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InterviewSolution
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.
| 1651. |
A train starts from rest and accelerates uniformly at 100 m "minute"^(-2) for 10 minutes. Find the velocity acquired by the train. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph and use it to find : (i) the retardation in the last 5 minutes, (ii) total distance travelled, and (iii) the average velocity of the train. |
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Answer» Solution :INITIAL velocity = 0, time interval = 10 minute, acceleration = 100 m `"minute"^(-2)`. Acceleration = `("Final velocity - Initial velocity ")/("Time interval ") ` `= ("Final velocity"-0)/("Time interval") ` or Final velocity = acceleration `xx` time interval = 100 m `"minute"^(-2) xx 10 ` minute = 1000 m `"minute"^(-1)` `:.` The final velocity acquired = 1000 m `"minute"^(-1)` The velocity - time graph is shown in Fig 2.28.  (i) Retardation in the last 5 minutes = - slope of the line BC. ` = - (BE)/(EC) = - ((0-1000)"m minute"^(-1))/((35-30)"minute" )` `= - (-1000 "m minute"^(-1))/(5 "minute") = 200m"minute"^(-2)` (ii) Total distance travelled = Area of trapezium OABC `= (1)/(2) (OC + AB) xx AD` `= (1)/(2) (35 +20) ` minute `xx1000` m `"minute"^(-1)` `= 55 xx500` m = 27500m ( or 27.5 km ) . (III) Average velocity = `("Total distance travelled ") /("Total time of travel")` `= (27500m)/(35 "minute ")=785.7 "m minute"^(-1)` . |
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| 1652. |
How will you test whether a given rod is made of iron or copper? |
| Answer» Solution :Iron ROD gets magnetised when placed near a BAR MAGNET by magnetic induction, while COPPER rod does not get magnetised | |
| 1653. |
A sprinter completes a round on a circular path of circumference 400 m, what is his displacement ? |
| Answer» Solution :As the FINAL position and INITIAL position of the SPRINTER is the same, his displacemt would be zero. | |
| 1654. |
The mass of a cyclist together with the bicycle is 90 kg. Calculate the work done by cyclist if the speed increases from6km/h to 12 km/h. |
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Answer» Solution :Mass of cyclist together with bike, m = 90 kg. Initial velocity, U = 6km/h = `6x(5//18)` ` = 5//3` m/s Final velocity, `v = 12km//h = 12xx(5//18)` ` = 10//3` m./s Initial kinetic energy `K.E._((i)) = (1)/(2) mu^(2)` ` = (1)/(2)(90)(5//3)^(2)` ` = (1)/(2) (90) (5//3) (5//3)` = 125 J Final kinetic energy `K.E._((f)) = (1)/(2) mv^(2)` `= (1)/(2) (90) (10//3)^(2)` ` = (1)/(2) (90) (10//3) (10// 3)` ` = 500J` THe workdone by the cyclist = CHANGE in kinetic energy = `K.E_((f)) - K.E_((i))` ` = 500J -125 J = 375J` |
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| 1655. |
A boy gets into a floating boat ? (a ) What happens to the boat ? (b) What happens to the weight of water displaced ? What happens to the buyount force on the boat ? |
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| 1656. |
The speed-time graph of a car is shown in figure given below Speed-time graph] (a) Find how far does the car travel in the first 4 second. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car ? |
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Answer» Solution :(a) On the time axis 5 DIVISIONS = 2s On the speed axis 3 divisions `= 2 ms ^(-1) therefore 5 xx 3 = 15 ` small squares `= 2 xx 2 ms ^(-1) = 4m` Total area under the speed-time graph for the FIRST 4 seconds = 60 small squares `therefore` Distance covered by the car in first 4 seconds `= (60 xx 4)/(15) = 16 m.` (B) After 6 second the graph REPRESENTS UNIFORM motion of the car. |
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| 1657. |
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^(-1) strikes a stationary wooden block and comes to rest in 09.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. |
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Answer» Solution :Initial velocity u=150 m/s Find velocity v=o (since the bullet finally comes to rest) Time taken to come to rest t=0.03 s . According to the first equation of motion V=u+at Acceleration of the bullet ,a `O=150+(axx0.03s)a=-150//0.03 =-5000m//s^2` (Negative sign indicates that the velocity of the bullet is decreasing) According to the third equatino of motion `v^2=u^2+2as` `0=(150)^2+2(-5000)` =22500/10000 =2.25 m Hence the DISTANCE of penetration of the bullet into the block is 2.25 m. From Newton.s second LAW of motion. Force=Massx Acceleration Mass of the bullet m=10 G =0.01 KG Acceleration of the bullet , `a=5000 m//s^2` F =ma =0.1 x 5000 =50N Hence the magnitude of force exerted by the wooden BLCOK on the bullet is 50N . |
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| 1659. |
A constant force of 1 N is applied on an object. The body displaces 1 m in the direction of the force. Then work done is ……………. J |
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| 1660. |
Figure X shows a trace of a sound wave produced by a particular tuning fork: (a) On the graph paper given in Figure Y, draw a trace of the sound wave which has a higher frequency than that shown in Figure X. (b) On the graph paper shown in Figure Z, draw a trace of the sound waves which has a large amplitude than that shown in Figure X. |
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| 1661. |
An object weigh less in water 500 grams in air .This object is then fully immersed in water .State whether this will float or sink in water of more in water of more in water .Give reason for your answer. |
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| 1662. |
Motion is a relative concept. |
| Answer» Solution :The description of motion of an object DEPENDS on the PLACE from which the object is OBSERVED. If the place of the observer is changed, the state of motion of the object is found to be changed. So it can be said clearly that motion is a RELATIVE CONCEPT. | |
| 1663. |
S.P. Balasubrahmanyam is conducting a musical night in an open auditorium in New York . Taking into account , two persons, one who is sitting in the auditorium at a distance of 1 km from the stage and the other who is watching the live program on a television set sitting in front on it Hyderabad , who will hear him first ? Explain . |
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Answer» Solution :(i) Difference between velocities of mechanical and electromagnetic waves. (II) Is it the mechanical or electromagnetic form of sound that is RECEIVED by the observer (a) in the AUDITORIUM ? (B) watching the live PROGRAM ? Which of electromagnetic and mechanical waves have larger velocity ? (iii) Person in Hyderabad |
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| 1664. |
What does the slope of a displacement-time graph represent? |
| Answer» SOLUTION :ACCELERATION | |
| 1665. |
Which of the following measures a small length to a high accuracy: metre rule, vernier callipers, screw gauge ? |
| Answer» SOLUTION :SCREW GAUGE | |
| 1666. |
A block of wood of weight 200 g floats on the surface of water. If the volume of block is 300 cm^(3)calculate the upthrust due to water. |
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Answer» SOLUTION :Upthrust of floating OBJECT = WEIGHT of the water DISPLACED Weight = mg ` = 0.200 KG xx 10m//s^(2)` ` 2 N ` |
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| 1667. |
PendulumA isx cmshorterthan asecondspendulum and pendulum B is x cm longerthan the secondspendulum . Theratio oftheirtimeperiods is 3: 4 . Thelength of the pendulum with higherfrequency is |
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Answer» 86 cm work`xx` volume `xx` speed `=(N m) xx (m^(3)) xx m s^(-1)` `=N m^(5) s^(-1)` `("work "xx" VELOCIT ")/("volume ")= ((N m ) xx (m s^(-1))/(m^(3)) =NM^(-1)s^(-1)` `(" work" xx "velocity")/(" rateof change of volume") =((Nm)xx(ms^(-1)))/(m^(3)) =N m^(-1)s^(-1)` `("work")/("time") xx (" volume")/("velocity") =(Nm)/(s^(-1)) xx (m^(3))/(ms^(-1)) =N m^(3) s^(2)` |
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| 1668. |
An object travels 25 m in 4s on a linear path and then another 50 m in 6 s. The average speed of this object is .........ms^(-1). |
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Answer» Solution :Total distance `=25 m+ 50 m = 75 m ` Total time ` = 4S + 6 s = 10 s ` Average speed `= ("Total distance")/("Total time") = (75m)/(10s) = 7.5 ms ^(-1)` |
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| 1669. |
A ball is projected vertically up with a speed of 50 m/s. Find the maximum height , the time to reach the maximum height, and the speed at the maximum height (g=10 m//s^(2)) (AS_(1)) |
| Answer» SOLUTION :`125 m, 5S,` ZERO | |
| 1670. |
When a carpet is beaten with astick, dust comes out of it. Explain. |
| Answer» Solution :When the carpet is beaten, it is suddenly SET into motion . The DUST particles TEND to remainat rest due to to inertia of rest, THEREFORE the dust COMES out of it. | |
| 1671. |
While on a gaint wheel ride, a person experiencesloss of weight on the descent. Why? |
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| 1673. |
A mat of mass 1kg and length 1m is placed on the floor. One end of the mat is pulled with a constant speed of 1m/s towards the other end till the other end comes in to motion (till the mat is reverse). How much force is required to do this ? |
Answer» Solution : As shown in figure, a mat is being pulled with a coonstant speed of `v = 1m//s,` so that the mass of the part of the mat is continuously increasing. Hence here the mass is a variable. The time required for bringing the entire mat in motion is given by `Delta t =("DISTANCE covered by the end")/("speed")` `= (2m)/(1m//s) =2s` (Distance covered by the end =1m + 1m =2m) From Newton.s second law of motion, `F _(n et) = (Delta p)/(Delta t) = (Delta (mv))/(Delta t)` Here v is constnat, so we get `F _( n et) =v (Deltam)/(Delta t)` Where `Deltam` is the cahrge of mass in `Delta t` time. THge charge of mass in 2s is equal to entire mass of mat. ` F _(n et) = ((1m //s) X (1kg))/(2s)` `=1/2N` In the horizontal direction only ONE FORCE is acting. Hence the required force is `1//2N` |
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| 1674. |
Which of the following converts electrical energy into mechanical energy? |
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Answer» motor |
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| 1675. |
Suppose a spring balance with a body suspended from it is allowed to fall. What will be the reading shown by the balance? |
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| 1676. |
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelength of sound waves in sir corresponding to these two frequencies ? Take the speed of sound in air as 344 m//s. |
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Answer» Solution :(i) In the FIRST case: Speed, `v = 344 ms^(-1)` Frequency, `F = 20 Hz` And, Wavelength, `lambda = ?` (To be calculated) Now, `v = f xx lambda` So, `344 = 20 xx lambda` `lambda = (344)/(20)m` Wavelength, `lambda = 17.2m`...(1) (ii) In the second case: Speed, `v = 344 ms^(-1)` Frequency, `f = 20 KHZ` `= 20 xx 1000 Hz` (Because `1kHz = 1000 Hz)` `= 20000 Hz` And, Wavelength, `lambda = ?` (To be calculated) Now, `v = f xx lambda` So, `344 = 2000 xx lambda` `lambda = (344)/(20000)` Wavelength, `lambda = 0.0172m` ...(2) Thus, the wavelengths of SOUND in air corresponding to the frequencies of 20Hz and 20 kHz are 17.2m and 0.0172 m respectively. |
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| 1677. |
A stone is thrown vertically upwards with an initial velocity of 40 m s^(-1). Taking g = 10 m s^(-2), draw the velocity-time graph of the motion of stone till it comes back on the ground. (i) Use graph to find the maximum height reached by the stone. (ii) What is the net displacement and total distance covered by the stone ? |
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Answer» Solution :Given u = 40 `m s^(-1)` g =10 m `s^(-2)`. As the stone rises up, the velocity decreases at the rate of 10 m `s^(-2)`. When the velocity becomes zero, the stone is at its highest position. Then it begins to fall and its velocity increases at a rate of 10 m `s^(-2)`. The velocity of stone at DIFFERENT instants is shown in the following table (the upward direction is taken positive).  Fig 2.29 SHOWS the velocity - time graph .  Maximum height reached by the stone = Area of `Delta OAB` `= (1)/(2) OBxx OA ` `= (1)/(2) xx4s xx 40 m s^(-1)= 80 `m. (ii) Net displacement = Area of `Delta `OAB- Area of `Delta BDC` `= (1)/(2) OBxxOA - (1)/(2) BD xx DC` `= ((1)/(2) xx 4S xx 40 m s^(-1)) - ((1)/(2) xx 4s xx 40 m s^(-1))=0` Total distance covered = Area of `Delta` OAB + Area of `Delta` BDC = 80 m + 80 m = 160 m |
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| 1678. |
The speed of the light in a diamong is 1, 24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS_(1)) |
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| 1679. |
A rocket is moving in upward direction with the velocity v, suddenly its velocity becomes three times, then find the ratio of its initial and final kinetic energy. |
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Answer» `E_(k)=(1)/(2)mv^(2),E_(k)propv^(2)` |
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| 1680. |
A body of mass 1.5kg moving with a velocity of 10m//s encounters another body of mass 2.5kg moving in opposite direction with a velocity of 16m//s, and they stick to eachother. What is the velocity of combination? |
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Answer» Solution :Here, `m_(1)= 1.5kg, u_(1)= 10m//s` `m_(2)= 2.5kg, u_(2)=16m//s` `v=?` Applying the principle of CONSERVATION of linear momentum, `(m_(1)+m_(2))v=m_(1)u_(1)+m_(2)u_(2)` `v=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))= (1.5xx10+2.5(-16))/(1.5+2.5)= (-25)/(4.0)= -6.25m//s` NEGATIVE sign indicates that motion of combination of BODIES is along the DIRECTION of motion of second BODY. |
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| 1681. |
As the angle between force and displacement increases, the magnitude of work …………. (remains constant, increases, decreases) |
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| 1682. |
Find out the reasons for the following. When a body was placed in a liquid it remained in the same position. |
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| 1684. |
Two wheel and axle systems P and Q areconnected as shown in the figure. The radii of wheels and axles of P and Q are 20 cm, 27 cm, 3 cm and 5 cm, respectively. If L = 540 kgwt, find E. |
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Answer» Solution :(i) Mechanicaladvantage of a wheel and AXLE (ii) Relate the load of Q to the EFFORT of P. (III) Mechanicaladvantage of a wheel andaxle is equal to the ratio of theradius of the wheel to that of its axle. (vi) E = 15 kgf |
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| 1685. |
A body with an initial velocity of 18 km h^(-1) accelerates uniformly at the rate of 9 cm s^(-2) over a distance of 200 m. Calculate : (i) the acceleration in m s^(-2) (ii) its final velocity in m s^(-1). |
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Answer» Solution :(i) ACCELERATION =9 cm `s^(-2) = (9)/(100) m s^(-2)` `=0.09 m s^(-2)` (II) Given initial velocity u = 18 km `h^(-1)` `= (18000m)/(60 xx60 s)= 5 m s^(-1)` Acceleration a = 0.09 `m s^(-2) ` and distance S = 200 m From equation of MOTION `v^(2)= u^(2) + 2A S ` `v^(2) = (5)^(2) +2xx0.09 xx200` or ` v^(2)= 25 + 36 = 61 ` `:.` Final velocity v = `sqrt(61) = 7.81 m.s^(-1)` |
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| 1686. |
An electric bulb of 60 W is used for 6 h per day. Calculate the 'units' of energy consumed in one day by the bulb. |
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Answer» Solution :POWER of electric BULB = `60W=0.06kW` TIME used `t=6h` ENERGY = power `xx` time taken `=0.06kWxx6h` `=0.36kWh=0.36` .units. The energy consumed by the bulb is 0.36 .units.. |
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| 1687. |
Theminimum distance require to hear distinct echo is _______. |
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| 1688. |
How high does the mercury barometer stand on a day when atmospheric pressure is 98.6 kPa ? |
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Answer» <P> Solution :`H = P_("ATM")/P_(Hg) = 98.6 XX 10 xx 10 xx 10 `` = ((N//m)^(2))/(13.6 xx 10 xx 10 xx 10 Kg//m^(3))` ` = 9.8 m//s^(2) = 740 MM` |
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| 1689. |
In a uniform magnetic field, the field lines are : |
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Answer» CURVED |
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| 1690. |
The total energy of a body falling freely towards the earth …………… (increases, decreases, remains constant) |
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| 1691. |
Find the velocity of a body of mass 100 g having kinetic energy 20 J. |
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Answer» `E_(k)=(1)/(2)MV^(2)` `therefore 20J=(1)/(2)xx0.1xxv^(2) therefore v^(2)=4 therefore v=2ms^(-1)` |
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| 1692. |
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car : |
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Answer» does not change |
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| 1693. |
The speed of sound in air is 330 ms^(-1) and in water is 1650 ms^(-1). It takes 2 s for sound to reach a certain distance from the source placed in air. How much time will it take for sound to reach the same distance when the source is in water ? |
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Answer» Solution :In water, V = 1650 m `s^(-1)` , d = 660 m `therefore ` TIME taken by sound to travel the DISTANCE d in water will be `t= d/V = (660 m)/(1650 ms^(-1)) = 0.4 s` |
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| 1694. |
The speed of sound in air is 330 ms^(-1) and in water is 1650 ms^(-1). It takes 2 s for sound to reach a certain distance from the source placed in air. Find the distance. |
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Answer» Solution :Given in AIR V= `330 m s^(-1) , t = 2S` From relation `V= d/t` Distance travelled by sound in air, `d= V xx t = 330 xx 2 = 660`m |
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| 1695. |
A scooterist travels at 30 km h ^(-1) along a stright path for 20 min. How much distance has he travelled ? |
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Answer» SOLUTION :Distance s = SPEED `xx` Time `= (30 km//h) xx (20 min)` `= (( 30km)/(60 km)) xx(20 min) =10 km` |
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| 1696. |
A simple machine |
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Answer» acts as a FORCE multiplier |
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| 1697. |
A batsaman hits a cricket ball which then rolls on a level ground. After covering a short distance , the ball comes to rest. The ball slows to a stop because. (a) The batsman did not hit the ball hard enough. (b) Velocity is proportional to the force exerted on the ball. (c ) There is a force on the ball opposing the motion. (d) There is no unbalanced force on the ball, so the ball would want to come to rest. |
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Answer» SOLUTION :The ball slows down and comes to rest due to opposing forces of air RESISTANCE and frictional force on the ball opposing its motion. Therefore the CHOICE (C ) there is a force on the bal opposing the motion is correct. |
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| 1699. |
Explain the working and application of a sonar. |
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Answer» Solution :Working SONAR standsfor Sound NAVIGATION and Ranging. It consists of transmitter and detector , transmitter transmit electrical signal into ultrasonic sound signal and detector CONVERTS ultrasonic sound into electrical signal. .d. is the total distance , t - total time taken , v - VELOCITY of the ultrasonic sound waves. APPLICATIONS : (a) Depth of the ocean can be determined (b) Location of submarines. |
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| 1700. |
Think of situation when an obejct gets displaced in the absence of a force acting on it. |
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Answer» SOLUTION :Yes. If the object moves with a constant velocity, then no force ACTS on it. (Newton.s first law of motion.) Consider a puck on a horizontal air table (a frictionless table) : When an impulse is given to the puck, it moves (GETS displaced) with constant velocity in the absence of a force. An object observed from a moving reference frame : If you are sitting in a moving train, a tree or a pillar appears to be moving in the opposite DIRECTION. |
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