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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.
| 14451. |
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds. |
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Answer» SOLUTION :`a=omega^(2)x,a=v,A=3cm,x=2cm` `v=omegasqrt(A^(2)-x^(2))` `omega^(2)x=omegasqrt(A^(2)-x^(2))` `((2pi)/(T))2=sqrt(3^(2)-2^(2)),(4PI)/(T)=sqrt(5)` `T=(4pi)/(sqrt(5))impliesT=5.6sec` |
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| 14452. |
Read each statement below carefully and state with reasons and examples, if it is true or false, A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. |
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Answer» Solution :True, E.g. : If OBJECT is thrown UPWARDS, then speed will be zero at maximum height. But acceleration g = 9.8 `ms^(-2)` in downward direction. (b) False. Because speed is magnitude of velocity. Hence speed is zero then velocity is also zero. (c) True. When particle MOVES with constant speed in straight line, then velocity is constant. So the change in velocity is zero and hence acceleration is zero. (d) False. The speed of object thrown upwards decreases. But if the speed and acceleration are in same direction, then its speed increases. |
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| 14453. |
Two masses m_(1) and m_(2) are attached to a spring balance S as shown in Figure. If m_(1)gt m_(2) then the reading of spring balance will be |
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Answer» `(m_(1)-m_(2))` From FBD of `m_(2), T_(2)-m_(2)g=m_(2)a` TAKE `T_(2)=T_(2)`, solving the above eq.s, we get .a. |
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| 14454. |
Consider a string in a guitar whose length is 80 cm and a mass of 0.32 g with tension 80 N is plucked . Compute the first four lowest frequencies produced when it is plucked . |
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Answer» SOLUTION :The velocity of the wave , `v = SQRT((T)/(mu))` The length of the string , L = 80 cm = 0.8 m . The mass of the string , `m = 0.32` g = `0.32 xx 10^(-3)` KG Therefore , the linear mass DENSITY , `mu = (0.32 xx 10^(-3))/(0.8) = 0.4 xx 10^(-3) kg m^(-1)` The tension in the string , T = 80 N `v = sqrt((80)/(0.4 xx 10^(-3))) = 447.2 ms^(-1)` The wavelength corresponding to the fundamental frequency `f_(1)` is `lambda_(1) = 2 L = 2 xx 0.8 = 1.6` m The fundamental frequency `f_(1)` corresponding to the wavelength `lambda_(1)` `f_(1) = (v)/(lambda_(1)) = (447.2)/(1.6) = 279.5` Hz Similarly , the frequency corresponding to the second harmonics , third harmonics and fourthharmonics are `f_(2) = 2f_(1) = 559` Hz `f_(3) = 3f_(1) = 838.5`Hz `f_(4) = 4f_(1) = 1118` Hz |
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| 14455. |
In the system shown in the figure all surfaces are smooth, pulley and string are massless. The string between the two pulleys and between pulley and block of mass 5 m is parallel to the incline surface of the block of mass 4 m. The system is released from rest. Find the acceleration of the block of mass 4 m . ["tan" 37^(@) = (3)/(4)] |
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| 14456. |
A prism can produce a minimum deviation delta in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is |
| Answer» ANSWER :B | |
| 14457. |
The pressure of one mole of an ideal gas varies according to the law p = p_0 - alpha V^2where p_0 and alphaare positive constants. The highest temperature that the gas may attain is |
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Answer» `(2p_0)/(3R) ((P_0)/(3alpha))^(1//2)` |
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| 14458. |
Column I shows different charge distributions and short electric dipole at a distance x from the charge distributions. Column II gives the dependence of force acting on the dipole as of function of x. |
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Answer» <P> Charge thin INFINITE disc `F_("net")` on DIPOLE `=0` (Q) Electric filed due to uniformaly `=sigma_(x)/epsi_(0)` charged infinite `F_("net")` on dipole `=(rho ql)/epsi_(0)` (R) Electric field due to uniform `=(2kl)/(x)` infinite line of charge `F_("net")` on.dipole `=(-2kl)/(x(x+l)) ~~(-2k lambda l)/x^(2)` (S) Electric field due to uniformly `=(KQ)/x^(2)` charged sphere `F_("net")` on dipole `=(-KQq(2xl))/(x^(2)(x+l)^(2))~~(-KqQ(2xl))/x^(4)` `~~(-2KqQl)/x^(3)` |
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| 14459. |
When an spacecraft goes from earth to moon along a straight path, what changes occurs in weight? |
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Answer» Solution :(i) As an spacecraft goes away from the earth toward moon it weight decreases. (ii) At NEUTRAL point on the LINE joiningbetween earth and moon its weight become zero. (III) In further journey its weight increases andits weight on the moon will be `(MG)/6` |
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| 14460. |
A man walks on a straight road from his home to a market2.5 km away with a speed of5 km //h. Finding the market closed, he instantly turns and walks back with a speed of 7.5 km//h. What is the (a) magnitude of average velocity and (b) average speed of the man, over the interval of time (i)0 to30 min . (ii) 0 to 50 min (iii) 0 to 40 min ? |
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Answer» 0,0 Time taken by the boy to return back from the market to his home, `t_2= (2.5 km)/(7.5 km h^(-1))= 1/3` h `therefore` Total time taken = `t_1 + t_2 = 1/2 h + 1/3 h = 5/6 h = 50` min In t = 0 to 50 min. Total DISTANCE TRAVELLED = 2.5 km + 2.5 km = 5 km) Displacement = 0 (As the boy returns back home) `therefore` Average speed `= ("Distance travelled") /("Time taken")`=`(5 "km")/(5/6h) = 6 km h^(-1)` Average velocity `= ("Displacement")/("Time taken") = 0` |
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| 14461. |
A simple pendulum is of length 50cm. Find its time period and frequency of oscillation. (g = 9.8 m//s^(2)). |
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Answer» 2.419 SEC, 0.8045Hz |
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| 14462. |
Two particles A and B locat at (0,0) and (4,4) respectively start moving simu lataneou ly with velocities v_(A)=-4hati and V_(B)=-4hatj: |
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Answer» the shortest distance between them is `4sqrt(2)m` `AB=4sqrt(2)m` Because the distance b/w them increases from START. 
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| 14463. |
A spring balance shows wrong readings after using for a long time. Why ? |
| Answer» Solution :When a spring balance has been USED for a long time. It develops an elastic fatigue, the spring of such a balance take LONGER time to recover its original configuration and therefore it does not GIVE CORRECT measurement. | |
| 14464. |
The equation of state corresponding to 8g of O_2 is |
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Answer» PV=RT |
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| 14465. |
The kinetic energy of a body becomes four times its intial value , the new linear momentum will be ……….. |
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Answer» four times the INITIAL value `K = (p^(2))/(2m)` when m = mass of BODY `:. (p^(2))/K =2m ` constant ` :. (p_(1)^(2))/(K_(1))=(p_(2)^(2))/(K_(2))` But `K_(2) =4K_(1)` ` :. (p_(1)^(2))/(p_(2)^(2)) =(K_(1))/(4K_(2))` But ` ((p_(1))/(p_(2)))^(2) =1/4` ` :. (p_(1))/(p_(2))= 1/2 ` ` :. p_(2) = 2p_(1)` |
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| 14466. |
A layer of glycerine of thickness 1 mm is present between a large surface area and surface area of 0.1 m^2. With what force (in N) the small surface is to be pulled. So that it can move witha velocity of 1m. S^-1 ? (Given the coefficient of viscosity =0.07 kg. m^-1. s^-1.) |
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| 14467. |
Define simple harmonic motion and explain it. |
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Answer» Solution :Simple Harmonic Motion : The periodic motion about a fixed point on a linear path under the influence of the force acting towards the fixed point and PROPORTIONAL to displacement of the BODY from a fixed point is called a Simple Harmonic Motion. Or ..Simple harmonic motion is not any periodic motion but one in which displacement is a sinusoidal function of time... In a oscillatory motion restoring force is EXERTED towards mean position from any point hence in simple harmonic motion displacement of particle from the ORIGIN varies with time as `therefore x(t) = A cos (omega t+ phi)` where `A, omega" and "phi` are constants  In figure a particle oscillating back and forth about the origin of an X-axis between the limits +A and -A is SHOWN. |
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| 14468. |
If x = a t + bt^(2), where x is the distance travelled by the body in kilometer while t is the time in second, then the unit of b is |
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Answer» km/s |
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| 14469. |
A wire under a load suffers an elongation x. The load is then completely immersed in water. If the load is made of brass (sp. gravity 8), what will be the elongation? |
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| 14470. |
A body is just sliding dwon an inclinedplane due to its own weight what is the relation between angle of inclination and angle of repose |
| Answer» SOLUTION :Angleof INCLINATION=- Angleof REPOSE. | |
| 14471. |
Can an object have an eastward velocity while experiencing a westward acceleration ? |
| Answer» Solution :YES, A pendulum oscillating in east-wast direction will have eastward VELOCITY and westward acceleration in HALF CYCLE of its oscillation . | |
| 14472. |
Three vectors vecP,vecQ and vecR such that |vecQ|=Asqrt(2) and the angles between vecP and vecQ,vecQ and vecR,vecR and vecP are 90^(@),150^(@),120^(@) respectivlely. Find the value of |vecP|= |
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Answer» `(A)/SQRT(2)` `VECR` 
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| 14473. |
A spring balance has a scale that reads 0 to 20 kg. The length of the scale is 10 cm. A body suspended from this balance, when displaced and released, oscillates with period of (pis)/(10) What is the mass of the body |
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Answer» 4.9 KG |
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| 14474. |
A simple pendulum suspended from the ceiling of a stationary van, has time period T. If then van stats moving with a uniform velocity, the periodic of the pendulum will be……….. |
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Answer» less than T |
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| 14475. |
Two small rings, each of mass 'm', are connected to the block of same mass 'm' through an inextensible massless string of length 'l'. Rings are constrained to move over smooth rod AB. Initially, the system is held at rest as shown in Fig.Let a and v be the velocities of ring and block, respectively when string makes an angle 60^(@) with the vertical. |
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Answer» `u=SQRT((GL)/5)` From conservation of energy `mglcostheta=2(1/2"mu"^(2))+1/2mv^(2)` `u=sqrt((2glcostheta)/(2+tan^(2)THETA))=sqrt((gl)/5)` USING `theta=60^(@)` `v=sqrt((3gl)/5)` 
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| 14476. |
What are dimensions of E/B? |
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Answer» `[LT^(-1)]` `:. E/B=[m/s]=[L/T]=[LT^(-1)]` |
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| 14477. |
A gas is governed by an equationV=(aT^(3))/(P)where P,V and T are pressure, volume and temperature of the gas respectively and .a. is a constant. If the temperature of the gas is doubled at constant pressure then the workdone by the gas is |
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Answer» `6AT^(3)` |
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| 14478. |
When 'n' numbers of particles each of mass 'm' are at distances x_(1)=a, x_(2)=ar, x_(3)=ar^(2)….x_(n)=ar^(n) units from origin on the x-axis then find the distance of their centre of mass from origin. |
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Answer» SOLUTION :`x_(cm)=(ma+m(ar)+m(ar^(2))+…..+m(ar^(N)))/(m+m+m+…..+m(n"TERMS"))` `x_(cm)=(m(a+ar+ar^(2)+…+ar^(n)))/(MN)` If `R gt 1` then `x_(cm)=(1)/(n)[(a(r^(n)-1))/(r-1)]=(a(r^(n-1)))/(n(r-1))` If `r lt 1` then `x_(cm)=(1)/(n)[(a(1-r^(n)))/(1-r)]=(a(1-r^(n)))/(n(1-r))` |
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| 14479. |
A uniform capillary tube of length l and inner radius r with its end sealed is submerged vertically into water. The outside pressure is p_0 and surface tension of water is gamma. Whena length x of the capillary is submerged into water . It is found that water levels inside and outside the capillary coincide. The Value of x is |
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Answer» `(L)/((1+(p_0r)/(4 gamma)))` `p_0(lA)=p.(l-x)A` or, `p.=(p_0l)/(l-x)""…(1)` As the water levels inside and outside the capillary coincide, so `p.-p_0=(2gamma)/(r )""...(2)` Solving equation (1) and (2) we get, `x=(l)/(1+(p_0r)/(2gamma))` |
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| 14480. |
A bullet of mass 5 g moving with a velocity of 300 ms strikes a plank of wood of mass 1.995 kg suspended by a thread and gets embedded in it. Find the velocity of the plank with bullet immediately after collision and vertical height through which the combined mass rises. |
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| 14481. |
The work done in lifting a body of mass 20 kg and specific gravity 3.2 to a height of 8m in water is, (g=10 ms^(-2)) |
| Answer» Answer :D | |
| 14482. |
Relation between prepssure P and average kinetic energy E per unit volume of a gas is |
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Answer» <P>`P=(2/3)E` |
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| 14483. |
A particle is performing a linear simple harmonic motion. If the acceleration and the corresponding velocity of the particle are a and v respectively, which of the following graphs is correct? |
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| 14484. |
A small body of mass m moving with velocity v_(0) on rough horizontal surface, finally stops due to friction. Find, the mean power developed by the friction force during the motion of the body, if the frictional coefficient mu=0.27,m=1 kg and v_(0)=1.5 ms^(-1). |
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Answer» Solution :The RETARDATION due to friction is, `a=("force of friction")/("mass")=(mu mg)/(m)=-mug` Further,`"" v_(0)`=at Therefore, `""t=(v_(0))/(a)=(v_(0))/(mug)`………..(i) From work energy theorem, work done by force of friction =change in kinetic energy or `"" W=(1)/(2)mv_(0)^(2)`..............(ii) Mean POWER`=(W)/(t)` From Eqs. (i) and (ii), we get `P_(mean)=(1)/(2) mu mgv_(0)` Substituting the values, we have `P_(mean)=(1)/(2)xx0.27xx1.0xx9.8xx1.5~~2 W` |
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| 14485. |
Explain that you can't make a decent cup of tea up on a mountain ? |
| Answer» Solution :The air on mountainis thin . Hence ,PRESSURE is low . Boiling point of WATER decreases at low pressure of `100^(@)C`. As result , TEMPERATURE is not sufficient for boiling TEA , sowe can.t make a decent cupof tea. | |
| 14486. |
The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is (Take g = 10 ms^(-2)) |
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Answer» `40 ms^(-1)` ` thereforev_(max ) = sqrt(mu g r) = sqrt(0.25xx 10 xx 40) = 10ms^(-1)` |
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| 14487. |
A cylindrical rod of mass .m.length L and radius .R. has two cords wound around it whose ends are attached to a rigid support. The rod is held horozontally with the two cords vertical. When the rod is released, the cords unwind and rod rotates. Find the tension in the cords as they unwound and determine the linear acceleration of the cylinder as it falls. |
Answer» Solution :The rod ROTATES and translates while falling down.  `Mg-2T=Ma` (translational motion) `2TR=I ALPHA` (Rotational motion) Since `I=(MR^(2))/(2) and alpha=(a)/(R )` on solving the above equation we get `a=(2g)/(3) and T=(mg)/(6)` |
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| 14488. |
Abody starts moving unidirectionally under the influence of a sourceof constant power . Which one of thegraph correctly shows the variation of displacement (s) with time (t) ? |
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Answer»
=increase in KINETIC energy per second = COSTANT So, velocity of the body INCREASES with time. Therefore , slope `((ds)/(dt)=V)` of the s-t graph increases with time. |
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| 14489. |
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts ofheat involved in these steps are Q_1 = 5960 J, Q_2 = -5585 J, Q_3 = -2980 J and Q_4 = 3635 Jrespectively. The corresponding works involved are W_1 = 2200 J, W_2 = -825 J, W_3 = -1100 J, and W_4respectively What is the efficiency of the cycle? |
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Answer» 0.11 |
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| 14490. |
A thin rod AB of length a has variable mass per unit length (rho_(0)(a+x))/(a), where x is the distance measured from A and p_(0) is a constant. The rod is pivoted at A and is hanging vertically in equilibrium, when it is struck by a horizontal impulse of magnitude J at the point B. Show that if B passes through a point vertically above A, then J gt(sqrt(10xa^(3)rho_(0)^(2)g))/(6) where 'x' is |
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| 14491. |
N moles of an ideal diatomic gas is contained in a cylinder at temperature T. On supplying some .heat to cylinder, N/3 moles of gas dissociated into atoms while temperature remains constant. Heat supplied to the gas is |
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Answer» `(NRT)/(3)` |
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| 14492. |
Two black bodies at temperature 327^(@)C and 427^(@)C are kept in an evacuated chamber at 27^(@)C. The ratio fo their rates of loss of heat is |
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Answer» `((6)/(7))` `:. (E_(1))/(E_(2)) = (T_(1)^(4) - T_(S)^(4))/(T_(2)^(4) - T_(S)^(4)) = ((327 + 273)^(4)- (27+ 273)^(4))/((427 +273)^(4)- (27 + 273)^(4)) ` `=((600)^(4) - (300)^(4))/((700)^(4)-(300)^(4)) = (10^(8) (6^(4)-3^(4)))/(10^(8) (7^(4)-3^(4)))= (((6^(2))^(2) - (3^(2))^(2)))/(((7^(2))^(2)- (3^(2))^(2)))` `= ((6^(2) -3^(2)) (6^(2) + 3^(2)))/((7^(2) - 3^(2)) (7^(2) + 3^(2))) [ :. (a^(2)-b^(2)) = (a +b) (a-b)]` `=((36-9) (36+ 9))/((49-9) (49+ 9)) = ((27) (45))/((40) (58)) = (243)/(464)` |
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| 14493. |
The uncertainty contained in any measurement is……………. |
| Answer» Solution :error | |
| 14494. |
State the nature of the graph drawn between pressure and temperature at constant volume. When the graph is extrapolated, at what temperature it meet the temperature axis ? |
| Answer» Solution :A graph is a straight line, when a graph is extra polated it meets TEMPERATURE AXIS at `-273^(@)C`, this indicates that at constant volume. The pressure of the given MASS of gas REDUCES to zero at `-273^(@)C`. | |
| 14495. |
A circular coil of radius 5 picm having atotal of the magnetic meridian plane (i.e. thevertical plane in thenorth-south direction) . It is rotated about itsverticaldiameter by 45^(@)and a currentof sqrt(2A) is passed through it. A magnetic needle placed at thecentre of this coil points west toeast. The horizontalcopmonent of the earth'smagnetic field is |
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Answer» `20 xx 10^(-5)T` |
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| 14496. |
Gravitational force is ……………and ………………is a…………….. . |
| Answer» SOLUTION :CONSERVATIVE FORCE `, `FRICTIONAL force `, `non-conservative force | |
| 14497. |
P-V diagram of an ideal gas for a process ABC is as shown in the figure. Change in internal energy of the gas during the process ABC |
| Answer» SOLUTION :`DELTA U_(ABC ) =0` | |
| 14498. |
A tennis ballis dropped on to the floor from a height of 4m.It rebounds to a height of 2m.If the ball is in contact with the floor for 12xx10^(-3)s, its average acceleration during |
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Answer» 0 As u=0 , HENCE `V^(2)= 2as` or `V = sqrt2as` When the tennis ball is DROPPED the velocity gained by it just before colliding with the floor is : `u= sqrt2xx9.8xx4 = 8.85ms^(-1)` (upwords) `=- 8.85ms^(-1)` (upwords) time of contact of the ball with the floor (GIVEN `t = 12xx10^(-3)S` Hence , average acceleration `a= (v-u)/(t) = (6.26-(-8.85)/(12xx10^(-3)` `=+1260ms^(-2)` |
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| 14499. |
A spring is cut into two pieces in such a way that one piece is double the length of the other. If the force constant of main spring is k then force constant of the longer part is |
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Answer» `2/3`K |
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| 14500. |
A particle of mass 10 g is placed in a potential field given by v=(50x^(2)+100)erg/g. Calculate the frequency of oscillation. |
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