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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.
| 13351. |
Viscous force similar to friction in solids, but viscous force a) is independent of area but friction depends on area b) is temperature dependent while friction force between solids depends upon normal reaction c) is velocity dependent while friction is not d) is velocity independent while friction is velocity dependent |
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Answer» a, B, C are correct |
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| 13352. |
In the figure, if m_(1) is at rest, find the relation among m_(1), m_(2) and m_(3) ? |
Answer» Solution :`m_(1)` is at rest `rArr` point B does not move, `m_(2)` and `m_(3)` move with ACCELERATION `a=((m_(3)-m_(2))/(m_(2)+m_(3)))G, m_(3)GT m_(2)` `T=(2m_(2)m_(3)g)/(m_(2)+m_(3)), T^(1)=2T=(4m_(2)m_(3)g)/(m_(2)+m_(3))` `m_(1)cancel(g)=(4m_(2)m_(3)cancel(g))/(m_(2)+m_(3)) "" [(4)/(m_(1))=(1)/(m_(2))+(1)/(m_(3))]` |
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| 13353. |
Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter? |
| Answer» SOLUTION :Sphere of smaller DENSITY will have larger MOMENT of inertia. | |
| 13354. |
Impulse of force when F force act on body for Deltat. Similar impulse will provided when 2F force act on body for …... time. |
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Answer» `F_1Deltat_1=F_2Deltat_2` `:. FDeltat=2FxxDeltat_2` `:. Deltat_2=(Deltat)/2` |
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| 13355. |
A student plucks at the centre of a stretched string and observes the wave pattern produced.What type of wave is produced in the string |
| Answer» SOLUTION :Tranverse STATIONARY WAVE | |
| 13357. |
A spring balance gives erroneous readings if it is used frequently over a long period of time. Explain. |
| Answer» Solution :On USING a spring BALANCE frequently over a long period of time, its elastic property degrades and it deforms permanently. As a result, the elongation of the spring is more than the elongation it should suffer for a given load SUSPENDED at it free end. So, we GET a WRONG reading. | |
| 13358. |
Refer to the plot of temperature versus time figure showing the changes in the state of ice on heating (not to scale). Which of the following is correct ? |
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Answer» The region AB represents ice and WATER in THERMAL equilibrium. Temperature of water becomes `0^(@)C` to `100^(@)C` [it is REPRESENTED by to C, BC] Now, Cd part shows the equilibrium of water and steam. |
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| 13359. |
A satellite orbits the earth at a height of 1000 km. Find its orbital speed ? |
| Answer» SOLUTION :`7.354 KMS^(-1)` | |
| 13360. |
(A) : Dimensions of constant of proportionalities can be derived from dimensional method. (R) : Numerical value of constant of proportionality can be found from experiments only. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 13361. |
Explain which properties are necessary to understand the speed of mechanical waves. |
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Answer» Solution :Once disturbance is passed through the MEAN position, restoring FORCE and elasticity are required for string particles to come back to their original position. Oscillated particle is displacement depends on the inertia of medium. THUS, for PROPAGATION of mechanical waves, elasticity and inertia are NEEDFUL. Thus, by using dimensional analysis based on two properties wave speed can be obtained. |
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| 13362. |
Matching |
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Answer» (i) G, (ii) H, (iii) C, (IV) B, (v) C |
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| 13363. |
which distance is represented by a fermi ? |
| Answer» Answer :B | |
| 13365. |
A long spring when strectched by 'x' cm has a potential energy 'v'. On increaseing the stretching to 'nx' cm the potential energy stored in the spring will be |
| Answer» ANSWER :C | |
| 13366. |
Identify the increasing order of angular velocities of following (a) Earth rotating about its own axis (b) Hour.s hand of clock ( c) Seconds hand of clock (d) Fly wheel of radius 2m making 300 r.p.m. |
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Answer» a, B, C, d |
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| 13367. |
Discuss strings stretched between fixed points. |
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Answer» Solution :(i) LET a spring be stretched between two POINTS . Vibrations will set up in this string. When it is disturbed , say from middle . (ii) The figure given below shows that vibration set up in this string appears to be similar to a half wave having wave length `(lamda)/(2)` `lamda=2l`  Length of the spring `l=(lamda)/(2)` If T is TENSION in the string and M is mass PER unit length of the string, then velocity of disturbance is given by , `v=sqrt((T)/(M))(Mrarr`Linear Density) `v=("wavelength")/("TIME")="wavelength"xx"frequency"=lamdaf` `v=2lxxf` `f=(v)/(2l) =1/(2l) sqrt((T)/(M))""[:'v=sqrt((T)/(M))]` |
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| 13369. |
A loop of a string of mass length mu and redius R is rotated about an axis passing through centre perpendicular to the plane with an angular velocity omega.A small disturbance is created in the loop having the same sense of rotation. The linear speed of the disturbance observer is |
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Answer» `OMEGAR` |
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| 13370. |
A body of mass 6kg travelling with a velocity 10m/s collides head -on and elastically with a body of mass 4kg travelling at a speed 5m/s in opposite direction. The velocity of the second body after the collision is |
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Answer» 0m/s |
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| 13371. |
A sample of an ideal gas occupies a volume 'V' at pressure 'P' and absoulle temperature 'T'. The mass of each molecule is m. The expression for the density of gas is |
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Answer» MKT |
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| 13372. |
What are the derived units? |
| Answer» Solution :UNITS which are expressed in terms of fundamental units are called derived units. For EXAMPLE, speed is distance/time and is therefore expressed in `MS^(-1)` and is a derived units. | |
| 13373. |
Volume of mercury in the bulb of a thermo-meter is 10^(-6)m^(3). Area of cross section of capillary tube is 2 xx 10^(-7) m^(2). If the temperature is raised by 100°C, the length of mercury column is ( gamma_(Hg) = 18 xx 10^(-5)//""^(@) C) |
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Answer» 9 cm |
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| 13374. |
An incandescent light bulb has a tungsten filament that is heated to a temperature 3xx10^(3)K when an electric current passes through it. If the surface area of the filament is approximately 10^(-4)m^(2) and it has an emissivity of 0.32, the power radiated by the bulb is |
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Answer» 150 W |
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| 13375. |
A clock with a brass pendulum keeps correct time at 30^@C . How many seconds will it loose or gain if the temperature falls by 10^@C ? Take, alpha_("brass") = 0.000019^@C^(-1) |
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Answer» Solution :Time period of pendulum ` T_30 = 2S` The new temperature will be `(30 - 10)^@C = 20^@C`and LET `T_20`be the time period at this temperature. Then, `l_30 = l_0 (1+ 30 alpha)` `l_20 = l_0 (1+20 alpha)` where `l_30, l_20` and `l_0`are the lengths of the pendulum at `30^@C, 20^@C` and `0^@C` , respectively. ` T_30 = 2pi sqrt( (l_30)/(g) ) = 2pi sqrt( (l_0 (1+ 30 alpha) )/(g) )` ` T_20 = 2pi sqrt( (l_20)/(g) ) = 2pi sqrt( (l_0 (1+ 20 alpha) )/(g) )` ` therefore (T_20)/(T_30) = sqrt( (1+ 20 alpha)/(1 + 30 alpha)) = (1+ 20 alpha)^(1//2) (1+ 30 alpha)^(-1//2)` `= (1+ 1/2 xx 20 alpha) (1- 1/2 xx 30 alpha)` `= (1 + 10 alpha )(1 - 15 alpha) = (1 -5 alpha)` ` therefore T_10 = T_30 (1 - 5alpha)` ` = 2 (1 - 5 xx 1.9 xx 10^(-5) )s ` ` = (2- 1.9 xx 10^(-4) ) s` SINCE `T_20 lt T_30` , the clock will gain time when the temperature falls to `20^@C` . Gain in time for 2 seconds = `T_30 - T_20 = 1.9 xx 10^(-4) s` Gain in time for 1 day ` = (1.9 xx 10^(-4) xx 24 xx 3600)/(2)` = 8.21 s |
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| 13376. |
A hollow and a solid sphere of same material and identical outer surface are heated to the same temperature |
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Answer» in the beginning both will EMIT EQUAL amount of RADIATION per unit time |
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| 13377. |
Why does sound travel faster on a rainy day than on a dry day? |
| Answer» Solution :The amount of water vapours present in the atmosphere is much higher on a rainy day than on a dry day. As the water vapours are lighter than dry air, density of wet air becomes LESS than of dry air. Because the speed of sound is inversely PROPORTIONAL to the square root of the density, sound TRAVELS FASTER on a, rainy day than on a dry day. | |
| 13378. |
An astromical telescope has an angular magnification of magnitude 5 for distant objects. 36cm and the final image is formed at infinity. Determine the focal length of objective and eye piece. |
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Answer» SOLUTION :For final IMAGE at infinity `M_infty=f_0/f_e and L_infty=f_0+f_e therefore 5=f_0/f_e`……(i) and `36=f_0+f_e`…….(II) Solving these two EQUATIONS, we have `f_0=30cm and f_e=6cm` |
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| 13379. |
The angle friction thetais given by |
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Answer» `TAN mu_s` |
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| 13380. |
A body covers a distanceof 4 m si3rd second and 12 m in 5 th second. If the motion 25 ms ^(-1) and after 8 second ,it is uniformly accelerted ,how far will it travel in the next 3 second ? |
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Answer» |
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| 13381. |
A manwalkseasrwithcertainvelocity.A caris travellingalonga roadwhichis30^@westofnorth. Whilea busistravellingin anotherroadwhichis60^@southof west. Findthe anglebetweenvelocityvectorof( a)man and car(b)car and man(c )busand man |
Answer» Solution : From the DIAGRAM the angle between velocity vector of CAR and BUS is `60^(@) + 60^(@) = 120^(@)` The angle between velocity of bus and MAN is `30^(@) + 90^(@) = 120^(@)` |
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| 13382. |
The quantity which is conserved in an inelastic collision of two bodies is …….. |
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Answer» TOTAL KINETIC energy |
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| 13383. |
A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has a pressure P and volume 5Vand the other part has pressure 8P and volume V, the piston is now left free. Find the new pressure and volume for the isothermal and adiabatic process. |
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Answer» Solution :Final pressure will be same on both sides. Let it be `P^(.)` , with VOLUME `V^(.)` , in the left side and `(6V-V^(.))` in the right side. (A) If the CHANGE is isothermal : For the gas enclosed in the left chamber, `8PxxV=P^(.)(6V-V)^(.) "…….." (2)` Solving these for `V^(.) and P^(.)`, we get `V^(.)=30/13V and P^(.) = 13/6 P and (6V-V^(.))=48/13V`  (B) If the change is adiabatic : For the gas in the left chamber. `P(5V)^(gamma)=P^(.)(V^(.))^(gamma) "............" (3)` and for the gas in the right chamber, `8P(V)^(gamma)=P^(.)(6V-V^(.))^(gamma) "........" (4)` Dividing (4) by (3), `[(6V-V^(.))/V^(.)]^(3//2)=8/5^(3//2)or (6V)/V^(.)=1+4/5 [:. gamma=3/2]` i.e., `V^(.)=10/3V` Substituting it in EQN. (3), `P^(.)=P[(5Vxx3)/(10V)]^(3//2)=(3sqrt3)/(2sqrt2)P=1.84P` So `P^(.) = 1.84P, V^(.)=10/3 VAND (6V-V^(.))=8/3V` . |
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| 13384. |
What will be the force of contact it the same force is applied on the mass m_(2) in the above problem . |
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Answer» |
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| 13385. |
Assertion: A carpet is wrapped like a cylinder and is pushed on a horizontal surface to unwrap. As the carpet unwraps its angular speed is found to increase. Reason: Angular momentum of rolling part remains conserved |
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Answer» If both assertion and reason are CORRECT and reason is the correct explanation of the assertion. |
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| 13386. |
At the highest point of oblique projection, which of the following is correct? |
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Answer» Solution :This projectile motiontakes place when the intial velocity is not horizontal, but at some angle with the verticl , as shown in FIGURE (Oblique projectiom). Example, Water ejected out of a hose pipe held obliquely. Cannon fired in a battle ground, consinder an object thrown with intial velocity at an angle `theta` with the horizontal. then ` vecu=u_(x)hati+u_(y)hatJ` Where `u_(x)=u costheta` is th horizontal componant and `u_(y)=u sintheta`, the vertical component of velocity. since the accelereration due to GRAVITY is in he DIRECTION opposite to te directionopposite to the direction of vertical component `u_(y)` this component will gradually reduce to zero at the MAXIMUMHEIGHT of the projectile . At this MAXIMUM height the same garvitational force will oush the projectile to move downward and fall to the ground. There is no acceleration along the x directionthroughout the motion , So the horizonral component of the velocity `(u_(x)=u cos theta)` remains the same till the object reches the ground.  Hence after the time t, the velocity along horizontal motion ` v_(x)=u_(x)+a_(x)t=u_(x)=u costheta`  Thuis , `x=u cos theta. t or t =(x)/(u costheta)` Here `u_(y)=usin theta, a_(y)=-g` (acceleration due to gravity acts opposite to the motion). Thus `v_(y)=u sin theta-g t` The vertical distance travlled by the projectile in the same time t is Here , `s_(y)=y, u_(y)=u sin theta, a_(x)=-g` Substitute the value of t from equation (1) in equation (2), we have `y=u sin theta (x)/(u costheta)-(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)` `y=xtantheta -(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)` |
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| 13387. |
A particle is subjected to two mutually perpendicualr simple harmonic motions such that its x and y-coordinates are given by x=sinomegat, y=2cosomegat The path of the particle will be : |
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Answer» an ellipse |
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| 13388. |
If energy (E), Velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be |
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Answer» `[E^(-2) V^(-1) T^(-3) ]` |
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| 13389. |
The efficiency of a heat engine if thetemperature of the source is 100^(@)C and sink is 27^(@)Cis nearly |
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Answer» 0.1 |
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| 13390. |
Which of the following is a dimensional variable ? |
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Answer» FORCE |
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| 13391. |
One mole of an ideal gas with heat capacity at constant pressure C_p undergoes the process T=T_0 + alpha V, where T_0and ' alpha ' are constant. Find the amount of heat transferred to the gas, if its volume increased from V_1 and V_2: |
| Answer» SOLUTION :`a(V_2 -V_1) C_p +RT_0 LOG _e ((V_2 )/(V_1))` | |
| 13392. |
One mole of an ideal gas with heat capacity at constant pressure C_p undergoes the process T=T_0 + alpha V, where T_0and ' alpha ' are constant. Find heat capacity of the gas as function of its volume, |
| Answer» SOLUTION :`C_p +(RT_0)/(alphaV) ` | |
| 13393. |
From arod fixed on a vertical wall, a simple pendulum is suspended. Its bob is pulled away from the wall to a horizontal position and then released. If the coefficient of restitution is 2/sqrt(5), what would be the minimum number of impacts of the bob with the wall, after which the amplitude of the pendulum becomes less than 60^@? |
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Answer» |
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| 13394. |
A block of mass m sliding on a smooth horizontal surface with a velocity vecV meets a long horizontal spring fixed at one end and having spring constant k as shown in figure. Find the maximum compression of the spring? |
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Answer» `Vsqrt(k/m)` |
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| 13395. |
A ball is dropped from a building. It takes 4s to reach the ground. The height of the building is (use g = 10 m//s^(2) ) |
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Answer» Solution :From equation of motion. `s = UT + 1/2 at^(2)`s = h : g = a : u = o `h = 1/2"gt"^(2)` `h = 1/2 xx 10 xx (4)^(2)` , h = 80 m |
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| 13396. |
Two bodies of masses m_(1) and m_2 moving on the same direction with velociities u_(1) and u_2 collide. The velocities after collision are V_(1)and V_2. If each sphere loses the same amount of kinetic energy, then: |
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Answer» `u_(1) + u_(2) + V_(1)-V_(2)=0` |
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| 13397. |
While measuring the thermal conductivity of a liquid we keep the upper part hot and the lower cool so that: |
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Answer» Convection may be stopped |
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| 13398. |
A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 27^(@)C, it weighs 30 g. When the temperature of the liquid is raised to 42^(@)C, the metal piece weighs 30.5 g. Specific gravity of liquid at 42^(@)C is 1.2. Calculate the coefficient of linear expansion of the metal, |
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Answer» Solution :Loss of weight of a body at `27^(@)C = 46 - 30 =16 g` Loss of weight of body at `42^(@)C = 46 - 30.5 = 15.5 g ` If `V _(27)` is the bolume of body at `27^(@)c,,` VOLUME at `42^(@)C = V _(27) (1 + gamma _(s) t ) = V _(27) (1 + 15 gamma _(s)) = V _(42),` where `gamma _(s)` is the volume coefficient of expansion of solid. SINCE loss of weight of body = weight of the liquid displaced `= V d _(1) g,` We have ` 16 = V _(27) d _(27) implies 15.5 = V _(42) d _(42)` `therefore (16)/(15.5) = (V _(27))/( V _(42)) . (d _(27))/(d _(42))(or) (32)/(31) = ( V _(27))/( V _(42)) . ( d _(27))/( d _(42)) (or) (32)/(31) = (V _(27))/( V _(27) {1 + 16 gamma _(s) })xx (124)/(120) (or) (32)/(31) = (1)/(1 + 15 gamma_(s)) xx (31)/(120)` `therefore 1 + 15 gamma _(s) = (31)/(30) xx (31)/(32) = (961)/(960) therefore 15 gamma _(s) = (961)/(960) -1 = (1)/( 960) therefore gamma _(s) = (1)/(960) xx (1)/(15) = 3 alpha _(s),` where `alpha _(s)` is the linear coefficient of expansion of solid . `therefore alpha _(s) = (1)/( 960 xx 45) = 0.00024 = 2.4 cc xx 10 ^(5) //K` |
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| 13399. |
Explain main type of force with suitable example. |
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Answer» Solution :FORCEIS REQUIREDTO bringto bodyat rest.Tonecessaryto be incontactof the body.Hencethere aremainlytwotypeof force . In contactforcethe is physical interationbodystateof motinof restof anobjectcan be changed Herecontactforectact on bothobject. Whenan OBJECTIS ingravitationfieldforcedue tofield. Anobjectfallingfreelyfrom TOPOF abuildinghasacceleratedmotiondue togarvitationfieldof the arth.Whennailis placenearbarmagnetit isattracted dueto magneticfieldof bar MAGNET. |
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| 13400. |
The velocity-time graph of a particle moving along a straight line is given as below. The displacement time curve for the particle is given by : |
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Answer»
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