along the y-axis? |
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Answer» 20J |
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| 12410. |
It is known that density rho of air decreases with heighty as -y//y_(0)rho=rho_(o)e where rho_(o)=1.25kgm^(-3) is the density at sea level and y_(o) is a constant . Thisdensity variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions).Also assume that the value of g remains constant. (b) A large He balloon of volume 1425m^(3) is used to lift apayload of 400 kg . Assume that the balloon maintains constant radius as it rises . How high does it rise ? (y_(o)=8000mandrho_(He)=0.18kgm^(-3)). |
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Answer» Solution :The rate of decrease of DENSITY `rho` of air is directly proportional to the height y. It is given as , `-(drho)/(dy)proprhoor-(drho)/(dy)=-krho` where K= CONSTANT and negative sign indicates that density decreases as height increases. From height 0 to y density becomes `rho_(o)` to `rho` . By integrating above equation, `int_(rho_(o))^(rho)(1)/(rho)=-kint_(o)^(y)dy` `[Inrho]_(rho_(o))^(rho)=-k[y]_(o)^(y)` `[In rho-Inrho_(o)]=-k[y=0]` ln `((rho)/(rho_(o)))=-ky` `log_(e)((rho)/(rho_(o)))=-ky` `therefore(rho)/(rho_(o))=e^(-ky)` `thereforerho=rho_(o)^(e-ky)` TAKING constant `k=(1)/(y_(o))` `rho=rho_(o)e^((y)/(y_(0)))` The balloon will rise upto a height where density of air equal to the density of balloon Volume of ballon `V=1425m^(3)` MASS of He has in balloon, `V=1425xx0.18=256.5kg` Total mass of balloon (with payload), `M=400+256.5` `=656.5kg` Density of balloon `rho=(M)/(V)=(656.5)/(1425)` `rho=0.46kg//m^(3)` `y_(o)=8000m` `rho_(o)=1.25kg//m^(3)` `rho=0.46kg//m^(3)` `rho=rho_(o)e^((y)/(y_(o)))` `therefore0.46=1.25_(e)^((-y)/(8000))` `thereforee^((y)/(8000))=(1.25)/(0.46)=2.7` `thereforey=8000loge^(2.7)` `=8000xx1.025` `=8200m` `=8.2km` |
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| 12411. |
The energy stored in an electric device known as capacitor is given by U=(q^(2))/(2C) where U= energy stored in capacitor C= capacity ofcapacitor q= charge on capacitor Find the dimensions of capacity of the capacitor |
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Answer» `[A^(2)M^(-1)L^(-3)T^(4)]` :.C=(q^(2))/(2U) :.[C]=([AT]^(2))/([ML^(2)T^(-2)])=[A^(2)M^(-1)L^(-2)T^(4)]` |
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| 12412. |
A very flexible chain of length L and mass M is vertically suspended with its lower end just touching the table. If it is released so that each link strikes the table and comes to rest, what force the chain will exert on table at the moment 'y' part of length falls on table ? |
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Answer» Solution :Since chain is uniform, the MASS of y part of chain will be `((M)/(L)y)`. When this part reaches the table, its total force EXERTED MUST be EQUAL to weight of y part RESTING on table + Force due to momentum imparted `=(M)/(L)yg+(((M)/(L)dy)sqrt(2gy))/(dt)=(Mg)/(L)y+(M)/(L)v.sqrt(2gy)` `("as"(dy)/(dt)=v)=(Mg)/(L)y +(M)/(L)sqrt(2gy).sqrt(2gy)=3(My)/(L)g`. |
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| 12413. |
A uniform circular disc of radius R is rolling on a horizontal surface. If v be velocity of its CM and omega be its angular velocity about CM such that v gt R omega, then which of the following are correct ? |
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Answer» No point on the disc will have velocity less than `v-R omega` |
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| 12414. |
(A) : A block of mass .m. rests on a wedge of mass M, which, in turn, is at rest on a horizontal table. All the surfaces are friction less. The block starts from rest. The position of c.m of the system will change only in vertical direction.(R ) : A block of mass .m. rests on a wedge of mass M, which, in turn, is at rest on a horizontal table. All the surfaces are friction less. The system is free from external force along horizontal direction. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 12415. |
The refractive index of a material of a prism of angles 45^@-45^@-90^@ is 1.5. The path of the ray of light incident normally on the hypotenuse side is shown in |
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Answer»
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| 12416. |
The equation of simple harmonic motion y = a sin(2pit+alpha) then its phase at time t = 0s is |
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Answer» `2pin t` |
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| 12417. |
Angle (in rad) made by the vector sqrt(3)hat(i)+hat(j) with the x- axis is |
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Answer» `pi//6` |
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| 12418. |
what percentage of length of a wire will increase by applying the stress of1kg.weight/mm^(2) on it ?(Y = 10^11 N/m^2), 1 Kg Wt =9.8 N |
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Answer» `0.0078%` |
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| 12419. |
In an experiment using post-office box, the resistance of a wire is found to be (64 pm 1) ohm. The length and radius of the wire are measured to be (156.0 pm 0.1) cm and (0.26 pm 0.001) cm. Calculate specific resistance (or resistivity) of the material of the given wire, within the limits of percentage error. |
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Answer» Solution :Specific resistance, `rho=(pir^(2)R)/(L)=8.7xx10^(-4)` ohm metre `(DELTA rho)/rhoxx100=[2((Deltar)/r)+(DeltaR)/R+(DELTAL)/l]xx100=2%` |
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| 12420. |
There is a fixed half cylinder of radius R on a horizontal table. A uniform rod of length 2Rleans against it as shown. At the instant shown,theta = 30^(@) and the right end of the rod is sliding with velocity v. (a) Calculate the angular speed of the rod at this instant. (b) Find the vertical component of the velocity of the centre of the rod at this instant. |
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Answer» (B) `v_(y) = (v)/(4) (DARR)` |
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| 12421. |
Suppose the Sun expands so that its radius becomes 100 times its presents radius and its surface temperature becomes half to its present value. The total energy emitted by it then will increase by a factor of. . . . .. . |
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Answer» Solution :From Stefan-Boltzmanns law, we have `E=esigmaAT^(4)` `:.EpropAT^(4)""(e sigma" = CONSTANT")` `:.E prop(4piR^(2))T^(4)""(becauseA=4piR^(2))` `:.E propR^(2)T^(4)""(4pi" is constant")` `(E_(2))/(E_(1))=((R_(2))/(R_(1)))^(2)((T_(2))/(T_(1)))^(4)` `=((1000R_(1))/(R_(1)))^(2)((T_(1))/(2T_(1)))^(4)` `=(100xx100)/(16)=625` `:.E_(2)=625E_(1)` |
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| 12422. |
The system shown in figure is in equilibrium. Pulley, springs and the strings are massless. The three blocks A, B and C have equal masses. x_(1) and x_(2) are extensions in the spring 1 and spring 2 respectively. (a) Find the value of |(d^(2)x^(2))/(dt^(2))| immediately after spring 1 is cut . (b)Find the value of |(d^(2)x^(1))/(dt^(2))| and |(d^(2)x^(2))/(dt^(2))| immediately after string AB is cut. (c)Find the value of |(d^(2)x^(1))/(dt^(2))| and |(d^(2)x^(2))/(dt^(2))| immediately after spring 2 is cut . |
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Answer» `|(d^(2)x_(1))/(dt^(2))|= 2g ;|(d^(2)x_(2))/(dt^(2))| = 2g` (C)`|(d^(2)x_(1))/(dt^(2))| = (G)/(2) ;|(d^(2)x_(2))/(dt^(2))| = (3g)/(2)` |
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| 12423. |
A pendulum is hung in a very high building oscillates to and fro motion freely like a simple harmonic oscillator . If the acceleration of the harmonic oscillator. Itthe acceleration of the bob is 16ms^(-2) at distance of 4 from the mean position , the the time period is |
| Answer» ANSWER :D | |
| 12424. |
When light of wavelength lamda_0 in vaccum travels through same thickness t in glass and water, the difference in the number of waves is _____. (Refractive indicates of glass and water are mu_g and mu_w respectively). |
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Answer» Solution :We KNOW NUMBER of waves of a given light in a MEDIUM of REFRACTIVE index `mu` is `(t mu)/lamda_0` `therefore` Difference in number of waves `=t/ mu_0 (m_g-mu_w)` where `mu_g and mu_w` are the refractive indicies of glass and WATER respectively. |
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| 12425. |
An insect trapped ina circular groove of radius 12cm moves along he groove steadily and completes 7 revolutions in 100s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a contant vector? What is its magnitude? |
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Answer» Solution :This is an example of uniform circular MOTION. Here R = 12 CM . The angular speed `OMEGA` is GIVEN by `omega = 2pi//T = 2pixx7//100=0.44rad//s` |
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| 12426. |
State law of floatation. |
| Answer» Solution :The law of floatation STATES that a body when float in a liquid if the WEIGHT of the liquid displaced by the immersed part of the body EQUALS the weight of the body. | |
| 12427. |
A boy weighing 42 kg eats bananas whose energy is 980 calories. If this energy is used to go to height h fing the value of h. |
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Answer» SOLUTION :Energy gained by the boy in EATING bananas = `980 "calorie" = 980 xx 4.2 J` . If m is the mass of the boy, the potential energy gained by the boy in going up through a height h is mgh. In SI, energy = potential energy `980 xx4.2 = "mgh" , 980 xx 4.2 = 42 xx9.8 xx h` ` h= (980xx4.2)/(42xx9.8) = 10m` |
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| 12428. |
Match list I with list II{:(,"List - I",,"List - II"),((a),"Ratio of angular velocities of hours hand of a clock and self rotation of the earth",(e ),12:1),((b),"Ratio of angular velocities of seconds hand to minutes hand of a clock",(f),60:1),((c ),"Ratio of angular velocities of seconds hand to hours hand of a clock",(g),2:1),((d),"Ratio of angular velocities of minutes hand to hours hand of a clock",(h),720 : 1):} |
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Answer» a-g, b-f, C-e, d-h |
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| 12429. |
State whether elasticity of a metallic substance increases or decreases with the rise in temperature. |
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Answer» |
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| 12430. |
Show that the oscillations due to a spring are simple harmonic oscillations and obtain the expression of periodic time. |
Answer» Solution :According to figure a block of mass m fixed to a spring, which in turn is fixed to a rigid wall.  The block is placed on a friction less horizontal surface. If the block is pulled on one side and is releasedit then executs a to and fro motion about a mean POSITION. Let x=0, indicate the position of the centre of the block when the spring is in equilibrium. the positions -A and +A indicate the maximum displacement to the left and the right of the mean position. For spring Robert Hooke LAW, ..Spring when deformed, is subject to a restoring force, the magnitude of which is proportional to the deformation or the displacement and acts is opposite direction... Let any time t, if the displacement of the block form its mean position is x, the restoring force F acting on the block it `F(x) = -kx"""............"(1)` Where k is constant of proportionality and is called the spring constant or spring force constant. Equation (1) is same as the force law for SHM and therefore the system executes a simple hormonic motion. But `F(x) = ma(x)` `therefore ma(x) = kx` `therefore m (-omega^(2)x)= -kx ""[therefore a(x)= -omega^(2) x]` `therefore omega = sqrt((k)/(m))` `therefore (2pi)/(T)= sqrt((k)/(m))` `therefore T = 2pi sqrt((k)/(m))` is the period of block ATTACHED with spring having small oscillations `therefore T propto (1)/(sqrt(k))` (for a given block) and `T propto sqrt(m)` (for a given spring) and frequency `v= (1)/(T) = (1)/(2pi) sqrt((m)/(k))` Stiff spring have high value of k and period is small and frequency of OSCILLATION is large. So the oscillation becomes rapid and for soft spring k is small and HENCE period is large and frequency of oscillation is small so oscillation becomes slowly. |
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| 12431. |
The flow rate of water from a tap of diameter 1.25 cm is 0.481L/min. The coefficient of viscosity of water is 10^(-3)Pas (b) After sometimes the flow rate is increased to 3L/min. Characterise the flow for both the flow rates. |
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Answer» Solution :Let the speed of the flow be v and the diameter of thet tap be d=1.25 cm. the VOLUME of the WATER flowing out per second is `Q=vxxpid^(2)//4""v=4Q//d^(2)pi` We then estimate the REYNOLDS number to be `R_(e)=4rhoQ//pi d eta` `=4xx10^(3)kgm^(-3)xxQ//(3.14xx1.25xx10^(2)mxx10^(-3)Pas)` `=1.019xx10^(8)m^(-3)SQ` Since initially (a) `Q=0.48L//"MIN"=8cm^(3)//s=8xx10^(-6)xx10^(-6)m^(3)s^(-1)`, we obtain `R_(e)=815` Since this is below 1000, the flowis steady After some time (b) when `Q=3L//"min"=50cm^(3)//s=5xx10^(-5)m^(3)s^(-1)` we obtain`R_(e)=5095` The flow will be turbulent |
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| 12432. |
Is the work done by an external agency against an opposing force positive or negative ? |
| Answer» SOLUTION :POSITIVE | |
| 12433. |
Several games such as billiards, marbles or carrom involve collisions. When to objects collide, after collision they could move together, the collision is ……… (elastic, completelly elastic, inelastic, completely inelastic) |
| Answer» SOLUTION :COMPLETELY INELASTIC | |
| 12434. |
The diameter of the vertical tubes of a U-tube are 10 mm and 4mm. The tubes are filled with water. What wil be the difference of heights of water coloumns in the tubes? (angle of contact theta=0^(@)) |
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Answer» Solution :Given `d_(1)=10mm, r_(1)=5MM, d_(2)=4mm, r_(2)=2mm` `Deltah=h_(2)-h_(1)` `=(2T cos THETA)/(rhog) [(1)/(r_(2))-(1)/(r_(1))]=(2 xx 0.072 xx 1)/(10^(3) xx 9.8 xx 10^(-3)) [1/2-1/5]=0.01469 xx ((5-2)/(10))` `Deltah=4.4 xx 10^(-3)m` This difference of HEIGHTS of WATER columns in the tubes. |
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| 12435. |
A pilot of mass 70 kg in a jet aircraft moves in a vertical circle while executing the loop-the-loop at a constant speed of 200 m/s. If the radius of the circle is 2.5 km what is the force exerted by the seat on the pilot at (a) the top of the loop and (b) the bottom of the loop. |
Answer» Solution : Speed = v=200 m/s Radius = `r= 2.5 KM = 2.5 xx 10^(3)` m At the top [The pilot will be UPSIDE down in his seat] Centripetal FORCE = `(mv^(2))/r = N_(1) + mg` `N_(1) = (mv^(2))/r - mg` `N_(1) = mg(v^(2)/(rg)-1) = 70 xx 9.8 ((200 xx 200)/(2.5 xx 10^(3) xx 9.8)-1)` `=70 xx 9.8 xx 0.63` =432.18 N At the bottom Centripetal force = `(mv^(2))/r = N_(2)-mg` `N_(2) = (mv^(2))/r + mg = mg (v^(2)/(rg) +1)` `=70 xx 9.8(1.632 +1)` `=1805.6` N |
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| 12436. |
A ship of mass 3 xx 10^7 kg initially at rest is pulled by a force of 5 xx 10^4 N through a distance of 3 m. Assuming that the resistance due to water is negligible, what will be the speed of the ship? |
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Answer» `0.1ms^(-1)` |
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| 12437. |
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11kms^(-1) , the escape velocity from the surface of the planet would be ....... |
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Answer» `1.1 KMS^(-1)` Escape speed on planet `v_p = sqrt((2GM_p)/R_p)` but `M_p = 10 M_e and R_p = R_e/10` `:. v_p =sqrt((2G(10M_e))/(R_e/10))= sqrt((100xx2GM_e)/R_e)` `:. v_p = 10 XX v_e = 10 xx 11 = 110 kms^(-2)` |
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| 12438. |
On a two-lane road, car A is travelling with a speed of 36 km h^(-1). Two cars B and C approach car A in opposite directions with a speed of 54 km h^(-1) each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ? |
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Answer» `1m//s^2` |
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| 12439. |
If the distance between the centre of gravity and point of suspension of a compound pendulum is l and the radius of gyration about the axis passing through its centre of gravity is k, its time period will be infinite if |
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Answer» `l = 0` |
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| 12440. |
To avoid random errors, what should wo do? |
| Answer» SOLUTION :Random ERRORS can be AVOIDED by taking the MEASUREMENTS a number of TIMES and then finding the arithmetic mean. | |
| 12441. |
Displacement of a body is (5hati + 3hatj - 4hatk) m when a force (6hati + 6hatj + 4hatk) N acts for 5 sec. The power in watt is |
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Answer» 16 |
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| 12442. |
A car moves from X to Y with a uniform speed V_(u) and returns to Y with a uniform speed V_(d) The average speed for this round trip is |
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Answer» `SQRT(v_(u)v_(d))` |
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| 12443. |
Figure shows top view of an airplane blown off course by wind in various directions. Assume the magnitude of the velocity of the airplane relative to the wind and the magnitude of the velocity of the wind to be the same each case. vecv_(A//w)=velocity of the airplane relative to the wind, vecv_(w//g)= velocity of the wind in ground frame |
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Answer» Air plane travels fastest ACROSS the ground in case d Net velocity of the airplane is the resultant of two given velocities. The resultant is maximum in case d and minimum in case c. Lateral velocity (velocity perpendicular to `vecv_(A//w)` is maximum in case a. hence lateral displacement is maximum in case a. In each case `vecv_(A//w)` is towards north, so `vecv_(w//A)` will be towards south. |
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| 12444. |
A wreckage of mass m present in space collides with a satellite of mass 10m and orbital radius R as shown. As a result of collision, the wreckage sticks to the satellite and the satellite is transferred to an orbit whose minimum distance from the planet is R/2. If mass of planet is M, with what velocity the wreckage collided the satellite? |
| Answer» SOLUTION :`SQRT((58GM)/( R))` | |
| 12445. |
The speed with which a bullet can be fired is 150cm^(-1). Calculate the greatest distance to which it can be projected and also the maximum height to which it would rise. |
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Answer» Solution :The GREATEST horizontal range is achieved at an angle of projectionn of `45^(@)` COMPONENTS of INITIAL velocity `=150cos45^(@)=106.06ms^(-1)` Now, if T is the time of FLIGHT, then considering the vertical motionof the bullet, `u=106.06 ms^(-1),a=-9.8ms^(-1),s=0,t=T` using `s=ut+1/2at^(2)` we get `0.106.06T-1/2 9.8T^(2)impliesT=(106.06xx2)/9.8=21.64` sec `:.` Maximum Horizontal range = horizontal component of velocity `xx` total time of flight. `=106.06xx21.64=2295.14m` Again if `H_("max")` be the maximum height of when the bullet rises, then `u=106.06ms^(-1),a=-9.8ms^(-1)` `v=0,s=H_("max")=(u^(2))/(2)impliesH_("max")=((106.06)^(2))/19.6=573.91m` |
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| 12446. |
A pendulum has time period T in air. When it is made to oscillate in water, it acquireda time period T= sqrt(2)T. The density of the pendulum bob is equal to (density of water = 1) |
| Answer» ANSWER :B | |
| 12447. |
A uniform solid brass sphere of radius b_(0) and mass m is set spinning with angular speed omega_(0) about a diameter. If its temperature is increased by Deltatheta, what will be its new angular speed? |
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Answer» |
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| 12448. |
How angle of contact depend on temperature ? |
| Answer» SOLUTION :As TEMPERATURE INCREASES , ANGLE of CONTACT increases. | |
| 12449. |
Find the number of significant figures in the following numbers. 6729""4""(ii)0.024""2 (iii)6.0023""5""(iv)2.520xx10^(7)""4 (v)0.08240""4""(v)4200""2 (vii)4.57xx10^(8)""3""(viii)91.00""5 |
| Answer» SOLUTION :SIGNIFICANT FIGURES | |
| 12450. |
Dimensional formula for Intensity of Magnetization is |
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Answer» IL |
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